groups-rings-fields

[spiral wolf]

I think i devised a method using gaussian elimination where the columns are from the powers of 2^1/4+i

[spiral wolf]

We got it

Thank you very much you are a lifesaver

[celest furnace]

Awesome great job!

[crystal vale]

In ring homomorphism is it necessary that 1->1 ?

[void cosmos]

no

[crystal vale]

But in many books given that homomorphism should follow this property

[void cosmos]

yes , it is most of the time assumed in commutative algebra ig

but in like total generality , it is not necessary

[crystal vale]

Okay

[viral tiger]

How do i determine the center of GL(2, R)

[crystal vale]

Hint: by using elementary matrices

[celest furnace]

Can anyone give me a (small) hint on how to prove this -- tried a lot of things none of which have worked

Was able to get it for n = 2 but no clue how to generalize

[cobalt heath]

Which direction are you struggling with?

[celest furnace]

Norm => of that form

[cobalt heath]

IIRC you have to utilize some polynomial for the opposite direction

Wait, no

[cobalt heath]

Did you prove that if norm is 1, then alpha = beta / sigma(beta) ?

[celest furnace]

No thats what im struggling on

[cobalt heath]

Ah sorry, I misread

Yeah so you have to 'construct' beta

So that sigma(b) = (1/a) b.

[cobalt heath]

You can approach it as solving this equation to compute b

(Indeed, not easy)

[celest furnace]

Can I have a slightly bigger hint

[cobalt heath]

My memory is lacking, but you need something like c_1 a + c_2 sigma(a) + c_3 sigma^2(a) + ...

Ah, might be multiplications instead of '+'

[crystal vale]

2 hint

[celest furnace]

4ab = (a+b)^2-(a-b)^2

[boreal inlet]

@cobalt heath Hey, I was wondering, if the problem you led me to the solution on, if I take any field of characteristic 0 and then start the whole thing, would the result be the same?

[cobalt heath]

Yeah, I think so

[boreal inlet]

Though, how would we take care of the root business

I could claim the root is in C because C is algebraically closed

[cobalt heath]

You need to work around not being algebraically closed, but will work

Yea so

[boreal inlet]

Let's say the field is Q

[cobalt heath]

You can find roots of f(x) / g(x) = f(0) / g(0)

There should be at least one root, 0.

[stark hearth]

this is Hilbert 90

[boreal inlet]

But only in Q. I think if the root exists in Q, that would only contribute 0.

If does not, now we have a lot more cases instead of just the constant polynomials. Because irreducibles.

[boreal inlet]

Oh right

[cobalt heath]

Then you can use similar arguments

[boreal inlet]

So that would be the field itself?

[cobalt heath]

Yeah

[boreal inlet]

Thanks, I'll be trying to write this out and see if I get stuck

[cobalt heath]

Good luck!

[boreal inlet]

I didn't get stuck

I worked with K closure and the same argument worked

But I see the proof fails immediately when the field is of characteristic p

[crystal vale]

In construction of ring of fraction, how can I show that every element of D is unit in its rings of fraction?

[coral spindle]

What have you tried so far?

[glossy crag]

Can you have Galois extensions in characteristic p with Galois group Z/p x Z/p? I figure something like the composite of two distinct Artin-Schreier extensions of F_p(t), e.g. for t and t-1, ought to work, no?

[crystal vale]

Define equivalence classes and then define the set of all equivalence classes be ring

[pliant forge]

Let E be an algebraic extension of F and $\sigma\colon E \to E$ be an embedding of E into itself over F. Then we want to show that $\sigma$ is an automorphism.

For a $e \in E$, there is a minimal polynomial $p(x) \in F[x]$ with root $e$. Set $E’$ to be thesubfield of E generated over F by all roots of $p(x)$ that are in E. Then $E’$ is a finite algebraic extension of F. Every root gets sent to a root by $\sigma$. So $\sigma$ maps $E’$ to itself, hence $\sigma(E’) \cong E’$.

Now my question is why does the author continue with showing that this implies $[\sigma(E’):F]=[E’:F]$, and since $\sigma(E’)$ is a subspace of $E’$, $\sigma(E’) = E’$. Hence, there is a $b \in E’$ that gets sent to $a$

[cloud walrus]

Eso

[pliant forge]

isnt the fact that its isomorphic to E’ enough that there is some element b that gets sent to a?

and what does the equality on the last paragraph mean? are the roots required to be sent to the same root?

[south patrol]

I agree with you yeah

Slightly more concisely: E is a union of vanishing sets V of irreducible polynomials, and sigma sends each V -> V injectively, hence bijectively

So that sigma is surjective overall

[coral spindle]

So you haven't actually tried to find inverses yet. Try just writing down what you think the inverse of an element might be and proving it works. Just try.

[crystal vale]

Got it thank you

[rocky cloak]

Let F be GF(p) and K= GF(p^p) be the degree p extension. Consider F(t) inside K(t)[x]/(x^p - x - t).

The Galois group is generated by the Frobenius map on coefficients and the map given by x |-> x+1. These have order p and commute, so the Galois group is CpxCp

[crystal vale]

Not get it

[cloud walrus]

smidgin

[south patrol]

Good question, and I think yes it is possible

[coral spindle]

Not sure what you mean by this. The isomorphisms do have to be isomorphisms, which means that in particular they satisfy the usual conditions. Maybe you can find a clever way to avoid doing hard work for some of these, but in the end you will prove this.

[south patrol]

Basically you can view V_4 as having presentation <a,b,c | stuff>. So any bijection {a,b,c} -> {a,b,c} will induce a unique map V_4 -> V_4

Like the point is that the presentation is symmetric in a,b,c

Hm

[coral spindle]

I mean, yes?

Often?

But this is a strange question because it's so granular

Yes, you will need to verify that the things you claim are isomorphisms are indeed isomorphisms

[south patrol]

Often you will have a nice handle on your group which puts restrictions on which bijections can be homomorphisms

For example, it must preserve orders of elements

[coral spindle]

Oh I see yes

[south patrol]

Well it is so small that you have a complete description

[coral spindle]

Automorphisms send sets of generators to sets of generators. If you identify a nice set of generators, you only need to think about where they get sent. That's a nice way to narrow things down.

A good example of where it's very easy to find the automorphisms is Aut(Z/m), precisely because of this generator observation.

[chilly ocean]

Take u and v s.t u+v = 1 and then take a high enough power of it

[crystal vale]

How it will help me?

[chilly ocean]

Because it will be a sum of elements in I^2 + J^2 which is 1

[south patrol]

This notation for ideals is hurting my brain lol

Another nice way is to consider radicals

Or is that what you had in mind nine

[chilly ocean]

Nah what I had in mind was using the binomial theorem

[south patrol]

Like || sqrt( I^2 + J^2) contains I and J, so it is the whole thing. So 1^n in I^2 + J^2 for some n lol||

[chilly ocean]

Yeahhh but it seems too advanced for that exercise tbh

[south patrol]

Sure

[chilly ocean]

u mean M_1 and M_2 tho right

😭

not I and J

[crystal vale]

Is I^2=I?

[chilly ocean]

No

[crystal vale]

Why?

[chilly ocean]

Take (2)

[crystal vale]

Okay

[chilly ocean]

You could also look at u+v=1, then square and get u^2 + 2uv + v^2 = 1 and then multiply by u..... (and do the same for v)

[coral spindle]

I'm not aware of any easy way to find all automorphisms of a group

[dreamy granite]

hey, can someone maybe help me in terms of mathematics on call or something?

[coral spindle]

I'd recommend just asking your question

[flat tree]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[crystal vale]

Is it like that u^3+ 3u^2v +3v^2v+v^3=1 now we can say that u^3+3u^2v is element of I^2 and 3uv^2 + v^3 is element of J^2 hence then we can say that I^2+J^2= R ?

[formal ermine]

if only this would be true 🙏

[chilly ocean]

Yes

[crystal vale]

Thank you

[flint cave]

Is the the word problem for groups equivalent to being able to write a program to tell if a word is trivial or not from its presentation?

Today my professor was telling me 'with a grain of salt' that it is slightly different and having solvable word problem doesnt mean being able to implement it as code

More specifically an amalgam of two groups have solvable word problem when the groups have solvable word problem and also when the subgroup along which we amalgamate has to be countable/countable index or something right? (since otherwise the algorithm for unique reduction will involve some sort of choice)

[coral spindle]

Yes. I imagine your professor was referring to the fact that Turing machines are infinite and our computers are, sadly, only finite.

Perhaps also your professor was saying that the fact that the word problem is computable doesn't necessarily mean that we know an algorithm to compute it, but merely that it exists.

[stark helm]

for the first question, is it correct to show it by using division algorithm?

i mean by saying every element in Z3[x] can be expressed as q(x)*(x^3+x62+1)+r(x)?

[glossy crag]

Yes

[sturdy yew]

a few questions:

1. if order of a group is taken as a function, does ord(G/N) (G/N being a quotient group) imply ord(G)/ord(N) = ord(G/N)?
2. is a "reverse order" function well-defined, in that if you have a natural number n which divides evenly by m, ord^-1(m/n) exists such that ord^-1(m/n) = G/N, G being a group of order m and N being a normal subgroup of order n? or would this be an abuse of notation? i presumed that if 1. holds, this reverse order function would similarly hold and be well-defined
3. from this, is there a symbolic derivation of cauchy's theorem, meaning going from ord(G) * 1/p = m (p being a prime and m being a natural number), you can go to ord(G) * 1/m = p, and apply a reverse order function ord^-1 such that the "reverse order" of the quotient is simply G and a normal subgroup N of order m, and p becomes a quotient group of prime order? this would, of course, imply cauchy's theorem if it were to hold

[flint cave]

Can you give an example of an algorithm computable by a Turing machine but not by a computer?

[crystal vale]

In Modules , if R is act on group M , then what is operation of M, under addition or any binary operation?

[delicate orchid]

Yes, basic counting
No, this is far from well defined. You’d have to map into classes groups of groups up to order. By Lagrange m/n must be an integer anyway so I don’t get the point.

[sturdy yew]

pardon? i made the point clear, mind you, just an attempt to see if cauchy's theorem can be implied symbolically, again with the example "derivation" (again, i'm asking if this is well defined, and so i don't know if it's correct), from the formula cauchy's theorem is implied, also what does "classes groups of groups up to order" mean, i don't know if i'm reading this wrong or something

if you were to define a reverse order function it would map any natural number to a group with that order

for cyclic groups, this can be pretty easily defined: it maps any natural number n to a cyclic group of order n, and of course regardless of the actual group obtained there still exists an isomorphism between that and any other cyclic group of order n

the issue isn't at all with the definition of reverse order, it's with the idea of taking the reverse order of a quotient, would taking the reverse order of this quotient, that is, for a quotient m/n, ord^-1(m/n) would this be equal to ord^-1(m)/ord^-1(n)? the reason this is such a big deal is because would it be an abuse of notation to presume that if ord^-1(m) is a group G and ord^-1(n) is a group N, G/N is then a quotient group equivalent to ord^-1(m/n)?

the issue with the definition is that it holds perfectly for natural numbers, but you run into an issue when you obtain from it a group divided by another, logic tells that this should simply be a quotient group (and it almost makes sense, the quotient group of a group and a normal subgroup can be of prime order GIVEN THAT a prime number is a divisor of g, and you derive this former from the latter) but i genuinely just don't understand how this works, it's as if it reifies everything i formerly knew about quotient groups

[mighty kiln]

For each element in the domain of a function, there exists a unique output

Even when n is a divisor of ord(G) there is no guarantee for a normal subgroup of order n to exist

And there are also many cases where you have many different normal subgroups of size n, so even if the quotient exists it may not be unique

[sturdy yew]

oh ok thank you

[crystal vale]

In this definition if I remove that hypothesis f(R) is contained in center of A and then if I define r•a=f(r)a then is it R-Module?

[mighty kiln]

Yes

[crystal vale]

And I have one doubt there is given Klein 4 is vector space over F2 so here R=F2 and M is Klein 4 so here operation of M is not addition right?

[mighty kiln]

The vector space addition is the Klein 4 group operation

[crystal vale]

Okay

Is vector space addition is exactly the Group operation on which we act

[mighty kiln]

?

It is the Abelian group structure of the module M yes

The Klein 4 group is isomorphic to {(0,0), (0,1), (1,0), (1,1)} under pointwise addition in F2

[crystal vale]

Yes

[mighty kiln]

You add component-wise multiplication by F2 and it becomes a vector space

[crystal vale]

Okay

And any ring with identity is a Z-algebra? I think ring must be commutative

[mighty kiln]

Any ring with identity is a Z-algebra

Since image of Z must commute with everything

[crystal vale]

How can I show this image of Z must commute with every element of R

[mighty kiln]

Center is a subring

And f(1) is in it

[crystal vale]

Not get it🥲

[crystal vale]

Is it like if I denote identity of R is 1 then m let be in image of f then for any s in R, m•s=(1+1+.......+1 - m times)•s = s+s+s+.......+s m-times then it is equal s•(1+1+1.............+1 m times) =s•m thus m•s=s•m ?

[mighty kiln]

Yes

More generally, if f: G → H is a morphism of groups, S generates G, and f(S) is in the subgroup H' ⊂ H, then f(G) is in the subgroup H'

So you can take G = Z, S = {1}, H' = center

[crystal vale]

Oh because S generates G so f(G) contain in H'

Thank you

[crystal vale]

If G is infinite cyclic group so how many generators there are?

[dull marsh]

You just said that G is cyclic

Or you mean which elements can be the generator?

[crystal vale]

Which element can generate the group?

[dull marsh]

There's just one

Infinite cyclic groups are isomorphic to Z so let's talk about Z

The only generator of Z is 1

[summer brook]

I have a map from group G = < x,y : r > to a group H = < X, Y > with phi(x) = X and phi(y) = Y. I'm trying to prove it's a homomorphism but I'm stuck because I don't know how to find what arbitrary g in G maps to, without assuming it's a homomorphism. Algebraically that is all I know. Can I do this algebraically? Or do I need to use other things as well?

I seem to be going in circles, so want to know if I'm missing something

[coral spindle]

This is not true; -1 also generates Z.

[dull marsh]

Oh whoops my bad

1 and -1 yeah

[crystal vale]

Yes

[pliant forge]

is it right to say that the root $\alpha = \frac{1}{10} (-1 \pm 3\sqrt 11 i)$ of the polynomial $p(x) = 5x^2 +x+5 \in \bZ [x]$ is an example such that $\bZ(\alpha) = \bZ(\alpha^2)$ and $[\bZ(\alpha) : \bZ]$ ($=2$) is not odd?

Professor is hinting to cyclotomic extensions, but this seems to already fulfill the requirements?

Edit: it should be Q not Z as its required to be a field, but id assume this root still makes this valid, unless im missing something?

[cloud walrus]

Eso

[crystal vale]

Is there any proper subgroup of Q which is not cyclic

[mighty kiln]

The dyadic rationals

[crystal turtle]

Many, in fact. There are countably many cyclic subgroups of Q, but uncountably many subgroups of Q.

[dull marsh]

How can that be shown?

[pliant forge]

ah ic i guess what ive come up with is basically the hint given

[crystal vale]

Is it like any cyclic Subgroup of Q is form of (n) where n is integer so integer set is countable

[crystal turtle]

||Injection from the power set of primes to the set of subgroups||

[dull marsh]

Ah

[crystal vale]

Is it correct?

[south patrol]

Not quite

You have subgroups generated by any rational and indeed these are always distinct (except q and -q generating the same ones)

So from this it is clear that there are exactly countably many such subgroups

[south patrol]

Nice

V cool

[crystal vale]

Oh I actually thought as ideal my mistake

[south patrol]

Oh lol so Ryx is it like

||for each set S of primsa consider set of rational with denominators formed from S||

[crystal vale]

Ryx?

[crystal turtle]

Yeah

[south patrol]

Yes, the person who helped u

[crystal vale]

In this mapping how can I show 1 map to 1(R) hint

[dull marsh]

Q is a field so every ideal is either the zero ideal or Q itself

[crystal vale]

Yeah that's my mistake I actually thought subgroups as ideal

[south patrol]

Sure, virtually any degree 2 poly will work

It's just it is way easier to consider quadratics with easy roots

[crystal vale]

By using given property r•(ab)=(r•a)b=a(r•b) ?

[dull marsh]

Yeah wait shouldn't it follow from the definition?

[elder wave]

it follows because all elements except 0 are units

[dull marsh]

I mean the 1_R getting mapped to 1_A

[cloud walrus]

smidgin

[coral spindle]

It looks OK at a glance but it is much easier if you first prove the lemma $aH = bH$ iff $b^{-1}a \in H$.

[cloud walrus]

Boytjie

[coral spindle]

Well, I suppose in any case it just follows once you realise this is an equivalence relation, but maybe you aren't familiar with that.

Sure

[vocal pebble]

Yeah

[crystal vale]

?

[stark helm]

is there any recommended way to figure out the inverse of elements in this question? I can only see that x^3+x^2=x(x^2+1)=x^2*(x+1), which is the inverse of x^2 and x^2+1.

[rustic crown]

in general you would do euclidean algorithm to find f and g such that
(x^3+x^2-1) * f(x) + (ax^2 + bx + c) * g(x) = 1
so modulo the ideal, g(x) is the inverse.

another way is to just plug g(x) = a'x^2 +b'x + c' and solve for a', b', c'

[stark helm]

do you think we need to use ax^2+bx+c or x^2+bx+c in this equation?

[rustic crown]

idk can't say without actually carrying out the computation

[rustic crown]

doesn't really matter that much.

if a is non-zero then (ax^2+bx+c)^-1 = a^-1 * (x^2 + b/a x + c/a)^-1

[stark helm]

does g(x) mean g(x)*(ax^2+bx+c)=1?

[rustic crown]

yea but modulo I

[cloud solar]

Find n>=2 s.t. there is a field K with n elements with the property that for every x in K, there is y in K with x^2=y^3.

n=p^t with p prime and t>=1

[analog hemlock]

Considering the group of f(x) = 1/x, g(x) = 1-x and their compositions:

• What are its elements?
• What subgroups are there?
• What is a partial ordering on the subgroups?

[cloud solar]

Then if p=2, the function from K-{0}->K-{0}, f(x)=x^2 is bijective so the extension to K is bijective. Now we can rewtrite the problem as: for every x in K there is y in K s.t. y^3=x. That means the function g:K->K, g(x)=x^3 is surjective so bijective because K finite. and that means also the function x^3 from K-{0}->K-{0} is bijective. So 3 and 2^t-1 are coprime. So t must be odd right?

For p>2: the number of squares in the field K is (n+1)/2

[cloud solar]

How to deal with this case

[rocky cloak]

Are you aware that the multiplicative group of a finite field is cyclic?

[cloud solar]

Yes

[rocky cloak]

Right, so then the question is just for which n do we have that
2Z/(n-1) is contained in 3Z/(n-1)

||which is just equivalent to 2 being in 3Z/(n-1)||
||Which means 3-2 = 1 is in 3Z/(n-1)||
||Which means n-1 relatively prime to 3||
No case work required.

[rocky cloak]

Try to compute f(g(x)), and the order of f, g and fg in the group. Does it look like a group you're familiar with?

[cloud solar]

Ok so n is 3t or 3t+2

n=p^s. If n=3t then n=3^s. If n=3t+2 then 3t+2=p^s. 3t+2 has a prime factor of the form 3q+2. So n=p^s with p=3q+2 prime

[stark helm]

How can I prove that Ic is a non-empty set? I am confused because I can not guarantee that in the set R, we must have an element c such that f(c)=0.

although it can map to real numbers, but I don't know if that would be convincing to argue that Ic must be non-empty

[rose prism]

consider the constant function f(x) = 0?

or the polynomial f(x) = x-c?

or…

what

[static glen]

This is the kernel of some ring homomorphism and hence non-empty

[stark helm]

do you mean we can say Ic is non-empty because f(x)=0 exist in R? so we can pick it in Ic?

[celest furnace]

What

[static glen]

The map $\phi:R\to\mathbb R, f\mapsto f(c)$ is obviously a ring homomorphism with $I_c=ker\phi$ by definition

[cloud walrus]

Max

[static glen]

and kernels always contain at least 0

this here is even larger

it has even codimension 1 as a vector subspace of R

[stark helm]

I mean f(x)=0 must exist in the set R, so we can quote it to say Ic is non-empty?

[celest furnace]

Yes

[tulip glacier]

Hello! I am having trouble understanding how there exists an infinite cyclic group - what power of the generating element would the identity element be ? There is no such power, thus not all elements of the set are generated from a single element, thus no infinite group can be cyclic. What am I getting wrong ?

[prisma ibex]

take the group of integers Z under addition

it's generated by a single element (e.g. by 1)

so it's cyclic

[tulip glacier]

How would 0 be generated ?

[prisma ibex]

what

[celest furnace]

0 * 1

In general if G = <a> then e = a^0

[tulip glacier]

Does that really work ? I know it makes sense if we multiply this way, but theoretically, at this level we're still only talking about adding 1 to itself a finite amount of times.

[prisma ibex]

yes lol

[celest furnace]

0 is finite right

[tulip glacier]

I know what you mean, but it still feels like "cheating"... we're not actually composing 1 with 1 any amount of times.

Like, you literally have to do 1 + 1 + 1 + 1 + ..., starting from 1, no ?

[prisma ibex]

remember you also have inverses

a*a^{-1}=e

[tulip glacier]

Right, these also have to be generated somehow from 1 + 1 + 1 + 1 + ..., and they're not, causing further problems.

[prisma ibex]

no

you should review how group presentations are defined

[tulip glacier]

I agree that the whole numbers are generated by {-1, 1}, but it is not a single element.

[prisma ibex]

I think you're confusing monoids with groups

for any element g of any group G, g^n is an element of G for every integer n

[celest furnace]

<a> is by definition {a^n : n in Z}

In particular it contains a^0

[coral spindle]

You clearly need to review definitions

[tulip glacier]

Okay, I see, I see.

[coral spindle]

A much more straightforward definition of <X> is the smallest subgroup containing the set X, so this should clear confusions. Every other result mentioned here is a consequence

[prisma ibex]

actually group presentations sort of overcomplicate things here, this is literally just an immediate consequence of the definition of a group

[tulip glacier]

Yeah, my problem was that I thought the power had to be strictly positive, and not a whole number. Thank you all!

[stark helm]

for part(b), does anyone think that Ic1 uinon Ic2 will be an ideal? I consider Ic1={f in R|f(c1)=0} and Ic2={g in R| g(c2)=0}. Then Ic1 union Ic2={f,g in R| f(c1)=0, g(c2)=0}. Then it seems that f-g in R, f-g(c1)=f(c1)-g(c1) might not be valid since we don't know value of g(c1), can someone help me check if I am correct?

[narrow marsh]

you want to show that for any such f in R, if x is in the union so is fx, ie fx(c1)=0 or fx(c2)=0

does that make sense

[rocky cloak]

It might be easier to prove the more general fact that if I and J are ideals such that neither contains the other, then the union is not an ideal.

You may also replace ideal with subgroup.

[narrow marsh]

of course, you would have to first verify that the union is closed wrt addition :)

[rocky cloak]

Or verify that it isn't 😉

[stark helm]

so you mean the union of Ic1 and Ic2 are not an ideal?

[rocky cloak]

That's right, just like you were onto when noticing that g(c1) could be anything

[stark helm]

Therefore, I can just say that since g(c1) can be anything, I can suppose that g(c1)=1 in R, then f(c1)-g(c1)=0-1=-1, which can make union not an ideal?

[rocky cloak]

Yeah, or you also need to make f(c2) - g(c2) nonzero. But once you have that you see that f-g is not in the union.

[stark helm]

Another question is about when we have ring homomorphism phi: F(field) to R(ring), and we have proven that phi is an injection, then can I directly say that there is an injection from F to im(phi)?

[summer brook]

I'm still a bit stuck here, trying to figure out if I can do this algebraically or if I need other stuff. Anyone have ideas?

[rocky cloak]

Yes, if a function f:X -> Y is injective, then also the function X -> Im(f) will be

[rocky cloak]

If all you know about the map is that phi(x) = X and phi(y) = Y, and you have no information about what phi does elsewhere, then there is no way to tell if phi is a homomorphisms.

But I suspect you're asking the wrong question. A better question would be "does there exist a homomorphism with maping x to X and y to Y?" Or even better: "is there a unique such homomorphism?"

[summer brook]

Right okay. I'm given that map and asked to prove it's a homomorphism, so I guess I need to look more into the structure. I have geometric structure to work with, maybe that will help. Just would be easier if it was purely algebraic

[rocky cloak]

Well it is purely algebraic

It's just ||phi(xy) have to map to phi(x)phi(y) etc, so phi is uniquely determined||, and ||this defines a well defined homomorphism iff phi(r) is the identity.||

[summer brook]

Makes sense. In general if ||phi(x)=X and phi(y)=Y, do you have phi(x)phi(y)=phi(xy) (where x,y generate the first group and X,Y generate the second)?|| If that doesn't hold, I can't do anything with the relation (that I'm aware of)

[wooden ember]

i'm guessing here they mean k_D = (0 sqrt(7); sqrt(7),0)?

[coral spindle]

Yeah, this is sparse matrix notation and it's very standard.

[wooden ember]

no but i mean it shouldnt be -sqrt(7)

but sqrt(7)

for k_D

[coral spindle]

Oh I see

I think you are right

[glossy crag]

Jesus what's with all the typos.

[sturdy yew]

sorry to come back to this but of course even though this is generally true i feel as if in the case that the order of G divides by a prime number p, there also exists an integer m such that whenever the order of G divides by p, it implies G has a normal subgroup of order m (whether this implies that this normal subgroup is G itself or another subgroup) and the quotient group G/N has an order p

and so if the order of G divides by a prime to produce an integer m it implies the existence of a normal subgroup of G which is of order m

i'm not sure if this is true or not but i can't think of any counterexamples off the top of my head

given the converse of lagrange's theorem isn't true (again, issue with defining a reverse order function) i would presume that this is not generally true

[coral spindle]

A_11 has no normal subgroup of order 10!/2.

[sturdy yew]

right

and so what are the conditions where the converse of cauchy's theorem (or lagrange's theorem) holds (that is, if the order of a group G divides by a prime p to get an integer m, there exists a normal subgroup of G of order m and a quotient group of order p)

[coral spindle]

I am not aware of any universal conditions for this

[nova holly]

Sylow theorems give some similar stuff

[sturdy yew]

i'll look into it

thank you

[tough raven]

Doesn't Cauchy's theorem state that if p | |G|, then there is a subgroup of order p?

In general, I feel like finite groups tend to have a lot fewer normal subgroups than just subgroups.

[sonic coral]

if you have an abelian group, then there is a subgroup for every divisor of its order

all of which are normal

[crystal vale]

If I have group {1,a,b,c} and it is given that no of them has order 4 so without using Lagrange theorem, is there other way to show that every element has order 2?

[tough raven]

Bash it out: if a has order 3, then 1, a, a² are distinct. Say a² = b. Then a1 = a, aa = b, ab = 1 which forces ac = c by (left) cancellativity, which contradicts (right) cancellativity since 1c = c.

In fact, one can show "directly" that any group {1, a, b, c} of order 4 with no element of order 4 has multiplication given by ab=ba=c, bc=cb=a, ca=ac=b, aa=bb=cc=1.

Is there a reasonably large class of (say commutative for simplicity) rings in which
x^n - 1 is invertible for all n >= 1 => x lies in the Jacobson radical?

I can think of local rings with residue field of positive characteristic p and algebraic over F_p.

Actually, I think that classifies all local rings with this property.

[crystal vale]

Okay thank you

[crystal vale]

In this question, is it enough that if we let a=s and b=sr and then write r in terms of a and b then show that sr=r^(-1)s holds so this generates D(order of 2n) ?

[dim widget]

groups generally don’t have many normal subgroups, you’re unlikely to get any general way of constructing them for an abstract group G

[narrow wagon]

Is it possible to say when will multiplicative group be cyclic?

[solemn garden]

A sufficient condition is when it is that of a finite field

[solemn garden]

Yes, after you show that they have the right orders, and also do the same starting from a and b

[solemn garden]

Is this commutative algebra

[narrow wagon]

Ie it’s enough to check every element is invertable modulo n?

i’m given the unitary group: U(n)

I need to check if it’s a cyclic group

[tough raven]

You can interpret it as either commutative algebra or a question about rings; it fits both.

[narrow wagon]

I suppose U(n) is already a field

[tough raven]

You should probably find the order of the group and look for a single element with that order.

[solemn garden]

You might want to ask in #advanced-algebra

[tough raven]

?? How?

[solemn garden]

Observe that a cyclic group is always abelian

Is this abelian?

[narrow wagon]

yeah U(n) is abelian

converse may not be true

[narrow wagon]

mb

[solemn garden]

Do you mean this, a group of matrices? https://en.wikipedia.org/wiki/Unitary_group

or do you mean the group of units of a ring

[narrow wagon]

Yes, I think it’s the only group under modular multiplication

[solemn garden]

So, which one?

[narrow wagon]

U(n)={x: x<n and x is a relatively prime to n}

[tough raven]

Oh

[narrow wagon]

essentially collect all the invertable integers of Zn this will be a group

[tough raven]

That's … not what "unitary group" usually means.

[narrow wagon]

The notation in my textbook is U(n)

[solemn garden]

So you mean the group of units

[narrow wagon]

It can be seen as the group of units of the ring Zn

[narrow wagon]

yes

[solemn garden]

I see, then the term you are looking for is “primitive root”

[narrow wagon]

yes

primitive root modulo n

[tough raven]

So this is actually a fairly non-trivial question in elementary number theory; you can't just immediately answer it by general group-theoretic considerations.

You can get somewhere by using CRT if n is not a prime power.

[narrow wagon]

i thought if the order is some thing of the form p^2 or idk, then this maybe be cyclic

like groups of certain order is cyclic.

Definitely groups of prime order is cyclic but U(n) can never be odd as euler totient function is even

[narrow wagon]

can we say something like this?

[tough raven]

Yes, it's sometimes cyclic and sometimes not cyclic.

[narrow wagon]

We know any group of prime order is cyclic right?

is there something similar to this ?

like any group of ____ order is cyclic.

I hope you get what im looking for

[static glen]

The Sylos theorems maybe are what you are looking for( idk if ig is really since it is no statement that directly speaks of cyclic groups but at least assures the existence of cyclic subgroups of some order)

[delicate orchid]

There’s a notion of a “nilpotent number”, which guarantees that all groups of that order are nilpotent. So if that number is also square free then your groups are guaranteed to be cyclic

I can’t recall if the implication is an if and only if

[static glen]

This is probably not true, if $n$ is not prime, we might factor $n=km$ with $k,m>1$. The group $\mathbb Z/k\mathbb Z \times \mathbb Z/m\mathbb Z$ is not cyclic and of order $n$

[cloud walrus]

Max

[static glen]

Is this not cyclic, please correct me if I‘m wrong

[delicate orchid]

Nah you’re correct, just set n = 4 lol

A number being nilpotent + square free is a very strong condition, each prime has to appear only once (duh) and cannot be congruent to 1 mod any of the other primes in the factorisation

So the classic example is that groups of order pq with q > p and p not dividing q-1 are cyclic

[static glen]

No i guess I‘m wrong the group Z/kZ times Z/mZ can be cyclic by the chinese remainder theorem

[narrow wagon]

I think for U(n) the only nilpotent is 0

[delicate orchid]

Well yeah if k and m are coprime

[static glen]

Indeed

[delicate orchid]

Considering that group is abelian for every number, no

[narrow wagon]

maybe im talking about something else, if I consider known cyclic unitary group, U(11)={1,2,3,4,…,10}

No element is nilponent here, ie reaches zero

ie so, then it wouldn’t generate the group U(11) ig

[delicate orchid]

Well 0 isn’t in U(11) so if one of them reached 0 U(11) wouldn’t be a group

[narrow wagon]

yes

so what’s the notion of nilpotent number here?

[mighty kiln]

Is it be this? https://www.jstor.org/stable/2589118

[narrow wagon]

It’s written every group of order 15 is cyclic

nice to see that

I wonder why tho

[delicate orchid]

#groups-rings-fields message up here

[rocky cloak]

By the sylow theorems, a group of order 15 will have a subgroup of order 5 and one of order 3 and they will both be normal. Hence any group of order 15 is the direct product of C5 and C3. I.e. the cyclic group of order 15

The same argument extends to nilpotent square free numbers, like wew mentioned

[narrow wagon]

Makes sense

Thanks for sharing, i have to look more on it

Suppose U(n) is cyclic for some n

so is there some similar characterization exist for finding the primitive root?

[rocky cloak]

So if n is prime you can show that U(n) is cyclic with a little field theory.

If n = p1 * p2 * ... pm is a product of distinct primes, then
U(n) = U(p1) x U(p2) x ... U(pm)
Which is not too hard to see when will be cyclic if you know the classification of finite abelian groups.

Showing that U(p^k) is cyclic for odd p, I think you can do with a little fideling and induction.

[crystal vale]

Okay thank you

[solemn garden]

For vector spaces, Hom(V, W) is isomorphic to V* x W, so is there an analogous relation for modules over commutative rings and abelian groups?

[static glen]

That's interesting, how is $Hom(V,W)\equiv V^*\times W$? in finite dimension even the dimensions don't match i would say

[solemn garden]

Oh I mean the tensor product

[cloud walrus]

Max

[solemn garden]

Don’t know how to type it

[static glen]

\otimes

[crystal vale]

Btw in polynomial ring we take commutative ring why? Because of indeterminate variable?

[static glen]

First of all there's no problem in defining polynomial rings over noncommutative rings

[solemn garden]

Ok, $$ \text{Hom}(V, W) \cong V^* \otimes W $$

[cloud walrus]

Maxime Baudin

[static glen]

i always confuse \cong and \equiv

[solemn garden]

Same, I edit it until it works

[crystal vale]

I don't know in many book they take commutative ring

[rocky cloak]

If V is finitely generated projective then
Hom(V, W) = Hom(V, R)(x)W
That works over any ring.

[solemn garden]

Does this work for all abelian groups?

(finitely generated)

[rocky cloak]

The only finitely generated projective abelian groups are the free ones

So it works, but it's not so interesting

[solemn garden]

So it only works for homsets with domain as free abelian groups

[mighty kiln]

It fails for Ab(Q, Q), for instance

Since Q* is 0

[rocky cloak]

Yeah in general Hom(G, Z) is usually 0.

[solemn garden]

Then is there a relation for homsets between any abelian groups

[mighty kiln]

This probably ^

I'm not aware of better ones

[rocky cloak]

Like if you restrict yourself to finite abelian groups then Hom(A, B) = A(x)B

[solemn garden]

But Hom(Z/7Z, Z/8Z) \cong 0

[mighty kiln]

Indeed, Z/7Z ⊗ Z/8Z = 0

[rocky cloak]

Yeah, for cyclic groups you just get Z/n (x) Z/m = Z/gcd(m,n)

[solemn garden]

How do tensor products work on abelian groups?

[rocky cloak]

In the usual way, not sure what you're asking

[mighty kiln]

Ab = Z-Mod

[solemn garden]

Oh

[solemn garden]

Thanks! @rocky cloak @mighty kiln

[rocky cloak]

If you're working over an Artin algebra or in another setting, where you have an injective cogenerator DR that induces a duality.

Then if you define DM as Hom(M, DR)
You have
Hom(M, N) = D(M(x)DN)

If you do something similar for abelian groups (having DR be Q/Z) you get an injective map

Hom(A, B) -> D(A(x)DB)
f |-> (a(x)g |-> g(f(a)) )

Which becomes an isomorphism when A and B are finite.

[solemn garden]

I don’t know what Artin algebras or cogenerators are

Is that like a generator of the dual

[rocky cloak]

A module M is a cogenerator if for any other module X and nonzero element x in X, there exists a homomorphism f: X -> M such that f(x) is nonzero.

For example Q/Z is a cogenerator in the category of abelian groups. Any nonzero vector space is a cogenerator in the category of vector spaces.

And an Artin algebra is a finite algebra over a commutative artinian ring. Typical examples being finite dimensional algebras

[pine crow]

If A is a commutative ring and I a subring of A. We know A/I is a field. I know A/I being a field is equivilant to I being a maximal ideal if I is an ideal. But let's say we don't know if I is an Ideal but know A/I is a field. can we deduce out of those informations that I is an ideal?

[barren sierra]

A/I isn't a well defined structure if I isn't an ideal

[coral spindle]

Do you mean to ask if we can deduce that I is a maximal ideal?

Oh no, you don't.

[barren sierra]

It's a very classical proof, A/I is a ideal iff I is an ideal

Also idk what you mean by A/I is a field if I is a subring

Also a good proof is that A/I is a field iff I is maximal

[cloud walrus]

smidgin

[coral spindle]

Hint: check that (xN)(yN) = (xy)N is well-defined.

[mighty kiln]

y^-1Ny ⊂ N for all y implies the other direction

[coral spindle]

Hint 2: ||you can specify xyN to be the identity||

[mighty kiln]

I think they're proving the converse?

[coral spindle]

Yes, the point being that you simply do the same in reverse.

[mighty kiln]

Oh

[pine crow]

I learned math in a different language so bear with me.

Korollar:
If A is a commutative Ring with 1 and I an Ideal of A then

A/I is a field equivilant to I is a maximal Ideal

Situation: It is known that A/I is a field but I is not defined as Ideal. Can we deduce that I is an maximal ideal out of that or no?

[mighty kiln]

How would you define A/I?

If not an ideal

[coral spindle]

Spamakin already provided an answer

No.

Well-definedness is a property of a relation, meaning that it defines a function.

The classic example of this is with equivalence relations

If I try to define a function f([x]) = x, where [x] is the equivalence class of x, typically this will not actually define a function, since [x] may contain some other element y.

Then we'd have y = f([y]) = f([x]) = x

So we say that "f is not well defined" in this case.

Well-definedness of the product in a quotient group is saying that if xN = x'N and yN = y'N, then (xN)(yN) = (x'N)(y'N).

Yes.

[mighty kiln]

That should translate to the same thing in this case tho?

[coral spindle]

Yes.

[coral spindle]

Sorry, I made a mistake here. I've edited it. Maybe you saw it already

Maybe? Still overcomplicated

[pine crow]

ye mb. skimmed through my algebra notes and noticed that I misremembered the definition of A/I

[mighty kiln]

It might be helpful to know some calculus for inclusions of subsets too

gA ⊂ B is equivalent to A ⊂ g^-1B

[coral spindle]

Spoilered solution based on this hint, when you're ready. ||Let g in G and n in N. Then (gng^-1)N = (gnN)(g^-1N) = (gN)(g^-1N) = N by well-definedness, so gng^-1 in N. So gNg^-1 subset N. QED.||

[mighty kiln]

In particular, y^-1Ny ⊂ N yields N ⊂ yNy^-1

[stark helm]

If we can not use maximal, and want to show that Z2[x]/(x^3+x+1) is a field, is it the only way to prove that thisn set is a domain and all elements in this ring has multiplicative inverse?

[coral spindle]

That would be the first thing that comes to mind

[rustic crown]

finite domains are already fields, so you don't need to explicity find inverses

[coral spindle]

Oh right, Z2[x]/(...) not Z[x]/(...)

[glossy crag]

Where's Wew when you need him, dude was here 24/7 up till recently.

[cloud solar]

,rotate

[cloud solar]

I know the answer is 2^(n+2)-1

Using the viete sums

But I want to find it using the companion matrix of P

It should be Tr(A^(n+2))

Is there a formula to find the trace of a power of the companion matrix?

[glossy crag]

Over C, the trace of A^k is \sum a^k , where a are the eigenvalues.

[cloud solar]

Yes I know that but finding the sum a^k directly is just using viete sums for the polynomial P

And I mind if I can find that sum from the trace

[glossy crag]

What's the question then.

[glossy crag]

Ah, do you mean how to express tr(A^k) in terms of tr(A)?

[cloud solar]

Yes

[glossy crag]

Well you probably can't.

[cloud solar]

I know there is some formula to find the power of the companion matrix

So there should be some formula to find the trace of the power

[glossy crag]

Try asking in #linear-algebra.

[barren thorn]

Suppose we have a ring of integers modulo n. What can we infer about this ring if its order is divisble by a prime number?

[patent girder]

There's some nice things: $n = p_1^{e_1} ... p_n^{e_n}$ means $Z/nZ = A \cong Z/p_1^{e_1} Z \oplus \ldots\oplus Z/p_n^{e_n}Z =: A_1 \oplus ... \oplus A_n$. But this means that the prime ideals are just take a maximal ideal $I_i=(p_i)$ of $A_i$ and consider $ A_1 \oplus ... \oplus I_i \oplus ...\oplus A_n.$ You can show some more nice things about what things can exist, but these are some of the basic ones that come to mind.

[cloud walrus]

Zander

[patent girder]

And that decomposition follows from the Chinese Remainder Theorem

[crystal vale]

Hint? How can I show that S is infinte group?

[next obsidian]

Just come up with infinitely many automorphisms

There’s a ton and they’re easy to write down

[crystal vale]

Any contradiction way?

[crystal turtle]

Show it directly

No need for contradiction

[next obsidian]

Yeah the contradiction way is

Assume it’s finite

Here’s infinitely many

So that was a contradiction

Okay there’s infinitely many

[crystal vale]

Okay thank you

For every positive integer there are permutation (1 n) .....is it correct?

[next obsidian]

Yah

[chilly ocean]

What is a group action?

Why does this foreshadow a group action?

[next obsidian]

Group acting as bijection of a set in a way that composes the way a group does

An example is something like GL_n acting on the space of nxn matrices via multiplication

[mighty kiln]

You may have seen groups visualized as "acting" on some object

E.g. the dihedral group permuting the vertices of an n-gon

The symmetric group permuting n elements

The free group on {U, D, F, B, L, R} permuting the 26 pieces of a Rubik's cube

A group action is when you let elements of a group G permute some set S

That is, you have a homomorphism G → Sym(S) (the group of permutations of S)

[chilly ocean]

Cool thank you 👍

[crystal vale]

When R is commutative ring, I want to prove that Hom(R,M) and M are isomorphic as left R-Module so for first I define:

an Homomorphism from R to M such that r->rm so it is R-Module homomorphism(name as π(m))

Now I map M->Hom(R,M) such that m->π(m)
So for injective if I let π(m)=0 that's mean for all r belong to R π(m)(r)=0 which is equal to rm=0 and thus for when r=1 it implies that m=0

Is it correct for injective?

[mighty kiln]

Yes

[velvet lake]

I just realized the ring of symmetric functions $\Lambda_{n} := \mathbb{Z} [x_{1},....x_{n}]^{S_{n}}$ has dimension $\sum_{k=0}^{n} p(k)$ where $ p(k) $ is the k-th partition number, over $ \Z$ as a $ \mathbb{Z} $ algebra. Yet $ \mathbb{Z}[x_{1},....,x_{n}] $ has dimension only $ n $ so for $ n = 3 $ say the subring $ \Lambda_{n} $ of $ \mathbb{Z}[x_{1},x_{2},x_{3}] $ has dimension $ p(0) + p(1) + p(2) + p(3) = 1 + 1 + 2 + 3 = 7 $ over $ \mathbb{Z} $ but $ \mathbb{Z}[x_{1},x_{2},x_{3}] $ only has dimension 3 even though it contains $ \Lambda_{n} $ as a subring.

My question is that naively, I expected the subring to have a smaller dimension over $ Z $ then the ring it is contained in, but that's not true. What is going on here? In general is there some way to relate the dimensions of rings and their subrings?

[cloud walrus]

spikey1421
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[crystal vale]

How can it show that surjective?

[mighty kiln]

Show that every R-module map R -> M is determined uniquely by where it maps 1

[crystal vale]

Can you please tell me what is the concept of determined uniquely ? How it will help us

[mighty kiln]

*Given any R-module map R -> M, show that it is given by the map r -> rm for some m

[crystal vale]

Yes but that's a problem how can I show that

[mighty kiln]

Consider where 1 in R is mapped to

[crystal vale]

For this 1 map to m (for some m)

[mighty kiln]

yes

[crystal vale]

So for any R module homomorphism we know that 1 map to some element of M so let say f(1) so then we simply take π(f(1)) , is it?

[mighty kiln]

Yup

[tough raven]

In the (more commonly seen) polynomial ring construction, the variable commutes with the ring of coefficients. Hence you can't "evaluate" polynomials except at inputs that do commute with the ring of coefficients.

Here's an example of why that matters: in the polynomial ring over the quaternions, we have the factorisation
X² + 1 = (X + i)(X - i),
which you would think forces the only zeroes of the LHS to be ±i (since multiplication is cancellative in the quaternions), but that's not true because
j² + 1 =/= (j + i)(j - i).
In fact, there are uncountably many zeroes of X² + 1 in the quaternions even though its degree is just two and multiplication is cancellative in the quaternions.

(This does not mean that these polynomial rings are useless. For one example, there is an equivalence between R-modules M equipped with an R-linear endomorphism A : M -> M and R[X]-modules (the R-linearity of A ensures that it “commutes” with R), which does not require that R be commutative.)

[tough raven]

Is this ring not infinite-dimensional over Z?

Assuming that by dimension over Z you mean rank as a free Z-module.

[rocky cloak]

I don't think there is any nice relationship between the dimension of a ring and of a subring no.

For example if R is an integral domain, then R is a subring of its field of fractions, which is 0-dimensional. But the dimension of R can be arbitrary.

[rocky cloak]

Assume the size is n, show that Sn is a subgroup. |Sn| > n, contradiction.

[crystal vale]

Thank you that's what I wanted

[crystal vale]

Thank you I will look into it

[hidden wind]

so true paolo

[fading field]

i love aluffi

[low wyvern]

Quick question; does the cancellation property hold for rings (in general) the same way it does for groups?

[scarlet estuary]

depends whether you're talking about the ring's + or * operation

a ring's + operation is an abelian group, so of course the same properties hold

[low wyvern]

Well, I assume addition does due to the first axiom

so x?

[scarlet estuary]

a ring's * operation does not necessarily form a group, so to try and construct a counterexample, consider a ring where it is not a group

for example, Z/4Z

can you find a case where a * c = b * c mod 4, but a ≠ b mod 4?

[coral spindle]

I daresay that there is only a single ring for which the * operation forms a group...

[scarlet estuary]

well, you know what i mean

[coral spindle]

Hopefully Kernal will see

[scarlet estuary]

where it lacks the group properties used in the proof that groups have cancellation

but fair enough, i was imprecise

[coral spindle]

It's worth seeing that too

[scarlet estuary]

||that said, if you know your ring has no zero divisors - or even if you're able to consider a subsemigroup with no zero divisors - you do actually get the cancellation property even if your ring has no 1 element!||

||it might be a useful exercise to convince yourself of this, using the example i gave as inspiration||

[wooden fulcrum]

what does this mean

[coral spindle]

I suggest you think about it if it's not clear

If you're asking what a ring/group is, then you're a little stuffed

[wooden fulcrum]

no I mean what do you mean there's only a single ring for which the multiplication forms a group

or do you mean for each set

[scarlet estuary]

0

[crystal turtle]

Up to isomorphism

[dull marsh]

Up to isomorphism there's only one ring

[crystal turtle]

All rings are 0 sotrue

[wooden fulcrum]

per set?

[coral spindle]

Wym per set

[dull marsh]

I.e., all rings that forms a group under * are isomorphic (the same)

[scarlet estuary]

up to isomorphism means we dont care about the underlying set

we are only considering the structure

and can "relabel" the set if we wish

[wooden fulcrum]

wtf so Q and R are isomorphic?

[coral spindle]

No

[scarlet estuary]

no since they have different structure

[coral spindle]

They are not.

[wooden fulcrum]

they're both fields no?

[dull marsh]

No

[coral spindle]

Not all fields are isomorphic.

[dull marsh]

That doesn't provide an isomorphism

[wooden fulcrum]

yes I know

[scarlet estuary]

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

[wooden fulcrum]

I don't think they;re isomorphic

but I don't get the statement then

[coral spindle]

The field R does not form a group under multiplication

Why?

Because there's 0!

[wooden fulcrum]

why is 1 a problme

ha ha

I still don't get the statement

[coral spindle]

Idk what else there is to explain; I think we've covered every possible misunderstanding.

[scarlet estuary]

let's say you naively tried to add "division" (inverses) to the multiplicative structure of a ring

when you do this, you get a field, right?

but fields explicitly have 1 element you cant divide by

[wooden fulcrum]

oh you mean like

[scarlet estuary]

0

[wooden fulcrum]

if you consider R withou removing the zero

ah ye

kinda pedantic but...

[coral spindle]

It needed to be pointed out earlier.

[scarlet estuary]

eh its important to clarify, i was imprecise

[wooden fulcrum]

yeah thanks for clarification also

[dull marsh]

Being isomorphic yields an equivalence relation, what's meant by "there are n groups/rings/fields satisfying this up to isomorphism" means that there are n equivalence classes where every group/ring/field satisfies "this"

[low wyvern]

Having worked through the example I see, but I don't know what a subsemigroup is lmao. This was beyond galilean

[coral spindle]

The confusion was much simpler than that

It was simply confusing F\{0} with F.

[dull marsh]

Ah, I see

[coral spindle]

FB just did not communicate that at all

[wooden fulcrum]

sorry I just got used to conflating the 2

like talking about whether * forms a group just automatically not considering 0

[scarlet estuary]

in this case, just some particular subset that preserves closure and associativity

[low wyvern]

Ok, thanks 🙂

[scarlet estuary]

of *

[scarlet estuary]

clearly we need to remove 0 from all rings to avoid this issue in the future

[wooden fulcrum]

so true

make R + just a commutative semigroup with division and everything will be so much easier

[scarlet estuary]

i present a modified number line for consideration

[wooden fulcrum]

I mean have you ever actually encountered 0 of an object?

didn't think so

[crystal turtle]

It looks like something is missing this doesn't seem very natural

[low wyvern]

Just a quick question on this point then, does this imply that the cancellation property (for *) only holds for that single ring?

[scarlet estuary]

no, again i was imprecise

[coral spindle]

Yes, but as we have just spent quite a while elaborating on, you really want to consider R\{0} when talking about this to avoid this triviality.

[scarlet estuary]

as long as you consider the ring without 0 its fine (as in, you get both examples and counterexamples)

but obviously 2 * 0 = 0 * 0 does not imply 2 = 0

actually, you can be slightly stronger and just require you not cancel out 0 specifically

this is, in fact, what we typically do when we work with ℝ

[wooden fulcrum]

in a ring you can prove that 0*anything = 0, where 0 is the (R, +) neutral element, so if you want there to be an inverse 0^-1 for 0 such that 0 * 0^-1 = 1 where 1 is the (R, *) neutral element then the neutral element has to be 0 so the whole ring is just a single element

[scarlet estuary]

we just say that division by 0 is disallowed, but dividing 0 by things still works fine and still exhibits cancellation

0 * a = 0 * b implies 0 = 0

[low wyvern]

Got it 👍

[scarlet estuary]

just not a = b

this is a very fancy way to say "grade school mathematics works"

[allegedly]

[vernal needle]

anybody else fail this class 🕺

[crystal vale]

I am not sure about the F(A) notation.... can someone help me to understand the meaning of F(A) ?

[tender wharf]

the last line in your picture gives some motivation

but basically it's just saying free R module with basis A

[crystal vale]

So why do they use this F(A) notation? Any specific reason?

[chilly ocean]

Free

[crystal vale]

Thank you

[static glen]

and it depends on A, for other A this will be another module

[cloud solar]

Let K be a finite field with char≠2. Show that every element in K is a sum of four cubes

I dont think that char≠2 implies that every element is a sum of two cubes

[rocky cloak]

(x+1)^3 + (x-1)^3 -x^3 - x^3 = 6x

So if the characteristic is not 2 or 3, this map is surjective.

If the characteristic is 3, then every element is a cube.

You're also right that two cubes is not enough. Look at the field with 7 elements for example.

[hollow mica]

How would you finish the proof of the fwd direction

I know it can go like this

But I'm specifically interested in the fragment of a proof I have above

[coral spindle]

It looks like you want to prove that ah = b for any h in H. This just isn't true, you can't write the proof like this.

[hollow mica]

oh sorry

the proof is incomplete

[coral spindle]

And there's something wrong written in it, so I can't really complete it

[hollow mica]

but basically what I intended to convey is that I want ah to equal b*(something in H) through some series of substitutions

but I don't know how to do that

[coral spindle]

Well indeed b = a(a^-1b) so

[hollow mica]

great so then I can write bh as a*(something in H)

but for the other way

idk

[coral spindle]

Do you agree that aH contains a

[hollow mica]

yea

[coral spindle]

So a is in bH

[hollow mica]

ye

wait

[coral spindle]

Well, now I regret the order I did this, but you should see the end now.

[hollow mica]

So before we showed that bH is a subset of aH, and now you're saying that aH containing a implies that a is in bH ?

[coral spindle]

Are you not asking about starting from aH = bH and concluding a^-1b in H?

[hollow mica]

other direction, sorry

That's why I started my proof with "suppose a^{-1} b \in H"

[coral spindle]

So you meant the other inclusion, not the other implication. OK.

[hollow mica]

Yes

[coral spindle]

So we've showed that bH \subseteq aH.

Now note H is a subgroup, and a^-1b in H.

So (a^-1 b)^-1 is in H too.

[hollow mica]

Yea

[coral spindle]

You finish the rest.

[hollow mica]

Lmao that's straight up what I originally had

[coral spindle]

You made a mistake there.

(a^-1 b)^-1 is not ba^-1.

[hollow mica]

Oh oops

[coral spindle]

Now do you see it

[hollow mica]

Yeah

I think

[coral spindle]

Explain how this finishes things

[hollow mica]

ah = b(b^-1a)h = b(b^-1ah) = bh'

nice

Ty

ok so exactly the same argument

[coral spindle]

Yes, it's the exact same argument as before but with a and b flipped.

[hollow mica]

👍

[cloud walrus]

spikey1421

[opal osprey]

I'm having a bit of trouble understanding this definition of \epsilon-quadratic forms as certain coinvariants

what does this mean in practice?

appearenly, coinvariant is synonymous with orbit under a group action

but i don't see how in this case this leads to the intuitive definition of what a quadratic form should be (say for * the trivial involution and \epsilon = 1)

[boreal inlet]

... should I open my own thread? I'm currently reading Field Theory so it might be helpful to organise everything in one place.

[boreal inlet]

Thank you

Riku's Field Theory Shenanigans

[hearty ledge]

Am I right to say that $s_3$ is not isomorphic to any of the other groups because $s_3$ is of order 2 and all the other groups are of order 6?

[cloud walrus]

Bizcuits

[dull marsh]

How come S3 is of order 2?

[boreal inlet]

What..... what do you mean by order 2

[hearty ledge]

because evry element is an inverse of its self

i might just be misunderstanding order

[dull marsh]

Well, what's the inverse of (1 2 3)?

[dull marsh]

Order of a group is the number of elements in it

[hearty ledge]

[dull marsh]

There are 3! = 6 distinct ways to permute 3 objects, making the order of S3 equal to 6

[hearty ledge]

ah you right

i am thinking of period

[hearty ledge]

You right. This is why I can't be rushing this shii

(1 2 3) is not it's own inverse 💀

[cobalt heath]

Permutation difficulty

[hidden wind]

you should convince yourself (1 2 3) generates a subgroup isomorphic to Z3

[indigo ridge]

interesting did not know that

[hidden wind]

S3 is the same thing as D3

[dull marsh]

In general,
Any element of finite order n generates a subgroup isomorphic to Zn
Any element of infinite order generates a subgroup isomorphic to Z

[coral spindle]

This is perhaps even a workable definition of being of a certain order!

[indigo ridge]

ah ofcourse, still cool but kind loses some magic lol

[hidden wind]

in which you should quickly recognize (1 2 3) as a simple rotation

[cobalt heath]

Why does directed limit exist in the case of "algebraic" structures? Also does the functor onto directed limit preserve exactness?

(Please let me know if I am unclear here)

[dim widget]

Because algebraic structures have a left adjoint to the forgetful functor

[cobalt heath]

Ohh?!
(Ah sorry for overreaction)

So for forgetful C -> Set, we have Set -> C which is a left-adjoint?

[dim widget]

Yes it’s the free functor

I.e. free group free ring free algebra etc.

[cobalt heath]

Ah, I see. Sorry I am dumb

I need to think of free functors more often

[dim widget]

Ah wait

Actually I’m not answering your question

The point is that the free and forgetful functor provide a monadic adjunction between algebraic structures and sets

This is why filtered colimits always exist in categories of algebraic structures

The argument I was giving was for why limits are well behaved for algebraic structures

[cobalt heath]

Hmm, directed limit is colimit right

Ah, thus being filtered colimit

[dim widget]

Yes

A directed limit is a filtered colimit

[cobalt heath]

I see, thanks a lot for good intuition!

[dim widget]

Actually I’m no longer sure this is a good justification

I’m too tired

I think there should be a beautiful categorical explanation

But essentially if you have a finitary algebraic structure, then to define it on an increasing union it suffices to just define it on each step in the union by some compactness argument

So this is why filtered colimits exist so ubiquitously in categories like rings groups etc

Where every object is an increasing union of finitely presented objects

But I’m not sure I have the most beautiful or simple way of understanding this

[cobalt heath]

Hmmm

I need to think more about this then

[crystal vale]

If I am correct then the following statement is true.

Consider R-Module M where R= F(Real number) and M= F × F (cartesian product of F × F) then if I consider the set A ={ (1,2) , (1,0) , (2,3) ,(5,0) } then so R-Module M is not free on A.
Is it correct?

[coral spindle]

Indeed that set A is not a basis for M.

[crystal vale]

Yes

But there is no introduction of basis, they first introduce a free then introduce basis

[crystal vale]

According to the last statement F(A)= Ra1 + Ra2 +.......+RaN is isomorphic to R^(n) so if I want to show that F(A) is free on A, for every non-zero element x of F(A) should have unique representation

How can I show that? Hint

[mighty kiln]

The representation is given by projecting to each summand

[delicate orchid]

oh right this is unrelated to this question

[mighty kiln]

If you have a finite direct sum ∏Ai = A1 ⊕ … ⊕ An, then you have projection maps ∏Ai → Ai for each i

That essentially give you the ith "coordinate" if you think of ∏Ai as Cartesian product

And this yields the coefficients of each ai for the above situation

[delicate orchid]

alternatively, if there are two different representations of some x, then try and construct a module such that the given triangle doesn't commute

[topaz solar]

For a less category brained version, see chapter 1 of tent and Ziegler describing via structures and existence of direct unions as structures. Looking further, many algebraic structures are universal or A_2 (= exactly the first order axiomatizable things closed under direct unions)

A_2 here meaning \forall x \exists y phi(x, y) type axioms

Like, say, fields

[cobalt heath]

Ah, I suspected this problem is a bit model-theoretical.

[delicate orchid]

sharp just makes everything model theoretical

[topaz solar]

Yeah, having a free functor is nice

But we don’t have that always

Look at complete lattices

Free complete lattice on 3 points contradicts choice

(I believe it works in ZF, but category theory kinda sucks there)

[topaz solar]

Yeah like how someone with topos theory in their name makes things categorical

In fact, tteg even says compactness argument

If you have a non-A_2 axiomatizable class, you may still have directed colimits which aren’t just taking unions

Since we remove objects under consideration

But, if we have free/forgetful we can say things by working in sets then quotienting algebras (supposing said object exists, which presumably you can see via the requisite quotient)

[topaz solar]

Cofinality of your directed set is probably infinite, so finite tuples are very bounded when going to the coordinate (and if it’s a finite cofinality then it’s kinda trivialized or way simpler). So just work in X_whatever and pass it up via the usual colimit diagram shenanigans or however you want to view it

Finitely presented isn’t necessary here

Cofinal set of net moment

Alternatively, in reverse, can view this as successively killing off arbitrarily large finite pieces of your directed set, then compactness type principles say why not do it all

[mighty kiln]

How does filteredness come in

[delicate orchid]

depending on how narrow your definition of "algebraic structrures" is, you can just straight up define the directed limit constructively

which is pretty much the only way I ever think about colimits

[topaz solar]

Yeah, which is kinda literally the A_2 axiomatizable way

Which is why it’s best

[delicate orchid]

whatever u say sweaty....

although viewing it via the adjunction is nice because (I think...) it generalises to any concrete category (over Set) with an adjoint

[topaz solar]

Filtered colimits relate to adjunctions

Ye, which axiomatizations can suck (or fail to exist if your category isn’t accessible, though I think that’s kinda hard to pop up w/ these adjunctions)

Or be very unnatural

[delicate orchid]

ermmm if my category isn't accessible then how am I supposed to use it 😹

[topaz solar]

There’s some things which suck for adjunctions, some things which suck for the model approach

Use anything that works

[delicate orchid]

do something else instead

[crystal vale]

What is the structure of elements of F(A)?
x= (r1,r2,......,rn) and I want to do that there exists unique s1,s2,.....,sn and a1,......,an, such that x=r1a1 + r2a2 +........+ rnan

[mighty kiln]

A direct sum of R-modules can be expressed as the Cartesian product with addition (c1, ..., cn) + (b1, ..., bn) = (c1+b1, ..., cn+bn) and scalar multiplication r(c1, ..., cn) = (rc1, ..., rcn)

Then you can identify a1 with (a1, 0, ..., 0), a2 with (0, a2, 0, ..., 0), etc

So (c1, ..., cn) = c1a1 + ... + cnan

[cobalt heath]

I wish I could understand this

[topaz solar]

“It’s directed so take the thing covering all your finite pieces”

[cobalt heath]

Sorry, I really have to go through the model theory book

[cloud walrus]

Riku

[boreal inlet]

I am so utterly lost in this one

Oh shit

For the last statement I meant f(a) = 0 implies f(b) = 0

[topaz solar]

Well

What does it mean for f(\bar a) = 0

[glossy crag]

You did write that though (with a caret).

Anyway, one direction of this is trivial, for the other you need to know about the isomorphism extension theorem for splitting fields, which you apply step by step to a tower of subfields.

[boreal inlet]

The conjugate to the f(a) implies part is the trivial one right?

Because the isomorphism induces another isomorphism in the polynomial ring

[glossy crag]

Yes

[delicate orchid]

mfs typing too quick these days...

[boreal inlet]

And you just send the evaluation at a through that induced map

Should go to f(b)

[glossy crag]

No low-hanging fruit for you, Wew

[glossy crag]

IMO that's overcomplicating things, just take any K-polynomial expression in the a_i that evaluates to 0, applying phi turns it into a K-polynomial expression in the b_i, end of conversation.

[boreal inlet]

Yeah I meant basically that

Thanks

[delicate orchid]

I just saw it as f(a hat) = f(a_1, ..., a_n) = 0 then u just BAM 👊 phi(f(a_1, ..., a_n)) = 0 BOOM 💣 f(phi(a_1), ... phi(a_n)) = 0 POW 💥 => f(b hat) = 0 OOSH 🤜

[boreal inlet]

Now for the other part

[boreal inlet]

ngl when i first saw the question this is exactly how I read it

[glossy crag]

Math Adam West-style.

[boreal inlet]

Wait what is isomorphism extension theorem again

I only know that any K-alg embedding can be extended to a L-alg embedding kinda

[delicate orchid]

yeah that's basically it

kind of

[glossy crag]

If f:K->K' is an isomorphism, L/K is a splitting field for a subset A<=K[x], and L'/K' is a splitting field for the subset A'=fA<=K'[x], then the isomorphism f extends to an isomorphism f':L->L'.

[delicate orchid]

_ _

oh my golly gosh it's just like wholesome brauer lifterinos

[glossy crag]

It's our time, old man.

[delicate orchid]

if u have... F_1 -> F_2 iso... and K/F_1.... then... iso... K -> "idfk what letter to use"/F_2 which is ur iso F_1 -> F_2 upon restriction

[boreal inlet]

Oh wait uh wjat is fA

Oh roght

Your isomorphism is f

Makes sense

[glossy crag]

What's a heckin Brauer lifterino?

[delicate orchid]

modular representation gibberish

[glossy crag]

Gotta dab on the undergrads, right.

[delicate orchid]

u seem to know more field theory than me lol

[boreal inlet]

lmao

[delicate orchid]

I wouldn't've been able to prove the reverse direction lol

[glossy crag]

Since you're here, wanted to ask you this the other day: good books on reprtheory? I know semisimplicity and Artin-Wedderburn, if that helps.

[south patrol]

Profinite

[delicate orchid]

pro-p, even

[boreal inlet]

Isn't this that topology on uh something something characteristic p

[delicate orchid]

bobby fusions part IV I always go to Liebeck's book

[south patrol]

Uh not really

[delicate orchid]

it's jsut something that can be made from finite things (usually via a limit/colimit)

[coral spindle]

James & Liebeck is a good introduction

[boreal inlet]

Oh

[glossy crag]

This?

[delicate orchid]

yus

[coral spindle]

Isaacs Character Theory is the bible for finite groups' characters

[delicate orchid]

Isaac's is good

[south patrol]

Oh nice, I hadn't heard of Liebeck

[delicate orchid]

but I'm not sure I'd recommed it to an undergrad as a first course

[coral spindle]

Serre's Rep theory is OK

[delicate orchid]

it's more of a reference text to me

[south patrol]

Another nice reference for characters imo is Serre's book.

[delicate orchid]

but whatever you do don't read fulton harris as a first course LMFAOOO

[south patrol]

but i don't think it's ideal for reps more generally e.g. not over C

[glossy crag]

I'm mightily annoyed by his right-side formalism

[coral spindle]

I find it harder to read than Isaacs

Meh

[delicate orchid]

I have bad news. They ALL do this

[coral spindle]

James and Liebeck does the same.

[south patrol]

Ah i think it's just our lecturer liked Serre so I used that book a lot for preparation lol

[glossy crag]

I know TRVE GROVP theorists do this, but I just can't get behind it.

[south patrol]

Wdym by this?

Or like are his reps like right-actions

[glossy crag]

Writing shit on the right.

[delicate orchid]

luckily he doesn't do it for characters, just maps

[delicate orchid]

xf = f(x)

[glossy crag]

What's worse, he writes half his shit on the left, half on the right.

[south patrol]

Oh bruh that is worse lol

[glossy crag]

This alone I'd be fine with, he likes to mix things up

[coral spindle]

Just swap the representationios in your hecking headerino

[delicate orchid]

^op ^op ^op, ^oppa gangnam style

[south patrol]

I was reading something and confused why the author kept using like $H\backslash G$ until in the next page or smth he wrote $(H \backslash G)//G$. double quotients scare me

[cloud walrus]

potato

[south patrol]

two sided stuff

[delicate orchid]

Boytjie what's ur favourite involution in the 2-endomorphism group of Cat and you can't say ^op

[boreal inlet]

This reminds me of herstein holy fuck that was pain to read

[south patrol]

Imagine if we taught double cosets etc in first year algebra

so that people just got cracked at it

Not really worth it probably but uh

[coral spindle]

What I actually dislike about Isaacs' book is that he uses kinda weird notation that is hard to look up. So e.g. he writes chi^G instead of Ind^G chi for induced characters – but this is a tame example.

[delicate orchid]

tame
he he, ho ho, ha ha

[coral spindle]

opFUCK

[south patrol]

I heard that in Copenhagen they ran a first lin alg course where matrices act on the right and all fields are now more generally division algebras

[boreal inlet]

Holy fuck...

[coral spindle]

Dang Danes

[delicate orchid]

oh double cosets? yeah the notation is weird

[south patrol]

Oh yeah iirc this was like, they wrote the first major lin alg textbook in Danish or smth

[delicate orchid]

you basically use them for Mackey's formula then stop caring/just use biset notation

[south patrol]

Skew fields is funny to me though because like

[coral spindle]

Um um actually 🤓

[south patrol]

you might as well just teach lin alg with fields so that people can use R and C etc

for intuition

and then just mention yeah the theorems will hold over division algebras

well virtually all of them

[coral spindle]

Double cosets are used in the Bruhat decomposition of algebraic groups and are actually quite helpful (not that I really understand this)

[delicate orchid]

ok yes that rings a vague bell

[south patrol]

In my case the point was that uh

[delicate orchid]

I think I've seen this for coexter groups

[south patrol]

We have G-spaces and then a functor which takes this quotient of them in a topological way

And then a natural G-set is just like H\G

so makes sense

But double cosets scare me.

[delicate orchid]

ermm please don't say natural unless there's a natural transformation involved 🤓

[south patrol]

Cringe.

[delicate orchid]

actually wait

[coral spindle]

ewwww undergrad category theorist everyone point and laugh

[delicate orchid]

taking burnside rings is functorial and they're generated by H\Gs

maybe you can get a natural transform lol

[south patrol]

Idk if you mean to say wew is undergrad lol

[delicate orchid]

yeah G -> H induces a natural transformation on double burnside ring functors A(G, -) -> A(H, -) ok whatever

[coral spindle]

I'm just being a silly little guy

[delicate orchid]

I just wanted to use the nerd emoji

[south patrol]

Or just cause this is an impression of an UGCT

fair heh

I might learn about the Burnside ring today actually

[coral spindle]

Nice nice

[south patrol]

Idk what it is

I am learning stuff in the wrong direction as ever

[delicate orchid]

I like Benson's exposition of it but that's very rep theory motivated

[coral spindle]

Uh Groth ring but for G-sets rather than G-spaces??

[south patrol]

I'm fine w that

[coral spindle]

Right?

[delicate orchid]

I don't actually recall where I learnt it from

[delicate orchid]

yeah literally

[south patrol]

which thing do you mean wew

[coral spindle]

OK nice

[south patrol]

Like is there a name I can look up I mean

Tbf i imagine googling what u gave is sufficient nvm

Thank

I don't know what a Groth ring is either

[coral spindle]

Grothendieck ring

[south patrol]

Oh ok

[delicate orchid]

u take a semiring and just fuckin shove in the minus sign

it's rigorous trust

[coral spindle]

Honestly once you've learned abt the Burnside ring you'll have a good intuition for the Groth ring, I won't bombard you now

[south patrol]

I know about Grothendieck rings etc dw

[delicate orchid]

not one mention of burnside rings in bobby fusions
he has failed me

[coral spindle]

Ok you're fine then lol

[south patrol]

Like for the representation ring and topological K-theory

When you say G dpaces

[delicate orchid]

actually fuck it potato I'll just exposit it now quickly

[south patrol]

Do you mean like G-vector spaces

[coral spindle]

Yeah, like kG-modules

[south patrol]

Sure

Just checking as I am working with G-spaces in the topological sense

[coral spindle]

So the Burnside ring is that but for the category of finite G-sets

[south patrol]

OK nice

So does taking free abelian groups pass to a map from rep ring to Burnside ring

Or at least a natural transformation i mean like

[delicate orchid]

by orbit stabiliser any set with a G action decomposes into a coproduct of G-sets which are isomorphic to G/H for H up to G-conjugacy. So u take the semiring spanned by these G/H with addition given by coproduct and u don't want to know how multiplication works. Then grothendieck completion

[coral spindle]

No I don't think so, surely you mean the other way?

[delicate orchid]

the useful map is going Burnside -> Rep via permutation reps

[coral spindle]

Like I can get a G-set and linearise it to get a kG-module

[south patrol]

Sorry yes thats what I mean

Sorry

Yes exactly

[coral spindle]

Yes then, the functor becomes a ring homomorphism

[south patrol]

Pog

[coral spindle]

V pog

[delicate orchid]

I nearly pooped my pants thinking about the grothendieck completion being contravariant there

[south patrol]

Ok I will think about this thanks

I haven't thought of that interpretation of orb stab

[delicate orchid]

the double burnside ring is where things get "epic"

[south patrol]

But makes sense lol like I guess if G acts on a set S transitively, is S iso to G/H

And then coprod over orbits

Makes sense

[delicate orchid]

oh shit sorry, it's spanned by the isomorphism classes

[south patrol]

I imagine multiplication is probably uniquely characterised by this

[delicate orchid]

not the G/H themselves

[south patrol]

Yeah dw I imagine so lol

[coral spindle]

Yes since it's an inclusion for k char 0

[topaz solar]

Set theorists about to complain

[south patrol]

Cool

[coral spindle]

And the whole lesson of elementary character theory is that this is a free Abelian group over the irreducible characters

[south patrol]

Well it's presumably fine se theoretically as these are explicitly quotients of G aha

Wait ig if you are taking coprods nvm

[delicate orchid]

no need to pass to characters my friend watch this

[coral spindle]

No I want my characters

[delicate orchid]

there's an analogue of characters for the burnside ring

[coral spindle]

Well yes it's characters of the permutation rep right

fixed 🅱️oints

[delicate orchid]

yeah but they have a really nice interpritation on the basis elements

[south patrol]

Based

[delicate orchid]

also no, not quite

they're evaluated on subgroups not elements

so they're not just straight up the permutation characters

anyway

[coral spindle]

:O