groups-rings-fields

[dim widget]

which part?

that \sigma(g) is the minimal polynomial of one of the roots of f?

[languid trellis]

Well I thought that g is by definition the minimal poly of some r* in E \ K, which means it can be the product of the roots of f, for example, so I'm not sure why is should divide f.

[dim widget]

Well in what we’re doing we want r^* to be one of the roots of f

that’s the point

[languid trellis]

Ok. Then i see why g should divide f

[dim widget]

Since the information we have only tells us that the roots of f generate E we should use that

[languid trellis]

Why sigma(g) should divide f is less clear

[dim widget]

okay well \sigma(g) divides \sigma(f) no?

now do you see how to conclude?

[languid trellis]

Yes, I agree, because we can apply sigma to f(x) = g(x)h(x) for some h(x) \in F[x].

And no I don't see how to conclude

Presumably, sigma(f) divides f, but I am only saying this because it gets me the conclusion I need, not because I actually believe it

[dim widget]

\sigma(f) is f!

Because \sigma is assumed to be an F-homomorphism

[languid trellis]

I can't lie to you, I'm quite embarrassed to say that I still don't get it

By F-homomorphism I'm understanding that sigma is a homomorphism of the F-vector spaces K and E?

And perhaps we have that linear maps map 0 to 0?

I've spent like 3 hours today thinking about field theory stuff my brain is so broken

[tough raven]

sigma is a F-homomorphism means that sigma is the identity on F.

[languid trellis]

Oh my god

obviously

sigma(f) being f is clear

Ok well that solves it

@tough raven @dim widget I appreciate how patient both of you have been with me. Thank you both for the help ❤️

[void cosmos]

yw

[tardy hedge]

Whats an example of a free R-module with a submodule that is not free?

[rocky cloak]

R = Z/4 for example

Or R = k[x, y] if you want a domain

[tardy hedge]

Oh its free as a Z/4-module

Like any ring

The rank of an R-module with R a domain is always well-defined, like its unique?

Im not sure if dummit and foote proved this. It feels really subtle now with all the slight variation

[rocky cloak]

For a domain you can just tensor with the field of fractions, so then it's just a question about dimension

And dimension is indeed well defined

[tardy hedge]

When u do that it becomes a vector space over the field of fractions (extension of scalars kind of idea right?) But why is this dimension the same as the original rank

Actually I did see a similar thing like this in the procedeeing proof

where they embedded into field of fractions and then used a vector space argument

When you have an relation in this vector space, they "cleared denominators" to get a relation in the original R-module

[rocky cloak]

Yeah, linearly independent sets map to linearly independent sets, and conversely you can clear denominators

[languid trellis]

Can someone give me a small hint here? I'm struggling to get started. Also, the definition of "seperable" being used is "A polynomial in F[x] is called seperable if its irreducible factors have distinct roots." So, for example x^-p -t in (Z/(p))(t) isn't separable, because it is not coprime to its formal derivative in E[x], E the splitting field of x^p-t over (Z/(p))(t), whereas x^2-2 in Q[x] is seperable. (im explaining mostly to myself)

We are only given the information of f in the question, so my only candidates for g so far have been f itself and (f(x))^p^e

[dim widget]

do you know why every irreducible polynomial in characteristic zero is separable?

[dim widget]

(f(x))^{p^e} and f(x) don't have the right degree

[languid trellis]

If f is irreducible, char F = 0, then if (f',f) neq 1, degree considerations imply that f' divides f, so f' =0, but f' =0 iff f is in F. So, f' is 1, and so every root is simple if f is irreducible

[languid trellis]

🤦‍♂️ I didn't even consider this

[dim widget]

okay! so what does that tell you about f'(x) if f(x) is not separable

trying to apply the same reasoning

[languid trellis]

if char F = p, (f,f') \neq 1, f(x) = a0 + a1x + ... + anx^n, f'(x) = \sum i * ai * x^{i-1}. Degree considerations again lead us to say that if (f,f') \neq 1, then f | f', so f' = 0. f'(x) = 0 iff i * ai =0 for all 1 \leq i \leq n. Then either i =0 or ai =0. If ai \neq 0, then i=0, which means that i is a multiple of p, so f(x) = b0 + b1x^p + ... + bm* x^mp = g(x^p) for some p.

To be clear: I ripped this from the book I'm reading with some small modifications lol

But I see the argument now

I need to run to a lecture now. Thank you for the guidance TTEG

I'll continue thinking along these lines in about an hour

[dim widget]

ok

[somber sleet]

Can someone tell me, why if V is infinite dimensional, then V and its dual space V* are not isomorphic?

[dim widget]

when V is infinite dimensional the cardinality of the dual space is larger

so no chances

you should think about why!

In terms of cardinal arithmetic the general statement is something like: Card(F) \cdot dim(V) = Card(V), but Card(F)^{dim(V)} = Card(V*)

and these only happen to agree when V is ft dim.

[dull ginkgo]

This is sorta a direct sum vs direct product sorta deal.

A direct sum is basically where elements are of the form (a, b, c…) but only finitely many are nonzero so we can basically write elements as a sum of scaled basis elements (0,0… 1, 0… 0).

Direct products on the other hand are where there isn’t that only finitely many nonzero restriction, but then you can’t really write elements of it as sums over the basis ones above.

V of dim K is iso to a direct sum of F K-many times. On the other hand, consider the dual V*, functionals from V to F. Then we have to consider that functional \phi can map the basis vectors literally to any value of F, so there isn’t that “finitely many nonzero” restriction. We can still add them and shit and it would behave like a direct product where the components are the images of the basis vectors in the original V

For the finite case, this isn’t a problem

[inner needle]

I love this question.
This answer is essentially the same explanation as the one above, but worded a bit differently.
Okay, in the finite case, what do you define as the isomorphism? If ${v_k}$ is a basis of $V$, and ${f_k}$ is the corresponding dual basis for $V^{\ast}$ , then an isomorphism is given by $ v_k \rightarrow f_k$ right? But for $\text{dim} \ V = \infty$, this is just an injective, but not a surjective linear transformation. In other words, if $v_\alpha$ are a basis for V, then not not every $f \in V^{\ast}$ is a linear combination of $f_\alpha$'s. Try to see why that is the case! A hint is to just consider an $f$ with each $f(v_k)$ nonzero.

[cloud walrus]

Transience

[inner needle]

What this shows is that the image of the aforementioned injective linear transformation $V \rightarrow V^$ is a strict subspace of $V^$, hence $dim V < dim V^{\ast}$

[cloud walrus]

Transience

[inner needle]

In defining the linear transformation, I've assumed V has a countably infinite basis ${v_{k}}$ but can you see why I haven't lose any generality? If this holds for the countable case, it surely holds for the uncountable one. Alternatively, just define the map as $v_\alpha \rightarrow f_\alpha$, where $(v_\alpha)$ are a basis for V with $(f_\alpha)$ the corresponding dual basis.

[cloud walrus]

Transience

[inner needle]

Sorry, this is actually not the right argument. in the infinite dimensional case even if $W$ is a strict subspace of $U$ is does not follow that $\text{dim } W < \text{dim } U$,

[cloud walrus]

Transience

[inner needle]

It seems some sort of cardinality argument will be required for the proof, I'll let others take over.

[inner needle]

Certainly, this is a better way to look at the actual reason why.

[dim widget]

I think your original idea or intuition is pretty spot on, the point is that one can view V^* (fixing a basis B of V) as the set of F-valued functions on B. There is a subspace (coming from the choice of basis) isomorphic to V consisting of the set of functions with finite support.

this inclusion is obviously strict, but as you say a little more set theory is required to show they're not isomorphic

[inner needle]

Exactly! All we have to prove is that they are isomorphic if and only if(emphasis on the only if part), V is finite dimensional

But the proof comes from some amount of set theory involving cardinals

[dim widget]

yes you do at some point need to show that |F|^{dim(V)} > |V|

[slim kayak]

The "information" of a functional is completely described by the values the functional takes on each basis vector. You then get an infinite list with values in your field.

Your vector space consists of all finite linear combinations of basis vectors.

It might make sort of sense that the former collection should be much larger in terms of cardinality and using more set theory you can show that this trend must continue to the dimension using cardinal arithmetic.

(Ngl my logician department issued topology course had a special lecture for cardinal arithmetic but I think I forgot the actual proofs like a week after, just googling them is fine)

[next obsidian]

This is simple enough

[inner needle]

In fact, when you put it this way, this kind of a way of looking at things is useful in studying, say representation theory and in other fields of algebra

[next obsidian]

If alpha <= beta then alpha^kappa <= beta^kappa

And then apply the proof to show kappa isn’t bijective to P(kappa) = 2^kappa

[inner needle]

Yeah, it seems only the most elementary of arguments will be used.

As long as we simply work with the cardinalities of the basis, that is

[bitter rover]

The cardinal arithmetic fact causing things to break down "at infinity" is that $$\kappa \cdot \nu = \operatorname{max}\left{\kappa, \nu\right}$$ if either $\kappa$ or $\nu$ is an infinite cardinal.

[cloud walrus]

Cufflink

[slim kayak]

Why is |F| < |F|^(dim V) for |F|> dim V?

[wraith cargo]

what's F?
Some finite field?
What's the context?

[slim kayak]

The same as the above context

[wraith cargo]

oh lmfao

[slim kayak]

For F finite there are standard results, so for F infinite is what I am wondering rn

[inner needle]

At this point, you can just replace this statement with arbitrary sets/cardinalities, right?

[dim widget]

if dim(V) is finite it's not obviously

[slim kayak]

Well duh

It's a bit unsatisfying to only get that some dual spaces are too large to have the same dimension for cardinality reasons. Does the cardinality argument not extend for all infinite dimensional V?

[dim widget]

what?

the cardinality argument is totally general

it applies to literally any vectorspace

but it seems you're asking why |X| \cdot |Y| < |X|^{|Y|} for X, Y two sets?

[slim kayak]

Okay so why does this hold (for dim V infinite) is my question

[inner needle]

Thinking about it, your very first answer was already the perfect one. But then again, the idea is equally as important as the proof

[slim kayak]

Well, |F| <= |V| <= dim V * |F| = |F| holds in the case |F| > dim V, so this is what the inequality comes down to

But I can't think of the argument here rn

Like, simpler question. Is the cardinality of R^infty really larger than R and why?

[dim widget]

hmmm actually I'm no longer sure it is...

[inner needle]

I think you don't really mean R^infty but rather R^aleph_0

[dim widget]

I think if |X| <= 2^{|Y|} then one can show that |X^Y| = 2^|Y|

[slim kayak]

(okay, for context I am going off the wiki article only providing the property for the base finite. I guess I'll consult a proper source if they have it for the base infinite as well)

[inner needle]

The former is actually the sequences with finite support

[dim widget]

so then it would seem that if X = P(Y) for instance that this argument doesn't work

or indeed if X is anything between Y and P(Y)

[slim kayak]

Okay, after reading some unis PDF there is a corollary that kappa < kappa^mu if the cofinality of kappa is less than or equal to mu

[dim widget]

what is the cofinality of a cardinal?

[slim kayak]

Something that reminds me I don't mess with logicians and set theorists

Gimme a sec

[inner needle]

Honestly, I'm getting all kinds of flashbacks reading this whole discussion

Activating mental muscles that I've not used for a while

[slim kayak]

https://www.ucl.ac.uk/~ucahcjm/ast/ast_notes_4.pdf
Okay, here is just the entire thing I am reading rn

Section 4 has the cofinality, the corollary mentioned before isn't too far below that

[dim widget]

interesting

well anyway it is much weaker than GCH to see that 2^a = b^a if 2 <= b <= 2^a

but according to this pdf you have some impenetrable stronger iff under AC + GCH, but either way the claim I thought was "obvious" is actually just wrong

[slim kayak]

So anyways, who is gonna take one for the team and well-order the reals real quick to check if N is cofinal here

[dim widget]

I think again for your specific question we just need to note that a*a = a (a infinite)

and use standard rules of cardinal arithmetic

to show that R^{N} is not actually any bigger than R

Because 2^|N| <= |R^N| <= |(2^N)^N| = 2^{|N| \cdot |N|} = 2^N = |R|

[slim kayak]

Yeah true that, |R|=2^|N|=2^(|N| * |N|) = R^|N|

[dim widget]

anyway but it is true that the dimensions are different always!

and that's just from |X| < 2^|X|

[slim kayak]

Yeah, for infinite dim VS that always true.

Hm, how?

I thought the aleph_0 dimensional case for real vector spaces leads to both having the exact same cardinality. I mean, there is an inclusion of said vector space into its dual space, and the latter has cardinality R^|N|=R

[dim widget]

well by the argument given above we know that |V^*| = |F|^{dim(V)} and certainly dim(V) < |F|^{dim(V)} by the above

I considered repeating the argument, but it's just: fix a basis, then V^* is F-valued functions on this basis

this breaks down for the field with one element but otherwise we should be fine 🙃

[slim kayak]

Absolute geometry enjoyers are absolutely furious rn

[dim widget]

and it's not hard to see that $|V^| = dim(V^) \cdot |F| = dim(V^)$ since one can construct a linearly independent set of $V^$ of size $|F|$

I think

[cloud walrus]

Math_Discord_Final_Girl

[dim widget]

okay I am thinking this is hard to do

[slim kayak]

Can we? My gut feeling is that we cant have more than 2^dim V linearly independent elements

Or rather, if we switched to a bigger field I dont really get why we should find more independent elements

[dim widget]

Yeah it is hard to see I think

but I think already you can see that F^N contains at least |F| independent elements

[slim kayak]

Yeah, you are in a tough spot since you can only multiply the entire list and you only get to add finitely many of those

[dim widget]

okay one way to see this is that there is no nontrivial solution (set of a's) to the equations \sum_{i \in I} a_i x_i^j = 0 for all j ranging from 0 to |I| - 1 for I a finite set and all x_i are distinct

so one can take the linear functionals F(e_i) = x^i

for x some element of F

[hidden wind]

i’m not sure if i want to know what this is

[dim widget]

it was a joke

there is a concept called the field with one element but it's probably not literally a set with one element

[dim widget]

^^^ this works

[hidden wind]

not a field and not with one element, i imagine

[sly crescent]

Correct

[next obsidian]

It’s the set {}

[dim widget]

also Chmonkey I led you astray

[next obsidian]

Omg

[dim widget]

if you read the above conversation it turns out this thing about duals is much harder than I had made it out to be

in particular the claim that ab < a^b for a, b infinite cardinals is just wrong

[slim kayak]

is too powerful, it fulfills =/=

[next obsidian]

Right if b is larger than a then the both just become b right?

[dim widget]

no

[next obsidian]

Or like, the left is b, maybe the right is also b?

[dim widget]

the right will typically be the successor cardinal to b

[slim kayak]

Uh, a=2?

[dim widget]

or larger

[next obsidian]

Isn’t the left always the max?

Wait no there’s something fucked here

I remember multiplication isn’t even commutative or something

[slim kayak]

multiplication is doing fine here

[next obsidian]

Well okay, the proof I had was this

[slim kayak]

the product in set doesnt care too much about ordering

[next obsidian]

k^(+)alpha has cardinality max{k,alpha} I think.

If alpha <= k, then you use the fact that a union of <= k sets of size k is size k

Which I guess is nontrivial

And if alpha > k I think you can see its size is alpha… hmmm

Anyway, then Hom(k^(+)alpha,k) = k^alpha by an easy argument blah blah finite support

And then k^alpha >= 2^alpha > alpha

[next obsidian]

But I guess the set theory really comes in here

The other stuff is simple I think

[slim kayak]

Anyways, while a < b^a tends to be pretty straightforward, a < a^b isnt. It is if b is finite, but the case b infinite seems a bit tricky.

[dim widget]

even if b = {*}

[next obsidian]

You mean {}

[dim widget]

sorry, yes, I made a typo

[slim kayak]

And yeah,
dim V < |F|^ ( dim V) = | Hom(V,k) | = |dim Hom(V,k) | |F| = max( dim , |F|)

so it comes down to showing that F cant ever be bigger than the dimensions, which is where you can finally have some fun

[dim widget]

the hard claim is that dim(V^*) >= |F| and after that everything is easy

although kerr and I kind of realized this backwards

as in that was the last thing we worked out

anyway, interesting that it was much harder than I remembered!

[slim kayak]

took the scenic route, as you might say

[rotund aurora]

I didn't read the above but if b>=a the inequality is clear, otherwise you could take a=N^N and b=N, then a^b=(N^N)^N=N^(N x N)=N^N=a?

[opaque finch]

I am struggling with an idea. The problem is as follows
Let $g$ be a generator of $(\mathbb{F}{p^n})^{*}$. Then the minimal polynomial of $g$ over $\mathbb{F}p$ is $\prod{i = 0}^{n - 1}(x - g^{p^i})$.
Of course $g$ is a root since $(\mathbb{F}{p^n})^{*}$ is a cyclic group with generator $g$ and has order $p - 1$, so any element is of the form $g^{p^i}$ for the $i$ in given range. All I got to prove is that this is an irreducible polynomial over $\mathbb{F}_p$. Any ideas on how to do that ?

[cloud walrus]

mycroftholmes1703

[opaque finch]

Sorry for the typo, the indexing starts from $0$

[cloud walrus]

mycroftholmes1703

[opaque finch]

So $g$ is indeed a root.

[cloud walrus]

mycroftholmes1703

[rotund aurora]

irreducibility follows from minimality

[opaque finch]

I have to prove that it is indeed the minimal polynomial. I have proved $g$ is a root. To prove it is minimal, I have to show irredicubility. Irreducibility is not given, but I have to prove.

[cloud walrus]

mycroftholmes1703

[rotund aurora]

Say f is the minimal polynomial over F_p. f(g)=0. f(g^p)=0, etc...

[opaque finch]

why should we have $f(g^p) = 0$ and for successive powers ?

[cloud walrus]

mycroftholmes1703

[rotund aurora]

x-->x^p fixes F_p. This is just Galois theory

[opaque finch]

uhhhh true true

Thanks @rotund aurora

[rotund aurora]

np

[slim kayak]

I mean circumstances where b is actually smaller. The only real theorem I could find uses some weird order theoretic property that shows that the inequality holds as long as the "cofinality of a" is smaller than b.

For R^N there is a trick to leverage the multiplication of cardinals, but for a's larger than 2^N it gets less clear

Must've been the wind

[rotund aurora]

sorry it was me, I made a mistake

Does it follow from GCH that any uncountable cardinal is a power of 2?

nvm, no

[slim kayak]

I mean, yeah. Thats what it says

[rotund aurora]

nope

https://math.stackexchange.com/questions/965769/does-sf-gch-imply-that-every-uncountable-cardinal-is-of-the-form-2-kappa

[slim kayak]

Oh, sure

[rotund aurora]

If a=2^A, b=2^B then a<b^a holds and follows simply from A<2^A the inequality was a<a^b duh

[slim kayak]

That one holds whenever b isnt empty or a singleton though

[rotund aurora]

oh yeah nvm me

[slim kayak]

For the more interesting one it comes down to whether b is bigger than A="a-1", since that evaluates to 2^max(A,b)

Well, for successor cardinals. But who tf is studying vector spaces with limit cardinal cardinality

[rotund aurora]

Okay sorry, i was thinking of a<a^b. In the case a and b are powers of 2 it reduces to A<A2^B which holds if and only if B>=A, iff b>=a

[slim kayak]

You really wouldn't want to express b as a power of two here tbh

Oh shit, I got called chronically online by a bot (emeritus -> active)

$ 2^{A 2^{B}} $ just isn't a very friendly expression

Uh hello

$ e^x $

[rotund aurora]

isn't k-->2^k strictly monotonic?

[slim kayak]

$\text{ I have been forsaken}$

[cloud walrus]

Kerr

[slim kayak]

Oh, you meant it like that. Yeah, A < 2^B so A <= B

[rocky cloak]

I think that's independent of ZFC

[rotund aurora]

Yeah, you are right

https://math.stackexchange.com/questions/3935960/is-cardinal-exponentiation-strictly-monotone-in-the-exponent

[hidden wind]

so it’s consistent with zfc that there’s a largest cardinal?

[slim kayak]

Set theorists really eating crumbs at this rate

I figure it's again the limit cardinals bere

[rotund aurora]

The power set axiom says otherwise. It's not difficult to prove |X|<|P(X)|, it doesn't use choice or anything fancy

[rotund aurora]

Well no, some answer says 2^omega=2^(omega_1) with some Cohen trick. I assume omega_1 is the successor of omega, but I could be wrong

[slim kayak]

Under GCH the strict monoticity reduces k -> k+1 being strictly monotonic

[hidden wind]

i thought 2^k was just a fancy way to talk about the cardinality of P(X) for some X with cardinality k, i guess i should go read up on basic cardinal aritmetic

[slim kayak]

Which doesn't have to hold for non-successor cardinals

[rotund aurora]

Yes, 2^|X|=|P(X)|, that's the definition.

I think there's a confusion here. k-->2^k being strictly monotonic means that l<k implies 2^l<2^k.

[hidden wind]

guh thanks yeah i always mix up increasing, fsr with the property “x < f(x) for all x”

which i forget if has a name

[slim kayak]

Is that one common in analysis stuff

?

[opaque finch]

I doubt that stuff has a name

[hidden wind]

it’s just what my brain goes to but for no good reason as far as i am aware

[slim kayak]

Oh

[rotund aurora]

Suppose a<a^b and c<a. Does it follow that c<c^b?

[chilly ocean]

So I'm trying to find the dimension of a galois extension over Q and a basis and I'm really lost. I have a polynomial and it's splitting field, I've deduced that the polynomial is reducible over Q and found the factors. I've also found the roots of the polynomial. I've done examples where the given polynomial is irreducible over Q but now that it isn't I don't really know what to do. I'm also still very confused on how finding a basis works, sometimes it's obvious but other times like now I just don't know how many of the roots I should be extending Q by.

The polynomial is x^4-x^3-x^2+3 and I found it is equal to (x^2+2+2)(x^2-3x+3)

The roots are 1+-i and (3+-isqrt(3))/2

I figure as a basis just using one of the terms that has square roots in it would do the trick (using its powers as basis elements)

But I still have no idea how many as I don't know the dimension

[rocky cloak]

If you know how the irreducible case works, what you can do is take one of the irreducible factors and take it's splitting field K/Q, then factor the rest of the polynomial over K and take it's splitting field E/K.

Then if you have a basis for K over Q and a basis for E over K you get a basis for E over Q simply by taking all possible products of the basis vectors from the two bases.

[chilly ocean]

Oh so I can just treat each factor individually and then add it all up in the end?

Assuming they're irreducible over Q ofc

[rocky cloak]

Well, it's not really about being irreducible over Q. Something that is irreducible over Q can become reducible over K, so you would have to factor out an irreducible factor at each step.

[chilly ocean]

I see, so it's probably best to choose a factor that makes showing irreducibility easy

[rocky cloak]

For example
(x^2 + 1)(x^2 + 4)
has splitting field Q(i)

[chilly ocean]

That makes sense

Thanks for the help

[dim widget]

under GCH + AC yes

[rotund aurora]

How does GCH work without AC

[dim widget]

what do you mean?

[rotund aurora]

You don't have cardinals. But yeah, you can formulate it. GCH + ZF implies AC https://math.stackexchange.com/questions/3646072/the-meaning-of-gch-in-zf

[dim widget]

Yes it does

[dim widget]

I don’t think you’re correct about the first statement

I just think the usual definition in terms of ordinals doesn’t make sense

But you can just take equivalence classes of sets under bijection

Anyway unsurprising…

I guess depending on what one wants from a theory of cardinals one might want to take canonical representatives in each equivalence class? But for this there is the usual trick https://en.m.wikipedia.org/wiki/Scott's_trick

anyway not obvious what you were trying to say

[rotund aurora]

This question?

It just surprised me that you also included AC explicitly. ZF is always assumed here, right?

[dim widget]

Ah I see what you mean

[dim widget]

yes just GCH, I guess I was thinking about the fact that under AC one can show that given a lower bound on a cardinal b (depending on a) one has that a < a^b, but adding GCH the bound becomes an iff

the bound just being b >= cofinality(a)

[static mauve]

There is a problem in my math book that asks the following:

"Let a and b be elements of a group G. Show that if ab has a finite order n, then ba also has order n."

I thought that order applied only to groups (since it is essentially the cardinality of a group). Wouldnt ab just be an element of G?

[bitter rover]

Your book almost certainly has the definition of the order of an element.

What book are you using?

[static mauve]

Fraleigh (a first course in abstra algebra)

Maybe I totally overlooked said definition!

[bitter rover]

to the index batman

I'll race you

Downloading textbook now

you better hurry

[static mauve]

yeah, i guess it does have this

[bitter rover]

🙂

Anyhow, it's the smallest number n such that g^n = e.

Every element in a finite group has a finite order (by the -hole principle).

[static mauve]

Yeah, that totally makes sense

[bitter rover]

An element in an infinite group might not have a finite order, e.g., any non-identity in the additive group of the integers

In any case, you can write, e.g,. $$(ba)^5 = b(abababab)a = b (ab)^4 a$$

[cloud walrus]

Cufflink

[bitter rover]

bababababa

[grizzled crane]

Is this proof correct?

The first line follows because every subfield of C has a characteristics zero, right?

[slim kayak]

That feels possibly circular

One way to define Z would be as the cyclic group generated by 1 under addition, from which inclusion follows

[south patrol]

Well you need to mentions there are no relations i.e. the subgroup generated by 1 is free

And also there is the ring structure on Z to consider

[grizzled crane]

What does it mean for a subgroup to be free?

[crystal vale]

How many group homomorphisms are there from Z/3Z×Z/4Z×Z/9Z to Z/18Z?

I guess { (1,0,0), (0,1,0), (0,0,1)} generate Z/3Z×Z/4Z×Z/9Z, so for each there are 18 choice, isn't 18×18×18 homomorphism, I know this is not a answer but where I am wrong in counting?

[hidden wind]

not all of those will be homomorphisms

for example if you send each generator to 1 in Z/18Z, that’s not a hom, as it doesn’t preserve the identity

[crystal vale]

Why?

[hidden wind]

we need φ(0) = 0

oop nvm me i’m too sleepy and not thinking straight

[crystal vale]

Oh wait

Actually order of image matters here

Got it

[hidden wind]

if a,b,c are the generators, we need φ(xa+yb+zc) = xφ(a) + yφ(b) + zφ(c), where do we go from there?

i don’t see it i need my coffee

also is the number of homomorphisms 81?

[crystal vale]

No it is || 54 ||

[hidden wind]

ooh that’s a lovely number

[crystal vale]

See we have to consider the order of the image of generators

[lone niche]

You can split this into two questions

[crystal vale]

So for the first generator we have only 3 choice

[lone niche]

number of homomorphisms Z_n -> Z_m

[crystal vale]

For second there is 2 and for third there is 9

So we get 54

[crystal vale]

Yes

[lone niche]

and computing the number of homomorphisms from a product groups, given the number of homomorphisms on each factor

putting both things together gives you the answer

[crystal vale]

Yes

[glad osprey]

Is the question how many surjective group homs are there to Z/18Z?

[crystal vale]

No but yes this is an interesting one

[hidden wind]

right i’ve been meaning to get to this one woops

[glad osprey]

hmm, but then why do you need to consider the order of the image of generators? Like, why are there only 3 choices for the first generator?

[crystal vale]

Great

[crystal vale]

Because say (1,0,0) maps to a then order of a must divide 3

So we have to consider those elements which have order which divides 3

So in Z/18Z there are 3 element whose order divides 3

[serene dune]

for the lemma i have hand waved

[crystal vale]

You can easily prove that

It is just on a fact where your generator maps

[glad osprey]

I see, thanks

[serene dune]

linked the very thing

i wrote a few weeks ago

[hidden wind]

this definition is just plain wrong, right?

[lone niche]

Maybe they are assuming R is a ufd?

But also I think they also missed non-constantunit?

[little shuttle]

Is there a way to show that the conjugation of an element in the inner automorphism group of S_4 is also in the inner automorphism group of S_4

WITHOUT using cycle structure

or order

I thought I could maybe try showing that since all elements of the inner automorphism are products of 2 disjoint transpositions/identity element that (distninct abcd) (a b)(c d) is mapped to (a' b')(c' d') which are distinct and therefor ein the inner automorphism since conjugation is bijective

[long swan]

that’s my guess too yeah

although for some reason the way they’ve written it seems as if they’re treating the terms prime and irreducible synonymously rather than saying they’re distinct terms but coincide if R is a UFD

Is there like a disagreement in convention for that terminology or something

[wicked patio]

I would just say like n>0 = 1+1+1+...+1 n times and then n<0 = -(-n) or something for Z

[hidden wind]

yeah i think they should qualify that it is not a product of lower positive degree polynomials

[tardy hedge]

A torsion module tensor divisible module would always be 0?

[tardy hedge]

FEIN FEIN FEIN FEI FEI FEIN FEIN

[dim widget]

Yes

[forest night]

I have seen unimaginable horrors as algebra enthusiast, I saw how number theorists prove little Fermat's theorem...

[rapid junco]

How else would you prove it

[elfin wraith]

The only two proofs I can think of are just directly via group theory or as a corollary of Eulers theorem but these are both just algebra. What are you getting at?

Elementary number theory is generally just algebra, the majority of the ideas generalise, think Chinese remainder theorem etc

[south patrol]

Well there is a cheap proof that uses the fact that the groups involved is abelian

As in, fix a prime p, fix x in F_p^x, and let A be the product of all elements of F_p^x. Now multiply all the elements of F_p^x by x before taking the product and since this is a bijection we get A = Ax^(p-1)

So x^(p-1) = 1

(Here we critically used commutativity at the last step to put all the x terms at the back of the product)

This sidesteps using Lagrange's theorem

This proof should work to prove Lagrange's theorem for any finite abelian group

But saying number theorists "prove flt" in a certain way is sort of eh cause this is a very basic result and they will be very likely aware of all these different proofs lol

[rocky cloak]

I guess "number theorists" here really means "first year course in elementary number theory"

[south patrol]

Yeah that is what I assume lol

Idk Ig it is worth pointing out it is funny pitting algebra against number theory

[tardy hedge]

If you have a quotient like M/L = N, are elements of M given by pairs of elements (l,n) in L and N?

Oh yea it does right

an n specifies the coset, and the l then chooses an element in that coset

Hom functor being exact makes more sense now. Pairs of homomorphisms determining a hom

There was something in dummit foote saying that but i never understood it

[spice whale]

Struggling on this problem

It seems like some (even finitely generated) infinite groups don't even have subgroups of finite index

[tardy hedge]

If you have an R-module M and create a free R-module F(M) with basis M, what does the kernel of the projection F(M) -> M look like?

[rocky cloak]

It looks absolutely huge and complicated.

[spice whale]

actually maybe they do

i'm really not sure where even to start here

[tardy hedge]

Oh damn really??

Theres not a way to describe what it looks like easily?

[rocky cloak]

I mean if you can the generators f_m. Then for every equations

sum r_i m_i = 0

you'll have an element
sum r_i f_m_i
in the kernel.

So that's an easy description, but not that enlightening

[rocky cloak]

There are such groups yes.

[spice whale]

I thought of Q

but it's not finitely generated

[rocky cloak]

There's these
https://en.m.wikipedia.org/wiki/Tarski_monster_group

[spice whale]

yikes

[tiny jolt]

Is the book looking for a geometric argument?

[spice whale]

no?

how would you even do a geometric argument

also not a book

[tiny jolt]

yeah i just meant whatever context this problem is from

[spice whale]

i mean

it's a problem sheet for an intro group theory course

[tiny jolt]

oh lol oops

[elfin wraith]

Intro group theory why is the course at my uni so bad

[rocky cloak]

A hint could be that if H < G is a subgroup, then G acts on G/H and H is the stabilizer group of the coset H

[spice whale]

ok thanks

i'll try orbit stabiliser from there

[wicked patio]

What about the group <r,s | sr=r^-1s>

[spice whale]

what about it?

[wicked patio]

finitely generated but has infinitely many groups of index 2: all groups of the form <sr^n>

oh

that's order

nvm

[spice whale]

the problem is i don't even know why G being finitely generated is important

[bitter rover]

A homomorphism is determined by it's action on the generators. If a group is finitely generated then there are only finitely many homomporphisms into any finite group.

[spice whale]

oh and Sym(G/H) is finite

thanks!

wait

something like that anyway

[dim widget]

Have you thought about a way to rephrase being finitely generated by a set of cardinality X in a more universal way? It means there is a surjection of groups from something to G.. this helps a lot not just with this problem but as a general idea

[spice whale]

free group

i thought about this but they haven't been covered in the course

so i assumed they were off limits

[dim widget]

I see, well you end up eventually just saying the same thing as in cufflink’s hint

[spice whale]

I'm not sure where the finite group is

like sure there's Sym(G/H)

[dim widget]

Here, it’s finite and independent of H!

[spice whale]

oh yeah

it's S_n

but like

hm

I don't see how this helps

I can't think of a way to associate morphisms G→S_n to subgroups

or like

[dim widget]

well, there is certainly always the kernel, but you should think of what process would reasonably recover H

[spice whale]

at least inject the set of index n subgroups into Hom(G,S_n)

the kernel has the wrong index

[dim widget]

yes it’s not the kernel because H doesn’t have to be normal

[spice whale]

and is also necessarily normal

yeah

[rocky cloak]

Well what you want is to associate subgroups to morphisms (in an injective way)

That way showing the are only finitely many subgroups

[bitter rover]

gotta find something s t a b l e

[warm dove]

what's the deal with this proof

I've proved through part f

but part g seems like nonsense?

I think |\mathcal{O}_T| is actually [G:H] not |G|

by orbit-stabilizer (H is the stabilizer of T)

[crystal vale]

Why?

[crystal vale]

Yes

[chilly ocean]

is this proof good

[crystal vale]

Why do you take R[x] = { a_0 +a_1x +a_2x^2 +... | a_i in R }?

I think it is a finite sum not infinite

[chilly ocean]

You're right. My carelessness qwq

[crystal vale]

I think it is correct

[chilly ocean]

[crystal vale]

But A is a principal ideal generated by x not <x>

And you used the fact A is <x>, which is principal ideal generated by x therefore A is an ideal but did you prove that <a> = {ar | r in R} where R is commutative ring with unity is an ideal?

[chilly ocean]

I see

[crystal vale]

Can you tell me what the definition is used for < a > in your textbook?

[chilly ocean]

Sure

(?)

[crystal vale]

Gallian, right?

[chilly ocean]

Yes 💯

[crystal vale]

Okay now prove that <a> is an ideal

[errant wedge]

How could I count the number of subgroups of Z x Z of index 3?

[lone niche]

If H is a subgroup of Z x Z of index 3, then we know that | Z x Z / H | = 3.

Then we would want to look at surjective morphisms Z x Z - > Z x Z / H and count them.

[slim kayak]

The question was how many such H there are tho

[lone niche]

This can be further reduced to counting morphisms Z - > Z x Z / H and then squaring that.

[slim kayak]

Do you know how many index 3 subgroups Z has?

[lone niche]

Yup I am translating the problem into a problem of counting homomorphisms.

[errant wedge]

Ah ty

im dumb lol

[slim kayak]

Wait, wouldn't you want to count the surjective homomorphisms into Z/3Z?

[errant wedge]

Z x Z is abelian so all subgroups are normal so this is the same as counting the number of quotients of size 3

and Z/3Z is the unique finite abelian group of size 3

[errant wedge]

its equiv

[errant wedge]

tyvm ^^

[slim kayak]

Yeah, just worded a bit odd for my taste

[slim kayak]

Want some further tips, specifically one that makes the calculation easier?

[errant wedge]

oh ty, i got it tho i think (if 4 is right)

[warm dove]

Well then, how can we finish the proof of par g if the statement is incorrect

[slim kayak]

I would get 8

[warm dove]

I don’t see why the index is bounded by p^k

[slim kayak]

But more generally, I meant the knowledge that maps out of finite products of groups G_i are determined by tuples of functions G_i -> H.
In the special case G_i = A this comes down to considering choices for where 1 gets mapped to.

Into Z_3 we get 3 choices, any besides mapping to 0 would be surjective already. If considering pairs it is thus sufficient to any pair where not both are the zero maps

So from 9= 3 x 3 functions subtract one to get 8

[rocky cloak]

You forgot to account for isomorphisms of Z/3

[slim kayak]

Wdym

[rocky cloak]

There are only 4 subgroups of index 3. For example the map that sends
(1, 0) to 1 + 3Z and (0, 1) to 0
and the map that sends
(1, 0) to 2 + 3Z and (0, 1) to 0
has the same kernel.

In general if you compose a map with an isomorphism it still has the same kernel

[slim kayak]

Makes sense

[chilly ocean]

What does this question mean
Prove that f (x) = x^4 + 4x + 1 is irreducible over Q

[slim kayak]

That you can't factor it

[lone niche]

I just realized the ending of this argument was b.s. which came up with the right answer by chance. Really embarrassed sorry.

[tardy hedge]

Broski don't be so embarrassed we all make mistakes

[bitter rover]

No, be more embarrassed.

Let your embarrassment fuel your success.

[tardy hedge]

I really disagree with that sentiment but ok

I don't think embarrassment is the right emotion to focus on here

A5 contains all 3-cycles meaning every element of A5 is a 3-cycle?

or, if its just every 3-cycle is in A5 is easy isnt it because every (a,b,c) is an even permutation

[bitter rover]

Every 3-cycle is in A_5. In fact, A_5 is generated by the 3-cycles.

[tardy hedge]

Meaning if you take an arbitrary permutation in A5, you can write it as a product of 3-cycles?

[bitter rover]

https://en.wikipedia.org/wiki/Generating_set_of_a_group

[tardy hedge]

Do you think that first question is asking to show that?

A5 obviously contains 3-cycles because (a,b,c) = (a,c)(a,b)

[barren sierra]

Yes

[barren sierra]

And yes the question is asking you to show exactly this

That's basically the whole proof of (a)

[bitter rover]

What would (a) have to be saying for (a) and (b) together to imply that A_5 is simple? That's what it must be saying.

[slim kayak]

from jagr's comment it seems that you can still count the set of surjective homomorphisms, or rather the number of orbits under the action of automorphisms of Z/3

Extending that further, it would also give a cute little argument as to why p-1 divides p^n-1 when generalized :3

[barren sierra]

Yes (a) & (b) together show that A5 is simple

Once you add the fact that two permutations of the same cycle type are conjugate to each other

[bitter rover]

Right, but @tardy hedge was asking what (a) meant, and rather than interpreting it for them I wanted them to think about which interpretations combine with (b) to give the result. 🙂

[barren sierra]

Ah my bad

[bitter rover]

Or whether the difference in interpretation matters at all.

I think their two interpretations were "A_5 is exactly the set of all 3-cycles" vs. "A_5 contains every 3-cycle, and is actually generated by them".

But with a little thought I was hoping they'd see the former couldn't possibly be the case.

[tardy hedge]

But a) doesnt say A5 is generated by 3-cycles it just asks that it contains the 3-cycles

Which is why im kind of confused by how this question is being asked

[bitter rover]

It'd take less time to prove A_5 is generated by all 3-cycles than you've already spent trying to figure out whether that's what (a) is asking!

So if you can see how "generated by" gets you where you want then just prove that.

Maybe there's a way to use only the fact that A_5 contains every 3-cycle without also proving it's generated by them.

(There is, but I don't think it matters from the perspective of proving A_5 is simple.)

[crystal vale]

In general Artin used this proof, so basically up to e we developed our candidate Sylow p-subgroup H and then we will prove that |H| has the form p^m, then using |G| = u|H|, we can conclude that |H| = p^k since p does not divide u.

But in your problem, they use a different approach first, proving f and then g, maybe g is correct.

Which book?

[warm dove]

This is Judson’s book

[crystal vale]

| H | ≠ | O_T | but we have to prove | H | ≤ p^k

[warm dove]

Yeah, basically this is what I need to prove

[crystal vale]

It is clear that statement is wrong

[crystal vale]

Okay let's do it

[warm dove]

I know p^m | H and p^m < H

Yeah I just need to know H <= p^m

[crystal vale]

What is m here?

[warm dove]

oh

I meant k

my bad

[crystal vale]

Since | H | = | G |/ u

Now p doesn't divide u

So let | G | = p^ka, where p doesn't divide a

[warm dove]

then |H| = p^k (a/u)

but how do we show a/u = 1

[crystal vale]

So |H | ≤p^k, right?

[crystal vale]

Use f

in f you showed that p^k ≤ | H |

[warm dove]

well, doesn't this give H >= p^k?

since u < a

so this just gives f again I would think

[crystal vale]

Oh yes

[warm dove]

oh wait

I think I see a way

the |H| = |O_X| makes no sense

but what we can do is for x \in X, consider the coset xH

this must be contained in X (since H is by construction the stabilizer of X and so for x \in X, hx \in X), so |Hx| < |X|?

or not X

T is the letter I used earlier

but basically this coset is contained within the subset of size p^k

[dull ginkgo]

MARLIN DOING GROUP THEORY

FUCK YEAH

[dull ginkgo]

Is this sigh love theorem stuff

[warm dove]

Yeah

Idk what algebra book to read next

[dull ginkgo]

JACOBSO

I love the Sylow proof

you’re using the Stabilizer of subsets

Method?

[tardy hedge]

FUCK YEA MARLIN DOIN GROUP THEORY WOOOHOOO WE LOVE MARLIN

Thank you for making my life better Marlin ❤️

[dull ginkgo]

Yap machine time:

Assume we have a finite group |G|. We can allow G to act on its own subsets (i.e on P(G), the power set of G) via the images of sets by multiplication from the left (or right, just stay consistent) i.e g(S) = gS. Obviously this is a natural group action, and note that multiplication by each g is of course a bijection, which means subset cardinality is preserved, e.g |gS| = |S|

Then, naturally the next question is analyzing the Stabilizer subgroups. An element of G stabilizing a subset S is essentially saying that g maps S to S, therefore we can restrict/localize to the action of Stab(S) on S. Partitioning S into orbits, they take the form of Stab(S)x i.e right-cosets of Stab(S), implying S is a (disjoint) union of right-cosets of Stab(S). Therefore, this implies that |Stab(S)| divides |S|.

So lets assume that |G| = mp^k and p doesn't divide m. Easy combinatorics says that there are nCr(mp^k, p^k) subsets of |G| of size p^k. If |S| = p^k, then |Stab(S)| must be a power of p. We can partition the family of subsets of size p^k into subfamilies based on the log_p of their stabilizer size.

Now, lets assume that we don't have a subgroup of G of size p^k, then no set can have a stabilizer of size p^k obviously, so that family is empty. Now let |Stab(S)| = p^n, n < k.
Then all the sets in Orb(S) have cardinality p^k too and take the form of gS for some g. Stab(gS) = {h : hgS = gS <=> ghg^-1 S = S}, i.e Stab(gS) is a conjugate of Stab(S), so has the same cardinality. That means that Orb(S) is entirely within the log_p subfamily, so that family is a disjoint union of orbits, and |Orb(S)| = p^(k - m) and is thus divisible by p, implying so is the cardinality of the family.

Since the set of all p^k-sized subsets of G is a disjoint union of these families with cardinalities each divisible by p, that means nCr(mp^k, p^k) is divisible by p.

But nCr(mp^k, p^k) is not divisible by p

[naive lance]

Why is f = 16Y^8 -8Y^4X + X^2 + Y^2 + 1 in C[X,Y] irreducible
Maybe i am blind, just see the binomial not anything helpful for any criteria

[warm dove]

Yeah this is basically the argument I was doing in the exercises

Idk

I was thinking about Jacobson but idk if I’m ready + exercises look too hard or smth

Idk I was thinking between artin Aluffi and d&f

[warm dove]

Do I know you

[elfin wraith]

D&F is nice but dear god is it long

[bitter rover]

Makes for a great reference after the fact!

[tardy hedge]

No broski you dont 🥰♥️😍

[dull ginkgo]

I gave my copy of D&F to a friend

[warm dove]

Idk

I think what’s most likely is artin because I really need to get good at computation (dear god Judson is horrible for computation)

[tardy hedge]

Half of my d/f ripped off the damn book seal

Damn annoying

Just happened yestersy

[dull ginkgo]

Jacobson isn’t terrible

[tardy hedge]

Whyd they have to make such a thicc delicious book

[warm dove]

Ok bro

[dull ginkgo]

The one good thing about softcovers is that it won’t implode like hardbacks do

[warm dove]

Idk

[dull ginkgo]

I paid less than $20 for both Jacobson I and II paperback lol

[warm dove]

I think I’ll probably read artin

And then either Aluffi or Jacobson

Probably Aluffi

[dull ginkgo]

Jacobson is just dated

And uses row basis vector convention which is vile

[warm dove]

Aluffi it is

[dull ginkgo]

It’s only used for some basic lin alg crap

In all seriousness I only used Jacobson because I got cheap physical copies

[naive lance]

would b nice if sb sees something in this polynomial

[warm dove]

Try Eisenstein in C[Y][x]? Just a thought

[naive lance]

tryed it but the 1 is annoying and i straight up dont see a random prime element deviding it

[dim widget]

One way to see this is with the Lefschetz hyperplane theorem, but surely there is a simpler way.

In general any affine curve in C^2 cut out by a single equation is connected, and your curve is nonsingular so it must be irreducible

There is probably a simpler way to prove the result in the first paragraph, and then connected + nonsingular implies irreducible

[naive lance]

i ll take it xD thank u
this saved me from crap like trying a primary decompo

[tardy hedge]

true

[naive lance]

öhm i just looked over the polynom but isnt (0,0) a singular point?

[dull ginkgo]

Maybe try a small argument shift in the X by some P(Y)
X^2 + (-8Y^4)X + (16Y^8 + Y^2 + 1)
set X to X + P(Y)?

The binary nums there make me feel like that’s the route

[dim widget]

(0, 0) doesn’t lie on the curve because of the constant term

[dull ginkgo]

@naive lance X to X - 2Y^2 actually knocks out the middle X term:

(X + 4Y^2)^2 + (-8Y^4)(X - 4Y^2) + (16Y^8 + Y^2 + 1) = X^2 + (16Y^8 + 32Y^6 + 16Y^4 + Y^2 + 1) which of the form X^2 + K(Y)

Then 16Y^8 + 32Y^6 + 16Y^4 + Y^2 + 1 is a square if it’s reducible

Observing the roots of 16x^4 + 32x^3 + 16x^2 + x + 1

There is a single root at x = -1

But not multiplicity 2 because the remainder is 1 after dividing by (x+1) again

Implying it’s not a square so the polynomial is irreducible

[naive lance]

thank u<3

[warm dove]

anyone have good first course in abstract algebra ring theory exams?

I really need to practice and not bomb my next one

so

💀

[cloud blaze]

Here is the final from the intro class I took. I don’t have a version with the solutions

[dull ginkgo]

I have not done any two variable irreducibility exercises before and that was throwing pasta at the wall to see if it sticks I’m glad it did

[rocky cloak]

Are the nZ's supposed to be Z/nZ or do they actually have questions about the ring structure of nZ

[cold sparrow]

How to prove the isomorphism of π? I understand homomorphism and injection, but I don't understand surjection's proof.

[dull ginkgo]

Group morphism? Consider the image of 1, the generator of Q lol

Maps are unique up to the image of generators

And there is obviously a map sending 1 to any x in Q*, and it’s an automorphism. Boom, surjective

[cold sparrow]

Thank you!

[dull ginkgo]

Welcome

[dull ginkgo]

Same thing works for all the prime fields btw

[cold sparrow]

but why 1 is generator, how can i get 1/2 for example?

[dull ginkgo]

1 generates the prime ring Z, of which Q is its field of fractions

[dull ginkgo]

in other words it's additive. f(n) = f(1 + 1 + 1... + 1) = f(1) + f(1) + f(1) ... f(1) = nf(1). Think of what happens to 1/2... f(1/2 + 1/2) = 2f(1/2) = f(1)

[cold sparrow]

yeah, understood. thx

[chilly ocean]

as jagr point out that it is not UFD so it cannot be pid so it cannot be Euclidean domain

[cinder fox]

What's the biggest set $A \subset \mathbb{C}$ s.t. $arg(z)$ is a group homomorphism from $(A, .)$ to $((-\pi,\pi],+)$?

[cloud walrus]

VirtualCode

[coral spindle]

Any subgroup of C\{0} works, because a homomorphism f : G -> G' always restricts to subgroups of G.

So I think this answers the question if you're willing to interpret this a bit

[glad osprey]

I'm a bit confused by this paragraph. Why does A^2 = 0 imply that A is not a simple A-module?

[coral spindle]

What is "trivial multiplication" supposed to mean? Are they really talking about rngs here?

[glad osprey]

Yes, rings are rngs in this book (it's a pretty weird book)

[coral spindle]

Baffling

What's their definition of a simple module then?

I can't imagine what on earth they mean

[glad osprey]

Hmm, I think it's just a module without submodules

[coral spindle]

Then it's completely unclear to me since the submodules are just subgroups, and indeed a group of order p has no nontrivial subgroups

I suspect they may have some special rung definition for simple modules

[coral spindle]

Ah. There you go

If RM =/= (0).

[dim widget]

well okay this is tautological then

[coral spindle]

Seems like a bullshit requirement lmao

[glad osprey]

Ah, A^2 = 0 is referring to the RM != (0) requirement

[coral spindle]

Yeah

Why did they call it (0)

weird

very weird

[glad osprey]

We're following this book, but our course requires rings to have unity, so I think I can ignore a few things in this book

another weird thing in this book: integral domains are not necessarily commutative

[lone niche]

What book is it? From the font I thought it was dummit and foote

[glad osprey]

Basic Abstract Algebra by Bhattacharya, Jain and Nagpaul

[spice whale]

isn't this normal

like

I don't see why you'd need that in the definition

[glad osprey]

every other book I've read require integral domains to be commutative, while domains are the non-commutative variant

[spice whale]

i see

I guess I've mainly only read stuff with an "all rings are assumed commutative unless otherwise stated" at the front

so the definition of integral domain would just be lack of nontrivial zero divisors

[glad osprey]

I see

I feel like most often you want integral domains to be commutative, plus "commutative integral domain" is way too long to type in every theorem

[chilly sparrow]

make "domains" commutative & call the old "domains" non-commutative domains 😇

and get rid of "Abelian" while we're at it

[void cosmos]

what

[elfin wraith]

Well you start with the most general thing and then specialise, not the other way around, so no?

[elfin wraith]

How old is the book? This all seems wild

[glad osprey]

First edition 1986, second (this) edition 1994

I love how a book can cause such an outrage in groups-rings-fields

[valid night]

Can someone help me with this? For the first I tried proving that $\mathbb{R}[X]/(f)$ is isomorphic to $\mathbb{R}[X]/(X^2 + 1)$, and for the second I tried proving that $\mathbb{R}[X]/(f)$ is isomorphic to $\mathbb{R}[X]/(X^2)$, both to no avail. For the third I have no idea where to even start

[bitter rover]

Isomorphic as what, vector spaces?

[valid night]

My bad, I meant R[X]/(f)

Let me edit it

[bitter rover]

Oh, I see. Uhh

Well, ok, morally what's going on

The discriminant says how many real roots there are.

[cloud walrus]

Kroros

[bitter rover]

If there are no real roots, it looks like C.

If there's one real root of multiplicity two, you get something where e^2 = 0 because f(x) = (x-a)^2

If there are two distinct real roots, map each to its own copy of R.

I suppose it means isomorphic as rings.

If there are no real roots then the two roots are conjugate to each other

[valid night]

Ah, so you're saying I should try mapping the roots to C, R[e] and R^2?

[coral spindle]

Well you're gonna struggle with R[e] because this is no longer a field

[bitter rover]

Right

[valid night]

Does epsilon not have an inverse?

Or is there another reason it's not a field?

[coral spindle]

Well let me show you a presentation for R[e]

R[x]/(x^2)

[bitter rover]

If you're curious, the set R[e] with e^2 = 0 are called the "dual numbers": https://en.wikipedia.org/wiki/Dual_number

They're actually quite handy. You can use them to implement automatic differentiation in a computer program: https://en.wikipedia.org/wiki/Automatic_differentiation

Anyways, no, it can't be a field because e is a zero divisor and fields don't have any zero divisors

[valid night]

Ah, I see, thank you

I'll try to figure the proof out

[gloomy mountain]

I didn’t know ChatGPT could explain like this
Rings on steroids 🔥

[vapid vale]

halfway house is insane

[bitter rover]

Can't say I see how this clarifies things. Haha.

[gloomy mountain]

Oh wait WHAT

I didn’t even know what it meant

[gloomy mountain]

You know what they say about stories: the more ridiculous, the more memorable

[bitter rover]

But what is being remembered??

Remember: algebras are like rings on steroids.

And therefore...what, exactly? Haha.

[gloomy mountain]

Yes scalars = steroids

[bitter rover]

Oh ok

[gloomy mountain]

So you are injecting scalars into the algebra

External aid

[bitter rover]

So how is an algebra distinct from a module?

[hidden wind]

never seen that, for me a domain is just short for integral domain (which may be either comm or non-comm)

[tardy hedge]

from my education thus far i have seen domain to just mean integral domains, and integral domains are always commutative

[hidden wind]

wikipedia agrees with this, for whatever that is worth

[glad osprey]

I can't remember if I've seen domain being used to mean non-commutative integral domain actually, but that might just be because I haven't done much non-commutative stuff

[spice whale]

that first one is very silly

like

yeah

[tardy hedge]

rings on steroids is crazy

someone should now chat gpt an image of rings on steroids

1. somehow personify the algebraic structure of a ring

2. make it on steroids

[dull ginkgo]

Associative algebras

[languid trellis]

I'm working on this exercise. The forward direction is simple enough apart for showing that roots have the same multiplicity (0 idea on how to approach). For the reverse implication, I again have no idea. I have worked through 3 examples, and have calculated the reverse implication if the splitting of F is E = F(r1,r2), and F is of char 2, by expanding f(x) = x(x+r1)(x+r1)(x+(r1+r2)). The examples I've worked through haven't been suggestive on how to approach this in general. Any suggestion would be great ( :

Perhaps something that could be useful is that if char F = p, a in F, then x^p-a is either irred in F[x] or is a pth power. Showing that if r1 is a root of f(x) in E the splitting field, then x^p -r1 divides f and is thus equal to (x-r1)^p

I'm just speculating now because I am completely lost.

But that lemma would only be useful if a_{m-1} neq 0, and a_m = 0

So perhaps there is a way to extend it?

[rocky cloak]

They have a multiplication / ring structure

[bitter rover]

I understand that. I meant in the metaphor of a "ring on steroids"

[rocky cloak]

The word ring is there

[bitter rover]

Ok, but the metaphor (apparently) is that scalar multiplication is steroids. So I guess it's just something else on steroids?

(I'm objecting to the metaphor. I'm not confused about the definitions involved.)

[rocky cloak]

An algebra is a ring with a scalar multiplication

So I don't see the injection

Objection*

[bitter rover]

🤷‍♂️

[rocky cloak]

Like
Module = abelian group on steroids
Algebra = ring on steroids
Makes perfect sense

[bitter rover]

We disagree on what makes for a helpful metaphor, then. That's fine.

[rocky cloak]

Idk if it's helpful, but if it's easier to remember "stereoids" than "scalar multiplication" then that'll do it

[bitter rover]

gets busy making flash cards, one side says steroids the other has the definition

[dull ginkgo]

If F has characteristic p then x |-> x^p is an endomorphism (The frobenius endomorphism), due to binomial theorem and the "middle" terms being killed off by the characteristic

x^p^m is this endomorphism iterated so yeah any sum of x^p^m's is an endomorphism i.e in the ring F[f], f = x^p

[dull ginkgo]

Ergo, ponder when the polynomial f(x) is an additive endomorphism, i.e f(x + y) = f(x) + f(y)

[glad osprey]

Med students somewhere: anabolic steroids are basically just scalar multiplication by a coefficient k > 1

[vapid vale]

finite dimensional basic unital associative algebra = ring on protein powder

[bitter rover]

The macronutritional theory of algebra

[rocky cloak]

= arrows on your back
More like it

[coral spindle]

Bringing new meaning to Sun Tzu's theorem

[tardy hedge]

Why is every module a quotient of a projective module?

I understand why every module is a quotient of a free module, since u can just take free module with basis M and then do a surjection

[wraith cargo]

free modules are projective

[tardy hedge]

Oh. Yeah.

Lol thx

For a field F, every F-module is projective, because every F-module is free?

[wraith cargo]

Yes

More generally this holds

[tardy hedge]

Free modules have no torsion elements, so no torsion R-module can be projective, because it could not be a submodule of a free module

Is that true?

[warm dove]

why is it natural or "obvious" to see the following:

A commutative ring with identity is a field if and only it has no nonzero proper ideals

basically, I guess, I'm struggling to see where the intuition for characterizing "fieldness" in terms of ideals comes from

[dull ginkgo]

The connection is actually between units and (PROPER) ideals

being a unit is actually, in a sense, being "outside" ideals

Actually this extends to the weaker "sided" units

[rocky cloak]

Is R a domain?

[bitter rover]

Think about what happens with an ideal if your commutative ring is a field.

And relatedly, what happens once your ideal is shown to contain the identity

[dull ginkgo]

It's a bit more obvious in that context.

A left ideal is an additive subgroup of R, G, that is closed under multiplication from the left by ANY element of r, i.e rG is in G for each r
A left unit is an element x where there is some rx = 1.

So what happens when x is in an ideal S?

[warm dove]

a unit gives rise to the entire ring

I think I understand how this works (with no nontrivial ideals, principal ideals give rise to the whole ring and therefore the unity)

[dull ginkgo]

In that way, a left unit can never be in a proper left ideal

[warm dove]

I guess my qustion is "what's the motivation for connecting the field structure to ideals?"

[dull ginkgo]

Or the set of units is the ring remove all the left ideals

[warm dove]

because this is also the thing with unique factorization

we talk in terms of ideals

and I guess the question is... why?

[dull ginkgo]

Well this is sorta extending the notion of "simplicity"

[warm dove]

they arise as kernels of homomorphisms

[warm dove]

aah

💀

that makes sense

[dull ginkgo]

A simple ring is where there is no nontrivial TWO-SIDED ideals, division rings are where there are no nontrivial one-sided ideals (if it's left-simple it is equivalently right-simple_

of course this is an equivalence when it commutes

but the picture is that quotienting by maximal ideals as modules (one sided) or rings (two sided) gives us "simple objects", like division rings

and a major strong point of these is that maps out of them can't have nontrivial kernels, i.e the map is a zero map or injective, no in between

furthermore for rings, we also get that if there isd no nontrivial left (right) ideals, then elements have multiplicative inverses since the genered left ideal Rr would have to be the whole ring for r neq 0

[dull ginkgo]

We often get fields out of quotienting by maximal ideals, which thankfully for most contexts we have some constructive one, or we use choice and zorn's lemma to assume they exist

[warm dove]

hmm

yeah ok I guess this makes sense

so the initial motivation comes out of practicality (how do we get fields?)

and then we get this bigger picture

[dull ginkgo]

yep, the fact that they are simple

[warm dove]

of fields exactly mean no nontriival idelas

[warm dove]

it's like how if we want a simple group we quotient by a maximal normal subgroup

hmm

[dull ginkgo]

exactly!

[warm dove]

ok

ty

[dull ginkgo]

similarly, you can consider generated subgroups, and how every element generates the group for the abelian case

[warm dove]

makes more sense

[warm dove]

ew fundamental theorem of finite abelian groups /j

my math teacher did the linear algebra proof in class instead of the "group-theoretic" approahc and it made no sense until I realized how obvious everything was

[dull ginkgo]

PID MODULE HORROR

[tardy hedge]

Idk, i wasnt thinking about that i guess

[dull ginkgo]

jk it's not terrible

[warm dove]

yeah but yk teaching that to a bunch of 15 year olds is cooked

the group theory test was terrible lmao

[dull ginkgo]

doom

[warm dove]

I dropped a B+ 💀

because I forgot how to compute cosets

and spent all my time trying to figure out S_4 / V (this is obvious in retrospect but)

[rocky cloak]

Well, what definition of torsion are you using in the non-domain case

[dull ginkgo]

Proving group representations of S_N via elements of the form (i i + 1) or (1 i) respectively were fucking miserable

surjectivity was the absolute worst

[warm dove]

ok this is harder than computing S_4/V

💀

[dull ginkgo]

what's V?

[warm dove]

kelin 4

[dull ginkgo]

Klien 4?

Ah

[warm dove]

Z_2 x Z_2

[dull ginkgo]

we both mispelled it \

lma

[warm dove]

yeah idk I knew it was S_3 but it took forever to write out two coset multipications to show non-abelian multiplication

and from there you can just go nonabelian of order 6

[dull ginkgo]

it gets a bit easier with the see love shit but still is nevertheless annoying

ring theory is a bit easier and revolves around less bullshit specific arguments

imo

[warm dove]

idk I mostly have intuition for number-theoretic rings bc I've done a lot of ENT

but I'm expecting that I'm gonna get cooked by matrix rings

and like M_2(Z)/M_2(kZ) or something

[dull ginkgo]

pain

[warm dove]

sk

group scare me

[dull ginkgo]

functorality thankfully is a bit nice

[warm dove]

What

[tardy hedge]

If R is a domain is what I said true?

[tardy hedge]

u a gym guy?

[tardy hedge]

If R is not a domain then every R-module would be a torsion module I guess right. Some r1r2 = 0 so for any m in M, if r2m is not already 0 then r1(r2m) is 0

in our class I do believe anytime torsion module was mentioned, R is an integral domain

[rocky cloak]

What you said is true for domains yeah.

A definition of torsion I like in general is module that doesn't have any homomorphisms to a flat module. But then it's just true by definition.

[tardy hedge]

Thanks. We didnt cover the definition of flat modules in our class, just covered how the tensor product functor would always be left exact

then moved to modules over pid

[rocky cloak]

Yeah, if you just use the naive definition, then the "torsion module" isn't necessarily a module in the non-domain case. So you may want to be a little careful

Possible pathces are to say that m is torsion if
rm = 0 for r not a zero divisor.

Or the thing with flat modules

[hidden wind]

something something injective

[glad osprey]

kind of I guess, not a steroid guy though

[tardy hedge]

cool me too

I mean like i just maintain now though. dont have the mental space required to make gains rn tbh.

thats the one nice thing of not being in school though. I had more mental space to focus on gains lol

[glad osprey]

Yeah, it can be hard to juggle studying and training (gains vs brains)

[vocal spindle]

With regards to isomorphic groups being equivalent, is it that they are precisely the same object written in a different way or are they just functionally the same similar to how a function with a different codomain but the same ‘rule’ can work the same way when applied but is still technically a different relation?

[bitter rover]

There are lots of perspectives on it. Check out When is one thing equal to some other thing? by Barry Mazur

https://people.math.osu.edu/cogdell.1/6112-Mazur-www.pdf

[vocal spindle]

Awesome thank you. I’ll give it a read. I thought the answer might not be so clear cut 😅

[bitter rover]

It's more a question of philosophy of math and only indirectly affects how mathematicians go about doing math.

Indirectly in that this-or-that perspective reflects someone's interests, what they pay attention to, what ideas occur to them and how they investigate those ideas, etc.

[vocal spindle]

So for all practical purposes I can call an isomorphism an equality?

[bitter rover]

One person's "practical" is another person's "theoretical"

[vocal spindle]

Fair enough. I’m quite new to group theory so probably won’t yet understand the nuance. I just don’t want to be stating falsehoods when I call groups equivalent.

[bitter rover]

Sometimes you need to be sensitive to it, sometimes you don't.

The thought of making "equivalence" equivalent to "equality" is called "univalence" and is part of the "third way" that Mazur describes in the paper I linked.

[hidden wind]

what a trip

[vocal spindle]

I’m reading through now. Thank you for your help

[rocky cloak]

If you just care about properties of a specific group I would say yes. But once you think about relationships between groups you have to be a little careful.

For example 2Z is a subgroup of Z, and 2Z and Z are isomorphic. But saying 2Z = Z would be confusing, because even though they're the same "as groups" they are quite different as "subgroups of Z"

[vocal spindle]

Got it

[tough raven]

Incidentally, to me this really emphasises that you could replace "non-zero divisors" with any multiplicatively closed subset S, and m is "S-torsion" iff it becomes 0 when we extend scalars to S^{-1}R.

[rocky cloak]

Yeah that makes sense. That's also fits with terminology like 2-torsion for example

[tough raven]

This is just 2ℤ and ℤ not being isomorphic objects in the coslice category of the category of groups over ℤ. 😤

[old hollow]

Let A be noetherian ring, and let M be an A-module. Why is it true that if M is generated by a single element, then M = A/a for some ideal a in A?

[tough raven]

Do you know that any cyclic group is of the form ℤ/nℤ for some non-negative integer n?

The idea is exactly the same.

Explicitly, ||let m be the single element and consider the map f: A → M: a ↦ am. By hypothesis, f is surjective, so by the first isomorphism theorem, it defines an isomorphism of A-modules from A/ker(f) to M||.

Incidentally, this doesn't depend on A being Noetherian.

[rich granite]

We know that $\overline{F_E}$ consists of elements of $E$ that are algebraic over $F$. However, I can't seem to grasp the idea of the algebraic closure of $F$, $\overline{F}$. By definition, it is said that $\overline{F}$ is an extension of $F$ whose polynomials in $\overline{F}[x]$ all have zeroes in $\overline{F}$. With this in mind, I have the following questions:

\begin{enumerate}
\item Is $\overline{F}$ the intersection of all such $\overline{F_{E_i}}$ for all $i$?
\item How is $E$ such that $F \le E$ contained in $\overline{F}$?
\end{enumerate}

Thank you!

[cloud walrus]

Inxi25

[old hollow]

this is not quite right

first, you shoud be saying "an algebraic closure is ...", because it might not be unique

(it turns out to be unique, but I assume you haven't proved this yet)

also, what is E_i? You haven't defined that yet

and wdym by "how is E such that F ≤ E contained in Fbar", I don't understand the grammar of that sentence

[old hollow]

Oh, suppose M is generated by the element m. Then we have a homomorphism A --> M sending 1 --> m. This is a homomorphism of A-modules. Since m generates M, it is surjective. Therefore its kernel is a submodule of A, that is, an ideal a in A. Therefore we have A/a = M by the first isomorphism theorem applied to the additive groups

damn that's not so bad

@tough raven~~ is the first iso theorem true for homomorphisms of modules?~~ sorry i couldve just looked this up lol

[slim kayak]

What on earth is that

Mfw when I am apply the categorical imperative to obtain a generating and cogenerating set of my category

[plucky dome]

actually he talks about the categorical imperative elsewhere

[slim kayak]

Oh uh

Mfw I am trying to learn field theory but accidentally end up at the existence of the categories of thought

[keen terrace]

Guys I have a test in a bit and I'm not sure about something.. can anyone help me solve a doubt?

If (G,+) is a group, and a,b ∈ G,
then <a,b> is {k*GCD(a,b):k∈Z}.
What about in (G,•)?

[coral spindle]

gcd(a,b) has no meaning in an arbitrary group G.

(G, .) could be just a completely different group, with zero relation to (G, +).

You mean G = Z, don't you? But you've forgotten that (Z, *) is not a group.

I think you need to adjust your question before you get a proper answer.

[keen terrace]

More like Z/nZ

This is the exercise I'm referring to

(It's italian)

[coral spindle]

(Z/nZ, .) is still not a group, so you still need to explain your question.

[keen terrace]

It isn't? I thought it was
I'm a bit confused rn

So (Z/nZ,+) is and (Z/nZ,•) isn't?

[lone niche]

Consider Z/4Z for example

What is 2x0 and 3x0? What can you conclude from that by cancellation?

[keen terrace]

0 and 0

I know that |Z/4Z|=4, so |2|=|3|=4
Is there something I'm missing?

[lone niche]

Can you elaborate on how you obtained that?

[keen terrace]

No wait I got confused, |2|=2, |3|=4

[lone niche]

Can you further elaborate on how you got that?

[keen terrace]

|a| = k:a•k≡0 mod n, so
|2|= k:2•k≡0 mod 4, so k=2
Same thing for 3

[lone niche]

So 0 is the identity in (Z/4Z, •)?

[keen terrace]

Yh

[lone niche]

Are you sure?

[keen terrace]

Wait in • is 1 sorry

in + is 0

[lone niche]

Okay so what is |a| again?

[keen terrace]

In + the above, in • is k: a^k ≡1 mod n

[lone niche]

What is |2| then?

[keen terrace]

Still 2

2²≡0 mod 4

Wait im panicking haha

If qe are on • the 2 doesn't have an order

[lone niche]

How is that possible?, given that every element of a finite group must have finite order.

[keen terrace]

Can't an element if a finite group be aperiodic?

0 can't be the period, and 1, 2 etc can give powers of 2, so it can't be ≡1

[lone niche]

Hmmm I guess you might not have proved this yet, but in a finite group every element will have a finite order.

Anyways, just take that for a fact for now, I will show you something more convincing later.

[lone niche]

What's the answer to this?

[keen terrace]

That Z/4Z isn't a group?

Maybe?

U(Z/nZ) is tho, right?

Just so i don't implode during the exam

[lone niche]

I assume that's the group of units then yes

[keen terrace]

The invertible elements

[lone niche]

Yup

[keen terrace]

Okk
Thx a lot

[lone niche]

The more convincing argument is here

You have 2x0=0=3x0, so that by cancellation you have 2=3

[rich granite]

all algebraic extensions of F (please correct if this idea is wrong)

[rich granite]

E, an algebraic extension of F, contained in Fbar

[crystal vale]

I have to show that Q under addition is not group isomorphic to R^+ under multiplication.

I know R^+ under multiplication is isomorphic to R, now any hint how can I show that R and Q is not isomorphic?

[coral spindle]

Hint: an isomorphism is, in particular, a bijection.

A different hint: for any x and y in Q, there are some integers a and b such that ax = by.

[crystal vale]

Got it

[crystal vale]

I got the previous hint but I am interested in this hint

[coral spindle]

Hm I should have restricted it a bit more

x, y must be nonzero; a, b are nonzero.

[crystal vale]

So if I let f be that isomorphism then af(x) = b f(y), but I think they tell us more than this

Why do we consider those are non-zero

Yes for zero maybe it is trivial

Oh no

Yes they have to be non-zero

Both

[coral spindle]

What does it mean for a number to be irrational?

[crystal vale]

You are great ❤️

May I get to know what the thought process is behind this? I thought something like that I didn't get it

[coral spindle]

I just thought of it, I can't say it came from anywhere in particular

Perhaps I have seen something like it before and remembered it without realising, idk

[crystal vale]

Cool

[coral spindle]

I disliked the first hint I gave because it wasn't algebraic, so I tried to come up with a more interesting reason, and lo I did

[crystal vale]

But the first one is easy but the second one is conceptual

[coral spindle]

It's easy only because you know a powerful fact about the real numbers

I would argue that the 2nd is more elementary, in some sense

[crystal vale]

Agree

Thank you

[hidden wind]

you’re so swag for that

[coral spindle]

I LOVE ALGEBRA !!!!!!!!!!!!!!!

[vagrant zinc]

Algebraic topology¿

[coral spindle]

No

Stinky

[crystal vale]

Suppose that there is a positive even integer n such that a^n = a for all elements of some ring. Show that -a = a for all a in the ring.

I conclude that (2^n -2)a = 0 for all a in R

But I don't think it is the correct way here

[dull ginkgo]

There has to be a minimum n for each x such that nx = 0, it's the additive order as shown

[crystal vale]

But a^n = a here it is ring multiplication

[dull ginkgo]

but this is the crutial part

each element has additive torsion so that order exists for each x,

[crystal vale]

Yes

[dull ginkgo]

ord(x) divides each n where nx = 0

But you've shown kx = (kx)^n = k^n x^n = k^n x implying (k^n - k)x = 0... so n ord(x) must divide k^n - k for each k

[dull ginkgo]

(note: k and k^n - 1 are coprime for each n)

[crystal vale]

So is it something we will conclude that only 1 and 2 is the possible candidate which divides k^n-k for each k

[crystal vale]

Yes

[dull ginkgo]

yes.

[crystal vale]

So order divides k or k^n -1 or both

[dull ginkgo]

k xor k^n - 1

(coprime)

ord(x) = 1 implies x = 0

[crystal vale]

Yes

I am not sure about how to show only ord can be 1 or 2

[dull ginkgo]

I haven't done this exercise in a long time

trying to remember the last step

[dull ginkgo]

@crystal vale I got it lol, assume prime p divides ord(x), p is naturally coprime to p - 1
then p | ord(x) | (p - 1) * ((p - 1)^(n - 1) - 1), so p | (p - 1)^(n - 1) - 1

reducing mod p: 1 = (p - 1)^(n - 1) = (-1)^(n - 1) mod p, but since n is even, n - 1 is odd, so 1 = (-1) mod p, and thus 2 = 0 mod p, implying p = 2

thus ord(x) is a power of 2, assume ord(x) = 2^k, then:
2^k | 2 ( 2^(n - 1) - 1) thus 2^(k - 1) | 2^(n - 1) - 1, which is coprime to 2 so k = 1, thus ord(x) = 2

2x = 0 for every x, thus x = -x

this probably isn't the intended way but aaaaaaa

[crystal vale]

Ah thank you @dull ginkgo

[dull ginkgo]

that was not obvious at all lol

[crystal vale]

Yes I know

Show that an element in a finite ring with 1 is a unit if it is not a zero divisor.

Let R = {a_1,..., a_n} be given ring.
Let a \in R be not zero divisor, then {aa_1,..., aa_n} = R because a is not zero divisor so aa_i ≠ aa_j when i ≠ j.

So aa_i = 1 for some i and similarly a_ja = 1 so a is a unit.

Did I miss any point?

[dull ginkgo]

nope, pretty good use of pigeonhole

[crystal vale]

Okay thank you

[dull ginkgo]

typically i've seen people use a_n = a^n and use pigeonhole, the way you did is a first for me lol