groups-rings-fields

[fluid kelp]

you can easily get 4 x 3^m x 5^n as the final order of the roots

which equals the degree of the original polynomial

Does this make sense

[rocky cloak]

Okay, I think I follow now.

So you're saying since the roots of this are primitive 3^n+1 5^m+1 roots of unity, their minimal polynomial should be the cyclotomic polynomial.

Unfortunately, the cyclotomic polynomials don't always remain irreducible modulo p.

Besides phi(3^m+1 5^n+1) = 8 * 3^m 5^n anyway.

[fluid kelp]

Yes

For the bottom part, wouldn't it be the lcm so it would be 4* 3^m 5^n

Or something of those sorts

Is there a way I can tweak my argument so that it can work? (if it doesn't)

[rocky cloak]

Pfff, maybe idk.

Like the 15th cyclotomic polynomial splits mod 2 as a product of two degree 4 polynomials, so if you could argue something similar for the 3^n 5^m cyclotomic polynomial it would go through.

But that seems harder than the original problem...

[fluid kelp]

The main reason that I asked was because I saw a post of stack exchange that was for x^2n + x^n + 1

But now I realize that it is cyclotomic

I don't really understand the proof that you said earlier 😅

[fluid kelp]

Wait also x^4 + x + 1 is not cyclotomic is it?

[rocky cloak]

No, it's one of the factors that the 15th cyclotomic polynomial splits into mod 2

[elfin wraith]

Still struggling with this

I get that the projection map M->P has kernel N basically by definition, and we’re told that the only homomorphisms which are zero are the ones N->P vice versa. So taking the projection of End(M)->End(P) our kernel will just be End(N) and similarly the other way, trivial intersection by the N->P fact. But I feel like that’s quite hand wavey.

I’m also still struggling to show that f(N)<N, I get that it’s basically cross terms getting sent to 0 but idk how to actually write it up

Can we like compose with the projection/restrict f in some nice way?

Not sure what’s catching me out with this one so much

[rocky cloak]

Think about what it means for f(N) to be in N.

It means that if you compose f with the inclusion of N into M, the image lands in N.

Now think about the hint, the kernel of M -> P is N

[rocky cloak]

And just to comment on what your saying here, in general "taking the projection End(M) -> End(P)" isn't something you can do. The existence of a ring homomorphism End(M) -> End(P) hinges on the fact that f(P) < P

[elfin wraith]

Yeah I didn’t feel great about saying that, it didn’t feel like it should generally be possible

I’ll keep thinking about this for a bit and see where I get, this really isn’t clear to me for some reason

[rocky cloak]

Maybe it helps to draw a diagram:

N -> M -f-> M -> P

what can be said about this composition?

[elfin wraith]

It should be 0 I think

Well then again I’m not sure, I don’t think it’s exact anywhere because I’m not sure what I can actually say about f

[rocky cloak]

Right, because it's a map N -> P

[elfin wraith]

Oh yeah of course we’re assuming maps N->P are 0

[rocky cloak]

So if you compose a map with M -> P and get 0, what does that tell you about the image of that map?

[elfin wraith]

Its the kernel of the other map

[rocky cloak]

Right, so then f(N) < N, success!

[elfin wraith]

Thank you, that was a struggle lol

Should maybe actually learn a lesson from the homological algebra I’ve been doing and draw some sequences!

[fluid kelp]

I assume this is the conjugate of y with x rather than inverse of xyx right

[grave sedge]

Yeah (it's more or less the same because x has order 2 though)

[rocky cloak]

Both work actually, so that's kinda fun

[fluid kelp]

Why is it that the conjugate works?

Oh wait apologies wrong ping

[rocky cloak]

Well, the conjugate is sort of what you would expect work, since it's supposed to be a normal subgroup and all

[fluid kelp]

That's actually so cool!

[cloud walrus]

mycroftholmes1703

[opaque finch]

amazingly one has the formal derivative of this polynomial over $\mathbb{F}_5$ as $0$ !

[cloud walrus]

mycroftholmes1703

[vast stump]

conveniently, 250 and 375 are both multiples of 125

[slim kayak]

For f(N)<N:
As the coproduct of N and P the map M-> M corresponds exactly to a pair of maps N-> M and P ->M. I figure you can see the rest of the argument here

[elfin wraith]

I worked it out but thank you!

[slim kayak]

Oh, missed that 😅

But it's good to keep your products/coproducts close at hand.

I had one seminar on adjunctions and their continuity once and it's still paying off, especially with the hom set properties with limits/colimits

[elfin wraith]

I really don’t know what products and coproducts are concretely in this context tbh, I only really know it in the context of catagories generally and even then I’m not feeling great about it

[slim kayak]

With that nuke the End_R disintegrate into the products of end(N,N) x end(N,P) x end(P,N) x end(P,P) = end(N,N) x end(P,P).
Came in handy surprisingly often with central associative algebras

[slim kayak]

Well, with finite products you will often find they are the same as finite coproducts.

Them being distinguished by having projections/injections and getting induced maps into them or out of them.

Also, in a lot categories they tend to look essentially the same as their simpler bethren in Ab or Vec

[tough raven]

(when morphisms can be added)

[fluid kelp]

How do you find the order of a root in F2?

[slim kayak]

The field with two elements?

[rapid junco]

If I have this criterion, I want to take the intersection of the series.

Proving normality is trivial but what tells me that $[A^{i + 1} \cap B^{i + 1} : A^i \cap B^i] = p?

[next obsidian]

I don’t think it is my guy, I think you might need to refine it more. If the index is prime, you can’t wedge any other subgroup between them, but A^i+1\cap B^i sits between them, possibly not equal to either

[crystal vale]

let G is a p-group and H is proper subgroup then N(H) > H. any hint, i think i did this before

[rocky cloak]

If you consider the action of H on G/H (by left multiplication), then the fixed points are exactly N(H)/H

[crystal vale]

in this cardinality of set of fixed points is a multiple of p and there is an one already therefore N(H) > H, correct?

how did you think about action? How this idea comes?

[rocky cloak]

Well, I've seen this quite a few times.

I guess you have a lot of strong theorems about actions of p-groups, so you want to wedge an action in there somehow.

From there there's really only 4 options, either action on G/H or conjugation, with either H or G acting.

So just see whichever gives you something related to N(H)

[crystal vale]

interesting thank you

[opaque finch]

can you tell me what book is that ?

[crystal vale]

Show that for each positive integer k there exists a positive integer N such that there are at least k non-isomorphic groups of order N.

I am thinking to use Fundamental theorem of abelian group to construct at least k non-isomorphic groups, if I have k then I will take the N = p^k, where p is a prime and since number of partition of k is at least k except when k = 2,3 , so there are at least non-isomorphic groups.

Does this idea work?

[tough raven]

Yes.

[crystal vale]

Thank you

[mystic ether]

Is there an Eckmann Hilton argument for group actions? IE, suppose I have defined two different looking group actions of a group G on a set X. Can I show they commute with each other or some other condition and magically the group actions end up being the same?

obviously the actions commuting isn't enough, eg left and right action of a group on itself

[languid trellis]

I don't see how we get from $g(r) = 0$ to $\bar{g}(\zeta(r)) = 0$ rather than just $\zeta (g(r)) = 0$

[cloud walrus]

swifteeee

[languid trellis]

Any advice? ( :

[rocky cloak]

Zeta extends eta and the coefficients of g are in F, so applying zeta to them gives gbar

[languid trellis]

Oh right, of course.

Thank you

[toxic zephyr]

sorry this is so late. what is a localization and why does it help that S^c is multiplicatively closed?

[rocky cloak]

Localization is when you invert a multiplicatively closed set.

The magical thing is that prime ideals of the localized ring are exactly the prime ideals that don't intersect the set your localizing.

But if you don't already know that I guess you need a different methode...

[toxic zephyr]

yeah this is theoretically a basic group/ring theory class (grad level tho)

[toxic zephyr]

is that sort of like the field of fractions?

[rocky cloak]

I mean you're discussing modules and prime ideals, so it's not like localization is at a different level.

And yes, field of fractions is an example of localization

[toxic zephyr]

we haven't technically learned modules yet i just know them and it's easier for me to think in terms of them 😆

but thats interesting

this is from ch3 sec 2 of hungerford (ex. 15).
so i'm guessing there's something more elementary?

i'm gonna think to see if there's a nice ideal contained in S which would quotient to an integral domain ig

[rocky cloak]

I mean unpacking the standard proof for how primes in the localization works could be a thing:

Like, prove that there is an ideal in S maximal among ideals contained in S. Prove that such an ideal is prime.

These two steps are both pretty elementary, though you'll need to use something like Zorn for the first one.

[toxic zephyr]

so i'm not quite seeing the big picture, but it sounds like this guaranteed prime ideal is pretty much going to be this maximal contained ideal from the localization thingy. and there isn't indeed something simpler for this very general question?

i mean unless like... just the ideal generated by S itself lol
that wouldn't work would it? it can't be that simple

[uncut fox]

Hey I am trying to proof something that I think is pretty easy to show but I can't think of a good way to proof it, for now I have this

I don't know for sure how to show the other inclusion correctly. Does anyone have an idea how to cleanly show the other inclusion, or maybe even a better way than going over the fundamental theorem on homomorphisms?

Because it "kinda looks like pulling that (2)-Ideal in the factor into (Z/2Z)[X]" so I thought that it looks like some isomorphism theorem, but I couldn't figure it out

[toxic zephyr]

or wait, isn't S just the torsion submodule, which makes it an ideal? and if R/S=M/Tor(M) is torsion free, then doesn't that make R/S an integral domain and thus S is prime? i'm confusing myself i think

[opaque finch]

What do you want to prove @toxic zephyr?

[toxic zephyr]

my claim/idea is that this set S is itself a prime ideal

[opaque finch]

Ahh nicee.

[toxic zephyr]

thinking about it in terms of the more comfortable modules it seems that S is just the torsion submodule (i.e. an ideal?) and since R/S would be torsion free, then that would make it an integral domain (so S is prime)

[opaque finch]

Welll I can actually say much more, S is union of prime ideals. Try proving that

[surreal dagger]

Is this supposed to say (Z[x]/(x^7-1))/(2/(x^7-1))?

[toxic zephyr]

and a union of prime ideals is prime right? that would make some sense

but not sure

maybe not

[opaque finch]

Lemme say something in this regard. Union of ideals may not be an ideal even everytime.

Forget about primes

[toxic zephyr]

right yeah i know that. ex. 3ZU5Z is not an ideal

[opaque finch]

So you see it as well!

[uncut fox]

no

[toxic zephyr]

unless one is contained in the other, isn't it?

[opaque finch]

Right!!

[uncut fox]

it's supposed to say (Z[X]/(x^7-1))/2(Z[X]/(x^7-1))

[toxic zephyr]

i see. so the union you're speaking of is the chain of ideals that jagr was alluding to?

[opaque finch]

But lets see your problem like this. You just have to show that your S admits maximal elements and they are precisely the prime ideals. Then the rest is clear right?
Maximal elements are dealt with by zorn's lemma. The other case is that those maximal elements are prime ideals. Can you show this?

[toxic zephyr]

so just real quick, is my claim that S is itself a prime ideal (by all that torsion justification) wrong? i'm happy to try to work through this other way, but just wanna make sure i know if my initial idea is right or wrong.

[surreal dagger]

Is this the ideal generated by 2 in that quotient? Im not sure how to interpret the /2(Z[X]/(x^7-1))

[uncut fox]

yes, for R = Z[X]/(x^7-1), it's just R/2R, or R/(2)

[uncut fox]

it looks so close to being able to use the third isomorphism theorem but idk, if it even is possible to show it with it

[opaque finch]

Well S can be a prime ideal, I am not sure about that tbh, cause I didnt read your previous conversation. In that case that would be a pretty strong statement. But I should think that should be the case in general, gimme some time to read through it and get back to you on that.

What I am trying to say that for your case you just need the fact that I am stating which is weaker and easier to prove.

But your idea maybe legit just that I didnt go through it yet.

[toxic zephyr]

i would think this is a bit easier to just say S itself works... i mean the idea is basically that the zero divisors (and zero) is prime. if xy is a zero divisor, then x is a zero divisor or y is a zero divisor. isn't that true?

i was thinking about it in terms of the torsion of R as an R module.

say xy is a zero divisor and z is the nonzero annihilator. if x is not a zero divisor, then yz=0 means y is a zero divisor.

[opaque finch]

Can you tell me how you define torsion? For me I define it to be the set of all torsion elements. Torsion elements are such that when multiploed with non-zero divisors yeild 0

[toxic zephyr]

yeah

or wait

torsion is all elements x such that there is a nonzero scalar r such that rx=0

so anything that has a nonzero annihilator

i.e. zero divisors in a ring

[opaque finch]

Yes that is true

[opaque finch]

I think this is true only when it is an integral domain

Not generally tbh

[toxic zephyr]

but doesn't an integral domain have no nontrivial torsion?

[toxic zephyr]

does this not work?

shows that if xy is in S and x not in S, then y in S. i.e. S is prime

[opaque finch]

That is true.

[toxic zephyr]

and S is an ideal

(since torsion is a submodule)

[opaque finch]

Noo. I dont think so

Lemme try to understand the differences, based on what I understand and know. I may be wrong but hopefully not.

Let M be an R-module. An element m of M is called regular element if it is neither a left or right zero divisor. m is called torsion element if there exists such a regular element r such that rm = 0.

The reason I am putting stress over integral domains is because, it has no non-zero divisors and henxe ever element is regular and your definition of torsion works..

[next obsidian]

You want flat / free modules to always be torsion free

[opaque finch]

Exactly so.

[next obsidian]

So you have to define torsion only on regular elements because if R itself has zero divisors it would have torsion if you didn’t define it this way

And you get that flat modules are torsion free because if x is regular the multiplication map
R -•x-> R is injective and so stays injective when tensoring by a flat module M

And that becomes multiplication by x on M
M -•x-> M

[opaque finch]

Yes. Thats what I was aiming for. Thanks for making it clear.

So overall I think even though your arguments are absolutely correct @toxic zephyr, I dont believe the definition you are using is completely correct. Hence the conclusions doesnt actually want to correspond.

[opaque finch]

I would say no as well. Let R be an integral domain and M be a finitely generated module over R, then M is torsion free iff Ann(M)=0.

One can come with an examlle definitely with a non-finitely generated module consider direct products of Z/(p) where p ranges over all primes. Indeed they are integral domains.

[toxic zephyr]

what is a flat module? and in this case R is just a comm ring with identity so i'm just looking at it as some R module. if i'm defining the torsion to be anything with a nonzero annihilator (includes 0), then isn't that a submodule/ideal? and isn't the result that quotienting by the torsion gives something torsion free? and since that's a quotient ring with no zero divisors, wouldn't that be an integral domain which would make that submodule/ideal prime?

this is the original question, i'm just looking at it from a module perspective to say why i think the set itself is a prime ideal

[elfin wraith]

Flat modules preserve exact sequences under the tensor product

[next obsidian]

A flat module is one which preserves injections upon tensoring.

Torsion is an x in M for which there's a non-zero divisor r in R where rx = 0.

Quotienting by torsion would give you something torsion free, yes, but torsion lives inside of your submodule

But you can't quotient by torsion inside of R, because R will always have no torsion. You want R to be torsion-free, so you only define torsion to be stuff annihilated by a regular element of R

Here's another reason you need to define it in terms of regular elements

Say you defined torsion to be x in M for which there's non-zero r in R with rx = 0. How do you show this is closed under addition?

Take x in M with rx = 0, y in M with sy = 0, what kills x + y? The only good answer is rs, but if r and s are allowed to be zero-divisors then rs might be 0, and so you can't show x + y is torsion

[toxic zephyr]

ooof i see

so am i completely wrong that S is a prime ideal

[next obsidian]

What is S

[toxic zephyr]

what i'm calling the described set

[next obsidian]

the set of zero divisors and 0?

[toxic zephyr]

0 and 0 divisors

ye

[next obsidian]

yes, this is not prime

it isn't even an ideal

[rocky cloak]

If S is an ideal it will be prime, but usually it's not an ideal

[next obsidian]

in a say, reduced ring

it's the union of all minimal primes

In a Noetherian ring it's the union of all associated primes

You won't be able to show S is closed under addition

[toxic zephyr]

like in Z6, 2,3 in S but 2+3 not in S...

cry

[next obsidian]

There's a slick way to prove what it's asking

[toxic zephyr]

i'm listening...

[next obsidian]

Let T = R\S, this is a multiplicatively closed set, and doesn't contain any zero divisors. Thus T^-1R is a non-zero ring, in fact it's the "largest" localization of R and is called the total ring of fractions. Since T^-1R is non-zero it has a prime P, which corresponds to a prime ideal of R disjoint from T, which means P < S

[toxic zephyr]

people keep talking about localization and i've never heard of it before yesterday 😭
never brought up in class yet

[next obsidian]

gg

[void cosmos]

I did this!

in an exam

proving the intersection of prime ideals is the set of nilpotents

same argument kinda

was cool

[elfin wraith]

I did a pretty similar thing for non com rings yesterday too lol, showing the localisation at a prime ideal is the Jacobson radical and the unique maximal 2 sided ideal

[void cosmos]

yeah i see

localization is pretty cool honestly

whats cooler is that everything you write has deep geometric meaning behind of it

so theres that

[hidden wind]

everyone telling me GEOMETRY and i’m like omg i love geometry but WHERE ?!

[elfin wraith]

I don’t actually know what the geometry or anything behind it is in the noncom case, I asked my lecturer and she didn’t have a great reason, I think she’s just into rings for the love of the game largely

[void cosmos]

geometry is in the functorialty

(this is a joke idk what im saying)

[next obsidian]

I respect it

Geometry is when contravariant functor

[elfin wraith]

Honestly I do too, I’m becoming increasingly convinced I can just fall into a little rings rabbit hole and be quite content

[hidden wind]

tbh i’m enjoying rings quite a lot as well even though i’m not at all comfortable with them yet

[hidden wind]

between what

[tough raven]

Wait dang what

How do you even localise at a prime ideal?

[void cosmos]

take ur multiplicative set to be R-prime ideal

[elfin wraith]

Possibly not the right terminology but this is what I mean

Proof validity is yet to be seen

[tough raven]

Oh

Assuming it satisfies some Ore condition

I see

[chilly radish]

Jacobson radical is not the intersection of maximal left ideals generally

It is the intersection of colon ideals

It is a two sided ideal and the intersection is not a priori

[elfin wraith]

Yeah the localisation in a domain is left ore

[next obsidian]

Isn’t the Jacobson radical the intersection of all maximal two-sided ideals

At least you can define it that way or something

[chilly radish]

Only in the commutative case

[next obsidian]

Cuz I swear I did a hw problem which showed that you can use left or right

[chilly radish]

Such ignorance of our ways

[next obsidian]

And there’s something about 1 +xy and equivalently 1 + yx being invertible

[void cosmos]

no

[void cosmos]

left-quasiregular yeah

[elfin wraith]

Hmm, can’t say I know what colon ideals are

[chilly radish]

You prove this using the Colon ideals

[void cosmos]

an element is right quasi-regualr if there exists y such that x+y-xy =0

the jacobson radical is the unique right quasi-regular ideal

maximal

(quasi-regular idela meaning every element is quasi regular)

[tough raven]

Of all maximal left or all maximal right ideals, yes.

[next obsidian]

This was written by me

[chilly radish]

Oh I'm dumb

[tough raven]

Wait is it?

[chilly radish]

It is the intersection

My bad

I checked

[next obsidian]

Chmowned

[chilly radish]

The (I:R) thing is to show it's 2 sided

[elfin wraith]

We defined it to be the set of elements which annihilate all simple left ideals but then showed its equivalent to the intersection of all left and right ideals

[next obsidian]

Not needed

[chilly radish]

Shut up

[next obsidian]

See my proof

[chilly radish]

I know how to prove this

I had a brain fart

[tough raven]

Is it true that x ∈ J(R) iff 1-yx is right-invertible for all y ∈ R?

I know it's true with left or two-sided invertible.

[next obsidian]

You can define it for modules too, but let’s say we’re just in a ring. (I:J) = {x in R| xJ < I}

[elfin wraith]

Ok phew Ive not been lied to by my lecturer and I know what the Jacobson radical is haha

[void cosmos]

yeah

thats kinda what i said in the ocmmutative cae

about left/right quasiregular

[tough raven]

My question is all about the non-commutativity though.

[tough raven]

.

[elfin wraith]

It’s 1-axb is a unit for all a,b in R in the noncom case

[void cosmos]

yeah

its still true

[tough raven]

What is true exactly?

[chilly radish]

Yes it's symmetric

Nothing in the theory is side dependent

[tough raven]

Um

[elfin wraith]

The set of all those x is the Jacobson radical also, many such definitions

[tough raven]

Can I see a proof?

Because here's what I know: TFAE:

(i) 1-yx left invertible for all y (ii) 1-yx two-sided invertible (iii) 1-xy right-invertible (iv) 1-xy two-sided invertible

[chilly radish]

You can actually see it directly

Give me a few minutes

[tough raven]

It is symmetric overall, but it doesn't include 1-yx right-invertible for all y as sufficient.

[elfin wraith]

Just realised I missed the word maximal in the last part of this proof, if I drop a mark for that I’ll be so upset

I feel it’s quite obviously a typo but still

[next obsidian]

They won’t drop a mark, they’ll just kill you don’t worry

[elfin wraith]

I would honestly prefer that at this point

[chilly radish]

Do you believe the jacobson radical is 2 sided?

Also do you know that x is in the radical implies 1+x is left invertible for one definition and right invertible for the other

I can also prove this

[tough raven]

Yes.

[chilly radish]

It is not iff sorry

[chilly radish]

Well then you just apply this

So I can prove this

[tough raven]

1+yx left or 1+xy right for all y, yes

[chilly radish]

Ok do you want a proof prior to seeing that the "left" radical and "right" radical coincide?

Oh ok

I misread your question

My bad

[tough raven]

Nvm I misremembered how it works; #foundations message

[chilly radish]

Yes

That is more familiar

Your question follows since xy and yx are both in J

[tough raven]

1-xy li for all y => 1-yx li for all y => 1-yx ri for all y => 1-xy ri for all y => 1-xy li for all y

So they're all equivalent

[crystal vale]

If I have to think about a finite non-abelian simple group of even order then how should I think about the construction of the group?

[woven panther]

Can someone help with this?

Let $F$ be a field and $f(x), g(x) \in F[x]$ be polynomials such that every root of $g(x) \in \bar{F}$ is also a root of $f(x)$. Show that $g(x) \mid f(x)$.

Is this claim true as stated?

[cloud walrus]

Bean Man

[tough raven]

Try assuming both f and g are products of polynomials of the form (X-a) (may be with different a's or repeated a's) to begin with.

Well, it shouldn't be hard to find the invariant factors of a cyclic module.

[woven panther]

Then

$$g(x) = c\Pi_{i=1}^n (x-a_i)$$

and
$$f(x) = d\Pi_{j=1}^m (x-b_j)$$

Where each $a_i$ is a (possibly repeated) root of $g(x)$

I'm not sure where to go from here

[cloud walrus]

Bean Man

[tough raven]

So what are the roots of g and what are the roots of f?

What's your definition of the invariant factors of a module?

[woven panther]

all a_i are roots of g and all b_j are roots of f

[tough raven]

So what restriction is placed on the ai and bj by requiring every root of g to be a root of f?

Right. So in your question, aren't the modules already in this form?

[woven panther]

Each a_i has an equivalent b_j? How do I know that if an a_i is repeated then a b_j is repeated?

[tough raven]

https://tenor.com/view/thats-the-neat-part-you-dont-invincible-gif-27194608

[woven panther]

If g has repeated roots then doesn't f have to have the same repeated roots for g to divide f?

[tough raven]

It does.

Yep, so the invariant factors of O/(4) are 4 and the elementary divisors are (1+i)^2, (1-i)^2, and same approach for the other cases.

[static quarry]

How would you go about determining the greatest element order of $\mathrm{SL}(3,\mathbb{F}_7)$?

[cloud walrus]

DW0987

[static quarry]

It is easy to compute that $|\mathrm{SL}(3,\mathbb{F}_7)|=(7^3-1)(7^3-7)(7^3-7^2)=19\times7^3\times3^4\times2^6$, but I am not too sure how to proceed from here...

[cloud walrus]

DW0987

[late matrix]

is there a common term for the kernel of the conjugation action on subgroups of G, im trying to figure out what it would be equal to

clearly its a superset of the centre of G

its equal to the intersection of all the normalisers

uhhhh centralizer only fixes one subgroup

[static quarry]

One question: can we do any better, by showing that we can take the intersection of the normalizer of subgroups of a certain type?

[late matrix]

also i think centraliser is for conjugation of elements not conjugation of subgroups

[late matrix]

lemme think about this

[late matrix]

im not coming up with anything really...

sorry for pushing up ur question btw lol

ill link it back after

[static quarry]

I am still thinking too, but one fact that I think might be relevant is this: for a group $G$, indexed family ${H_i}_{i\in I}$ of subgroups of $G$, and a homomorphism $\phi$ on $G$, $$\phi(\langle H_i | i\in I\rangle)=\langle(\phi(H_i) | i \in I\rangle,$$. A motivation for this is that, for each $x\in G$, $c_x:G\rightarrow G$ such that $c_x(y)=xyx^{-1}$ for all $y\in G$ is (as you might have already learnt or realized) a homomorphism (and in fact an isomorphism) on $G$; so, if $x\in G$ such that $c_x(H_i)=H_i$ for each $i\in I$, then $c_x(\langle H_i | i \in I\rangle)=\langle c_x(H_i) | i \in I\rangle=\langle H_i | i \in I\rangle$. In other words, if $x\in G$ 'normalizes' each $H_i$, then it normalizes the 'join' of the $H_i$'s.

[cloud walrus]

DW0987

[opaque finch]

Kinda 😅 . That's what I was trying to describe last night.

But one sees the fact that it is indeed an union of prime ideals. And this is enough to prove your case. It is enough to show that any maximal ideal is prime and I think you can show that

[static quarry]

see if you can use the comment above to show that you can in fact take the intersection of the normalizer of subgroups of a 'smaller' class...

[crystal vale]

I don't get how the unitial ring of order p^2 is commutative

Yes Z(R), centre of ring has order p or p^2

And if it has order p^2 then we get the result

But what if the Z(R) has the order p?

[dull ginkgo]

The underlying additive structure of R is an abelian group, and Z(R) is a subgroup. As mentioned assume |Z(R)| = p, then |R/Z(R)| = p, so R/Z(R) is cyclic as an abelian group generated by any nontrivial r in R/Z(R).

Consider how that describes the elements of R off of Z(R) ||For any r in R\Z(R), r + Z(R) generates R/Z(R), so every x in R takes the form nr + c, for some c in Z(R). ||

[south patrol]

uwu

[dull ginkgo]

Alternatively, consider how Z maps into every unital ring via the unit

[hollow topaz]

problem, then solution. what does prescription mean here?

[coral spindle]

This is not a technical word. They are saying "definition" or "requirement" essentially.

[hollow topaz]

so the question is basically asking to show that multiplication and addition satisfy the axioms and stuff?

[coral spindle]

The question is asking you to show that those definitions actually work, meaning that if you choose different numbers a' and b' such that [a] = [a'] and [b] = [b'] that you still get the same result under multiplication.

There are no 'axioms' that are required in the questions.

[hollow topaz]

oh so its asking to show a mod n + b mod n = (a + b) mod n?

[ashen heron]

Its defining + on Z_n that way

[coral spindle]

In other words, it is asking you to show that multiplication and addition in Z_n is "well-defined"

[hollow topaz]

thanks! i dont really get it but i think thats just because i shouldnt be doing this yet.

[dim wagon]

I am not very familiar with the subject at all but I was writing some notes as my prof was talking but i'm not sure if they are correct and im having trouble tracing it online. If $G$ is a finite abelian group, then we can define the character/group homomorphism $\chi: G \rightarrow \mathbb{C}^\times$. We can extend this character onto the group ring $\mathbb{Z}[G]$, $\chi': \mathbb{Z}[G] \rightarrow \mathbb{C}$ defined by $\chi'(\sum_{g \in G} c_gg)=\chi'(\sum_{g \in G} c_g\chi(g))$.

Let $A,B\le G$, then they exists as group elements in the group ring and we denote them by $x_A,x_B$.

[cloud walrus]

somethingwrong

somethingwrong

[dim wagon]

could anyone verify whether these are correct?

[hidden wind]

ok so after having stared at the tensor product for a while, am i right to understand they’re basically just a fancy way to talk about arguments of multilinear maps?

(up to equality under any multilinear map)

[slim kayak]

Yeah, that's one way to think about them

[hidden wind]

whew, thanku

[slim kayak]

In my experience they have more interesting for their usage with R-algebras, extensions of scalars and it being adjoint

[hidden wind]

yeah i’m getting to alternating tensors next

[dim widget]

those are the arguments of alternating multilinear maps

[hidden wind]

thanks seems almost too simple

[dim widget]

it is pretty simple!

[hidden wind]

i’ve been hearing the word tensor for years and with no explanation, it sounded so scary

[dim widget]

it's just a nice language to box these things up

[dim widget]

they're just like some higher generalization of vectors and matrices really, nothing too crazy to imagine.

There is a certain logic to calculating with them or proving things with them that some people find counter-intuitive at first, but over fields everything is pretty down to earth

[tardy hedge]

Why is it true that if n divides the order of a finite abelian group then it has a subgroup of order n

[dim widget]

Because a finite abelian group is a product of its sylow subgroups

So then it reduces to the statement that any p-group has a subgroup of any size dividing the order of the group

And this you can prove by induction and cauchy’s theorem

(It’s also usually part of the sylow theorems)

[slim kayak]

How did I never realize this 😭

It's so obvious in hindsight

[dim widget]

a related statement is that a group is cyclic iff for every d|n there is only one subgroup of order d

[tardy hedge]

Ty

[toxic zephyr]

just proved in class that in a PID any ascending chain of ideals eventually stops growing. is this a way to prove that all PIDs contain a maximal ideal? or what's the "purpose" of this property?

[dim widget]

this property is called being Noetherian. The point is not to construct maximal ideals. The property is an important finiteness condition in commutative algebra. The basic point of the property is to capture some element of our intuition from thinking with the integers etc. One of the most common ways it's used is that in a Noetherian ring a submodule of a finitely generated module is finitely generated, but there are many more.

[toxic zephyr]

yeah I had just realized that was Noetherian interesting

does it mean all ideals are finitely generated?

[dim widget]

yes

[toxic zephyr]

yooo that's poggers

[dim widget]

since they are submodules of R which is finitely generated

in fact it's equivalent to all ideals being finitely generated.

[toxic zephyr]

that makes sense, actually.

[dim widget]

Yeah the proof is pretty similar to what you did to prove all PIDs are Noetherian

it's a small generalization

(the proof of the hard direction I mean)

[toxic zephyr]

so just to clarify, a fg module doesn't necessarily have fg submodules? but if over Noetherian, then it is?

[dim widget]

yeah for instance $k[x_1, \dots, x_n, \dots]$ is a finitely generated module over itself, but the ideal $(x_1, x_2, \dots)$ is a submodule which is infinitely generated

[cloud walrus]

Math_Discord_Final_Girl

[toxic zephyr]

oh wow that's so interesting

[dim widget]

but over a Noetherian ring one can show that this doesn't happen (and it's almost a tautology that it doesnt happen for submodules of R itself)

[toxic zephyr]

rings and modules are dope

[toxic zephyr]

one more clarification: you're saying submodules or R, is that because ideals are just submodules of a ring R as an R module?

[dim widget]

yep

that's what an ideal is

[toxic zephyr]

okay cool just making sure

yesterday I was trying to use submodule logic to reason about ideals (but I was doing it wrong lol)

[dim widget]

neat!

[next obsidian]

All infinite sized submodules are infinitely generated, your generating set comes free with your submodule

[languid trellis]

I believe the splitting field to be $\mathbb{Q}( \sqrt[^5]{2}, e^{\frac{i \pi}{5}})$ and the dimensionality to be $[\mathbb{Q}( \sqrt[^5]{2}, e^{\frac{i \pi}{5})} : \mathbb{Q}] = 5 \cdot 4 = 20$ as $x^5 -2$ is irred over $\mathbb{Q}$ and the other roots are given by $\sqrt[^5]{2} \cdot \zeta$ where $\zeta$ is any fith root of unity. $e^{\frac{i \pi}{5}}$ has minimum polynomial $x^4+x^3+x^2+x+1$, hence the dimensionality. Can someone verify?

[cloud walrus]

swifteeee

[dim widget]

Everything is right but can you justify why the degree of $\mathbb{Q}(\sqrt[5]{2}, e^{i\pi/5})$ over $\mathbb{Q}$ is 20? I feel like the argument you gave is unclear/insufficient. It's true that the degree of each of the subfields you mentioned are 5 and 4 respectively

[cloud walrus]

Math_Discord_Final_Girl

[languid trellis]

$\mathbb{Q}(\sqrt[5]{2}, e^{i\pi/5})$ is an extension field of $\mathbb{Q}(\sqrt[5]{2})$, which is in turn an extension field of $\mathbb{Q}$. So, we can apply dimensionality formula, $[\mathbb{Q}(\sqrt[5]{2}, e^{i\pi/5}): \mathbb{Q} ] = [\mathbb{Q}(\sqrt[5]{2}, e^{i\pi/5}): \mathbb{Q}(\sqrt[5]{2})] \cdot [\mathbb{Q}(\sqrt[5]{2}) : \mathbb{Q}] = 5 \cdot 4$

[cloud walrus]

swifteeee

[dim widget]

how do you know the first degree is 4 and not smaller?

that's the point of my question

i.e. why is $[\mathbb{Q}(\sqrt[5]{2}, e^{i\pi/5}): \mathbb{Q}(\sqrt[5]{2})] = 4$

[cloud walrus]

Math_Discord_Final_Girl

[dim widget]

you argued why $[\mathbb{Q}(\sqrt[5]{2}): \mathbb{Q}]$ is $4$ but didn't explain why that is still the minimal polynomial over the larger field

[cloud walrus]

Math_Discord_Final_Girl

[dim widget]

for instance over $\mathbb{Q}(\sqrt{5})$ the minimal polynomial of $e^{2\pi i /5}$ is smaller

[cloud walrus]

Math_Discord_Final_Girl

[dim widget]

also by the way you keep writing $e^{\pi i/5}$ but this is a $5$th root of $-1$ not a 5th root of $1$, I think you want the latter one.

[cloud walrus]

Math_Discord_Final_Girl

[languid trellis]

Yes I mean e^(2i *pi/5)

[languid trellis]

Give me a moment to digest why this is the case

Over $\mathbb{Q}(\sqrt{5})$ it suffices to adjoin $\sqrt{\frac{5}{8} \pm \frac{\sqrt{5}}{8}}$

[cloud walrus]

swifteeee

[languid trellis]

No it does not

We need to adjoin $i \sqrt{\frac{5}{8} \pm \frac{\sqrt{5}}{8}}$

[cloud walrus]

swifteeee

[languid trellis]

Which presumably has a minimum polynomial g over Q(\sqrt(5)) such that deg g < 4

[dim widget]

perhaps, but I like to think of it like, $z = e^{2\pi i/5}$ satisfies the quadratic $z^2 - z \cdot (1 + \sqrt{5})/2 + 1$

[cloud walrus]

Math_Discord_Final_Girl

[languid trellis]

Where does this come from?

[dim widget]

calculation

[languid trellis]

lmao. fairs

[dim widget]

the roots of this polynomial are $z$ and its complex conjugate

[cloud walrus]

Math_Discord_Final_Girl

[dim widget]

and one can notice that it has coefficients defined over $\mathbb{Q}(\sqrt{5})$

[cloud walrus]

Math_Discord_Final_Girl

[languid trellis]

Gotcha

[dim widget]

I arrived at this because I know that $\mathbb{Q}(\sqrt{5})$ is the largest totally real subfield of $\mathbb{Q}(z)$, so it was easy to figure out what the minimal polynomial should be (since the extension is galois with galois group generated by complex conjugation)

[cloud walrus]

Math_Discord_Final_Girl

[languid trellis]

I haven't defined anything galois yet, but I'll come back to that when I have

But I see, we need to be careful. Now i need to figure out a way of arguing that my polynomial (x^4+x^3+x^2+x+1) is the minimum poly over Q(\sqrt[^5]{2})

[dim widget]

there is an easy way around this using divisibility stuff and being clever btw

here is a nice general proposition: let L, K be fields of degrees m, n over some base. Then the degree of LK over the base is at least the lcm of m, n and at most mn.

[languid trellis]

What do you mean by LK?

Field generated by L and K?

[dim widget]

the field generated by L and K

the smallest subfield containing them in some (any) algebraic closure up to isomorphism

[languid trellis]

My first thought was to use that F[x] is a pid if F is a field, and we have unique factorisation up to units in a pid. Now that you've brought this up, we have lcm(4,5) = 20 \leq [Q(\sqrt^5, e^(2pi i/5)] \leq 45 =20. Because [Q(e^2pi * i/5) : Q] =4 by cyclotomic polynomial magic

Then the dimensionality formula implies that the degree of the minimum poly of e^2pi i/5 over Q(5th root of 2) is 4

[dim widget]

yep

and this shows more generally that if the degree of the minimal polynomial of z over $\mathbb{Q}$ is coprime to the degree of $L$ over $\mathbb{Q}$, then this is also the minimal polynomial over $L$.

[cloud walrus]

Math_Discord_Final_Girl

[languid trellis]

Yeah. Nice stuff

I just need to convince myself (and prove) the bounds

[dim widget]

great 🙂 good luck

[languid trellis]

Thanks a ton, appreciate it!

I will probably be back here tomorrow with another exercise haha

[vivid kestrel]

so if we talk about a field extension and a field containing another field, do we just mean we have a field homo/monomorphism from the smaller field into the larger field? I guess what im asking is, what does it mean for a field to contain another field

[next obsidian]

I mean, they’re kinda equivalent

But when you say L > F, you quite literally mean that F is a subset of L

But if F and L exist abstractly, this is basically giving a choice of map F -> L which gives an isomorphism of F to some subfirld of L

[vivid kestrel]

i dont know, there is a part of my brain that feels that is not a sensible definition

[next obsidian]

It’s sensible given context. Like we say R < C all the time, and this is interpreted as embedding the reals as the complex numbers with imaginary part 0

This is sensible, even though you can embed R into C a lot of different ways

[vivid kestrel]

at least the way field constructions are often done, taking a field and factoring out an irreducible polynomial over the polynomial ring, we start out with our field and then use it to construct the extension, how can we say a set is contained in a set that was used to construct it, even though there is a "canonical" way in which we can imagine the field ot be embedded

[vivid kestrel]

so when its phrased like K being a subset of extension L we always mean that there is a (canonical?) way to embed K into L?

[next obsidian]

It’s kinda like, already coming with one

[vivid kestrel]

of course given L you could just take a literal subfield and then call L the extension but that is at least not how I've seen it framed usually

[next obsidian]

Like idk, do you think there’s a meaningful different between the real numbers R and the set {a + 0i} < C?

I don’t think there really is, if you know what you’re doing you don’t ever run into trouble

We identify objects all the time, if I take a coordinate ball U inside of a manifold M I write stuff down as if U was literally R^n

Because I know all the things I’m gonna do will port over through the homeomorphism in the way I want

[vivid kestrel]

hm okay

[next obsidian]

This used to bother me, but then after working it out explicitly and being so annoyed I gained a sense for how to just identity things

[vivid kestrel]

i suppose not

[next obsidian]

And now I implicitly do it

I think grappling with these things is a point a lot of people come to at some point

But pretty much everyone realizes that these identifications are fine and basically necessary if you don’t want to remain super confused

And if you remain against identifications you go invent type theory or whatever

[vivid kestrel]

hmm, okay thank you

[glad spoke]

this may be stupid, but what is (x-sqrt(2), y-sqrt(2)) \cap Q[x,y] equal to, where the first ideal is generated over the polynomial ring in two variables in the algebraic closure of Q

is it just (x^2-2, y^2-2) generated as an ideal in Q[x,y]?

if so, how do i show that the intersection lies in (x^2-2, y^2-2)

[lilac sluice]

guys, a random question: suppose that a homomorphism A\to B is surjective. if I apply an action in A by a group G, the function A/G \to B is still surjective?

[coral spindle]

Assuming it is well-defined, of course -- it would be constant on the orbits. This is really not saying anything about homomorphisms though.

[glad spoke]

i’m starting to doubt this lol, how do i do this

[rocky cloak]

You're missing x-y.

Best way is probably to think about it as the kernel of the map Q[x,y] -> R that sends both x and y to sqrt(2)

[glad spoke]

you’re saying it is (x^2-2, y^2-2, x-y)?

[rocky cloak]

Yes, or just
(x^2 - 2, x - y)
Would be enough

But those are the same

[glad spoke]

oh, how do you see that

[rocky cloak]

R[x, y]/(x - sq2, y - sq2) = R
By sending x and y to sq2.

So restricting to Q you just have the kernel of the map Q[x,y] -> R mapping them to sq2.

Clearly x-y is in there, so modding that out you just need to compute the kernel of Q[x] -> R, where x is mapped to sq2.

And x^2 - 2 is the minimal polynomial of sq2

[glad spoke]

ohh, i see, thank y97

you*

what if it wasn’t sqrt(2) and sqrt(2), but one of them is sqrt(3), does the proof still apply?

it wouldn’t, right? since x-y wouldn’t automatically be in the kernel

how would you go on abt that then

[rocky cloak]

Would go about it in the exact same way.

Q[x, y] -> R

y^2 - 3 is in there so

Q(sq3)[x] -> R

minimal polynomial of sq2 over Q(sq3) is still x^2 - 2

So you get (x^2 - 2, y^2 - 3)

[glad spoke]

ah okay, i see now, this makes sense, thank you

[crystal vale]

got it, thank you

[dull ginkgo]

Epic

[crystal vale]

canonical mapping?

[dull ginkgo]

ye

[chilly ocean]

We have a result that Z[\sqrt{-5}] is not Euclidean domain under norm N(a+b\sqrt{-5}) = a^2 + 5b^2, but how can we show that it is not an Euclidean domain under any norm?

[rocky cloak]

Well, using that norm you can show that 2, 3, 1+sq(-5), 1-sq(-5) are all irreducible.

But 2*3 = (1+sq(-5))(1-sq(-5)) so it's not a UFD

[chilly ocean]

I see, thank you

[vivid kestrel]

Given a Galois extension F/K, if every Galois Automorphism fixes a certain element l of L, can we conclude that l must be in k

[slim kayak]

Well, for any element not in K you have a minimal polynomial of degree at least two. The automorphism of the field generated by its root has a non-trivial automorphism and this automorphism can be extended to one over all of F.

Or if you already had galois group stuff before, K is exactly the fixed field of all automorphism that fix K on F. But this is probably a circular justification

[vivid kestrel]

Hm, sorry I don't know much Galois theory, they came up in something im studying

[slim kayak]

Oh well

[vivid kestrel]

That's helpful though, thank you

[slim kayak]

You can ofc repeat the step above to adjoin even more stuff to K. You can then use a zorns lemma argument that if you didn't have an automorphism on all of F then you could extend it further

[rocky cloak]

This is sometimes taken as the definition of Galois extension.

[vivid kestrel]

its probably silly to see it as such, but would it also be a consequence of the fundamental theorem of galois theory, since by that the largest subgroup corresponds to the smallest intermediate field? sorry for the ping

[dim wagon]

do these look correct?

[rocky cloak]

I mean, claim 1 is technically correct, but the second case never happens

[dim wagon]

whats the technicality here

For claim 2 wouldn't $C_2 \times C_2 ={(1,1),(r,1),(1,s)(r,s)}$ and $C_2= {(1,1),(r,1)}$ mean that $(C_2 \times C_2)/C_2 \cong C_2$

[cloud walrus]

somethingwrong

[lone niche]

Have you seen a meme that goes like "Upvote if you like pizza or if you support hitler"?, jagr is referring to Claim 1. and what follows after the or.

[dim wagon]

Ohh okay I think know what you mean. I tried letting c_2 be yhe set containing (1,1) and (r^2,s^2) and the quotient of that seemed to be c_2 c_2c_2, I will try again

any thoughts on claim 2?

Ohh okay wait

He meant claim 1 is true but the or not happens

Gets

[slim kayak]

No that's exactly what I meant, but it might be circular as a justification

But if you are fine with blackboxing, yeah. Just cite the fundamental theorem and be done

[tardy hedge]

How can we show theyre irreducible using that norm?

Because every other element has an equal or higher norm or smth?

[tardy hedge]

Ive actually never understood this properly either

In dummit and foote they showed its not a euclidean domain by showing its not a pid. To show its not a pid they used that specific norm

So how does that generalize to any norm?

[languid trellis]

For this exercise I'd like to argue as follows. x^p^e -1 = 0 has one solution in Z/(p), x = 1. For a \neq 1 in Z/(p), the order of a divides p-1. So, for a \neq 1 to solve x^p^e -1 =0 we need p^e to be an integer multiple of p-1, which, intuitively, can't occur (I can't quite justify this step).

So now, we construct a splitting field by adjoining a p^e'th root of unity. So our splitting field is Z/(p)(e^{2pi *i/p^e})

Can someone verify my answer?

[rotund aurora]

I don't think your reasoning is correct. But recall (x+y)^p=x^p+y^p

In the way that you were trying to do it: the order should divide both p^e and p-1, but these are coprime

[late matrix]

can 2 non-isomorphic groups have the same composition series?

[next obsidian]

Yeah

Or uh, what do you mean by “same composition series”

They can have the same composition factors

But if you ask for the series themselves to be the same (up to isomorphism) then G = G_0 = H_0 = H so G and H are the same

[dim widget]

If it's not a PID it doesn't matter if you've used a norm to show it (though I don't see why they would use this), being a PID is a ring theoretic property and doesn't depend on any extra structure. Rings which are not PIDs can never be euclidean with any norm.

[late matrix]

sorry yh this is what i meant

whats an example of this btw

[next obsidian]

Z_8 and Q_8

[late matrix]

maybe like D_2n and Z/2nZ?

ah

[void cosmos]

any two p groups

also the composition series 1 <| Z_2 <| Z_2 x Z_2 <| A_4 <| S_4 has factors Z_2,Z_2,Z_2,

you can try to compose a similar series for some Z_n

[late matrix]

what even is the normal subgroup symbol on latex

[void cosmos]

idk honestly idk any latex

[late matrix]

$\trianglelefteq$

[cloud walrus]

ark

[void cosmos]

yeah cool

[late matrix]

well ig here not eq

[languid trellis]

Thanks

[rocky cloak]

Yes, that's right.

The norm is multiplicative so if 2 = xy, then N(2) = N(x)N(y).

N(2) = 4, so if you can show that no element has N(x) = 2 you're done.

You'll notice that this proof uses the norm, but that the conclusion (2 is irreducible) doesn't depend on the norm.

[tardy hedge]

Yeah, idk, I think my prof is being unclear on how she asked this homework question. The question just says "show if R is commutative with identity, M (x) N is an R-module and the middle linear maps become bilinear"

Im assuming she wants to show that if R is commutative we can make the standard (R,R) bimodule structure on R (with rm = mr) and then put a left R mod structure on tensor

but middle linear maps "becoming" bilinear is ?

[rocky cloak]

Like the rm = mr condition is reasonable, but you still need something that pulls the r out to get bilinearity.

Maybe "middle linear" means
rf(m, n) = f(rm, n)
f(mr, n) = f(m, rn)
f(m, nr) = f(m, n)r
?

It's not really a term I've seen before.

[tardy hedge]

It also comes from dummit and foote, they say for M a right R-mod and N a left R-mod and L an abelian group, a map M x N -> L is middle linear if:

f(m1+m2, n) = f(m1,n)+f(m2,n)

f(m, n1+n2) = f(m,n1)+f(m,n2)

f(mr, n) = f(m, rn)

[rocky cloak]

Maybe just a mistake on your profs part then

[tardy hedge]

In terms of what exactly

Yea i feel like something is weird but im not totally sure what

[rocky cloak]

In terms of the exercise not being true I mean

[tardy hedge]

then the next question says: why do we need R commutative? and then some sketch of like (m , rn) = (mr, n) neq r(m,n)

[rocky cloak]

Like the main idea is that if R is commutative, then M(x)N is an R-module. This is true, and when R is noncommutative you can find examples where M(x)N don't have any module structure.

Then I guess they tried to translate these statements into bilinear maps and didn't think it all the way through

[tardy hedge]

When u say M(x)N is an R-module, is it a left or right? Or both and the actions coincide because R is commutative? (so u do the natural one)

[rocky cloak]

The natural one yeah

[tardy hedge]

ok im guessing then for this first question she wants me to just show that when R is commutative both M and N can have natural bimodule structures and then M(x)N can be considered as both a left and right R-mod with the left and right multiplications all coinciding and working nicely

[rocky cloak]

I mean you can always ask her, but that seems like a reasonable thing to take away

[tardy hedge]

For M(x)N as an abelian group, why cant we just define a left R-action as r(m (x) n) = mr (x) n

im trying to understand why we need either M or N to be a bimodule

if they are not bimodules, can we just define an action like that?

[slim kayak]

What's (rs)(m (x) n)

[tardy hedge]

= (m(rs) (x) n ) ?

if R is commutative does stuff work out though

[rocky cloak]

If R is commutative it works. That's just the same as endowing M with the symmetric bimodule structure (rm = mr)

For noncommutative you'll notice
r(sm) = msr which may be different from mrs

[tardy hedge]

oh yeah

[tardy hedge]

if A x B is isomorphic to A x C as groups ( or i guess other structures too but im just thinking about groups rn), is it always true that B iso to C?

[dim widget]

no

[tardy hedge]

Im grading this group theory hw and it was A x B iso to A x C for A,B,C finite abelian groups, then B iso to C. A lot of ppl said basically something like there is an isomorphism from A x B to A x C and then they like restricted it to go from B -> C and i dont think thats a correct argument ...

maybe it is im not an expert on this

[dim widget]

consider Z^{\oplus \infty} = A and B = Z^n, C = Z^m for m \neq n

[tardy hedge]

Yeah many ppl did not use the fact theyre finite abelian groups and thats a key point to use here isnt it

if u dont use that then whatever proof they write should be wrong

[dim widget]

Yep

[tardy hedge]

Thanks

[next obsidian]

I think it’s true when things are fg

By essentially the same proof

Structure theorem

[wraith cargo]

I think the Krull-Schmidt theorem is the most general form of this

[tardy hedge]

what is the Z^{\oplus \infty}?

[dim widget]

just the infinite direct sum of Z's

[next obsidian]

But apparently Walker showed that if A x C ≈ B x C and only C is finitely generated then A ≈ B

[dim widget]

also it's true for finite nonabelian groups but it's pretty difficult

just an interesting fact

[opaque finch]

How can I show that a group of order 700 is solvable?

[dim widget]

sylow theorems?

[opaque finch]

I tried to reduce to that case even.

[rocky cloak]

I mean sylow-5 subgroup normal, mod that out, sylow 7 subgroup normal, done

[opaque finch]

Yesss. OMG so stupid of me. Thanks

[static quarry]

If $G$ and $E$ are groups, and $f:E\rightarrow G$ a non-surjective homomorphism, then does there always exist a group $H$ such that the pullback $\mathrm{Hom_\mathbf{Grp}}(G,H)\rightarrow\mathrm{Hom_\mathbf{Grp}}(E, H)$ along $f$ is non-injective?

If f(E) is contained in a proper normal subgroup N of G, then the answer is yes: the trivial map from G to G/N and the canonical projection of G onto G/N are distinct homomorphisms from G to G/N since N is a proper normal subgroup of G, but the pullback maps them both to the trivial map from E to G/N (and thus is non-injective).

But what if f(E) is not contained in any proper normal subgroup of G?

[cloud walrus]

DW0987

[static quarry]

In that case, is it true that, for all groups $H$, for all homomorphisms $g_1$ and $g_2$ from $G$ to $H$, if $g_1$ and $g_2$ agree on $f(E)$, then they are equal?

[cloud walrus]

DW0987

[tardy hedge]

If you have a middle linear (R-balanced) map phi, then u have phi = psi o i (universal property tensor product thing) where psi is a group hom from tensor product, i is inclusion

If you now put the R module structure on tensor product, i is now bilinear and psi can be a an r mod hom, does that mean phi can be extended to be bilinear?

i mean if psi is r mod hom and i bilinear then psi o i is bilinear

Im still just hung up on my prof saying “middle linear maps become bilinear”

Jagr was helping me with this b4

[rocky cloak]

Yes, this is true. In fancy words: "any epimorphism of groups is surjective".

What you can do is pick H to be the permutation group of the cosets of f(E) and one extra point.

Then you construct the maps by messing with the action of G on G/f(E).

[rocky cloak]

So if R is commutative, and psi is R-linear (with respect to the standard module structure on the tensor product), then phi will be bilinear.

If psi isn't R-linear, then phi won't be.

[warm dove]

how can I solve this?

the direction "if G has such a series, it is solvable" is easy

but the reverse direction has been stumping me hard

I can't see a way to do it without some notion of "finiteness" somewhere

if G is finite we can induct

but idk a way to do it for infinite G

[lapis latch]

you cant do it for infinite G, a counterexample is the rationals with addition

[warm dove]

the rationals are solvable?

[lapis latch]

they're abelian

[warm dove]

oh of course

💀

wait what

ok

I see

yeah they're certainly solvable bc the factor groups are abelian and we can construct SOME subnormal series

[glossy crag]

Can an algebraically closed field have finite transcendence degree over its prime subfield? I don't see what should stop that, but neither do I know of any examples.

[lapis latch]

how about the algebraic closure of a prime field

[warm dove]

question - does a solvable group necessarily have its factor groups finite abelian or just abelian?

my textbook doesn't specify, but some online sources say it does...

[rocky cloak]

Taking the algebraic closure doesn't change the transcendence degree.

So for example the algebraic closure of Q has transcendence degree 0

[abstract spear]

How are the order of cyclic groups always prime when I can give a simple counter example Zn for any n.

[vast stump]

the order of a cyclic group is not always prime

[abstract spear]

My teacher used that in a proof

So idk what’s going on

[rocky cloak]

Simple cyclic groups are prime, so maybe that's it

[rocky cloak]

I think definitions vary, but I would go with not necessarily finite.

[warm dove]

👍

Got it, thank you.

So I think the theorem only holds for finite G

[rocky cloak]

What theorem is that?

[warm dove]

@rocky cloak

exercise in my Abstract Algebra book

[rocky cloak]

Yeah, G must be finite then

[abstract spear]

I swear my prof used my statement in a proof but she probably meant to say the converse

prime order groups are cyclic

[rocky cloak]

That is true at least. What was the proof of?

[warm dove]

Ok, that makes so much more sense 😭 I was stuck for like a day. These composition series are kinda weird :(( Thanks for the help!

[warm dove]

how can I do this w/o invoking derived series or commutators?

I have the solvable if and only if factors are of prime order

but I don't think that helps very much...

it should be some kind of induction but I don't see a way to extend normality up the tower

[dim widget]

what is your definition of solvable then

[warm dove]

Subnormal series w/ factor groups abelian

[dim widget]

okay well then isn't it a tautology?

[warm dove]

Wdym?

I know somehow factor groups abelian relates to commutators but the exercise on it is meant to be done later and this exercise is meant to be done first

[dim widget]

okay it's possible to do this without that

have you tried the case of a three step filtration

[warm dove]

??

what's that

is this like a formal technique?

(this is a first semester of algebra w/no prereqs so not too much background)

[dim widget]

3 steps meaning you have 0 < H < K < G

for the subnormal series

[warm dove]

Then, H abelian and normal in K, we need to find a subgroup abelian normal in G. Then, is H \cap K abelian and normal in G isnce K is normal in G?

no H \cap K is just H again which defeats the purpose

[dim widget]

yep

[warm dove]

hmm

is HK normal? yeah but it's not abelian

[dim widget]

HK is just K

[warm dove]

Ok, we need to use K/H is abelian somehow.

[dim widget]

yep that is crucial

you should try to show that \cap_{g \in G} gHg^{-1} is normal in K with abelian quotient

[warm dove]

looking through old hw, I found this theorem - given N normal in G, G/N abelian if and only if N contains the commutator subgroup

[dim widget]

I thought you didn

't want to use commutators lol

[warm dove]

yeah true

I would rather not

[warm dove]

This is normal by construction since conjugating by anything gives another conjugation g'Hg'^{-1} so it's closed.

so abelian quotient...call this N, then we have aN, bN in the quotient group

aNbN = abN = \cap_{g \in G} (ab) gHg^{-1}, but taking g^{-1}, this is equal to the right coset N (b^{_1}a)^{-1}

abN = \cap_{g \in G} (ab) gHg^{-1}, but this abgHg^{-1} should be equal to the coset g H g^{-1} b^{-1} a^{-1} = (g a^{-1} b^{-1}) H (b^{-1} a^{-1} g^{-1}) b^{-1} a^{-1} ?

no that doesn't get us anywhere

[hot pebble]

Context: there is some homomorphism of commutative rings pho:R to S and some maximal ideal I of S s.t. preimage of I is not maximal in R

I have successfully produced the example of inclusion from Z to Q, where <0> is maximal in Q but not maximal in Z.
Question: what if I add the condition that the preimage cannot be trivial? Is the claim still true?
What i have tried is stuff from Z[x], since PIDs fails by "prime ideal iff maximal ideal" and "preimage of prime ideal is prime ideal". However no luck so far so any help would be appreciated

pho i meant phi

[mighty kiln]

Z × Q → Q × Q, maximal ideal {(0, q) : q ∈ Q}

[woven trout]

does anyone know how to do part b?

We need the subfield to be unique up to set equivalence

I dont see how thats possible

By the universal property, at least one subfield of E exists that is isomorphic to the field of fractions of R

I dont get why R being a subset of that subfield would be necessary though?

[rocky cloak]

I mean exactly the same example works for Z[x] -> Q[x], since (x) is a maximal ideal.

[woven trout]

I should say, I see why at least one subfield F of E exists which contains R and is isomorphic to R's fraction field. I just dont see why its unique

[rocky cloak]

The statement isn't simply that it is isomorphic to a field of fractions of R. You should also use the inclusion R < F

[woven trout]

Is the question not true without that?

I interpereted it more as: If R is a subring of E, then its ring of fractions F can be embedded in E, and moreover this e,bedding is unique and contains R

[rocky cloak]

Consider R equal to Q[x1, x2, x3, ...] inside E = Q(x1, x2, ..., y1, y2, ...) for example

[rocky cloak]

If you also impose that the embedding restricts to the identity on R, then this is true yeah

[hot pebble]

thank you mr fox and jagr

[woven trout]

Youre saying the Fof of Q[x1,...] can be embedded into the subfield generated either by Q(x1,...,) or Q(y1,...)?

[rocky cloak]

Well, Q(y1, ...) doesn't contain R, but for example E and Q(x1, ...) are isomorphic

Their both isomorphic to the field of fractions of R

But only the latter is the "correct" one

[woven trout]

right

thanks

I was stuck so long lmao

if R has to be a subset makes way more sense

[glossy crag]

Do I have it correctly that |A[x_j:j\in J]|=max(|A|,|J|,aleph_0)? And the same holds for the rational function field.

[rotund aurora]

yes, except when J is empty

You are basically using that the set of all finite subsets of an infinite set K has the same cardinality as K

[glossy crag]

Yep, just double-checking.

[glossy crag]

Say L/K is a finite extension and x is K-transcendental, I want to show that |L(x):K(x)|=|L:K|. By breaking L/K into a tower we may assume that L=K(a) and since a certainly satisfies a K(x)-polynomial, we have |L(x):K(x)|\leq|L:K|, so it's enough to show {1,a,...,a^n-1} is linearly independent over K(x). If it were linearly dependent, then there would be non-zero polynomials f_j\in K[t] with 0=\sum_j f_j(x)a^j. Let F(t,s)=\sum_jf_j(t)s^j be the formal polynomial in K[t,s], then this is non-zero and its t-degree must be non-zero, else F(x,a)=0 would a K-linear relation for the powers of a, so we can write F(t,s)=\sum_i g_i(s)t^i with not all g_i zero. Now I'd like to conclude "but then \sum_i g_i(a)x^i=0 is a non-trivial L-polynomial relation for x, which is a contradiction since x remains transcendental over L", however I don't really know if g_i(a) are not all zero (e.g. what if each g_i is divisible by the minimal polynomial of a). What can I do at this point?

[rapid junco]

is the action on the set of all subroups of the same size by conjugation transtivie?

[rocky cloak]

Think about what happens with for example abelian groups

[rocky cloak]

But the gi all have degree less than the minimal polynomial of a, no?

[molten viper]

So, I want to make sure I'm on the right track for this problem.

Suppose $G$ is a group of order 75 with an element of order 25. I wish to show that $G$ must necessarily be cyclic.

Using Sylow Theory, I have concluded that there is exactly 1, or 25, subgroups of $G$ with order 3. We also conclude there is exactly 1 subgroup of order 25.

If there is exactly 1, I believe $G$ can be recognized as the direct product $C_{25} \times C_3$ (Both subgroups are normal)

If there are 25, I think this leads to a contradition. My idea is that if there were such a group, it would be a nontrivial product $C_{25} \rtimes C_3$, but there are no non-trivial homomorphisms $C_3 \rightarrow \operatorname{Aut}(C_{25})$.

Does this seem like it would work?

[cloud walrus]

Galstaff, Sorcerer of Light

[molten viper]

Update; it worked

[lime junco]

what chain can i hope to use here

the standard intersections i thought of dont work

and imkind of at a loss for what direction to go from here

[alpine island]

Consider A \cap B \subset A, try to just extend the chain of A further out to A \cap B.

Like you already have half a chain

[lime junco]

wdym

[lime junco]

im not sure what you mean by A and B here

[alpine island]

Where A, B are two arbitrary cool subgroups

[tough raven]

A more abstract (but probably the same when unfolded) proof: any basis of L over K is also a basis of L[X] (the abstract polynomial ring) over K[X], and in fact of L (⨯)_K K(X) = { p/q : p ∈ L[X], q in K[X] \ {0} } ⊆ L(X) over K(X). But in fact any non-zero polynomial in L[X] divides a non-zero polynomial in K[X] [proved at end], so L (⨯)_K K(X) = L(X), thus any basis of L over K is also a basis of L(X) over K(X). (Finally, since x transcendental over K implies x is transcendental over L, mapping X to x is an isomorphism from L(X) to L(x) restricting to an isomorphism from K(X) to K(x).)

Any non-zero polynomial in L[X] divides a non-zero polynomial in K[X] because L[X] is integral over K[X], since it is finitely generated as a K[X]-module.

[alpine island]

Let A > B > C be a tower of galois field extensions

and A > D > C some other intermediary field

is (A n D)/(B n D) galois?

[rocky cloak]

AnD is just D.

So just pick B=C

[alpine island]

True

Specifically this was for part b here

I thought the tower L_n cap K would work as a tower to show K was cool

[analog sedge]

What does Noether's theorem really mean? (also is Noether's theorem really in group theory? I'm not sure.)

Like, in Minecraft, ignoring numerical precision issues and chunk borders, there is a translational symmetry. However, the concept of momentum doesn't even exist in Minecraft. What does this actually mean?

[rocky cloak]

It does, but you'll have to use the hypothesis that the degrees are p to prove it.

[alpine island]

Yeah I'm hoping I can think of some way to do this.

I think it's trivial that deg (L_n cap K)/(L_n-1 cap K) is either p or 1

if it's 1 we're good and this gets culled out of the new tower

if it's p I need to find a theorem to throw at it

[rocky cloak]

So for a physical system if you take the kinetic energy minus the potential energy you get what is called the Lagrangian of the system.

Now the principal of least actions says that the system will developed so that the integral of the Lagrangian over time is locally minimized.

In that case you get a correspondence between (infinitesimal) symmetries, and preserved quantities.

As far as I'm aware Minecraft is not programmed to satisfy this, but idk how Minecraft physics work.

[analog sedge]

So, you do need a lagrangian mechanic for the Noether's theorem to work?

[rocky cloak]

That's my understanding of it anyway. But I'm not a physicist.

[alpine island]

I'm hoping to avoid dealing with minimal polynomials

I honestly don't have any leads on this

I bet this was one of the things covered in a lecture I missed, there were like 2 of those earlier in the term

unfortunately the prof's notes that he posts are near-unreadable

[analog sedge]

I thought Noether's theorem was a math theorem that applies to any "physical system" (formally defined somehow in math), and not just the world we happened to live in.

It's called a theorem, not a law.

[rocky cloak]

Well, you can think of it as a calculus of variations theorem. So anytime you're minimizing the integral of something you'll get something similar

[alpine island]

any hints on this?

[slim cipher]

Hi does anyone know how the highlighted portion is obtained? I think it somehow follows from Theorem 2.9.2 where K=E but Theorem 2.9.2 uses F^bar instead of E^bar, so am a little confused.

[glossy crag]

Hm, do they?

[dim widget]

If E is Galois then every embedding of E into an algebraic closure of F has the same image

So different embeddings just differ by an automorphism of E

[rocky cloak]

Well no power of s bigger than n-1 ever apears, so they have degree < n

[glossy crag]

Yeah, you're right, that was silly of me. Thanks!

[tough raven]

Noether's Theorem is about continuous symmetries, so I'm not sure you can apply it to Minecraft.

[chilly ocean]

Does f(x) being non scalar just mean it has degree > 0?

This is the first time my instructor has written this I just wish he kept the terminology consistent

[lusty marlin]

Yes

[chilly ocean]

I just got thrown for a loop because I've written down "of degree at least 1" about 10 times now and I have no idea why he would use different phrasing here

[warm dove]

Yes. That doesn’t mean there aren’t continuous conserved quantities or anything, but you can’t extract them from the Lagrangian. The easiest source of examples for this may be to look at non-Lagrangian Quantum Field Theories - I’m only vaguely informed on these (my research deals with transformations of the theory preserving the matter fields instead of conserved quantities within the theory).

Oh I didn’t respond

[warm dove]

For you

[elfin wraith]

It’s relatively standard wording, I think probably from linear algebra. Elements from your ground field are called scalars because they just scale vectors

[warm dove]

https://hypixel.net/threads/the-laws-of-minecraft-physics.5640116/

[elfin wraith]

Elements of degree >0 is just a bit more clunky I guess

[chilly ocean]

Yeah that makes sense, I've heard similar terminology before just not in this class

[warm dove]

Thread on Minecraft physics

Some interesting observations

[chilly ocean]

So to solve this I am guessing the minimal polynomial (my guess is (x^2-1)^4-6) and now I need to show it is irreducible over Q<sqrt(2)+sqrt(3)>. Is this the correct approach? If so, any tips on showing irreducibility over this field? I've tried looking at it by setting x = a+b(sqrt2+sqrt3) where a and b are in Q but I just get an absurdly awful equation I have no idea how to solve. Maybe by polynomial solution skills are just bad but I don't see any easy way to do this

This is a homework problem so I don't need a full solution obviously, I would just appreciate some hints/tips

I'm trying to use the linear independence of the normal a terms and the terms mixed with b(sqrt...) to get a system of equations but my problem is that they are all under this power of 4 so in order to do that I'd have to expand it all out and get a truly awful expression

[tardy hedge]

If I want to find all subgroups of order <= 4 in the group Z4 x Z4, what framework should I be thinking in about this problem?

I guess firstly you would know by lagranges that you only have to look at possible subgroups of order 2, 4.

Ok, I know <(2,2)> is a subgroup of order 2. How would I know if there are other subgroups of order 2?

I guess in this case do you just look manually for elements of order 2... ?

which there is only one

nevermind no

<(0,2)> and <(2,0)>

[molten viper]

I think one thing to note is that Z4 x Z4 only has elements of order 1, 2, and 4

It does have subgroups of order 8 though (2 of them)

[tardy hedge]

Right

[molten viper]

This may be a bit much to use, but every subgroup of Z4 x Z4 is of the form A x B, where A and B are subgroups of Z4

[tardy hedge]

How do we know tha

that

[molten viper]

its a consequence of Z4 x Z4 being abelian, but its also a hard fact to prove probably

yeah don't use that fact

[tardy hedge]

f.g abelian group thing?

[molten viper]

yknow I might be wrong

[tardy hedge]

Why did you think of it?

[molten viper]

So, yeah Z4 x Z4 is a finitely generated abelian group, so each of its subgroups are fg. abelian also. That means they are isomorphic to Zp1 x Zp2 x Zp3 x ... x Zpn, is part of the thought process

also you have some other facts

but a better question is like

elements of Z4 x Z4 are of the form (a,b), if (a,b) is in a subgroup of order 4 or 2, what values can a and b take

eg. can they both be odd?

what if a is odd?

probably work in cases

It may also help to know there are exactly 2 groups of order 4 up to isomorphism, and 1 group of order 2

[tough raven]

order((g, h)) = lcm(order(g), order(h)) is probably helpful.

[molten viper]

(In an abelian group)

[tardy hedge]

Why?

[coral spindle]

No this happens in any product of two groups, Abelian or otherwise.

[molten viper]

oh I see, I misread what ragh wrote

I blame dark mode

[molten viper]

its a good exercise to prove that fact (for any product of groups, to be clear) on your own

[tardy hedge]

yea i was asking about why u said only for abelian group

[molten viper]

I was wrong :P

[tardy hedge]

poop

[molten viper]

In an abelian group, you have the similar looking fact order(gh) = lcm(order(g),order(h)) iirc

[coral spindle]

Well that's also false

e.g. g = 1 and h = -1 in Z/nZ

[molten viper]

bah

Today is not my day huh

[coral spindle]

You get that order(gh) | lcm(order(g), order(h)) though.

[molten viper]

That's probably what I meant to say

idk

[sonic coral]

dark mode again

[molten viper]

true!

[tardy hedge]

Lol

[molten viper]

Well yeah, use the fact Ragh gave

I'm currently in semidirect product hell so my brain is not working

[tardy hedge]

yea ive just paused because im trying to understand orders in cyclic groups again. Im not gonna lie that shit has lowkey always haunted me

[molten viper]

I'm also doing that currently

kinda

[tardy hedge]

are you in a group theory class rn?

[molten viper]

Yeah

[tardy hedge]

cool

[molten viper]

though we've just finished groups and are doing rings

intro to

It's intro algebra, but the first sem is 90% groups

then 10% rings, and second sem is rings, modules, and fields as one might expect

[molten viper]

Really this is about understanding some number theory, I would say

[tardy hedge]

Yes

[fluid kelp]

How would one go about a prove capelli's lemma?

[hidden wind]

abelian groups my beloved

[tardy hedge]

Y

[hidden wind]

they are nice

[dim widget]

what is it?

[languid trellis]

Can someone give me a small hint on how to get started here?

[tardy hedge]

yea true

[tough raven]

Extend the monomorphism one element (i.e., simple extension) at a time, using the fact that E is normal.

You might need that E/K is normal.

[languid trellis]

What do you mean by normal?

[tough raven]

E/F is normal if every polynomial with coefficients in F that has a zero in E completely splits into linear factors over E.

[languid trellis]

I see

[languid trellis]

E is normal, so it contains all the roots of f(x), and is generated by F and the roots, r1,...rn. Suppose we have a monomorphism K/F -> E/F. K is a subfield of E, and it is an extension field of F, so it is generated by F and some of the roots of f(x).

Consider f(x) \in K[x]. As f has roots in K, it is reducible, so f(x) = (x-ri) ... (x-rj)g(x), for some irreducible g(x) in K[x]. Let r* be a root of g(x). E is an overfield, a fortiori overring of K, so we can extend our monomorphism K/F ->E/F to a homormophism K[x] -> E, where h(x) -> h(r*). Then the kernel contains (g(x)), so we have the induced map K[x]/(g(x)) -> E, and the map K[x]/(g(x)) - > K(r*). The latter map is an isomorphism of fields as g(x) is irreducible, so we may invert it and compose to get a monomorphism K(r*) -> E. We may then do the same procedure to K(r*), and this process will eventually terminate because we have a finite number of roots, at most deg f, which will give us an automorphism of E.

Some argument like this?

[elfin wraith]

Let $K$ be an algebraically closed field of char 0. Let $J$ be an ideal of $K[x,y]$ generated by $x^2+y^2-1$ and $x^2-y$. Show that $J =\sqrt{J}$

From the nullstellensatz we’ve got that $\sqrt{J}=I(V(J))$ and I’ve shown that $V(J) ={(\alpha_i,\alpha_i^2):i\in{1,\ldots,4}}$
with the $\alpha_i$s being the 4 distinct roots of $x^4+x^2-1$ but im kinda unsure how to go on from here

[cloud walrus]

Nope

[elfin wraith]

We’ve got that I(V(J)) is the intersection of <x_1-alpha_i,x_2-alpha_i^2> but I’m not sure if this does much for me

We’ve also got that K[x,y] is a UFD so maybe some irreducibility argument to get a prime ideal? (Since prime -> semiprime)

First time doing a problem with variety’s and stuff so not really seen many methods to go about this yet (comalg lecturer didn’t want to do anything geometric) so I’d appreciate a push in the right direction!

[tardy hedge]

I did not know that <(1,2)> = <(3,2)> in Z4 x Z4 at first glance

how could i tell? Is there any way besdies just writing out the elements and seeing they are the same?

[bitter rover]

Bezout's theorem

[tardy hedge]

what is that? heard of it

is it a gcd thing?

[bitter rover]

https://en.m.wikipedia.org/wiki/Bézout's_identity

[tardy hedge]

how can i use that here

[bitter rover]

That's sometimes called Bezout's theorem in this context, but there's an unrelated theorem called Bezout's theorem in algebraic geometry.

[bitter rover]

The second half, in particular.

[tough raven]

It's not true that K must be generated by some of the roots of f. It's also not true that if K is generated by some of the roots of f that f must factor as shown over K - intuitively, g could factor further without factoring all the way into linear factors. So there are some problems in the details, but I'm not sure whether the idea works or not yet.

[languid trellis]

I don't see how K could fail to be generated by some roots of f. Do you have an example?

[warm dove]

A perhaps easier way to see this is to notice that each component is kind of independent, I.e the group generated by <g,h> is <g> x <h> in the original group

1 and 3 both generate Z_4

2 generates Z_2 in both

[tough raven]

I think this is missing the key justification that the kernel is (g). This is actually false as stated: you need to map X to a root s* of sigma(g) rather than a root of g, where sigma is the monomorphism of K into E (why?). Only then will you be able to extend (by mapping r* to s*).

[tardy hedge]

Maybe silly question but is A x B isomorphic to B x A always?

you can just swap right

[bitter rover]

Prove that's an isomorphism. It's obviously bijective.

[languid trellis]

I've got a counterexample: Consider (x^2-2)(x^2-3) over Q, then the splitting field is Q(\sqrt2, \sqrt3), and we have Q(\sqrt(6)) which is a subfield of the splitting field, yet isn't generated by any of the roots of the poly

[tardy hedge]

not sure

[bitter rover]

Or you can just notice that (3,2) = 3*(1,2) modulo 4

[languid trellis]

I'm not quite sure why we need X to a root s* or sigma(g). This definitely feels more right for sure. I've become more convinced that my characterisation of g is also wrong - that f must factor in the way given over K, again, taking h(x) = (x^2-2)(x^2-3) over Q, then in Q(\sqrt{6}), h(x) is the same, it doesn't reduce further.

[tough raven]

Yep, that's the failure of K to be generated by roots of the same f that you used to generate E.

[tardy hedge]

Z4 x Z4 / <(2,2)> cant be cyclic because there isnt even an element of order 8 in Z4 x Z4 right?

[tough raven]

Check what it means for g to be in the kernel of your candidate homomorphism.

Don't forget that you want to extend sigma, so your map on K[X] should be sigma on K.

[languid trellis]

What do you mean by g? Some polynomial in K[x] that divides f(x)?

[tough raven]

Right.

Well, you want to extend sigma, so take any r* in E but not in K and let g be its minimal polynomial over K.

You want to pick some s* in E so that mapping r* to s* defines a valid extension of sigma to K(r*).

[languid trellis]

I'm slightly confused. Could you be more specific about how and why this approach goes wrong? I still don't see why we need X to map to a root s* of sigma(g), when sigma(g(x)) = g(r*) = 0 by construction.

We have a monomorphism sigma : K -> E. Let r* be in E, such that r* not in K, let g(x) be it's minimum polynomial in K[x]. We have a theorem where we can extend sigma to a hom sigma* :K[x]->E, such that the restriction to K of sigma* is equal to sigma, and such that x -> r*. Then, the kernel of this map is (g(x)), by definition, as g(x) -> g(r*) = 0. ( I really don't see what's going wrong here), so this induces a map K[x]/(g(x)) -> E which is an extension of sigma in the sense that we can identify K with a subfield of K[x]/(g(x)), namely the cosets k + (g(x)), k \in K.

[elfin wraith]

Just bumping this because I’m still having no ideas, I’ve really no clue how to proceed

[regal zodiac]

If i have a class function f from a group G -> C, is it true that f(abc) = f(acb)?

[coral spindle]

You are asking if abc and acb are always conjugate, and the answer is no.

[elfin wraith]

Ok so I think we have that J \subset I(V(J)) since the variety is specifically the common zeros of the polynomials that generate J, so we’d just need to show the other inclusion, I.e. that every polynomial which is 0 at the points in the variety are just some linear combination of x^2+y^2-1 and x^2-y

[coral spindle]

Now an example is annoying to find

[elfin wraith]

So do I need to do some horrible gröbner basis stuff?

But even then I’ve really no idea how to go about that

[elfin wraith]

Ok I’m not actually doubting this because I think that I’m implying every ideal is contained within its radical and that’s not true so I think I’m just utterly lost here

[elfin wraith]

Could I consider say K[x,y]/(x^2-y] \cong K[x] and then consider J to be the ideal <x^4+x^2-1>, because I think this is irreducible? And then if it is, well we’re in a UFD so irreducible iff prime and we must have that J is a prime ideal?

Wait no of course it’s not irreducible K is algebraically closed so it has roots

I guess I give up for now (and should have hours ago) but if anyone can point me in the right direction I’d definitely take the hint

[rocky cloak]

I think the easiest is just to calculate K[x, y]/J, then use that J is radical iff the quotient ring is reduced

[rocky cloak]

Right, that's exactly what you did here.

Then you see that x^4 + x^2 - 1 is seperable, so its all good

[elfin wraith]

I’m not familiar with separable polynomials, but I’m not immediately seeing how we know that we have no nilpotent elements in the factor ring

[rocky cloak]

Seperable means no repeated roots, so in this case it means the polynomial has no repeated prime factors

[elfin wraith]

Hmm ok, I think I’ll need to do some thinking on that, not come across those exact ideas yet but I’ll give it another go tomorrow

At least I’ve got an avenue of approach now, I wasn’t really getting anywhere with the variety thing

[rocky cloak]

Alternatively, by Chinese remainder theorem the quotient ring is just K^4

[tough raven]

The assertion that sigma* maps g to g(r*) is wrong. Write g = g_m X^m + ... + g_0 and compute its image under sigma* carefully.

[tough raven]

Take a = c^{-1} b^{-1} and d = 1 to see that f(1) = f(c^{-1} b^{-1} c b), whereas in general c^{-1} b^{-1} c b is not conjugate to 1 unless b, c already commute. So this is only true if G is already abelian.

(BTW, a function satisfies f(abcd) = f(acbd) for all a, b, c, d iff it factors through the quotient map G → G/[G, G], i.e., f(g) = f(h) whenever g^{-1}h is a product of commutators, whereas if g, h are conjugate, g^{-1} h must be a single commutator.)

[elfin wraith]

Is this just by the fact that we can write x^4+x^2-1 as the product of degree one factors, each of them is maximal? Then we have K[x]/(x-a) \cong K, field so maximal ideal hence prime hence semi prime?

Also im realising I don’t actually know that the 4 roots of the polynomial are distinct, I actually just assumed that. It makes some amount of sense to me because like the algebraic closure of Q should be the smallest algebraically closed field of char 0, and this is a polynomial with integer coefficients, but I’ve not taken galois or anything so I don’t actually know any field theory to justify that. Is there any like purely ring theoretic way to deduce that?

[rocky cloak]

So you could just solve this polynomial over Q like you said,

But in general a polynomial has a repeated root iff it has a common factor with its (formal) derivative.

So you can use Euclid's algorithm to determine it.

[elfin wraith]

Ah that makes sense, still though none of that is at all assumed knowledge for this course so feels like an oversight perhaps, unless the variety method leads to a solution where you only need to assume the existence of a root, but I don’t think that is the case

If we had repeated roots I’m not seeing exactly what would break here but i think something would

[elfin wraith]

Also I suppose just to clarify some places where I was getting confused, what is the issue here?

I don’t think that my logic for the inclusion is wrong, but I’m pretty sure the result is, so clearly I’m missing something

[rocky cloak]

Every ideal is contained in its radical yes. That's immediate from the definition

[rocky cloak]

Well just think about the polynomial x^2 for example.

Then x will be nilpotent in the quotient ring.

[elfin wraith]

Ok I guess that’s something I’ll need to prove tomorrow that wasn’t clear to me (but it’s also almost 3am so I might just be being dumb)

I mean I guess I have given a proof but by the definition

[rocky cloak]

Well the definition of the radical is just that x^n is in J for some n. Certainly x^1 is in J if x is in J

And also x^n for every other n for that matter

[elfin wraith]

Yeah no of course, I was way over complicating that

And just as a last thing, do you reckon showing the inclusion I(V(J)) in J is just a dead end or is there a way to go about that? I see the CRT proof and that’s very nice I think that’s by far the slickest way to do this, but just since we’ve not really done variety’s in the course I wanted to have a go using them

Not like an actual solution or anything but was the idea along the right lines do you think? I’ll probably keep thinking about it to see if I can do it that way too but I honestly don’t know if it’s just a total dead end

[slim cipher]

But these are embeddings into the algebraic closure of F and not the algebraic closure of E, or should that not make a difference?

[late matrix]

maybe im not understanding but i have 2 questions

1. how exactly would solving (i) and (ii) (see screenshot) classify all finite groups
2. how does JHT help, it only says that each group has a unique composition series but not the other way around (a composition series does not determine a unique group)

i still dont get why simple groups should be the "building blocks" of all finite groups or what "put together" even means

is that like direct and semidirect products

[late matrix]

apparently this is called the "extension problem", time to learn what an extension is i guess

ah so the extension problem is the converse of JHT sorta
is it basically asking, "given a composition series, find all groups with this composition series" and that is somehow solved using a thing called extensions which i havent fully learnt yet

[prisma ibex]

extensions are just short exact sequences of groups:

0->K->G->H->0

which just means that K->G is an injective homomorphism, G->H is a surjective homomorphism, and the composition K->G->H is zero

isomorphism classes of such extensions are classified by a group called Ext^1(H,K)

two such extensions 0->K->G->H->0 and 0->K->G'->H->0 are isomorphic if you have a group isomorphism G->G' yielding the following commutative diagram

the split extensions are exactly the semidirect products of groups

an extension 0->K->G->H->0 is split if you have a homomorphism s:H->G such that the composition H->G->H is the identity

so in general extensions are more general than semidirect products, not all extensions are split like this

every finite group can be written as an iterated extension of finite simple groups by composition series

so the problem of classifying finite groups comes down to classifying finite simple groups, and then understanding all the possible extensions between finite simple groups

[dim widget]

E contains F so an algebraic closure of E is also one for F

[slim cipher]

Oh ok thanks 👍

[languid trellis]

Our monomorphism K-> E need not be the inclusion. So sigma* as construced maps g= g_m X^m + ... + g_0 to σ(g_m)r*^m + ... + σ(g_1)r* + σ(g_0) = σ(g(r*)) (So (g) also need not be the kernel). So I need a root of σ(g(x)) in E, call it s*. Define a map F(r*) -> E by r* -> s*, and σ everywhere else.

[tough raven]

You do need to justify that s* exists. (Also nitpick: it's sigma on K, not everywhere else - for a general element of K(r*) written as a polynomial h(r*) in r*, it's mapped to sigma(h)(s*).)

[languid trellis]

Oh christ, I'm a little stumped now. What I can say is that sigma*(r*) will solve sigma(g)(x), but this relies on the existence of a mapping sigma* already. I have no ideas sorry

[tough raven]

You haven't yet used that E is normal.

[dim widget]

classic fresher

[languid trellis]

E being normal shows that f(x) splits in E[x], but this doesn't mean that sigma(g)(x) should split?

At least I'm not convinced that it should

[dim widget]

What’s the original statement you’re trying to prove

[languid trellis]

Soo uncalled for

[languid trellis]

#groups-rings-fields message

sigma : K -> F is the given monomorphism.

[dim widget]

okay seems like your approach is fine then

given a subfield K of E, E is the splitting field of a minimal poly of a root of f over K

so you have some g(x)|f(x) and can extend to K’ which is K(a) where a is a root of g

now repeat this process for all of the other roots of g

Finally you get E \to E

unless your definition of splitting field is not minimal

then you have to deal with the case that E is infinite and use Zorn’s lemma

[languid trellis]

This tacitly relies on the given monomorphism being the inclusion. Suppose it is not, but suppose we have extended our given monomorphism sigma. Then we have a map σ*: K(a) -> E, where a is mapped to a root of sigma(g)(x). That a root of sigma(g)(x) exists in E is the issue at hand

[dim widget]

There’s no such thing as "the inclusion"

any monomorphism is an inclusion

[tough raven]

E being normal implies that every polynomial with coefficients which has a root in E, completely splits over E. This isn't obvious at all but it is true, and I think you need to use it.

[languid trellis]

By "the inclusion" i mean the monomorphism that is the identity on K. as in σ:K -> E given by σ(k) = k

[dim widget]

that’s not well defined unless you’ve already fixed an embedding of K into E ahead of time, in which case it’s a tautology

[tough raven]

I don't think so - then E need not be normal, and the conclusion would be false. Also @languid trellis mentioned E being generated by the roots of f, so I assume that's part of their definition.

[dim widget]

I was assuming that the other definition of splitting field would just be a normal extension

Which splits f

[tough raven]

This one appears to be "field extension where f splits and is generated by the roots of f".

[dim widget]

that makes sense

okay then no axiom of choice

[languid trellis]

This is the definition given, to clarify the matter

[dim widget]

yeah so the smallest normal extension containing a root of f

[tardy hedge]

Im not sure why we can say, in particular n >= m

[dim widget]

seems like a long proof

[languid trellis]

I should be clearer. At this point I'm not even convinced that \sigma(g)(x) should have a root in E

[dim widget]

the fact that n>=m is a consequence of the fact that the b_I are part of a basis and all bases have the same cardinality

[dim widget]

What is g

[tough raven]

Yes, that's fair. You need (I think) to use the property of E being normal to find that root.

[languid trellis]

The approach is to do as you said, to "extend adjoining an element at a time", so we are choosing r* in E such that r* is not in K, and g is the minimum polynomial of r*.

[dim widget]

okay but then r* is a root by definition

[tough raven]

This implies in particular that if h2 | h1 are polynomials, and h1 has a root in E, then h2 also has a root in E. You should apply this with h2 = sigma(g).

[dim widget]

I think your problem arises from thinking of K as having one "privileged" embedding into E to begin with

[languid trellis]

Yes it is a root of g, but we are looking to extend sigma, so we are looking for roots of \sigma(g)(x)

[dim widget]

whereas you just have the one you’re given, and you consider E as a K algebra along that embedding

[tough raven]

I don't think so. There are two embeddings of K into E. So one of them is being used to identify K as a subfield of E, but the other one is then not an inclusion.

[dim widget]

Ah okay but then what is the point of the extra embedding? It’s irrelevant to the problem

[tough raven]

In terms of the pair of embeddings, you want an automorphism of E which forms a commutative triangle with the two embeddings.

[dim widget]

Ah I see

[tough raven]

In terms of the subfield-and-embedding formulation, you want to extend sigma: K → E to an automorphism of E along the inclusion of K in E.

[dim widget]

Okay well both are factors of f

So it’s clear that if g has a root so does sigma(g)

[tough raven]

Well that's much simpler than my approach.

[dim widget]

I think it’s he same now that I see

what you wrote

[tough raven]

No, I wanted to take some random r* in E \ K and use that E is normal to find the image. This way you can just use the fact that f splits directly (as opposed to any polynomial with a root in E splitting).

[tardy hedge]

They proved that as a corollary to this on the next page tho 🤷‍♂️

[dim widget]

well I guess regardless it follows from the statement

[languid trellis]

I really am struggling to see this. Take (x^2-2)(x^2-3) over Q[x]. The splitting field is Q(\sqrt{2}, \sqrt{3}). The splitting field contains \sqrt(6). The minimum polynomial of \sqrt(6) over Q[x] is x^2 -6, but x^2-6 doesn't divide (x^2-2)(x^2-3), in fact (x^2-2)(x^2-3) = (x^2-6)(x^2+1) + 12. What am I missing/misunderstanding?

[dim widget]

it’s implicitly implied that n >= m

[dim widget]

you’re just choosing two completely unrelated polynomials

[tardy hedge]

That statements shows we cant have a set of more than n lin independent elements ?

[dim widget]

yes because they all fit into a basis {b_1, …, b_m, a_{m+1},…,a_n}

so implicitly m <=n

[tardy hedge]

Ohh

Thanks

[dim widget]

in our case f is the minimal polynomial of some element and g is the minimal polynomial of the same element over some sub extension

so is \sigma(g)

[languid trellis]

I can't lie to you, I'm really struggling here. Would you be able to explain a little more fully why this is the case?