groups-rings-fields

[still dew]

Pls elaborate

[vivid tiger]

they're just saying what I said

[still dew]

You meant the trivial side

Oh

Okay the non trivial side is trouble

[south patrol]

I think, at least in my mind, the key thing is this:

[delicate orchid]

VANDERMONDE

[south patrol]

uhh

what was i saying

[still dew]

How

[vivid tiger]

So, is the fact that we have a polynomial ring actually important besides being a PID? I think not.

[still dew]

I guess so

[vivid tiger]

nope?

[delicate orchid]

Ok fine I’ll make an actual observation

Since it’s over a PID you can put it in smith normal form, which makes the eigenvalues obvious

[still dew]

Wtf is smith normal form

[delicate orchid]

That’s my one mildly productive contribution for the year done

[vivid tiger]

put what in smith normal form?

[still dew]

Oh it's just the normal form you get like a scalar matrix block

[south patrol]

also yeah what ca you put in snf

we're trying to find a matrix

(other direction is easier)

[vivid tiger]

as nads says, the <= is trivial

one may even say

obvious

clear

[rocky cloak]

Here's an idea:

Consider instead the element x = (p1, ..., pn) of R^n and the submodule M it generates. Now R^n / M is torsionfree, since for any element y in R^n, if ry = sx, then the gcd of the elements in ry equals s. So s is a multiple of r, hence y = (s/r)x.

Thus R^n/M is free, thus R^n = M (+) R^n-1.

Then there is an isomorphism R^n -> M (+) R^n-1 = R^n which we can think of as a matrix whose first column is x.

Since this is invertible, the determinant is in F. Rescaling one column we may assume the determinant is 1. Transposing the matrix gives the desired result.

[still dew]

I don't know modules 😔

I guess this excercise can't be done without them

[rocky cloak]

Oh, so you're just supposed to do some horrible calculations?

[still dew]

I suppose not

[rocky cloak]

Or I guess there might be a clever trick to construct a matrix

[still dew]

It's just a problem book

Maybe

But math is also for not so clever people I believe

So I will learn modules

And get back to this

Thanks for the solution

[south patrol]

Oh i think i may have an idea

Consider the set of all determinants of matrices whose first row is (p_1,...,p_n). This is an ideal containing p_1,...,p_n and thus 1 right?

Wait no it's not clear

[rocky cloak]

Is it obvious that it is an ideal?

[south patrol]

The main problem is addition lol

Rippies

Hmm

Okay, I'm tempted to say the following works - i'll write it out for n=3

Suppose we have $(p_1,p_2,p_3)$ = 1. Write $(p_1,p_2)=(p)$ where $(p,p_3) = (1)$. Then if $p = f_1 p_1 + f_2 p_2$, the matrix $A = \begin{pmatrix} p_1 & p_2 \ -f_2 & f_1 \end{pmatrix}$ has determinant $p$. If $1 = pa + p_3 b$ I think the matrix $\begin{pmatrix} p_1 & p_2 & p_3 \ -f_2 & f_1 & 0 \ \pm b & 0 & a \end{pmatrix}$ ought to be as desired

[cloud walrus]

Süßkartoffel

[lament bough]

what problem book is it?

[south patrol]

So a sort of inductive approach with block matrices

Eh okay doesn't quite work

Gr

Well this has nerd sniped me lol

[tough raven]

Essentially just the proof that torsion-free => free over a PID, right?

[rocky cloak]

I guess. But existence of Smith normal form is not that bad

Hmm, a more linear algebray idea:

Let w1 be the vector (p1, ..., pn) and v1 = (a1, ..., an) a vector such that w1v1 = 1.

Let K be the field of fractions of R, and consider the orthogonal complement of w1 in K^n. Pick a vector v2 in the orthogonal complement and rescale such that v2 is in R^n and gcd of coefficients is 1.

Pick w2 in R^n such that w2v2 = 1. Then w2v1 = r for some r. Replace w2 by w2 - rw1. Notice w2v1 = 0 and w2v2 = 1.

Pick v3 in the orthogonal complement of w1 and w2, and rescale as before. Pick w3, so that w3v3 = 1. w3v1 = r, w3v2 = s, replace w3 by w3 - rw1 - sw2.

Continuing you construct the matrix with rows wi and one with columns vj, that multiply together to the identity. Since both matrices have coefficients in R, the determinant is a unit in R, hence in F. Rescaling one row gives the desired result.

[still dew]

It's like grahm Schmidt lol

[still dew]

Ok cool thanks

[still dew]

Berkeley problems in maths

[dire siren]

@still dew I found this proof by induction: https://math.stackexchange.com/questions/1833917/if-gcd-a-1-ldots-a-n-1-then-theres-a-matrix-in-sl-n-mathbbz-with-fi
it's for integers, but that's not an issue; it works over any PID

[chilly ocean]

can somone explain this part where they say repeated use of the equation [y^2] = [xz] and conclude that [f] = [g(x,z) + yh(x,z)]?

[agile burrow]

Suppose I was just thinking about the ring K[x, y, z]. I'll view this as K[x, z][y], so elements of this ring are polynomials in y whose coefficients are polynomials in x and z with coefficients in K. That is, an element of degree 2 looks like f = g_0(x, z) + y g_1(x, z) + y^2 g_2(x, z) where the g_i are in K[x, z].

Now we have this same way of thinking in the quotient K[x, y, z] / (xz - y^2), but now we have the identification xz = y^2. Thus, in the above expression, we can rewrite f = g_0(x, z) + y g_1(x, z) + (xz) g_2(x, z) = h_0(x, z) + y g_1(x, z) where h_0 is a new polynomial in K[x, z], namely h_0 = g_0 + xz g_1.

Basically just do that repeatedly. So if I see a term of the form g(x, z) y^{2n}, rewrite it as g(x, z) (xz)^n, which is absorbed into the degree 0 term. If I see a term of the form g(x, z) y^{2n + 1}, rewrite it as g(x, z) (xz)^n y, which goes into the coefficient of the degree 1 term

[still dew]

Thanks for this

[tribal niche]

Can I get a hint on this?

Suppose M is an module over a Noetherian ring and that f : M to M is a surjective module homomorphism. Show that f is also injective.

I'm trying to get some contradiction by looking at some non-zero c for which f(c) = 0, and getting the existence of d such that f(d) = c, and so on. But I'm not sure how to use the Noetherian condition

[ocean charm]

Hi, is it true that if I have an integral domain A with field of fractions K, then the field of fractions of its normalization is also K?

[chilly ocean]

what is normalization?

[ocean charm]

integral closure in its field of fractions

i don't think it's true but i can't come up with anything

[coral spindle]

The integral closure of A is a ring containing A and contained in Frac(A). It is immediate that its field of fractions is also Frac(A).

[rocky cloak]

The kernel of f is a submodule of M.

Now, think about ||the kernel of f^n||

Also, I think you mean that M is a Noetherian module, not module over Noetherian ring. Because as stated it's not true

[solemn dew]

does anyone know why (x(x)y)+(y(x)x) is in Sym^2(V) for x,y in V?

[rustic crown]

hint: what's (x+y)⊗(x+y)?

[solemn dew]

can you really do that with tensors?

x⊗y + y⊗x = (x+y)⊗(x+y)?

arent they completely different things

[rustic crown]

yea they're different

but ⊗ is bilinear

so you can distribute it out

[solemn dew]

oh right

thank you!

[coral spindle]

Just making sure you understand: what you wrote here isn’t true; make sure you distribute and really understand det’s hint

[tribal niche]

I forgot to mention that M is finitely-generated, oops

But thanks for the hint - I think I've got it 🙂

[tough raven]

You didn't — a Noetherian module must be finitely generated.

[rustic crown]

(earlier the ring was assumed to be noe, not the module >.<)

[rocky cloak]

A finitely generated module over a Noetherian ring is indeed Noetherian though

But worth noting that the same proof works for any Noetherian module

Also, interestingly, it's true if you just assume M to be finitely generated, but then the proof is quite a bit harder.

At least if it's over a commutative ring, not sure if it holds in the noncommutative case....

[tough raven]

Oops 💀

[tough raven]

That a finitely generated module over a Noetherian ring is Noetherian?

It holds for the same reason — if R is Noetherian, R^n is as an extension of R^{n-1} by R (induction), then M is as a quotient of an R^n.

[rustic crown]

jagr prolly meant the original question. f : M --> M epi, M f.g. / comm ring A then f also mono.

since the standard proof of that requires nakayama

[south patrol]

in fact you can check that Sym^2 is exactly the symmetric things

like there's an action of C_2 on V (x) V which swaps factors and Sym^2 is the stuff which is fixed under that

[solemn dew]

yesyes
x(x)y + y(x)x is bilinear, and let it be a function f(x,y)
now f(x,y)=f(y,x) so it is symmetric

[south patrol]

good question

[rocky cloak]

It is indeed not true in the noncommutative case. Just let A be a ring with an element x such that x has a left inverse, but no right inverse.

Then if you take A = M, then multiplication by x on the right is surjective, but not injective.

[south patrol]

Nice

Classic example being endomorphisms of uh k^{\mathbf N} and the left shift operator x

[rustic crown]

or Q[x] and d/dx, ∫

[tough raven]

Is that Noetherian though?

[rocky cloak]

No, that's the point

If the ring was Noetherian then any finitely generated module would be Noetherian as well

If the module is Noetherian it doesn't matter if the ring is commutative or not

[south patrol]

Oh yeah maybe that was silly of me lol not quite an example for what you want

[rocky cloak]

Well, it's an example for what I wanted.

[rustic crown]

why is everybody so confoosed

[rocky cloak]

Probably people got too much blood to their head, when trying to read crysis' name

[summer path]

Confuwused

[tender wharf]

what does it mean for an element a \in M to be annihilated by I^k? Does it mean that for all i in I^k we have ia = 0?

[dire siren]

yes

[tender wharf]

thanks

[chilly ocean]

Annihilation of the wicked

[dull ginkgo]

is this subring being fixed pointwise (i.e dxd^-1 = x) or is it being fixed set-wise (dSd^-1 = S)?

The former is trivial, as dxd^-1 = x => dx = xd for all d in D so S is a subset of Z(D), else it must be all of D

It is the latter

[stark helm]

if I can show that any polynomial of degree n with real coefficient can be factored into product of linear or quadratic polynomials, can I argue that all quadratic polynomial as factors must have either one real and one complex or two complex roots?

[delicate orchid]

Good luck finding a quadratic with one real root and one complex root

[reef smelt]

it is not

[rocky cloak]

If a complex number is a root of a polynomial with real coefficients, then so is it's complex conjugate

[stark helm]

OK, i got it

Then my argument to say because all quadratic polynomials factors must have two complex roots, and it shows that all roots of polynomial of degree n with real coefficient must be in the field of complex numbers right?

[south patrol]

x^2 - 1 has a real root 1 and a complex root -1

:)

[chilly ocean]

x²-ix

[stark helm]

Prove that any non-constant polynomial p(z) with real
coefficients can be factored into a product of linear and quadratic polynomials, each with real coefficients. my solution is to suppose base case n=1 or 2, and suppose n=k is true. Then when degree is n=k+1, and suppose contradiction that there exist a factor with degree larger than 3, but the question is how can I argue that a polynomial with degree k+1 must have at least one linear factor or quadratic factor??

[rustic crown]

use that C is algebraically closed

[rustic crown]

then read this

[stark helm]

I think I can not use it at this time, here is the original question

[chilly ocean]

hmm if k+1 is odd then its easy

[rustic crown]

what all are you allowed to use?

[south patrol]

Eh you kinda have to assume C is algebraically closed to do this lol

[rustic crown]

if q is any polynomial in C[x], then q * bar(q) is in R[x]. and since quadratic formula is a thing, if you proved (a) that would imply C is algebraically closed.

[south patrol]

In the sense that these easily imply that

Damn, beaten to it lol

[rustic crown]

<

[south patrol]

Kinda funny how like, C being algebraically closed was smth we implicitly assumed on faith in high school

Like I wonder if, in a similar vein, anyone has ever asked a teacher why sqrt(2) exists

Because imo it's a good question lol

[rustic crown]

so (a) is more or less equivalent to C being algebraically closed. there are many proofs of this, using topology, galois theory, complex analysis. idk any that just uses induction

[south patrol]

the galois theoretic proof is beautiful

[rustic crown]

that's det's favorite too

[south patrol]

topological one is my least favourite ish

In the sense that the proof is sort of easy to visualise but slightly tedious to write down iirc

[rustic crown]

galois theory of covering spaces :giggwe:

i wonder if we can actually translate the proofs

[stark helm]

so part a is equivalent to prove that C is algebraically closed?

[rustic crown]

yep

[stark helm]

i am considering if C algebraically closed means we can factor exactly n roots for polynomials of degree n

[rustic crown]

yee

[stark helm]

Hence we can not reason that C algebracially closed here right?

since these two are same thm

[rustic crown]

yea if you used that, (b) would look stupid

[rustic crown]

i won't say it's any more tedious than the galois theory proof

<

if f(z) is never 0, then g(z) = f(z)/|f(z)| makes sense.

for tiny values of z, g hovers around g(0)

while for large values of z, g should make n loops

more precisely, h(z, t) = g(zt) from S^1 x [0, 1] --> S^1 is an uwu homotopy. similarly h'(z, t) = g(z/t) is also an uwu homotopy (it makes sense at t = 0 as well).

so constant g(0) is homotopic to (-)^n : S^1 --> S^1

<

[dull ginkgo]

oh god

Let x^n = y^m = 0

[hollow mica]

how does this follow from Hilbert's Nullstellensatz

[south patrol]

What is your form of the Nullstellensatz

There are some forms in which is part of the statement

Lol

Also feel free to ask about such things in advanced algebra (or even alg geo if it's towards varieties)

[hollow mica]

• k a field
• K an algebraically closed extension of k
• For any ideal of polynomials J in k[X1, ..., Xn], one has
• I(V(J)) = radical(J)

[south patrol]

Unfortunately the nullstellensatz is essentially a name for a family of theorems

Ah, okay.

[south patrol]

And are you asking specifically about the last line?

Or everything

[hollow mica]

everything

[south patrol]

For the first bit, it's easy to see that I(P) contains (X1 - a_1,... ) and that the latter is maximal since the quotient is just K again. So that is the first bit.

For the last bit about maximal ideals: this does follow from the characterisation you have if you upgrade it slightly to saying there's a correspondence between radical ideals of K[x1,...,xn]. and algebraic subvarieties of K^n. A maximal ideal corresponds to a minimal (non-empty) subvariety, which is necessarily just a point, and from the start you know the ideal corresponding to a point is of that form

For the second bit about the radical, well you can view that as a consequence of this correspondence in a similar way (any algebraic variety is a union of its points.)

You can also view it as saying that / as a consequence of the fact that k[x1,...,xn] is a Jacobson ring - that is, every prime ideal is the intersection of the maximal ideals which contain it. Since the radical of I is the intersection of all primes p containing I, and each such p is the intersection of the maximals containing it, in fact I is the intersection of the maximals containing it. And the third part characterises those maximal ideals

Sorry this is a lot of text, but I hope it's helpful and feel free to ask more!

Knowing about the correspondence between alg sets and ideals is super useful, and the nullstellensatz is the hard bit of this correspondence (everything else is much easier)

[dull ginkgo]

Assume $x^n = y^m = 0$, then $(x + y)^{nm} = \sum_{k = 0}^{nm}{\binom{mn}{k}x^{nm - k}y^{k}} = \sum_{k = 0}^{m - 1}{\binom{mn}{k}x^{nm - k}y^{k}} = 0$

[cloud walrus]

Request a new nickname

[dull ginkgo]

nm - k > nm - (m) > n(m-1) > n if m > 2

[south patrol]

Real

That works

Can just take like m + n tho I think

But yes

[topaz solar]

That indeed gets you sums of nilpotent are nilpotent (w/ commuting)

[hollow mica]

I am reading up on the correspondence between varieties and radical answer, thanks for the answer @south patrol

[rustic crown]

i'll compress a lil what potato said.

in general if X is any algebraic set, then I(X) = functions vanishing on X = {f such that f(p) = 0 for every p in X} = {f such that f in I(p) for every p in X} = intersection of I(p) for p in X.
so I(V(J)) = intersection of (x_1 - p_1, ...) over p in V(J) = radical(J)

and for the last part, if J is a maximal ideal, then radical(J) = J is a proper ideal. so since this is an intersection of I(p) for p in V(J), there should be at least one such p, else the empty intersection wouldn't give a proper ideal. This means J is contained in I(p) = (x_1 - p_1, ...) and by maximality, we get J = I(p) = (x_1 - p_1, ...)

[dull ginkgo]

Assume [x] in R/N is nilpotent, then for some n: forall r in [x] r^n = 0 (mod Nil(R)) implying r^n is in Nil(R), which implies (r^n)^m = 0 by definition, so r^nm = 0 thus r is in Nil(R)

thus [x] = 0

[south patrol]

Np

[south patrol]

Yes, though saying for all r in [x] is a little unnecessary given that you notation implicitly picks a representative

[hollow mica]

wait wow that all made sense

[dull ginkgo]

I am going to generalize this a bit. Assume z is in Z(U), and H is the subset of units x such that x = z (mod I)

x = z (mod I) => x - z is in I => u(x - z)u^-1 = uxu^-1 - z is in I => uxu^-1 = z (mod I) => uxu^-1 is in H. Thus H is normal in the group of units under multiplication

[hollow mica]

so is the correspondence between subvarieties of K^n and radical ideals just extra stuff?

[dull ginkgo]

@south patrol does this sound right?

[south patrol]

Wdym?

[south patrol]

Sure yes nice

[hollow mica]

Idk everything that det said made sense and didnt mention any correspondence

[south patrol]

Well he is using the correspondence implicitly / proving parts of it

The correspondence is basically the point of the nullstellensatz and gives the results you mentioned a natural interpretation

[dull ginkgo]

can I use the finite-rank crap of M_n(R)

if $e(n,m)$ is the matrix with a $1$ at $(n,m)$ and $0$ elsewhere, i'm pretty sure $e(m,m)Me(n,n) = M_{n,m}e(n,m)$

[cloud walrus]

Request a new nickname

[dull ginkgo]

$e(a,b)e(c,d) = \delta_{b,c}e(a,d)$ so $M = \sum_{n = 1}^{N}\sum_{m = 1}^{N}{M_{n,m} e(n,m)}$ \\

$e(k,k)M = e(k,k) \sum_{n = 1}^{N}\sum_{m = 1}^{N}{M_{n,m} e(n,m)} = \sum_{n = 1}^{N}\sum_{m = 1}^{N}{M_{n,m} e(k,k)e(n,m)} = \sum_{n = 1}^{N}{M_{n,k}e(k,m)}$ \\

$e(k,k)Me(h,h) = \sum_{n = 1}^{N}{M_{n,k}e(k,m)e(h,h)} = M_{h,k}e(k,h)$

[cloud walrus]

Request a new nickname

[dire siren]

Why do people write n before m?

[dull ginkgo]

arbitrary choice in this case as long as it's consistent

[dire siren]

usually when I see n written before m, I have to swap them in my head

[dull ginkgo]

Every I embeds into an ideal I via the constant-diagonal map

but idk how to go about this problem

[vivid tiger]

"number" and then "hey what's another letter that's like n but different..."

[dull ginkgo]

If R is a ring, R embeds straight into K = M^n(R) via the constant-diagonal matrix map

so for now I'm literally just considering R to be in M^n(R) for my sanity

actually gonna need to ponder how I'd do this formally

[south patrol]

Poor education.

||/s||

But yes lol I often go back and change n to m later so i can use n after

[south patrol]

Oof, interesting

[dull ginkgo]

If I can coerce some kind of uniqueness out of this it would be nice

[dire siren]

some thoughts: consider an ideal J of M_n(R) and let E be the set of entries of all the matrices in J
playing around with diagonal matrices, we conclude that <E>⊆E, so E=<E>
so J⊆M_n(<E>)

[dull ginkgo]

M_n(R) is a finitely generated module/algebra over R generated by the aforementioned all-but-one matricies, so xe(n,m) = ye(n,m) for x,y in R <=> x = y. If M is in J, then M_{m,n}e(m,n) is in J. Thus M_{n,m} is in E, but also in the image of E under the diagonal-matrix map back into itself

[cobalt heath]

I think considering it as End_R (R^n) would make it simpler

[dull ginkgo]

unfortunately we're not at modules yet

[cobalt heath]

Since it is more closely related with its algebra structure

[dull ginkgo]

I don't want to jump too far ahead, even though yes, it would be easier

and the fact that the double-indicies are reverse references the idea of the transpose

[cobalt heath]

I mean, you can still define R-linear maps R^n -> R^n

Dunno if this would be using module facts

[dull ginkgo]

that's a lot of work for a single problem

I really don't feel like going through all of that

[cobalt heath]

Ah. I thought dealing with e_ij would be of bigger hassle

[dull ginkgo]

i mean that's how jacobson gives it

[cobalt heath]

Ouch

[cobalt heath]

Oh, you solved it

[jovial tundra]

Can anybody visualize anything when it comes to algebraic groups? Because, for me, if the proofs doesn't flow into my mind, I'm going to be stuck for a while

[dull ginkgo]

Actually I think it just flat out comes from it being a finitely-generated module

[cobalt heath]

Hm I don't think it is obvious

[dull ginkgo]

It's obvious from the definition

[cobalt heath]

E.g. Z/2Z has less ideals than Z

[dull ginkgo]

just every way I try to approach it feels redundant via the definition as the "ideal of matricies with components in J"

[dire siren]

does this prove the claim? the fact that M_{n,m} is in E follows from the definition of E

we need the reversed inclusion

[dull ginkgo]

every way I try to do the reverse inclusion I run into this problem

[dire siren]

it's not redundant

[dull ginkgo]

Every element in J is a unique sum over the scaled "basis matricies"

[dire siren]

because M_n(I) means all the matrices with entries in I

but when I defined that set E, we only knew that J contains matrices with elements in E, but not necessarily all such matrices

the hard part is the reversed inclusion

[dull ginkgo]

It may be easier to start with an ideal of R, embed it into M^n(R) and consider the generated ideal

[cobalt heath]

I think it follows from like

E e_{ij} is included in J

[dull ginkgo]

This ideal BY DEFINITION is M_2(I)

...wait

unless I phrase it very particularly

[cobalt heath]

You still need to show the surjectivity

[dull ginkgo]

a matrix in M^n(R) is an element of M^n(I) if e(n,n)Me(m,m) = xe(n,m) for some element x of I for each n and m

[dire siren]

what exactly is E e_{ij}?

[dull ginkgo]

matrix where all entries are 0 except for a 1 at double index i,j

[cobalt heath]

Multiplied by ideal E (by E e(i, j) I mean)

[dull ginkgo]

I think the universal-multiplicative-closure of ideals in particular here is super important with that identity

Versus other finitely generated modules as an algebra

[dire siren]

@cobalt heath but do we really know that that set is included in J?

[cobalt heath]

By the kind of manipulation given here @dire siren

You can kind of "extract" certain coordinate of the matrix and move it to desired location

[dire siren]

yeah, but to what matrix you apply these manipulations?

[cobalt heath]

Of course, this only holds for both-sided ideals, but I think left ideals would be more involved anyway

[dire siren]

we don't really know how J looks like

[cobalt heath]

Uh

I mean

[dull ginkgo]

that's why I used that

to describe J.

and through the e(n,m) being a basis

[cobalt heath]

Some c \in E should appear somewhere in some matrix in J

Then the manipulation reveals the generators of J

[dull ginkgo]

M_n,m e(m,n) over all M in our ideal serve as an additive generating set over J

[cobalt heath]

Yeah
..wait that m, n is switched

[dull ginkgo]

of which correspond to x such that x = M_n,m for some M

due to e(n,m) being a basis

[dull ginkgo]

I'm chosing it to go n-m-m-n specifically

just due to the identity I showed

[cobalt heath]

Ouch, my bad

@dull ginkgo how far are you from modules?

[dull ginkgo]

I’m just trying to formalize it

[dull ginkgo]

Next chapter is modules over PIDs

[cobalt heath]

Ohh, that's great

[dull ginkgo]

Sorta cheating with this problem because I know a priori some module stuff and I’m using module theoretic ideas

While the textbook assumes you don’t

[cobalt heath]

I think it might be foreshadowing for modules

[dull ginkgo]

Sorta

[cobalt heath]

Giving ideas about how modules would go

[dire siren]

I see it now, thanks

I can barely think, it's 3 AM here

[cobalt heath]

Ah sorry, I innately assume everyone lives in the same timezone

I am usually in worse state at 3am..

[dull ginkgo]

Let me be formal with that problem.

M^n(R) admits a basis e(n,m) of which generates it as a unital associative R-algebra. It also has the identity that e(a,b)e(c,d) is 0 unless b = c, where it’s e(a,d). It also admits an isomorphism from R to it’s center

Thus let’s say we have an ideal J of M^n(R). from the above identity, e(n,n)Je(m,m) is a subset of Z(M^n(R))e(n,m) ~ Re(n,m) can be identified with a subset of R by bijection. The intersection of the image of all the e(n,n)Je(m,m) is an ideal of R because of the algebra-center structure.

This image is an ideal of R because Z(M^n(R)) ~= R, so it can be embedded back into R, and can be expanded through generation. It turns out this new ideal is the original J

in other words, $e(n,n)M^n(I)e(m,m) = Ie(n,m)$

[cloud walrus]

Request a new nickname

[dull ginkgo]

It has to do with the whole basis of the algebra shit and R being isomorphically it’s center AND the ring that the algebra is defined over

Which might actually classify matrix rings in retrospect alternatively

[cobalt heath]

R isn't center tho, is it

[dull ginkgo]

iso to the center

The subring of diagonal matrices

[cobalt heath]

What if R is not commutative

[dull ginkgo]

Oh shit, good point

[cobalt heath]

Ofc noncommutative rings also doesn't exist but /s

[dull ginkgo]

Proof still works, it just has to do with the embedding of R into M^n(R)

[cobalt heath]

I think it has to do with (semi)simplicity

But that involves more definitions 👀

[dull ginkgo]

@cobalt heath I had the horrid realization that M^n(R) is some god awful semigroup ring R[K] where K is the semigroup where e(n,m)e(k,h) = \delta_{m,k}e(n,h) with an annihilator 0

[cobalt heath]

Wh

[dull ginkgo]

I’m not wrong

[cobalt heath]

By semisimplicity, I mean specific properties of modules

[cobalt heath]

But that's.. interesting

[dull ginkgo]

You just ignore the annihilator in the semigroup ring

I’m still thinking of a way to formalize that goddamn exercise but it has to do with how you define the matrix ring in the first place

That semigroup ring def might literally be the most formal way to put it without calling it an endo ring explicitly

[cobalt heath]

Gotta speedrun to modules then

[dull ginkgo]

I have some stuff left in this chapter to do, it’s also a big chapter

[drifting mauve]

Hey everyone, I have this question: $\$If $H$ is a subgroup of index $2$ in a group $G$, and $g \in G$ has odd order, then $g \in H.$\\
What I've done to solve it is do it by contradiction and I'll try to summarize briefly what I did:\\
Suppose that $|G:H| = 2$ and g is odd and $g \in G - H\$
Then G/H = {gH, H}, thus $g \in gH$, also $g^2 \in gH\$
So for some $h \in H$, $g^2 = gh$, implying that $g = h$, so $\Rightarrow!\Leftarrow\\$
However, at no point during the proof did I use the fact that g is odd, so what have I done wrong because obviously this can't be true. I am by no means saying that this is the best or even correct way to solve this, just saying what I've tried.

[cloud walrus]

theaveragejoe6029

[topaz solar]

Why is g^2 in gH

[hollow mica]

I think the average joe thought that gH is a subgroup

[barren sierra]

Need sanity check

"same process" is just first iso theorem

but isn't this impossible? Z^3 / <n, n, n> is finite isn't it?

I may just be tired, friend texted me asking for help

but I'm not seeing it

nvm it isn't finite i'm dumb

I got it I'm dumb

[drifting mauve]

well, if g is in gH, shouldn't g^2 be in gH. Also since if g^2 is not in H, it has to be in gH.?? right?

[topaz solar]

Why would you expect g^2 in gH

If g has order 2, it definitely isn’t

[drifting mauve]

g doesn't have order 2 tho.

[topaz solar]

why is g^2 not in H

[drifting mauve]

by assumption for contradiction g is not in H, since H is a subgroup. g^2 also cannot be in H. I think thats right?

[topaz solar]

I’m saying your reasoning on g^2 not in H is the issue

Because why can’t g^2 be in H

Consider 2Z as a subgroup of Z

It has index 2, in fact

1 is not in Z, but 2 (i.e. 1^2) is

Just because g isn’t in H doesn’t mean g^2 isn’t

[drifting mauve]

oh okay. I get you, thanks.

Do you have any other suggestions on how to solve this question then?

[topaz solar]

Yes but I can’t just give the answer

What do you know about index 2 subgroups

(The thing being index 2 is important!)

[drifting mauve]

honestly not much. I'll have a read of some theorems.

hopefully you wont hear from me again.

thank you

[topaz solar]

But yeah, if you know much about index 2 subgroups and what nice property they must have

There’s a nice way to handle every element in G based on in or out of H

Theres ur hint

[drifting mauve]

Ah, just got it. Thank you 🙂

[boreal inlet]

Let b and d be rational numbers such that d is not a square in Q. How do we prove that x^2 - bx + d is irreducible over Q(√d)[x]?

Do we somehow apply Eisenstein

Also additional information : d(b^2 -4d) is also not a square in Q

and x^4 + bx^2 + d is irreducible in Q[x]

[If we need more context, I'm actually trying to determine the galois group of the splitting field of x^4 + bx^2 + d over Q, when d and d(b^2 - 4d) both aren't squares in Q. I ask the question because that is something would work to establish that the galois group is indeed D_8, as per Kappe-Warren's Lemma]

[south patrol]

Hint: quadratic formula

(Also minor terminological thing: I'd say irreducible over Q(sqrt(d)), or irreducible in Q(sqrt(d)][x], not irreducible over Q(sqrt(d))[x])

[boreal inlet]

Oop, got it

Also wait lemme check

[south patrol]

No worries lol

But hopefully my hint will help

[boreal inlet]

Hmm it's just sqrt(b^2 - 4d)

Now we do know that both d and d(b^2 -4d) aren't squares in Q

What if b^2 - 4d IS a square

[rocky cloak]

I guess it's worth noting that both of these additional informations are necessary

[boreal inlet]

And that also in Q(√d)

[rocky cloak]

You can use these to prove that b^2 - 4d is not a square in Q(sqrtd)

[boreal inlet]

hmmm...

b^2 - 4d = (b - 2√d)(b + 2√d) in Q(√d)

Those can't be in Q because d isn't a square

Of course both of these factors cannot be the same because otherwise d is 0 and it means d is a square in Q

So if this needs to be a square both these two need to be sqaures aswell

[rocky cloak]

It's possible for the product of two non-squares to be a square

Maybe try to instead think about what squares in Q(sqrtd) looks like

[boreal inlet]

(x + y√d)^2 = x^2 + y^2d + 2xy√d

It must be of this form right

Oh.

2xy must be zero for it, and that means either x or y is 0

[rocky cloak]

Yeah, now you're onto something

[boreal inlet]

One of x and y is zero

Because b^2 -4d is definitely rational

so if y is zero

Actually no.

I'm not sure what to do from there

I might be missing something very trivial it seems

[boreal inlet]

Because if one of x and y is zero, whole thing boils down to b^2 - 4d being a square in Q

How to know about that

Oh wait

b^2 - 4d = x^2 => b^2 - x^2 = 4d => (b - x)(b + x) = 4d

Oh fuck these aren't integers...

[lilac mango]

Is there any smart way to count subgroups of size p^k of (Zp)^n?

[lilac mango]

I was thinking counting linear subspaces of (Zp)^n as a Zp-v. space (as subgroups of (Zp)^n are subspaces of (Zp)^n and viceversa)

[lilac mango]

Maybe one can make this a question about counting some set of matrices

Which should be easier

[south patrol]

Yes exactly this

||assuming by Zp you mean the field w p elements||

[lilac mango]

Yes

But how would you do that?

Because I don't know how I would count that

[rocky cloak]

So you have the two cases, either x is 0 or y is 0.

If x is 0, then y^2d = b^2 - 4d, see how this contradicts your first but of information? Hint: ||multiply both sides by d||

If y is 0, then b^2 - 4d is a square in Q, see how this contradicts your second piece of information? Hint: ||How does x^4 + bx + d factor?||

[rocky cloak]

One thing you could do is start by counting the number of (ordered) sets of k linearly independent vectors. Think about how you might pick such a set one vector at a time.

Then do a similar calculation to determine how much you've overcounted

[lilac mango]

Oh ok so for the first one it would be something like:
(p^n-1)(p^n-p)...(p^n-p^(k-1)).
And each k-dimensional individual subspace should have
(p^k-1)(p^k-p)...(p^k-p^(k-1))
ordered bases.

So when counting linearly independent ordered sets of size k we are counting each k-dimensional space
(p^k-1)(p^k-p)...(p^k-p^(k-1))
times. So it would be the quotient of these two products, right?

[weary spade]

prove that in Z[sqrt(-5)] the number 2 - sqrt(-5) is irreducible

[dire siren]

use the norm

[lilac mango]

Thanks for your help (:

[dull ginkgo]

@dire siren that problem from last night, I’m pretty sure it can be proved through the following lemma:

If $I$ is an ideal of $R$, then $M^N(I)= \sum_{m = 0}^{N}\sum_{n = 0}^{N}{Ie_{n,m}}$ where $e_{n,m}$ are the matrices that are all 0 except for being 1 at double index $(n,m)$

[cloud walrus]

Request a new nickname

[dull ginkgo]

It’s not too hard to prove that fact and the correspondence that follows

[dire siren]

yes, after proving that E * e_{i,j} is included in J, we are done, because they generate M_n(<E>)

[dull ginkgo]

Yep

[sterile condor]

What’s a polynomial ring?

[rocky cloak]

A ring consisting of polynomials 🤓

[rocky cloak]

But more seriously, do you know what a polynomial is?

[sterile condor]

Yes, but is a polynomial ring just a ring that consists of polynomials?

[rocky cloak]

Basically.

The polynomial ring in n variables R[x1, ..., xn] is the ring consisting of all polynomials in n variables with coefficients in R.

When people say a polynomial ring, they usually mean the polynomial ring in some number of variables over some commutative ring.

So I guess it would be more appropriate to say it's the ring of all polynomials (in a certain number of variables / with certain coefficients)

[boreal inlet]

Yeah I actually solved the x = 0 part a few minutes ago just now, but I'm still confused on y = 0 part

OHHH WAIT

Let y = x^2.

Then x^4 + bx^2 + d = y^2 +by+d

Bruhhhh...

If that's irreducible this can't even work out because this quadratic factors like

(y -b/2 - sqrt(b^2 - 4d)/2) (y -b/2 + sqrt(b^2 - 4d)/2)

lol

If √b^2 - 4d is a square we ded

Btw, there's another problem. If we do assume that d(b^2 - 4d) is actually a square this time, does this change the Galois group?

(i mean if it doesn't then there's a problem with the question I'm trying to attempt)

[halcyon hull]

anyone here familiar with Banach-Lie groups?

[boreal inlet]

Nevermind, it takes similar arguments

Thanks a lot @rocky cloak @south patrol

[wide brook]

What would be the idHom_R_(R,M) where R is a ring and M is an R-module group?

[solemn dew]

does anyone have an idea why this is true? L is nilpotent and some 2x2 matrix:

[loud merlin]

binomial theorem?

[solemn dew]

tried it but don't see the pattern

[south patrol]

what is the (x) here? Like the hadamard product

[solemn dew]

tensorproduct

[south patrol]

do you mean nxn matrix? rather than 2x2

[solemn dew]

yea it works too

[south patrol]

Sure, then this looks like a basic induction

[solemn dew]

ok i don't know that this is true so i have to find this formula somehow

[south patrol]

well you have the formula and can just prove it holds by induction

[solemn dew]

no like i don't know how to come up with that formula

[celest furnace]

Well you don't have to you already have the formula

now you just have to show the n->n+1 step

[solemn dew]

i don't want to prove that it is true

i want to find how you would come up with that formula

like what is the reasoning that this works

maybe it has something to do with binominal theorem

ect ect

[rocky cloak]

Hom(R, M) is just equal to M

[rocky cloak]

If L is a nilpotent 2x2 matrix, then L^2 = 0, so most terms in the right are 0....

[solemn dew]

damn you are right

i am so stupid lol

thank you

[rocky cloak]

Still doesn't seem quite right to me though....

[wide brook]

Why though? That’s what I’m actually trying to prove that they’re equal

[solemn dew]

it is generalized for L^m=0

[rocky cloak]

You have an isomorphism sending f to f(1) and inverse sending m to
r |-> rm

[rocky cloak]

Where is this formula from?

[solemn dew]

It is from the solutions from Linear algebra class that I have
Question is:
Show that I⊗L - L⊗I is nilpotent if L is nilpotent, but that the reverse does not always hold

I is the identitymatrix, and L some linear function given by a matrix

[rocky cloak]

I see, well I think there are some binomial coefficients missing. But either way you get that it's nilpotent

[tropic spade]

Does anybody know of a good syllabus/course outline/study guide using dummit and foote for a kind of generic qual/prelim style course?

I'm planning to study from it over the summer.

I'm mostly looking for decent suggested sections with exercises/hw to do from it.

[crystal vale]

What is the definition of sub-semigroup?

[coral spindle]

A subset closed under the binary operation.

[delicate orchid]

a subobject in SemiGrp

[rocky cloak]

In general a sub-X is usually a subset so that the inclusion is a homomorphism of Xs

[chilly ocean]

In more general its a monomorphism

In even more general its an element of some distinguished class of monomorphisms (that is closed for composition maybe?)

often regular monomorphisms

[delicate orchid]

giving this answer but unironically

[crystal vale]

Okay thank you

[tribal niche]

I've come across the following statement:

If a ring R is a domain and there is an additive group isomorphism f : R -> Z^n for some n then R is Noetherian

could anyone explain why this is true? i dont think you can extend f to a ring homomorphism, so i'm not sure if you can say e.g. f(ideal in R) is an ideal in Z^n as a ring

[coral spindle]

Z^n is a noetherian Z-module, and any ideal of R maps to a sub(Z-)module of Z^n, so you get the ACC for free.

It's waaaaaay stronger than necessary tho

[south patrol]

Why is domain specified there?

I guess a more general statement is that if M is an S-module which becomes noetherian after restricting scalars then it's already noetherian

domain seems irrelevant

[winged void]

$Let ( \tau = (1\ 2\ 3\ \ldots\ n) \in S_n ) and ( \sigma \in S_n ). Prove that ( \sigma\tau = \tau\sigma ) if and only if ( \sigma = \tau^i ) for some ( i \in \mathbb{Z} ) with ( 0 \leq i < n ).$

[cloud walrus]

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

[winged void]

how to prove this ?

[south patrol]

There's a general formula for sigma tau sigma^-1 you may know

[winged void]

Which one

Can you maybe say it

[uneven island]

for a cycle $ \tau = (a_1\ \ldots\ a_k) $ and a permutation $ \sigma $, we have $ \sigma \tau \sigma^{-1} = (\sigma\ a_1\ \ldots\ \sigma a_k) $

[south patrol]

Basically note that like

Every element of {1,...n} is of the form σ(a) for some a in {1,...,n}

what is στσ^-1 applied to σ(a) ?

@winged void

[winged void]

Yes

Im here

[south patrol]

Yee tr this

[narrow wagon]

how to determine the factor ring $\mathbb{Z}[i] /<1-i>$?

[cloud walrus]

Dubs

[fading field]

why is your text orange

[delicate orchid]

write out an explicit isomorphism to Z

[fading field]

anyways you’ll do it by finding a ring hom with kernel (1-i)

[delicate orchid]

or that

[fading field]

you should know (and if you didn’t now you do) that kernels of ring homomorphisms are ideals and an analogous version of the first isomorphism theorem for groups also holds for rings

also why did you choose to make your text orange

[narrow wagon]

i haven't gone till there

this chapter just introduces ideals and factor ring

[dire siren]

@narrow wagon in the quotient ring you have 1=i
so any a+bi becomes a+b
and 2=(1-i)(1+i)=0, so the ring only contains 0 and 1

this can also be seen using the norm, but I guess this is not a suitable solution given that you just started with ideals: the fact that N(1-i)=2 tells us that the ring has only two elements

[narrow wagon]

you mean 1+ I = i + I?

[dire siren]

yes

[narrow wagon]

2=(1-i)(1+i)=0, so the ring only contains 0 and 1

how does this conclude

ring has only 2 elements

[dire siren]

so we proved that any element can be represented by an integer
and because 2=0 it follows that any even number is represented by 0, and any odd number is represented by 1

[narrow wagon]

so it's isomorphic to Z_2?

[dire siren]

yes

[narrow wagon]

so in general if $\mathbb{Z}[i] /<a-i>$

[cloud walrus]

Dubs

[narrow wagon]

this is isomorphic to Z_{a^2+1}?

as coset representative a and i are same

a^2=-1 ie a^2+1=0

[dire siren]

well, yes, but the proof is not that straightforward in this case

if k=0 in a ring we cannot necessarily conclude that the ring is isomorphic to Z_k

[narrow wagon]

maybe k is identity

not necessarily an element of order k too

[dire siren]

@narrow wagon basically to make it work, we should also ensure that the classes 0,1,2,...,a^2 are all distinct

that was actually also the case for the initial ring: the classes of 0 and 1 are distinct

for a=1 it's clear, but for arbitrary a is not that clear

[south patrol]

To be fair like Z[i] = Z[a-i]

So isnt this always Z

Or am I missing smth lol

Oh I'm thinking of i as a sumbol

But here it is actually a qrt of -1

[dire siren]

I guess it's still doable without morphisms

if k and l are in the same class, then k-l=(a-i)(x+yi) for some integers x,y
this is k-l=ax+y+(ay-x)i, so ay-x=0 and ax+y=k-l, so k-l=y(a^2+1)
but given the range of k and l, we must have k=l

anyway, the general result is that for a and b coprime we have Z[i]/<a+bi> ~ Z/(a^2+b^2)Z

[south patrol]

I would say like

Z[a]/(a-i) is Z[x]/(x^2 + 1, a -x) which is Z/(a^2 +1)

[dire siren]

that's a very clean solution

I just tried to find some elementary approach given that the guy who asked said that he was just introduced to ideals and quotient rings

[dreamy edge]

I have an assignment question where I need to describe all the homomorphisms from Z to Z_12, so my first thought was just phi(n)=kn where k is an integer between 1 and 12 inclusive, this mapping would map the generator of Z which is 1 to every possible element in Z_12, however, there are multiple ways to map 1 to lets say 2 for instance, 2*1, but also 2^1, the mapping phi(n)=2^n would still map 1 to 2 but other 3 would be mapped to 8 instead of 6, is this a different homomorphism or does it make sense to say the mappings are isomorphic to each other since the only thing that matters is the generator and phi(n)=kn are all the mappings up to isomorphism?

[chilly ocean]

what kind of homomorphisms? group homomorphisms? ring homomorphisms? something else?

[dreamy edge]

oh yeah sorry, group homo

with addition in both

[chilly ocean]

is phi(n)=2^n an homomorphism?

[dreamy edge]

I guess not because it doesn't preserve the group operation

[chilly ocean]

indeed it is not.

[dreamy edge]

thanks

[chilly ocean]

In ZF, is there a classification of dedekind-finitely generated abelian groups?

[crystal vale]

Hint, R/I is commutative but R not, and I don't want take I=R

[topaz solar]

What

Are you saying you want a noncommutative ring R so that you can quotient by an ideal I and get a commutative ideal?

Phrasing it as “hint” is odd

[crystal vale]

Yes

And I want an example such that a + I is nilpotent in R/I but not in R.

So if I take R= Z and I = 4Z.
If a = 2 then 2+4Z is nilpotent in R/I but not in R.

Is this correct?

[topaz solar]

Yes, 2^2 = 4, which is mapped to 0

So nilpotent

[crystal vale]

And similarly I want an example such that a is zero divisor in R but not in R/I.

So if I take R=Z_6, I is ideal generated by 3( element of Z_6) and a =2 then a is zero divisor in Z_6 but not in R/I.

Is it correct?

[topaz solar]

I believe that works

If by Z_6 you mean Z/6Z

[crystal vale]

Yes

Any hint for R/I is commutative but R not?

[topaz solar]

What have you tried

[crystal vale]

Similarly I want an example such that R/I has 1 but R has not

[crystal vale]

I thought if I take it I = R then it will work

And R is non- commutative

[topaz solar]

Well that does work, since you map to the 1 ring

[crystal vale]

Yes

But I want another example

[topaz solar]

Well this also works with R/I, as do others, but yeah

[crystal vale]

But I am looking for another example

[topaz solar]

If you have R = R_1 x R_2, can also quotient out R_2

[crystal vale]

And how can I prove that R/I = R then I = (0) ?

[crystal vale]

Got it, thank you

[topaz solar]

I’m not sure that one is true even so

[crystal vale]

It's given that

[topaz solar]

All you need is a surjective ring homomorphism R -> (something commutative) right?

[crystal vale]

If I take R = Z × 2Z and take I = Z × {0}, then it will work

[crystal vale]

I am not sure mapping terms

[topaz solar]

Like I said up here

[crystal vale]

Yes

[topaz solar]

This is commutative though

[crystal vale]

We can replace it with M( Z) × M(2Z) , right?

[topaz solar]

Well you want to say how big your matrices are

But yeah

[crystal vale]

2× 2 ?

[topaz solar]

Then if you quotient by M(2Z) you get M(Z) out

Which has a 1

[crystal vale]

Yes

[topaz solar]

So your ideal is {0} x M(2Z)

If you do that ideal in Z x M(2Z), it turns a non commutative into commutative too

[crystal vale]

Got it, thank you

[crystal vale]

I made a mistake here , I = {0} × 2Z

[topaz solar]

Yes

[winged void]

well i tried this but i did not really reach a specific solution

where i have power

im trying this --> way

[crystal vale]

If I take R= {0} then R is a field, right? Then R[X] is a field, right?

[coral spindle]

People often exclude the zero ring from being a field definitionally, but it's up to you. Indeed in this case, R[X] = R.

E.g. the statement "J is a maximal ideal of R iff R/J is a field" breaks if you allow the zero ring to be a field

[winged void]

this is what i have until now but idk what then

@south patrol

[crystal vale]

Yes

[south patrol]

Helo

[winged void]

Hello

[south patrol]

Okay sorry yes

[winged void]

Do not worry

[south patrol]

so yeah did you try uh what is στσ^-1 applied to σ(a)

[winged void]

I do not see it

[winged void]

Here is what I did

[south patrol]

It's just (στσ^-1)(σ(a)) = σ(τ(a))

[south patrol]

I'm not really sure what you're doing

[winged void]

How did you get that I do not get it

[south patrol]

Well just definition of composition

σ^-1(σ(a))=a

[winged void]

But how is that going to prove sigma = tau to the power of i

[south patrol]

I'll get that in a sec

the point is you can work out what στσ^-1 is

Then see what happens when it is equal to τ

So στσ^-1 sends σ(a) to σ(τ(a))

If we use the convention that like addition is taken mod n, then uh

στσ^-1 sends σ(a) to σ(a+1)

i.e. it's the cycle(σ(1), ..., σ(n))

so now we're saying (σ(1),...,σ(n)) = (1,...,n)

[winged void]

The last step I do not see how mod n

Will apply that sigma (a) will be send to sigma(a+1)

[south patrol]

τ(a)=a+1

so σ(τ(a)) = σ(a+1)

[winged void]

Sure I agree about that but we did not use mod here

Right

We use composition

Or am I saying something false

But how sigma (a+1) will apply that sigma = tau to the power i from here

[crystal vale]

In p-adic integer ring, non-units have form a/b where gcd(a,b) =1 and a is divisible by p, right?

[coral spindle]

No, the p-adic integers include irrational units.

The nonunits are those divisible by p.

[rustic crown]

local ring uwu

[south patrol]

Yeah there are a few ways to think about this. But note that e.g. if we view the p-adics as the set of sequences (a_i) where a_i is in Z/p^i and the a_i are compatible under the projections Z/p^i -> Z/p^{i-1} then (a_i) is a unit iff all the a_i are units, iff none of the a_i are multiples of p

[crystal vale]

But p-adic integer ring defines as { a/b | gcd(a,b ) =1 p does not divide b and a and b are integers} so how irrationals include, may be I am talking about other definition

[rustic crown]

that's Z localized at (p). the completion of that ring is the p-adics.

[coral spindle]

That is not the p-adic integer ring.

These are the p-adic rationals.

[rustic crown]

confoosing terminology

i would call Q_p padic rationals

[coral spindle]

They are the rational elements of Z_p

[rustic crown]

<

[coral spindle]

Whereas Q_p is usually referred to simply as the p-adic numbers

[crystal vale]

So if I take this definition then the non-unit form is a/b and p divides a , right?

[coral spindle]

I think this is really fairly standard terminology but you'll have to forgive me if I'm wrong, det

[rustic crown]

oh det doesn't know ><

[cobalt heath]

Ohhh

[rustic crown]

me can't figure out when the adjective is defining something new vs when it's restricting.

[cobalt heath]

I wonder how terminology ended up being like this

[rocky cloak]

I mean, it seems pretty reasonable.

You have Q_p and you call it p-adic numbers, because they're a new number system. Then you call [the closure of] it's ring of integers, the p-adic integers, makes sense.

The you split the p-adic integers into rational and irrational numbers, and call them p-adic rationals and p-adic irrationals.

[dull ginkgo]

Me when ostrowski’s theorem

[winged void]

i have just did what you have suggested

this is what you have suggested right ?

but then idk how to go further from here

so im not really sure how now

[dull ginkgo]

I’m pretty sure metrics on Q can be classified by the existence of a prime with n(p) < 1

[cedar path]

Hi. Sorry if this is not suitable for this channel. Would anyone help me to confirm this statement? It seems to say any G < GL(n,R) as n×n matrices could be written in block diagonal form, but I cannot see how this is true (the last sentence where it states the standard representation of GL(n, R) on extrrior algebra can be written as direct sum...)

[dire siren]

@cedar path this is a question for #advanced-algebra

[delicate orchid]

This is just decomposition into irreducibles

[rocky cloak]

Every matrix is can be written in block diagonal form by just having one nxn-block.

For some choices of G, there might be more blocks. Take G = {I} for a trivial example

[cedar path]

My algebra is a bit rusty, but iirc reducible and decomposable are not the same thing?

[rocky cloak]

I think the usage of the word is somewhat inconsistent.

But you're right that not every representation decomposes as a sum of simple representations

[delicate orchid]

are these not complex representations

tbf there's nothing in the image that says that they are so fair enough

[rocky cloak]

I don't see that it would matter

[delicate orchid]

the maschke's theorem in question:

where does the h go again

[rocky cloak]

Only for finite/compact groups though

[delicate orchid]

I'm leaving it there

[rocky cloak]

Like for G = Z = [1, Z; 0, 1], the canonical 2-d representation is indecomposable, but not simple

[delicate orchid]

the claim is that we can write it as a direct sum of irreducibles though, so now I'm confused

it's definitely not the case that every proper subgroup of GL(n,R) is compact

if it was just writing it as a sum of indecomposables then sure but that's not what it says

[cedar path]

I forgot to mention here G is assumed to be Lie group

[rocky cloak]

Well Z is a lie group, but maybe 0-dimensional groups are out

[delicate orchid]

ah it's only for the exterior powers ok

jagr does the character decomposition for \lambda^k V hold for these reps if we replace the characters with just trace functions

[rocky cloak]

I would guess so, but idk

[delicate orchid]

if so then yeah good luck actually seeing the jordan decomposition

[delicate orchid]

if you just think about it via schur functor nonsense then as long as V is finite dimensional nothing should go wrong

[rocky cloak]

What book is this?

[cedar path]

Riemannian Holomony Group and Calibrated Geometry by JD Lotay

[winged void]

@south patrol i found something but do not really understand it

what is here mod k suppose to mean

why like taking the reaminder of a_i+i / k

[rocky cloak]

Hmm, I can't seem to find that book

[cedar path]

Oh i didn't know Maschke is true for compact group. As holmony group are compact I think it would suffice for the purpose of this book. Thanks XD

[south patrol]

The proof works the same way just with Haar measure used to average

Well in fact it is just a generalisation if you note the original proof uses the Haar measure on a finite group

[south patrol]

bad book then lol

that's also notation i've never seen

[last spoke]

to be fair/pedantic the colon doesn't necessarily indicate a definition but merely an introduced topic. It seems to me that local rings are one way of introducing tho. I agree that the notation is strange

maybe further context does show that they mean this to be a definition

[delicate orchid]

oh brother this guy STINKS!

what is thaaattt

[chilly ocean]

I find it ironic how the so called "academics", when presented with ideas that differ from the mainstream, are always do quick to dismiss them and call them "bad". Its funny how despite academia being advertised as a place of inclusivity, knowledge and innovation, they still find a way to gatekeep opinions that don't suit the "norm". Academia has become an industry that sells fake knowledge and keeps away ideas that are too radical for their liking or political agenda. Perhaps the truth is that the brightest minds of our century are now not in the academia but independent thinkers. Ill let you reflect about this. Maybe if you try thinking for more than 5 seconds about it unlike your professors taught you to, youll find deeper meaning in the notation used by the book, and wont be so quick to judge a book by its cover next time.

[south patrol]

Can't tell if this is a shitpost

[delicate orchid]

everything carla says is a shitpost

[chilly ocean]

This shallow comment just proves my point..

[languid trellis]

we're just 2 lost worms swimming in worm bowls..... year after year.

[rocky cloak]

I'm not sure disliking what seems like intentionally misleading notation, should be seen as closed mindedness.

It's not like it's really differing ideas, it's just bad communication

[south patrol]

The main point was that they called it the p-adic integers when that has a very well established meaning

[chilly ocean]

This is the aame argument as used by religious dogmatics

[south patrol]

Saying that is an "idea that differs from the mainstream" is overselling it lol

[south patrol]

How lol

[languid trellis]

obvious troll is obvious

[chilly ocean]

The truth is school ans university today doesnt teach you to think for yourself

They just want you to think fast

[south patrol]

Lol

[chilly ocean]

Thats why exams are timed

[south patrol]

So I say smth is bad for using terminology at odds with everyone and now it feels like you are saying I never think for myself and shun open mindedness

Or have elements of that

[chilly ocean]

Quick thinking is useful in the workplace

[rocky cloak]

It's like if I told you I was a vegetarian that eat beef, you say that's not possible, and I reply: You're just closed minded about your definition of vegetarian.

[chilly ocean]

Quick thinking => more production (even if defective) + less complaints => rich people benefit from capitalism (basically modern equivalent of slavery disguised as "you can choose what you want to do"

[south patrol]

OK

[chilly ocean]

In math everything has definitions. The book uses a different way. But the book says it. Its not like the books changes the definition when it suits it

[delicate orchid]

stop trolling on the internet

[chilly ocean]

There are many different worlds of mathematics other than the mainstream one

With other axioms

[languid trellis]

https://tenor.com/view/troll-trolled-trollge-troll-success-gif-22597471

[rocky cloak]

But this book isn't defining some new revolutionary ideas. It's just taking about normal things with confusing terminology.

[south patrol]

I cannot find a single place online which uses p adic integers to mean anything else

[chilly ocean]

Sometimes the most revolutionary ideas are the simplest ones tho

[south patrol]

How is this going to be revolutionary lmfao

[crystal turtle]

Carla what the fuck are you talking about

[delicate orchid]

https://tenor.com/view/troll-gif-26178840

[crystal turtle]

Bad terminology and notation for something standard is bad. Big shocker.

[south patrol]

Let me rename force "pressure" and see how it revolutionises physics

[coral spindle]

Carla is delusional.*

* delusional I define to mean "a nice person who looks after their health and wellbeing"

Anyone in the mainstream who disagrees with this is being closed-minded about the definition of delusional.

[crystal turtle]

Also yeah that all around is unorthodox

[chilly ocean]

Hmm ig thatd be weird

[south patrol]

I hope it is also clear that I do not genuinely believe the whole book should be tossed out because of this one oddity

[delicate orchid]

I do

[chilly ocean]

maybe youre right

i was thinking more about axioms

And while technically definitions can be seen as axioms

They also serve purpose as communication devices

[coral spindle]

I think it's important to tell beginners when their textbook uses terminology that alienates them from the rest of the mathematical community. This happens quite often, unfortunately.

[south patrol]

That said, I do agree with much of what Carla has said

[chilly ocean]

Sorry for thr confusion

And thanks for clearing it up

[south patrol]

Np, sorry I got annoyed

[chilly ocean]

Is there any way i can make it up for you?

[south patrol]

You don't need to do anything, it's all good

[bitter tree]

[south patrol]

Lol the random reaction coming in late made me chuckle

[chilly ocean]

Happy valentines day, i dont rly care about april fools i was just saying random nonsense but if i say its april fools people will forgive me probably

Anyway

Is there a classification of dedekind-finitely generated abelian groups in ZF

[fading field]

you already asked this

[chilly ocean]

But noone answered

[night onyx]

that probably means nobody knows

[chilly ocean]

😭

[rocky cloak]

I'm curious. Any reason you're asking this besides general curiosity?

[chilly ocean]

What other reasons can there be to ask something in math, if not for curiosity?

[rocky cloak]

No, I mean, like does this come from something else you're interested in

[chilly ocean]

i'm interested in how things work without choice. and i'm interested in classification theorems. there's a really simple classification of finitely generated abelian groups in ZFC. and countable abelian groups are too complicated to classify. so dedekind-finitely generated in ZF is a candidate for a middle ground.

altho its not really a middle ground cuz infinite dedekind-finite sets are incomparable to countable ones, but still, would be interesting to me to know if they have a simple classification or not

[rocky cloak]

Well, I have no idea. But it's a little hard to imagine what such a classification would look like.

Like Dedekind-finite and finite being the same is compatible with ZF, so the classification would need to somehow invoice objects that don't really "exist" in ZFC. My intuition would say it's probably "harder" to classify dedekind finite groups than countable ones. But yeah, interesting question.

[dull ginkgo]

Hey chat I know some fun applications of CRT

But they're either number theory bs or linear independence of characters over domains

any other cool applications?

[delicate orchid]

characters would never be linear independ- :cereal2:

[dull ginkgo]

the one application of monoid rings

i think it holds for semigroups?

[rocky cloak]

Any commutative artinian ring is a product of local rings is a consequence

[dull ginkgo]

oh that's not immediate and kinda sick

[rocky cloak]

Or you can prove Artin-Wedderburn with it

[dull ginkgo]

is it a good exercise or take a lot of work

I know artinian ring in a commutative setting is every descending sequence of ideals terminates

[rocky cloak]

Well, I guess the key idea is really lifting idempotents modulo nilpotent ideals, but that would be a good exercise.

[dull ginkgo]

will try it in a bit

[rocky cloak]

Maybe it's really only half of A-W, but you can use CRT to prove that if M is an artinian module, such that the intersection of maximal submodules is 0, then M is semisimple.

I think that would be a nice exercise using CRT

[next obsidian]

It is easy

once you know some basic facts about Artinian rings

[dull ginkgo]

i know none

fun time

[opaque wind]

Hi, I'm trying to prove that cosets are equivalent i.e. either equal or disjoint. I have taken a look at a few example groups and I can accept intuitively that this is a true statement at least with respect to those particular groups, however I can't quite prove it myself for arbitrary groups. Here's the outline of my proof:

For all groups G, subgroups H of G and elements g and g' in G, gH=g'H \/ the intersection of gH and g'H is empty

Proof by case analysis on whether g and g' are in H.

g in H /\ g' in H => gH=g'H=H by the definition of a group

g in H xor g' in H => gH and g'H are disjoint by noticing that either g or g' is not in H which means that gh or g'h will produce an element not in H, for all h in H

~(g in H \/ g' in H) => Further case analysis is needed on whether g=g' or not.
g=g') gH and g'H are clearly equal.
g≠g') This is the case I'm a bit stuck on since cosets are not necessarily subgroups so you can't rely on any group facts. I guess the fact that neither g nor g' are in H and them being unequal implies that for all h in H, gh≠g' and g'h≠g, right? Is that all I need to finish the proof?

Thanks

[dull ginkgo]

by idempotents I am assuming you are talking about the central idempotents from the quotient being iso to a direct product

[rocky cloak]

Well, it's true for non-central idempotents too. But for a commutative ring all idempotents are central

[dull ginkgo]

i think this just flat out gives it away, sad face

if J is an ideal of artinian ring R, then J^n must stabilize

[rocky cloak]

Well, it's still proving the fact that you can lift idempotents. Or that an artinian ring with no nontrivial idempotents is local.

[dull ginkgo]

brushes over bezout lemma

[dull ginkgo]

I know how to do the first part (number of basises, i.e. order of the group) but at a glance the second one eludes me a bit

[opaque wind]

I think this last case can be proven by contradiction, because e.g. if there exists an h such that gh=g' that means h = g^{-1}g' which contradicts the assumption that neither g nor g' are in H. correct?

[dull ginkgo]

@rocky cloak for the second part, I don't think I can do much with the determinant or similarity of matricies unforunately

However that does assert that (A^q - 1)A^2 = 0, so A is a zero divisor

[rocky cloak]

Guess you can think about what the minimal polynomial of a non-invertible matrix could be

[dull ginkgo]

god

We haven't even TOUCHED Cayley-Hamilton

I can try to prove it for general rings

[south patrol]

Cayley Hamilton isn't needed for stuff w min poly

[rocky cloak]

Hmm, okay. So you want to do it without using any linear algebra?

[south patrol]

Well basic stuff

[dull ginkgo]

that's an insane amount of work for a single problem

[dull ginkgo]

?

[south patrol]

Lol idk what you want me to clarify

[dull ginkgo]

minimal poly?

[south patrol]

What about it

You don't need Cayley Hamilton to work out stuff about the min poly of a non invertible matrix

Is my point

[rocky cloak]

What's an insane amount of work?

[dull ginkgo]

doing a bunch of work proving the existence of minimal polynomial

[south patrol]

That's very easy

[dull ginkgo]

However, M^n(Z/nZ) is finite

[south patrol]

M_n(F) is a finite dim F vector space

Etc

[dull ginkgo]

I haven't seen the proof for the existence of a minimal polynomial

[rocky cloak]

I don't it would be much work, but if you really know nothing about minimal polynomials, then I guess you might aim for a different method

[dull ginkgo]

book hasn't covered anything about matrix rings besides like the basic algebraic definition and i don't know much about min polynomials of matricies, no

idk how he'd want it to be done

I seriously have no clue what to even try.

[delicate orchid]

do you have eigenvalues

[dull ginkgo]

We haven't touched anything eigenvalue related

Literally the section on the matrix ring itself was like 3 pages

and barely covered determinants

so we haven't done eigenvalues, polynomial rings, etc.

didn't explain it as an endomorphism ring, no modules yet

idk how the FUCK he expects us to solve it

or I'm just a fucking idiot

[delicate orchid]

uhhh

[languid trellis]

I'm guessing Linear algebra is assumed tbh

[delicate orchid]

nilpotent + diagonalisable decomp?

does that work here

[dull ginkgo]

idfk

[delicate orchid]

probs not our ring isn't alg closed

we'll get "invertible + nilpotent" at least

[dull ginkgo]

5 pages on matrix rings

that's it

[delicate orchid]

nah doesn't work (my idea at least, I'm pretty sure the decomposition is fine)

the +2 has to be something to do with the dimension

[rocky cloak]

Well, it does say that it assumes you're already familiar with linear algebra

[languid trellis]

woops wrong ping

[dull ginkgo]

i feel stupid half the time here having to ask for help for questions from "Basic algebra"

[dire siren]

@dull ginkgo if A is invertible, you are done because A^q=1 implies A^(q+2)=A^2
if A is not invertible, then det(A)=0, so A^2=tA, so a^q+2=t^qA^2
and t^q=1 pecause p-1 | q

t=tr(A)

[south patrol]

basic algebra is more algebra than what people usually do in basic algebra

ig

[dull ginkgo]

I need to prove existence of minimal polynomial and what form it takes

[dire siren]

@dull ginkgo I thought you wanted a proof without minimal polynomial

[dull ginkgo]

A^2 = tA

[dire siren]

no need of minimal polynomial for that

[dull ginkgo]

that's Cayley Hamilton, is it not

[dire siren]

it is, but you don't need it

[dull ginkgo]

i'm skipping this problem

[dire siren]

just write the matrix as
a b
c d
with ad-bc=0 and compute A^2

[rocky cloak]

But, so have you actually not taken any linear algebra before this?

[dull ginkgo]

I studied linear alg like 2-3 years ago and forgot a good amount of how to prove some of the stuff

and I don't really like using shit without understanding the mechanisms behind it

At least at the basic level

[rocky cloak]

Alright, just checking

[dull ginkgo]

rage

uses that hell problem I spent forever on

[rocky cloak]

Well this one should be pretty doable by just directly computing

[dull ginkgo]

The ideal part

but yeah

[delicate orchid]

yeah you just multiply by an arbitary element and see what happens

[languid trellis]

arent triangular matrices closed under matrix multi

[dull ginkgo]

yes

[dull ginkgo]

I don't get this step:

A^2=tA, so A^q+2=t^qA^2

[dire siren]

from A^2=tA you get
A^3=tA^2, then
A^4=tA^3=t^2A^2, etc.

small induction

[dull ginkgo]

OH

thanks man

struggling a bit still trying to determine the ideals.

Clearly it's a subspace.

Basically subspaces $L$ of $\mathbb{Z}^3$ such that if $(a,b,c) \in L$, then $(ax, bx + cy, cz) \in L$

[rocky cloak]

Maybe start with the principal ideals, then see how they sum together

[cloud walrus]

Request a new nickname

[dire siren]

did you try the idea from the other problem?

the one with ideals of M_n(R)

[dull ginkgo]

I could try but I'm hesitant because Z doesn't embed right into T

But I don't see why the proof of that one wouldn't work here still tbh

We're just... not including the basis matrix e(2,1)

The other three are still closed

Let $R$ be a unital ring, and let $M^N(R)$ be the matrix ring over $R$ generated as a finitely-generated module over the basis $e(n,m)$. \

For any subset $S$ of $M^N(R)$, define $S_{i,j} = {r \in R : \exists M \in S : M_{i,j} = r }$, then $S \subseteq \sum S_{i,j}e(i,j)$ \

Thus $J \subseteq \sum J_{i,j}e(i,j)$. However, for any matrix $m$, $e(a,b)me(c,d) = m_{b,c}e(a,d)$, thus $e(a,b)Je(c,d) = Je(a,d) \subseteq J$. However, this proves invariance of $J_{i,j}$, as $J_{b,c}e(x,y) = e(x,a)\left(e(a,b)Je(c,d)\right)e(d,y) \subseteq e(x,a)Je(d,y) = J_{a,d}e(x,y) \Rightarrow J_{b,c} \subseteq J_{a,d}$ and due to the universal quantification, the other direction $J_{a,d} \subseteq J_{b,c}$ holds, thus $J_{i,j} = I$ for some ideal $I$ of $R$. $J \subseteq \sum J_{i,j}e(i,j) = \sum Ie(i,j) = M^N(I)$, but each $J_{i,j}e(i,j)$ is contained in $J$ thus $M^N(I) \subseteq J$, implying $J = M^N(I)$

[cloud walrus]

Request a new nickname

[dire siren]

the idea works here as well by neglecting the (1,2) entry

[scenic saffron]

im kinda confused on this corollary, i understand using lagrange's theorem that the only divisors would be 1 and p so a subgroup would be the group itself, but why does that make G cyclic? maybe i forgot a lemma on cyclic groups but yea

[dire siren]

pick an element other than the identity; what is its order?

[scenic saffron]

less than p?

$$\leq p $$

[cloud walrus]

tim

[dire siren]

can it be smaller?

[scenic saffron]

i dont know

[dire siren]

you mentioned divisors of p

[scenic saffron]

yes

1 and p

[dire siren]

and the order of the element is a divisor of p

[scenic saffron]

oh is that true

[dire siren]

it follows from Lagrange

[scenic saffron]

omg

right

i was being fumb

dumb

hold on

right yeah so the order of G must be p but i just dont understand why that makes G itself cyclic

[dire siren]

you have an element of order p

[dire siren]

call the element a

[scenic saffron]

nvm this proposition

i get it now

forgot about this

[stark helm]

Suppose E=Q(sqrt(2), sqrt(3), sqrt(5)), then do you think Gal(E/Q)=8? Also, I write f(x)=(x^2-2)(x^2-3)(x^2-5), how will you find galois group of Gal(E/Q)?

First of all, I want to make sure if the degree of it is correct? second, I list sqrt(2), sqrt(3), sqrt(5), sqrt(6), sqrt(10), sqrt(15), sqrt(30) are all not in Q but in E, but have some difficulty to make sure that all groups can satisfy homomorphism.( except identity) and try to consider the case fixing sqrt(2), sqrt(3), sqrt(5) etc

[delicate orchid]

use the tower law to find the degree

[stark helm]

I find Q subset Q(sqrt(2) subset Q(sqrt(2), sqrt(3)) subset Q(sqrt(2), sqrt(3), sqrt(5))

[stark helm]

and then I use [E:Q]=[E:Q(sqrt(2), sqrt(3))][Q(sqrt(2), sqrt(3)): Q(sqrt(2)][Q(sqrt(2):Q]

[stark helm]

do you think degree is 8?

[delicate orchid]

indeed I do

Gal(E/Q) should be C_2^3

[stark helm]

but the question is how can we find all Gal(E/Q)

[stark helm]

list sqrt(2), sqrt(3), sqrt(5), sqrt(6), sqrt(10), sqrt(15), sqrt(30) are all not in Q but in E, but have some difficulty to make sure that all groups can satisfy homomorphism.( except identity) and try to consider the case fixing sqrt(2), sqrt(3), sqrt(5) etc

[delicate orchid]

the obvious automorphisms swap sqrt(n) with -sqrt(n)

composing them together you get 8 different automorphisms

and I'm 99.9% sure you get GL(E/Q) = C_2^3

[celest furnace]

Anyone have any tips on how to show the first part: pi(x^2) = -I

[topaz solar]

So x^4 = 1 should kinda have x as looking like i, and y as j (if you see Q_8 as 1,i,j,k & negations)

At the very least, pi(x^2)^2 = I

& your conditions on y while being 2x2

[celest furnace]

Yeah really struggling with this part i dont see how that last relation with x and y helps at all but i think its the key

[topaz solar]

Well, if you had some other matrix which squares to I and is invertible

Maybe looking at that conjugation condition?

[celest furnace]

Yeah im thinking of using that x^2 should be in the center of the image

[topaz solar]

(I do not know a full solution offhand so maybe I’m just wacky)

[delicate orchid]

centres are preserved under isomorphism so yes

[topaz solar]

Ye

[south patrol]

Because it's a functor ||/j||

[delicate orchid]

making me do rep theory by actually thinking about matrices rather than just throwing a frobenius-schur indicator at this and calling it a day

[delicate orchid]

[incredibly loud incorrect buzzer]

[south patrol]

Actually lol we were asked to prove in cat theory that uh

The assignment Z(-) on objects does not admit a refinement to a functor

:chad:

[delicate orchid]

trivial.... because I said so!

[south patrol]

it is tbf

you just factor the identity on Z/2 through S_3

por ejemplo

[delicate orchid]

stop saying "factor" like I know what that means. I don't want to infer two edges of a triangle

[south patrol]

yeah

[delicate orchid]

quotient and embedding but that's gross

stop making me think get out of my head get out of my head get out of my head get out of my head get out of my head

[celest furnace]

Any tips for showing that pi(x^2) might be a scalar matrix.. The setup i have is that pi(x^2) commutes with pi(y) which is not a power of pi(x) and is not itself scalar

Seems like this should be really strong but idk just not working

[delicate orchid]

honestly, that seems lame. I'd think about what the minimal polynomial of an element of order 4 has to be and then brute force what y has to be once you have x using the relations

or you could just try every matrix of order 2 in GL_n(R), there's not an obscene amount of them unless there's a family I've missed

[celest furnace]

So this is what im working with: Clearly pi(x^2) = T has the property that T^2 - 1 = 0. Thus minimal polynomial is either T-1 (cant be) T+1 (we want to show this) or T^2 - 1. In the final case it is diagonalizable and similar to {{-1, 0}, {1, 0}}. By commuting and being non-singular, we see that pi(y) also has eigenvalues 1 and -1, thus it is diagonalizable too. Simultaneously diagonalizing them shows that pi(y) = P{{-1, 0}, {1, 0}}P^(-1) for some P. But this would show that pi(y)^2 = I, a contradiction?

Okay this was nonsense. But i have tried something new: Again we see if the minimal polynomial can be $T^2-1$. This would mean you diagonalize as {{-1, 0}, {1, 0}} in some basis, so let pi(x) = P * (that matrix) * P^(-1). Then phi = P pi P^{-1} would be another embedding of Q8 into GL(2, R). Once again phi(x^2) is going to commute with phi(y): This tells you that

$$\begin{pmatrix} -1 & 0 \ 0 & 1 \end{pmatrix} \begin{pmatrix} a & b \ c & d \end{pmatrix} \begin{pmatrix} -1 & 0 \ 0 & 1 \end{pmatrix} = \begin{pmatrix} a & b \ c & d \end{pmatrix}$$

[cloud walrus]

Ultimate Chad

[celest furnace]

The LHS (after calculating) would show that b = c = 0, a contradiction since no fourth roots of 1 exist in R!

[delicate orchid]

Yeah this looks good, nice work homedawg

[dull ginkgo]

wew usually tells me I'm a dumbass

[celest furnace]

Lmao

[delicate orchid]

Oh brother…. This guy stinks!

[celest furnace]

You should see langs section on finite fields its nuts (Very well written)

[dull ginkgo]

I spent most of the time on this problem trying to remember the name of an invertible element

it's a unit, and I'm a fucking idiot

[next obsidian]

This is such a stupid exercise lmao

[dull ginkgo]

(Z/pZ)^x embeds into F^x as a goop via the prime-ring map, so by legrange

[celest furnace]

Lmao nice

[delicate orchid]

I dunno why mfs set problems like this which are trivial with the most basic of machinery

[next obsidian]

Any finite field of char p is a vector space over F_p and has order p^n

[delicate orchid]

But they don’t provide that machinery

[next obsidian]

from that it’s clear p-1 divides p^n - 1

[dull ginkgo]

THe jacobson lmao

[delicate orchid]

Yes that’s what I’d do too chmonkey

[dull ginkgo]

Then the next problems is the most awful shit

[next obsidian]

Like, I don’t get why they don’t just ask you to show it has size p^n

[dull ginkgo]

Jacobson exercises are either trivial or

fucking horrendous

[delicate orchid]

I still think u must have skill issued the matrix ideals one

[dull ginkgo]

that matrix ideal one vacuumed the life out of me and I am unable to do matrix ideals ever again