groups-rings-fields

[rocky cloak]

Just some thoughts:

2020 = 10*101

Since 101-1 = 100 is a multiple of 10, you should have a semidirect product C101 with C10 whose only normal subgroup is C101.

For 1990 you have the dihedral group of order 398 times C5, whose only normal subgroups are C5, C199 and C199xC5 and the dihedral group.

So then the product should have no normal subgroup of order 10

[storm kiln]

Ah, thanks. I was kinda surprised about this question since the ones before and after were way easier imo (prove or disprove: every group G is a normal subgroup of some other group, Z/4Z[x] is a domain).

(mock exam)

Why does that first semidirect product exist though? I get that Aut(C101) ~ C100, but why doesn't the product have more normal subgroups than C101?

I should probably get more comfortable with semidirect products

[tardy hedge]

For this one, does the prop. Keep on saying “at most” n, because this extension of degree n that contains “a” might be bigger than F(a), so the minimal polynomial of a might actually have degree less than this n?

[rocky cloak]

It has a couple more actually.

But all the nontrivial subgroups contain C101.

It just comes from looking at what the elements of the semidirect product look like. If you take some element not in C101 and multiply it by its inverse conjugate by something in C101 you should get a nontrivial element of C101.

(The other normal subgroups being the semidirect product of C101 with C2 and with C5 respectively)

[rocky cloak]

Yes, you can for example imagine a being in F.

[tardy hedge]

Right, then the min poly is just x-a

[golden turtle]

Any ideas for simplyifying this?

I am computing the 105th cyclotomic polynomial and I'm stuck at this stage

[rocky cloak]

long division works.

Or Wolfram alpha

[golden turtle]

yeah

I expanded the products

and started long division

it works but

it is degree 48

I'm supposed to compute it myself, I think

I know the end result

just stuck at the algebra at this stage

[rocky cloak]

Well you probably don't want to expand stuff you've already factored.

Like divide the x^10 thing by the x^2 thing to simplify further

But yeah, 70 is still a big number

[south rain]

Ah alright thanks, I didn't catch that at all. Was this just an educated observation or is there any method on finding the desired relation? Because I have no clue on what relation or condition I should obtain. A condition on s, a condition on r, no condition at all and just a relation.

[rocky cloak]

I just know that if you generalize Jacobson radical to nonunital rings then you replace the condition of 1+s having an inverse with
s + r + sr = 0
and recognized the formula.

But I guess putting myself in the shoes of someone who doesn't know that, the first thing to reasonable do would be to rewrite the equation
s + (1+s)r = 0

Then you see that if 1+s has an inverse you can solve for r.

Conversely if s=-1, then
s + r + sr = s =/= 0

[south rain]

Yeah I did arrive at your rewritten equation but was just lost on what I was looking for, thanks.
But since $D$ is a division ring and $1 + s \in D$, we must have that $1 + s$ is invertible except if s=-1, right? So is every element in a division ring left and right quasiregular except for -1?

[cloud walrus]

dellinger

[south rain]

Thanks!

[potent condor]

Extremely stupid question: if $M$ is a module and $S$ is a submodule such that $M \cong S\oplus M/S$, then must $S$ have a complement?

[cloud walrus]

EdgarAlnGrow

[south patrol]

Here's a silly example I came up with. Let $R = \mathbf Z$, $M = \mathbf Z \oplus (\mathbf Z/2)^{\oplus \mathbf N}$ where I mean $\mathbf N$ copies lol. Let $N$ be the submodule $2 \mathbf Z$. Clearly $N$ doesn't have a complement, but also $M/N \simeq (\mathbf Z/2)^{\oplus \mathbf N}$ and $N \simeq \mathbf Z$ so $M \simeq N \oplus M/N$

[cloud walrus]

Prismatic Potato

[south patrol]

There will probably be simple finitely generated counterexamples over e.g. some noncommutative rings, but this was the easiest thing I could come up with by just considering 2Z in Z and then using a "swindle" to make it work

@potent condor

[potent condor]

thanks!

[south patrol]

Np

[kind temple]

my fav fact recently: the xor operation forms a Z/2Z vector space over the integers

[thorn jay]

Even better, if you take \land as multiplication, you get a ring which is a Z/2Z-algebra

You can even go a step further and consider a set of propositions which contains TRUE, FALSE, and is closed under XOR and AND, and that will form a ring, a so-called boolean ring, which also forms a Z/2Z-algebra

Now, what is special about boolean rings is not that they are Z/2Z algebras, but that every element is idempotent (the AND of a proposition with itself is equivalent to itself), and in fact every ring where x • x = x for all x is isomorphic to such a set of propositions with XOR and AND

[storm kiln]

Dummit and Foote even defined them as idempotent rings and then had an exercise to show they were commutative

[tardy hedge]

A vector space with finitely many elements must be finite dimensional , obviously … right lol

The finitely many elements is a generating set and so a basis must be less than that

At most that

Anyway so then to show that finite field F is order p^n, can we say that F is n dimensional vspace over its prime subfield, so then F has p^n elements

[rotund aurora]

the cardinality of a basis is less than the cardinality of the vector space, so yes

[south patrol]

Ye

[thorn jay]

That's actually related to a problem in equational logic, finding a general proof for x^n ≈ x => xy ≈ yx

Lol

(≈ denotes and identity, i.e. that the equality holds for all values of x, y, ...)

[storm kiln]

For what n has it been solved already?

2 I guess, also are unital rings assumed?

[tardy hedge]

If [F(a) : F] is odd then F(a) = F(a^2)

[F(a) : F] = [F(a) : F(a^2)][F(a^2) : F]

a satisfies the polynomial x^2-a^2 over F(a^2), so F(a) over F(a^2) has degree at most 2, but then it must be 1 cuz fa over f is odd

Which shows degree of fa2 over f is same as fa over f

Does that work?

[thorn jay]

The problem is to find a general algorithm for all n (or n bigger than some N)

Because you can just write a program that gives a proof for some n, im sure

But who knows, maybe there is some N for which there exists a noncommutative ring satisfying x^N ≈ x

[lusty marlin]

Yes

[thorn jay]

Yes

I believe

And for me, nonunital "rings" are rngs anyways >.>

[rocky cloak]

Isn't it proven that if all x satisfies
x^n = x for some n (that can depend on x) then your ring is commutative?

https://math.stackexchange.com/a/2589300/306319

[thorn jay]

Yes, sorry, i should rephrase

We have not a general formal deduction of it

[rocky cloak]

Like you want some first order proof using like the ring axioms or something like that?

[thorn jay]

A formal deduction is a proof like
e • x ≈ x,
x • e' ≈ x,
e • e' ≈ e',
e • e' ≈ e,
e ≈ e'

[thorn jay]

Basically, im pretty sure?

Im not well-versed in logic, but a formal deduction in equational logic is a finite sequence of equations where every equation is either an assumed axiom, or a substitution/replacement of equations occuring in the sequence

[thorn jay]

So here e • x ≈ x, x • e' = x are assumed axioms, and when filling in e or e' for x, that is a substitution

[regal zodiac]

hello, does anybody have a hint for the first exercise, in the case where the field is Fp?

[rocky cloak]

So no existential quantifiers?

[rocky cloak]

Do you know what all the finite fields are?

[swift prawn]

this proof just shows that some n exists

it's one of the cool results of a noncomm alg course

sorry I've worded this wrong

this proof just has the stipulation that if your ring satisfies the property that for all x, x^n(x) = x for some n then R is commutative (which is wild btw)

what it doesn't tell you (iirc) is how to actually like

do it

[thorn jay]

Nope

This universal algebra, we work with universal quantifiers

Lol, but no, equational logic is the logic of algebraic identities

[thorn jay]

Ye

At least, it doesnt give a formal deduction like i described

But the fundamental theorem of equational logic says that one must exist for every n

[tardy hedge]

I like the group Q/Z. Cause somehow when you're first learning group theory, an infinite group with every element having finite order is unintuitive at first

[thorn jay]

It is goated fr

[coral spindle]

Me too

[south patrol]

I like \mu_p^oo(C) = elements with pth power order in C

Mostly cause it appears in geometry lol

[thorn jay]

So for groups these would be the silow subgroups?

If you extend that notion to groups from what i presume is modules

[south patrol]

Well if you replace C by a group A then these are only subgroups when A is abelian

But usually one applies this to commutative rings R to get a subgroup of R^x

[thorn jay]

Oh, i see

Where is that commonly used?

[south patrol]

Well it is a common/simple example of a formal group (in the sense of algebraic geometry)

[vernal ore]

I'm studying and applying Galois Theory in constructibility using ruler and compass, in specific I'm looking at regular polygon costructibility. I read that Gauss showed a proof of construction of the regular polygon of 17 sides solving cyclotomic polynomial of 17. Someone knows how he simplified this polynomial (x^16+x^15+...+x+1)?

[delicate bloom]

what do you mean by simplify? It's irreducible over Q but can be rewritten as (x^17 - 1)/(x-1) if that helps.

[vernal ore]

Yep, this polynomial has all 17-roots of unity which has as generator e^(2i*pi/17) = cos(2pi/17) + i sen(2pi/17) but he found out that

How he got this result?

[restive birch]

interesting puzzle i came up with:

let G be some finite group. if for any group H we are given |Hom(H, G)|, is it possible to classify G up to isomorphism?

[lusty marlin]

So G is a fixed finite group, and we have been given |Hom(H,G)| for all groups H?

[mint seal]

neat question

I'm too rusty on group theory to know how to attack that. No intuition for if it's true or false. But if I had to guess... false

knowing how many homomorphisms exist for any H to G sounds like it would give how many subgroups of different orders exist in G, maybe? But I don't think that's enough to determine G to isomorphism..?

[delicate bloom]

Can we solve any special cases of it? Either we find counterexamples that way or find strategies that might generalize to prove it.
I guess for the example of |Hom(H,G)|=1 for all H, I'm thinking that's enough to say G is the group with one element.

[mint seal]

that sounds right to me

[kind temple]

the only endomorphism would be the identity in that case

[mint seal]

Yeah, because if |G| > 1, then in case H = G, there's the identity map of G along with the constant map to 0

[lusty marlin]

Yes, if |Hom(H,G)|=1 for all H, then G can only be the trivial group

[chilly ocean]

Generally, if |G|>1, |Hom(G, G)| would have to be >1, but it isn’t

[kind temple]

left multiplication by elements of G is always an automorphism

what about |Hom(H,G)| = 2 for all H?

[mint seal]

That's not possible if H is the trivial group

But setting that case aside

[kind temple]

H non-trivial?

[chilly ocean]

This implies |Hom(H, H)| $\geq$ |H|, so |H| must be the cyclic group of order 2, since it isn’t the trivial group

[cloud walrus]

Larue’s #1 Lawyer

[kind temple]

nice

[chilly ocean]

I think we can get an upper bound on |Hom(H, H)| given |H| and the number of generators of H right?

Wait no idt we can nvm

[kind temple]

we certainly have a lower bound |G| <= |Hom(G,G)|

[chilly ocean]

Well

Every homomorphism is uniquely determined by where it sends the generators, right? So, take some set G that generates H.

|G|^{|S|} $\geq$ |Hom(H, G)|

Wait that’s backwards

[kind temple]

this is confusing

[chilly ocean]

lmao

[kind temple]

G is fixed

H varies

say S is a set of generators

[chilly ocean]

Good idea

[cloud walrus]

Larue’s #1 Lawyer
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[chilly ocean]

Shoot got H and G swapped

Doesn’t really matter

So for fixed G with generating set S

|G|^{|S|} $\geq$ |Hom(G, G)$\geq$ |G|

[kind temple]

is that true?
what is Hom(S_n,S_n)

[cloud walrus]

Larue’s #1 Lawyer
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[chilly ocean]

Oh darnit

[kind temple]

i was genuinely asking

i have no idea

like, generators of S_n, pick a transposition (i j) and an n-cycle with i and j adjacent

[chilly ocean]

Well, every element of a group G is a product of powers of elements of S, say, g=s•s, and determining every F(s) then determines the whole homomorphism, since, say, let k=s•h•j for s, h, j in S, then F(k)=F(s)•F(h)•F(j), hence we have an injection from the set of homomorphisms Hom(G, H) for some H and fixed G to H^S, the set of functions from S to H, right?

[kind temple]

agreed, if S generates H, then we have |G|^|S| >= |Hom(H,G)|

i believe this follows from the universal property of the free group

[south patrol]

Woah okay this is an interesting question hm

Are we allowing H to be infinite lol

[kind temple]

do you think that changes things?

[south patrol]

It eases some things but probably doesn't make a difference

[kind temple]

it wasn’t my question, but personally, i think relaxing it is fine

but G should likely remain finite

[kind temple]

original message

[flint cave]

What is a good sufficient condition for the a large set (size m>>n) of primitive vectors in Z^n to be a generating set, I want something better than "put the vectors into a matrix and see if you can do column operations to get I_n as an n x n minor". What I want to do is pick a large number of primitive vectors in Z^n and claim that they almost surely generate Z^n, the sampling is done based on a uniform measure on N (bound of coordinates of the vectors chosen) and then asymptotically increase N. I want a lower bound on the probability and argue that it tends to one.

[south patrol]

Ye

[delicate orchid]

Le Yoneda embedding has arrived

[south patrol]

I was waiting for someone to wrongly use Yoneda

[delicate orchid]

Your wish is my command

[flint cave]

How...

[south patrol]

Oh sorry that wasn't an answer lol

[rocky cloak]

Hmm, well you can at least figure out how many elements G has, and how many elements of each order. How many pairs of commuting elements there are.

There are examples of groups with the same number of elements of each order, but in the examples I know they don't have the same number of commuting elements.

[rotund aurora]

Isn't this false already if n=1?

[flint cave]

No? If you take k many elements, it should be easy to generate Z right?

[rotund aurora]

wait nvm yeah

[flint cave]

The whole set shouldn't have a common divisor

[vivid birch]

in this proof, why are tT and Tt closed?

[rotund aurora]

well you didn't specify that, what does that mean?

I mean in the Z^n case

[flint cave]

No, I was saying that if the m elements in Z do not have a common divisor, they generate (this is equivalent in fact)

[rotund aurora]

sure, but how does that condition translate in Z^n

[flint cave]

That restriction is automatic in Z^n cus I want my vectors to be primitive... the equivalent condition in Z^n is just that the matrix you get when you put the vector coordinates, you should be able to do column operations (apart from scaling) to get an n×n identity matrix as a "submatrix"

I keep forgetting the word, maybe it's that it should have an n×n minor with determinant 1

But this isn't amenable to a good probability estimate

[rotund aurora]

what does "cus I want my vectors to be primitive" mean? Why is {2,4} not a set of primitive vectors? Is {(1,1), (-1,1)} a set of primitive vectors?

[flint cave]

Primitive means coordinates have gcd 1

[rotund aurora]

of a single vector?

the vectors (2) and (4) are both primitive

for n=1, m=2, your problem fails. Maybe for n>1 the conclusion of the problem is still true idk, or for m larger

it's false for n=1 and any m I think

[rotund aurora]

also this is not right, as the example {2, 3} for n=1 shows. The thing is that while your set of vectors might generate all of Z^n there need not be a basis within that set.

But for m=n and any n your conjecture is false. There are about C N^{n^2} tuples of n primitive vectors with coordinates bounded by N, which you can view as points in R^(n^2). The condition is that their determinant is =+-1, this gives an equation of a curve. But curves are sparse I think (like in general a curve in R^n should have o(N^n) integral points with coordniates bounded by N, I think)

[south rain]

Here is the adapted proof of: Let {R_i | i \in I} be the set of (possible infinite) Rings R_i. Then the jacobson radical of the product is the product of the jacobson radical. Does this work now?

[rocky cloak]

gcd(2) = 2, so that's not primitive.

[rotund aurora]

According to his definition, it is?

If not I ask again, what is the condition on the vectors?

[rocky cloak]

This is the definition

[rotund aurora]

ah, so all coordinates of all vectors?

[south patrol]

ah lol

[south patrol]

ngl i thought croqueta was being a lil silly and then read the preceding message and now it makes sense

I was gonna joke about the story where a student was asked whether G and H were isomorphic and he said "G is but H isn't"

But yeah no there are two reasonable notions of primitivity here

[rocky cloak]

The way I interpret their question, they have a matrix with primitive columns, and they're asking for those columns to generate Z^n

[rocky cloak]

What's the other?

[rotund aurora]

he was talking about primitive vectors, so I interpreted it as each vector being primitive, individually

[rocky cloak]

Yes

Is that not how your supposed to interpret it?

[rotund aurora]

I mean the vectors in Z^n

[south patrol]

Oh that's not how i interpreted it

i interpreted it as like, you are given a collection of vectors (which you can put into a matrix M) and then you want M not to be like non-trivially an integer multiple of smth else

but then that would be implied by any of the individual vectors being primitive

hm

[rotund aurora]

I feel like you could have the first coordinate always be even, and the rest be anything else, doesn't this give a negative answer?

[rocky cloak]

I'm not sure what you mean by negative answer.

It's not a yes/no question?

[rotund aurora]

The question is whether the probability tends to 1 (picking m vectors in Z^n, m fixed, with coordinates bounded by N and showing that the probability they generate Z^n tends to 1 as N-->infty)

[rocky cloak]

Yeah, but then how can you fix the first coordinate if you're picking them at random?

[rotund aurora]

you can try showing that there's a "positive density" subset of tuples of m vectors in Z^n that have those properties but they don't generate Z^n

[wicked patio]

snow and i figured out |Hom(Z, G)| = |G|

[rocky cloak]

I see, m also fixed.

But wait. Was the question showing it has probability 1?

I thought it was just about finding a nice equivalent condition

[rotund aurora]

(#groups-rings-fields message) What I want to do is pick a large number of primitive vectors in Z^n and claim that they almost surely generate Z^n, the sampling is done based on a uniform measure on N (bound of coordinates of the vectors chosen) and then asymptotically increase N. I want a lower bound on the probability and argue that it tends to one.

ups sry didn't mean to ping again

[rocky cloak]

I see, yeah it's definitely not almost surely.

But what the probability is. That's a good question

[wicked patio]

[rotund aurora]

In order to avoid defining what primitive means, it's probably better to consider P_nm(N)={ v_1,..., v_m in Z^n : v_1,..., v_m have coordinates bounded by N and generate Z^n }. For example, if n=1, then P_1m(N)=2^m N^m/zeta(m)+o(N^m) as N-->infty (m fixed)

[wicked patio]

yeah for any fixed number of vectors the probability is going to be less than 1 unless the fixed number is like, more than half of the size of the allowed region

if you ignore the bounds

think about the index of the subgroup so far generated

that's basically the "probability" of choosing another dependent element

[tardy hedge]

Only ring endomorphism on R is identity. Hint is x <= y iff sqrt y-x is in R. Not totally sure where to go

[wicked patio]

what about the trivial one?

ah

probably 1 -> 1

nvm

[wicked patio]

when is sqrt(y-x) in R?

[south patrol]

I was so confused about the order for a minute until I realised this was the reals

[wicked patio]

same

[south patrol]

Anyway imo this is a sort of pointless hint

Maybe I am confusing it w smth else

[wicked patio]

that hint is like half the solution

oh

i see

[south patrol]

Well like just weirdly phrased lol

[rotund aurora]

It is actually true that if f(x)>=0 for x>=0 and f(x+y)=f(x)+f(y) then f(x)=f(1)x for all x

[south patrol]

Just say nonnegative numbers are squares

[wicked patio]

yeah it is

[south patrol]

Nice

[wicked patio]

so the hint is look at f(y)-f(x) ig

for y >= x

[rocky cloak]

Some further hints:

||Notice this gives you x<=y implies f(x)<=f(y)||

||f(1) = 1, what other elements must satisfy f(x) = x||

||Think about how R can be defined through Dedekind cuts||

[tardy hedge]

Thanks. Im gonna do this question later cause it seems more involved

[rocky cloak]

I think it's the kind of thing that feels obvious once you know the solution.

But yeah, might take some time and fideling the first time

[rotund aurora]

are there some conditions on n, m so that if v_1,...,v_m in Z^n generate Z^n then there exists a subset of m-1 vectors of the v_i that also generate Z^n?

like I'm sure that, for fixed n, and all m large enough, this is always true

[south patrol]

Lol I know how to solve this problem but not how|| it relates to dedekind cuts||

[rocky cloak]

Then I'm very curious how you're solving it

[rotund aurora]

||I think in the second line he is saying that f(x)=x for x rational, then approximate any real number by a monotonic sequence or something||

[rocky cloak]

||Dedekinds construction is exactly saying that a real number is determined by the rational numbers smaller than it||

[south patrol]

||okay so we have f increasing, and f fixes the rationals. Because it is monotone, the only discontinuities can be jumps, which are impossible since you hit all the rationals. Hence it is continuous and we're done||

[south patrol]

Oh lol sure

[rotund aurora]

I'm tempted to say the condition is m>2n, but not sure. I was just trying to see if you could do inclusion-exclusion for P_nm(N) but seems quite hopeless tbf

or perhaps you would want to keep track of the number of minimal generating sets of fixed size with coordinates bounded by N

[quaint mantle]

Hello I have an issue with the proposed solution of this exercise, would like to know if there is indeed more work to be done

to show their sigma is well-defined, they use that the intersection is generated by one element, but in general there are multiple possible generators of cyclic groups. So if a different generator was picked each time, it would contradict the solution, so does that mean it is necessary to do some axiom of choice shenanigans or am I being overly cautious and their solution works?

thanks in advance :)

[south patrol]

Yeah this certainly isn't well-defined as given

[regal zodiac]

https://tenor.com/view/cat-no-nonono-noooo-cat-no-gif-13597132752790194338

[quaint mantle]

thanks! do you have any clues on how to show it's well defined? I thought of constructing the sigma "explicitly": starting with one finite extention L, and then for each other finite extention L' extend it to make it generate the galois group. There are multiple subtle points like can you extend to a generator and the fact that we have to extend an infinite amount of times, but it's the only better idea I have 😅 \

we have normality of each extention so it's possible to extend to an automorphism fixing K each time, but showing that it can generate the galois group is trickier...

[south patrol]

Well you can translate it into the statement that a "procyclic group" (in the sense of being an inverse limit of cyclic groups) is topologically cyclic, but idk if you've seen that terminology

[quaint mantle]

have not seen that haha

[south patrol]

But yeah the point is like you have surjections Gal(L'/K) -> Gal(L/K) of cyclic groups and want to "lift" generators. In general I don't think you can just extend stuff cause it may not behave well e.g. if you consider Z/6 -> Z/3 then whilst 2 generates Z/3, its preimages don't

[rocky cloak]

Yeah, there is some work in making a coherent choice. But it's just the usual trick of applying Zorns lemma to all partial choices

[south patrol]

One way to get around this would be decomposing stuff into prime power order things, and then you actually can just consider extensions (i.e. preimages) i believe

oh yh good point jagr lol

[quaint mantle]

so assuming I start with a $\sigma_L$ with the correction terminology, and gjven another finite extention $L'$, I would consider the union all possible extentions of $\sigma_L$ and a ""maximum element" would necessarly generate the galois group?

[cloud walrus]

Mathologue

[quaint mantle]

I guess the order relation would be "ïf you are generated by another element"

is that a good start?

[rocky cloak]

Like what your interested in are Galois extensions L/k together with automorphism sigma such that sigma restricted to L' generates it's Galois group for each finite subextension.

Then you make this into a poset by saying (L, s) < (M, t) when L < M and t restricts to s.

Hmm, it's actually a little harder to show that a maximal element will be the algebraic closure than I first imagined...

[empty perch]

Hey folks, just a quick reminder that the Abstract Algebra Lecture series is meeting regularly again ^_^
We will meet to discuss the Sylow Theorems in about 40 minutes over in the ⁠#1055201711679082516. More information can be found in our thread: ⁠#1317307081535000606

[restive birch]

thanks, i was proud of it

[restive birch]

okay, interesting

[restive birch]

rough proof?

[coral spindle]

What is a map Z -> G? Just choose where 1 lands. How many such choices are there?

[restive birch]

oh, true

right

we can also determine how many elements of each prime order there are

by looking at Hom(Zp, G)

and i think once you have that you can look at composite order too?

is knowing the number of elements of each order sufficient for finite groups?

[sleek violet]

Given a Galois field ((F, +, \cdot)) with order 8. A group ({x^m \mid m \in \mathbb{Z}}) is constructed with element (x \in F), where (x^m) is calculated using the second operation (\cdot) in the Galois field (for example, (x^3 = x \cdot x \cdot x)).

How much is max x ({x^m \mid m \in \mathbb{Z}})?

[glad lance]

7

[sleek violet]

can you explain pls?

[glad lance]

I could try

Let me put them into steps

[sleek violet]

ok

[tribal wind]

Is there a name for the class of groups isomorphic to subsets of general linear groups over some field?

[vast stump]

linear groups?

[glad lance]

linear groups

[glad lance]

yeah

[rocky cloak]

It's not. For example the Heisenberg group of order 27 and (C3)^3 both have 26 elements of order 3, and one of order 1.

[restive birch]

right

is that the minimal counterexample?

[rocky cloak]

I think so, haven't checked

Might be some examples of order 16

[glad lance]

It's an order of 16

Wait

[vapid vale]

Q8 x Z2 and Z4 semidirect Z4 is a counterexample of order 16

[glad lance]

Don't mind me

[restive birch]

interesting

under the inversion thing?

im new to semidirect products sorry

[glad lance]

Same

[vapid vale]

whats the inversion thing

[rocky cloak]

Like Z/4 acts on Z/4 by multiplication by (-1)^x

[restive birch]

like you map the generator in Z4 to the inversion map on Z4

[vapid vale]

oh

[restive birch]

<x, y | x^4 = y^4 = 1, xy = yx^-1>?

[rocky cloak]

It is the only nontrivial action, so that would be it

[restive birch]

right

[restive birch]

okay so can we work through how we would distinguish these two using |Hom(H, G)|?

well first of all is |Hom(G, G)| for these two the same?

or like

like H = Q8 x Z2, K = Z4 semi Z4

what are |Hom(H, K)| and |Hom(H, H)|

the latter shouldnt be hard

its just 256 right? or not

wait sorry

not that

how would you calculate that

take H = <(i, 0), (j, 0), (0, 1)>?

but does each reasonable (order reasons) set of outputs for these three elements generate a homomorphism?

[rocky cloak]

According to groupprop there are 28 endomorphisms of Q8.

It's not so hard to see there are 2 homomorphism Z/2 -> Q8, 2 endomorphisms of Z/2 and 4 Q8 -> Z/2.

So that would give 448 in total

[restive birch]

huh

what about Hom(H, K)

or is Hom(K, H) easier to calculate

oh also another one:
given n ∈ N, how do you construct a group |G| = n to maximize |Hom(G, G)|?

[rocky cloak]

So their not isomorphic, so any map would end up in a proper subgroup. So for
Q8xC2 -> the semidirect product

that's either in C4xC2 or C2xC4, both abelian, so we can factor through the abelianization. Which is (C2)^3.

Okay, then the image will be in C2xC2, so that should be 64 possible maps.

And 64 is different from 448

[restive birch]

right

wait

so say you have some |G| = n

is it true that for any |H| = n, H not iso G, |Hom(H, K)| < |Hom(H, H)|?

that would solve the problem

[rocky cloak]

I would guess surely not. But I can try to cook up an example

[restive birch]

intuitively you lose all of the automorphisms

so i feel like its true

but maybe for cyclic groups its weird because they dont have many endomorphisms

wait yea isnt Hom(Z16, Z2^4) the same size as Hom(Z16, Z16)? or am i tripping

but then the question becomes if it holds if its <=

[rocky cloak]

Yes, they're both have size 16.

And same for Hom(C2^4, C16)

[restive birch]

yea makes sense

so my original statement is wrong but what about <=

actually is there any H with Hom(H, C16) \ne Hom(H, C2^4)?

[rocky cloak]

C4

Or just C2

[restive birch]

oh right but Hom(C2^4, C2^4) is bigger right

[mint seal]

According to a problem in Aluffi, C4 x C4 also works for this (ordinary product)

erm Z4, whichever notation

[rocky cloak]

Yeah, that one is pretty big

[restive birch]

16^4?

[rocky cloak]

Yeah

[restive birch]

= 65536?

okay

this problem is impenetrable

[rocky cloak]

Here's an idea:

For each finite simple group S, the number of injective homomorphisms S -> G is just |Hom(S, G)|-1.

Now consider a group H. The number of homomorphisms H -> G with kernel K is equal to the number of injective homomorphisms H/K -> G.

By induction on the length of H we can count the number of homomorphisms with nontrivial kernel.

Then we can determine the number of injective homomorphisms H -> G.

Combined with the knowledge that |H| = |G| we can determine when H and G are isomorphic.

So yeah, that just works.

[restive birch]

ooh, cool

we just range over all the quotients?

[rocky cloak]

Yeah

[restive birch]

thats very clever

[rocky cloak]

Argument doesn't work for infinite G though. But then I guess you probably can't distinguish Q and Q^2 anyway.

[restive birch]

i mean also the notion of |Hom(G, H)| becomes much less useful

[rocky cloak]

Yeah

[restive birch]

what about this

[mint seal]

Does length of H refer to Holder sequences? I need to review simple group / decomposition stuff

[rocky cloak]

Yes, length = length of longest subnormal series.

[thorn jay]

Thats so cool

I love that

[mint seal]

so if G1 = C4 x C4, and G2 = C16, can we find a group H with |hom(H,G1)| =/= |hom(H,G2)|?

we must be able to if the claim about classification is true

(I can’t see any issue with the idea above, I just don’t fully follow it)

[restive birch]

C8 right

i think

because if C8 = <x> then phi(x) can be any element of C4 x C4

but it can only be half of the elements of C16

[mint seal]

ah, because the odd numbers in C16 have order 16?

so they can’t be phi(x)?

that is good

[rocky cloak]

In general Cn just counts the number of elements with order dividing n.

So for C4 or C8, that would count everything in C4xC4, but not everything in C16.

C2 would also work, cause it counts 4 things in the former and only 2 in the latter

[empty perch]

Hello folks! The meeting today ran into technical troubles, so we'll be re-convening in a few hours at <t:1737338400:F>, still right in house at #1055201711679082516

[barren sierra]

This product should be from j = 0 to n - 1 right?

that's what I got and I can't find a mistake in my work nor do I see how the product from j = 1 to n is equal to the product from j = 0 to n - 1

[south patrol]

Very nice

[rocky cloak]

Plugging in j=n gives the same term as j=0

[barren sierra]

Ah incredible

[rocky cloak]

I mean omega^j only depends on j modulo n

[south patrol]

Nice question

I have not seen circulant matrices in years - , I wonder what their main use/feature is

[barren sierra]

That's sort of the point of the class lol

[south patrol]

Wdym

[rocky cloak]

I've never seen them before, but that is a nice formula for their determinant

[barren sierra]

The class is called "Problem Solving"

basically we are assigned problems and have to present them every week

I don't have to do all of them but I'll do probably 2 of them

but they're nice

[sly crescent]

Sounds interesting

[barren sierra]

it's basically "practice presenting math to people" class

[south patrol]

I guess this must have some nice method in terms of ||the eigenvalues somehow, like using the cyclic action on the matrix lol||

[barren sierra]

I bet these are the eigenvalues

I proved it in a different manner than eigenvalues

[south patrol]

Oh nice how did you do it

Or maybe I wanna think about it first aha

Also, I'm pretty sure 11 is the laplace equation lol

[rocky cloak]

Yeah, just look at the vector
(1, w^j, w^2j, ...) I guess

[barren sierra]

Close yea

[rocky cloak]

They are the eigenvectors

[barren sierra]

Nice

The downside to this class is that it's 4:10-6 PM on Wednesdays and the prof likes to go over time

[quaint mantle]

hello, I think there is an issue with the solution of this exercise:

the problem I have is that if $p^{s}-1$ is not square-free, then how do you express a primitive root of a power of $q$?

[cloud walrus]

Mathologue

[quaint mantle]

say a $q^2$ primitive root of unity

[cloud walrus]

Mathologue

[rocky cloak]

It could be that the problem was supposed to have q range over all prime powers.

It could be it's still true as stated though, but I'm not sure...

[quaint mantle]

Thanks, glad to know I didn’t miss something obvious

[rocky cloak]

Hmmm, so say e is the number such that Fp(mu) = Fp^e

Then the density of primes congruent to 1 modulo e is 1/phi(e), so non-zero.

Now for any such prime q, if p has order a multiple of e (for example if p is a primitive root), then
Fp(1^1/q) would contain Fp(mu).

It seems likely to me this would be the case, but I'm not quite sure what this second probability is, and these probably aren't independent so this wouldn't work as a proof. But it's some evidence that suggests it should be true

You don't even have to restrict to primes q, since you can use square free integers in general.

For example for p=2, then
F2(1^1/9) is contained in F2(1^1/21)

So actually you can have one prime q contribute to each prime factor in e.

So you would just need to show that for any prime, there are infinitely many primes q such that the order p mod q is divisible by that prime.

That very much seems like something that should be true

[glad osprey]

For a polynomial f(x), is the condition f'(x) != 0 equivalent to f(x) being non-constant? I feel like the answer is obviously yes, but I'm just not sure why this book doesn't just say "f(x) is non-constant"

[rocky cloak]

Not over fields of positive characteristic.

For example x^2 has derivative 0 over F2

There are definitely cases where this distinction is meaningful

[glad osprey]

Thanks I should've read the next line btw, they write "if char(F) = 0 then f'(x) != 0 for any non-constant f(x)"

sometimes when I get confused, I don't know whether to read on and hope they clear things up, or just stop and figure it out

the theorem I'm reading is "an irreducible polynomial f(x) is separable if and only if f'(x) != 0", so I bet the distinction is important

[rocky cloak]

I guess the strategy would be to skim ahead a little to see if it's likely it would be cleared up, then go back and think a little if not

[glad osprey]

yeah, that's probably a good idea

[rocky cloak]

Indeed, though your fields have to be a little funky to find interesting examples.

But if you take your field to be the rational functions over F2,
F2(t)

Then f(x) = x^2 + t is irreducible, but not seperable.

And indeed f'(x) = 0

[glad osprey]

I see, thanks for the example

[opaque finch]

I am reading some lecture notes on Representation of finite groups and came across this corollary of Schur's Lemma
If $U$ and $V$ are simple $\mathbb{F}G$-modules, then $<U,V> = 1$ if $V \cong U$ and $<U,V> = 0$ otherwise. Can someone please explain the notation $<U,V>$ ? It doesn't seem to defined anywhere

[cloud walrus]

mycroftholmes1703

[delicate orchid]

it's the dimension of the hom space between those modules

I've only ever seen it phrased in terms of characters, where it is $\langle \chi, \psi \rangle = \frac{1}{|G|} \sum_{g \in G}\chi(g)\overline{\psi(g)}$ for any $\chi, \psi \in R(G)$

[cloud walrus]

Wew Lads Tbh

[opaque finch]

I know it for characters

but I don't know this notation for modules or for group modules

[delicate orchid]

yeah and it corresponds to the dimension of the homspace between the corresponding C[G]-modules (C in this case, F in yours)

there's no analogous statement for Brauer characters unfortunately so this is the best I can give you over positive characteristic

[opaque finch]

hmmnn

thanks @delicate orchid

[delicate orchid]

np

[rocky cloak]

Is F algebraically closed?

[quaint mantle]

Maybe it’s because they are identifying the representation with a module structure (equivalent to define a representation), and simple modules are the equivalent of irreducible representations

And the bracket notation would be the inner product used with characters, giving 0 if the irréductibles are different due to Shur, and 1 if they are the same

[tardy hedge]

How should i approach 10?

The degree of an irreducible factor is the same as the degree of the extension when u adjoin a root

So probably use something like that right

[rocky cloak]

Yeah, notice a root of fg is an element x such that g(x) is a root of f.

What happens when you adjoin such an x

[tardy hedge]

If a is a root of fog then g(a) is an element of F that is a root of f. F < F(g(a)) < F(a) so [F(a) : F] = [F(a) : F(g(a))]*[F(g(a)) : F]

[rocky cloak]

Yes, so g(a) is not an element of F, but that is indeed the argument.

[tardy hedge]

Oh because a is coming from some extension of F.

What happens when you "adjoin" a root that is already in the base field. C[i] = C is isomorphic to C[x]/(x-i)?

[rocky cloak]

Yeah, not much happens.

In general the degree of the extension should be the degree of the minimal polynomial of what you adjoin

[coral spindle]

First isomorphism theorem my dude. f : C[x] -> C sending x to i has kernel (x-i)

[tardy hedge]

ok yes makes sense lul

[thorn jay]

Yes, one way to "intuitively" see this is that k(a) forms an algebra over k, where it's dimension is the degree of the minimal polynomial of a, and for any a in k, it's minimal polynomial is x-a, which is of degree 1, hence k(a) = k

[tardy hedge]

If u have F(a,b) where a and b are algebraic over F with degrees n and m, then its a finite extension so any element of F(a,b) is algebraic over F. What are the degrees of the other elements in F(a,b)?

[thorn jay]

Divides the degree of F < F(a, b)

I dont think there's much you can say on it

[rocky cloak]

They can be as high as mn, but definitely divides that.

[thorn jay]

Is there a way to compute which degrees occur without looking at every in-between extension?

[tardy hedge]

Thanks

[rocky cloak]

Most should appear in general, but I guess you would have to think about the Galois group.

[crystal vale]

8th one is good one

[thorn jay]

Ah, i was suspecting something along those lines

[tardy hedge]

yea i still havent done that one yet

last one to do

[thorn jay]

Your ordering is on some permutation of { 1, 2, ..., n } i see

[tardy hedge]

Am I using eisenstein wrong? x^2-1 is reducible over Z, but p = 2 doesnt divide 1, it does divide 0, and p^2 doesnt divide -1, so by eisenstein its irreducible?

[coral spindle]

By eisenstein it's irreducible over Q

Wait lmao sorry I got this backwards

Ah but

you forget that 2 has to divide a_0

The restriction is that p^2 doesn't divide a_0

Check the statement of Eisenstein one more time

[tardy hedge]

Oh yea thanjs

[young thorn]

im struggling on how to prove this

any tips

🙏

[rocky cloak]

Do you know orbit-stabilizer theorem?

[young thorn]

no

havent been taught it yet

[rocky cloak]

Hmmm, okay.

Well, you have a function
G -> conj_G(x)
given by
g |-> gxg^-1

[young thorn]

yes

[rocky cloak]

Maybe examine this function and it's relationship with G/C_G(x)

For example if two elements g and h have the same image, what does that say about them?

[young thorn]

that h^(-1)g is in C_G(x) ?

oh which would mean hC_G(x) = gC_G(x)

[young thorn]

but then once u show hC_G(x) = gC_G(x)

how can u assume that if g, h are elements of G s.t. they have different images, the cosets will be distinct?

cz all ive done so far is show that if h and g have the same imagine, then their left cosets are equal

[rocky cloak]

Well assume the cosets are equal. Can you then show the images will be the same?

[young thorn]

yes

ohh

big brain

because it's both ways then we can conclude that the distinct elements are distinct cosets

[rocky cloak]

So you have a bijection!

[young thorn]

so true

thxthx

[viscid pewter]

quick q, can you generate S_4 with two 4-cycles?

actually wait

i guess i should really be asking if you can in general generate S_n with two n-cycles

well ok no

that's trivially false for odd n because then n-cycles are even

but for even n...

[round zodiac]

Hey! I have this exercise from a professor's class notes. I started trying to prove it, but I was getting nowhere, so I looked for an example to get some intuition, and this is where I ended up. I'm not sure if I'm missing something about Sylow p-subgroups, quaternions, or the exercise itself; or if the propositions lack some stronger hypothesis.

Any feedback would be helpful!

Notation used: Syl_p(G) := { sylow p-subgroups of G} = {maximal p-subgroups of G} ; N(H) = Normalizer of H = { g in G : gH = Hg }

Edit: It is not written that H is also a Maximal subgroup, so it is a Sylow 2-subgroup

[rocky cloak]

Consider the cycles
s = (1 2 3 4 ... n-1 n)
t = (1 2 n n-1 ... 4 3)

Then st = (1 3 2)
x := s st s^- = (2 4 3)
st x = (1 3)(2 4)
(1 2 n n-1 ... 4 3)
y := t st x = (2 3) (n n-1 ... 5 4)

Now if n is even n-3 is odd, so y^n-3 = (2 3).

And we know Sn is generated by any n-cycle and any transposition.

[rocky cloak]

So following the hint assume x is in N(H).

Now H is normal in N(H) so we can consider the image of x in N(H)/H. What can the order of this element be?

As for your example notice that H is not a sylow subgroup as it has only order 4, while G has order 8

(the only sylow subgroup of G is just G itself, so the example is quite boring)

[round zodiac]

Oh!, so I'm not getting Sylow subgroups then!

thx!

I must ask why H is not a sylow p-subgroup

[south patrol]

Nice

[rocky cloak]

So if k is the largest number such that p^k divides the order of G.

Then the p-sylow subgroups of G are defined as the subgroups of order p^k

Alternatively you can define the sylow subgroups as the maximal p-subgroups

[round zodiac]

Ok, I have that definition. ¿The problem is that H is not a p-subgroup i guess?

[rocky cloak]

H is a 2-subgroup, but it's clearly not maximal, as G is bigger

Perhaps your confused about the competing terminology "maximal subgroup" which means a subgroup maximal among proper subgroups?

[round zodiac]

I thought maximal subgroups were those that had no subgroup different from G containing them, because that happens always.

[round zodiac]

ouh yes, I think it is the case my friend!

[rocky cloak]

Yes, but maximal p-subgroup doesn't mean "p-subgroup that is also a maximal subgroup"

[round zodiac]

OMG

[rocky cloak]

It means subgroup maximal among p-subgroups

[round zodiac]

So G is considered as just a regular p-subgroup, regardless of the fact that it is the 'base group' we are working with?

[rocky cloak]

Yes that's right

[round zodiac]

thx!

What an obscure way mathematicians name things

[rocky cloak]

And on the flipside in for example
Z/12Z, then 4Z/12Z is the 3-sylow subgroup, even though it's too small to be a maximal subgroup

[round zodiac]

Oh! Great.

[rocky cloak]

Really it's just
"maximal proper subgroup" that was shortened to "maximal subgroup" since they come up alot.

Otherwise "maximal X" usually means "thing maximal among Xs"

[round zodiac]

Okay, that´s nice

[viscid pewter]

that's hot, ty ty

[minor fulcrum]

Is there a word for a group action where the action is the identity only for the identity element

Is it something like faithful or something like that

Like that's kinda what a faithful representation is

[rocky cloak]

Faithful group action is the word yes

And a faithful representation is exactly a faithful (linear) group action

[minor fulcrum]

yeah makes sense

[acoustic igloo]

The nonzero elements of a field form a group under the multiplication in the field
i think this is true

am i right?

[next obsidian]

Yes

[acoustic igloo]

we need to show multiplication is closed, meaning you can't have elements whose product is 0

but yeah this seems to be true

thanks 🙏

let a,b not equal to 0 and assume ab=0
(a^-1) a b = (a^-1) 0
1b=0
b=0 -> contradiction
so multiplication is closed over the nonzero elements

[delicate bloom]

what was your definition of field going into this

[acoustic igloo]

"A field is a commutative division ring"

a division ring is a ring with unity 1 not equal to 0, where every nonzero element is a unit

it seems we can define a field as a ring whose nonzero elements form an abelian group under multiplication

[mint seal]

More generally, the units of a ring with 1 form a group under multiplication

where unit means invertible element. Which all the nonzero elements of a field are

[acoustic igloo]

oh nice

[sly crescent]

That isn’t the only instance of “maximal” being short for “maximal proper”

Really maximal sub-anything has an implied proper

[serene dune]

Krull's theorem (1929): Every nonzero unital ring has a maximal ideal.

well i haven't really looked up the proof but i got the information that it uses zorn's lemma
however is there anyway i can get any idea(maynot be the whole proof) of how and why this should be true

why am i asking this: i'm not very familiar with zorn's lemma

[crystal vale]

You know the definition of Zorn's lemma?

[night tartan]

suppose there does not exist a maximal ideal

that means every proper ideal has a proper ideal "bigger" than it

so let's start off with a proper ideal I_1

if there does not exist a maximal ideal then we can find ideals that are always bigger, i.e.

$I_1 \subset I_2 \subset I_3 \subset \cdots$

[cloud walrus]

LY

[night tartan]

recall that the union of a sequence of increasing proper ideals is a proper ideal

so we have another ideal $I_{\omega} = \bigcup_{n \in \mathbb{N}} I_n$

[cloud walrus]

LY

[night tartan]

but again, we've assumed no maximal ideal, so we can keep finding even bigger ideals

$I_{\omega} \subset I_{\omega+1} \subset I_{\omega+2} \subset \cdots$

[cloud walrus]

LY

[night tartan]

union them again to get another ideal $I_{\omega \cdot 2}$

[cloud walrus]

LY

[night tartan]

keep going until we get ideals $I_{\omega} \subset I_{\omega \cdot 2} \subset I_{\omega \cdot 3} \subset \cdots$

[cloud walrus]

LY

[night tartan]

union them again to get ideal $I_{\omega^2}$

[cloud walrus]

LY

[night tartan]

at this point, you might suspect that we get a contradiction if we're allowed to go on forever

because eventually "we run out of ideals"

this is what zorn's lemma formalises

anyway that's like some motivation why zorn's lemma & krull's thm should be true

[serene dune]

thanks for the input, in a bit i will go through this and will try to respond if my brain cant fathom anything

[crystal vale]

Let G be a group such that for any non-identity, a,b,c in G we have abc = cba then G is abelian.

I proved this, but I am now thinking about how I can use this exercise in future or what information I can get from this?

[thorn jay]

Probably not much, considering you almost never prove stuff for stuff besides the identity

Only thing i can think of is abelian heaps

A heap is an algebraic structure with a ternary term [x, y, z] satisfying the same identities as xy^-1z for groups

Then a heap is abelian if [x, y, z] = [z, y, x] for all x, y, z

Now, you can turn a heap into a group with for any element a defining
x • y = [x, a, y]
If [x, y, z] = xy^-1z in some group, then this induced group must be isomorphic to it

Nevermind, i thought you could do something but the "any nonidentity" fucks everything up lol

I think it's just an exercise with group operations

Nothing more

[tardy hedge]

That was only in 1929?

[rocky cloak]

When would you have expected it be proven?

The modern definition of a ring seems to be from 1921

Although varies families of rings where studied since the 1870s

[cloud walrus]

nastasya

[serene dune]

sounds ?

quite confident for the case of Z but not in this one

[serene dune]

emil artin and co

[rocky cloak]

Emily Noether. I guess she can fall under co there

[serene dune]

yeah true i just learnt about the artinian and noetherian modules a few days ago

also i dont know how number theory has to say so much about rings but i can kinda see where is it coming from !

[rocky cloak]

Number theory says alot about number rings at least.

That's where most of the motivation came from originally, and where the word "ideal" is from

[serene dune]

im trying to follow the borcherds' playlist too, he talks about some bizarre things from here and there which takes up a lot of time for me to grasp

[rocky cloak]

Number rings don't always have unique factorization into prime numbers, but they have unique factorization into prime ideals. So ideals are like an "ideal version of numbers" or how numbers should have been

[serene dune]

ohhh, i kinda get the insight now! as it feels

[serene dune]

in anyways, is there any flaw in this argument ?

[rocky cloak]

Seems fine.

Any automorphism is just given by multiplication by a rational number, and there are countable

[serene dune]

alr

can i get some motivation to think about R^op

it felt a bit intuitive but i dont wanna carry just the intuition

[rocky cloak]

You can notice that a right R-module M is the same as a ring homomorphism R^op -> End(M). In other words a right R module is a left R^op module.

This can be useful if you consider bimodules or objects with several different module structures, so you can swap between perspectives.

So if you have some sort of duality it might be useful to think of it through ring homomorphisms R -> R^op.

It also means that any general statement you prove about left modules will also be true for right modules.

In general I wouldn't say there's so much to "think about R^op", it's just convenient to have some object whenever you want to think about maps that reverses the order of multiplication.

[serene dune]

internet gone for some moment and got stuck into some other details, will go through this later

[terse crystal]

Two finite groups G and H. For any prime p, G, H have same number of Sylow p-groups, can we obtain G and H are isomorphic? And what about stronger condition: for any prime p , any r, G and H have same number of subgroups of order p^r? Or stronger any n, G and H have same number of subgroups of order n. Those conditions, is any one of them is sufficient for G and H being isomorphic?

[rocky cloak]

The first example I would look at would be to compare the dihedral group of order 16 and the semidihedral group (the semidirect product of C8 and C2 where C2 acts by x |-> x^3)

[terse crystal]

Thank you. Will exam this case.

[rocky cloak]

They appear to have a different number of elements of order 2

[terse crystal]

Oh..

[terse crystal]

Someone suggested me to consider Z/pZ rtimes Z/3Z for prime p=1 mod 3, I am examine this too

[rocky cloak]

That's only one group though. Like you need two

[terse crystal]

That person seems to mean that there will be same number of sylow groups for various semiproduct

Different rtimes

[rocky cloak]

Well, up to isomorphism there's only one besides the direct product

[terse crystal]

Oh…

[rocky cloak]

Something that might work though

Consider two primes p and q both congruent to 1 mod 3. And have C3 act on CpxCq.

Then you have a little room for different ways to act

[terse crystal]

Will try. Strange that I can’t find online resources discussing this.

[rocky cloak]

http://math.colorado.edu/~kearnes/Papers/group10.pdf

This paper seems to suggest such examples are possible

[terse crystal]

Shit. Same lattice, strongest condition I can even imagine, interesting. It certainly can answer all my questions

Thank you

Wait same lattice doesn’t necessarily mean same number of subgroups of some order right…

Like cyclic groups of prime order all have the same lattice, trivial subgroups

[terse crystal]

Z/2Z times Z/6Z, and Z/12Z both have only one sylow-2 group and one sylow 3-group.

[rocky cloak]

Yes, but they have a different amount of subgroups of order 2

So consider Z/7 x Z/13 and have C3 act by
(a, b) |-> (2a, 3b)
and
(a, b) |-> (2a, 9b)

The resulting semidirect products should be non-isomorphic (I haven't actually checked)

But both have 7*13 sylow-3 subgroups and only one for each of the other primes.

[rocky cloak]

In the paper they also talk about when the isomorphism of lattices preserves the order of the subgroups

[rocky cloak]

This example is in fact a special case of what the paper discusses in section 2. And according to what is claimed there, they will be non-isomorphic

[terse crystal]

Thanks a lot. Yes they are not isomorphic

https://math.stackexchange.com/questions/527800/when-are-two-semidirect-products-isomorphic

I used Nogard’s answer

[rocky cloak]

Yeah, that makes perfect sense

[tardy hedge]

If i know the degree of an algebraic element p over F is n , can i find the minimal polynomial from this info?

This will tell me that a0+a1p+a2p^2+..anp^n with ai in F is linearly dependent, but thats it i guess

[rocky cloak]

I mean any irreducible polynomial is the minimal polynomial of some element, so just knowing that the degree is n doesn't narrow it down the much

[tardy hedge]

Yeah

If i found some polynomial in F that has the algebraic element as a root, with the degree of the polynomial the degree of the element, must it be the minimal polynomial? I guess, must it be irreducible?

[void cosmos]

yeah

the minimal poly must divide it

[tardy hedge]

Right, thank you!

[void cosmos]

up to a constant then

[tardy hedge]

But if its monic then its the same right

[void cosmos]

u did not say it is

but yeah

[tardy hedge]

Minimal polynomial in my class and d/f is defined as monic

[void cosmos]

yeah but im talking about that other polynomial

[tardy hedge]

My bad

[void cosmos]

no its ifne ;D

[clear ingot]

Given a nxn F_q matrix C where q is a prime, how many ordered pairs (A,B) in Mat_n(F_q) satisfies AB=C?

[rocky cloak]

Depends on C. But if C is invertible, then it's the same as the amount of invertible matrices which is
(q^n - 1)(q^n - q)(q^n - q^2)...(q^n - q^n-1)

[tardy hedge]

Not sure, I guess I forgot the modern definitions are newer than I thought

[next obsidian]

Actually isn’t it two times that - # of idempotent matrices cuz it’s an ordered pair

No im dumb

🐒

[tardy hedge]

In Q10, what happens if we don't assume f(x) is irreducible?

[rocky cloak]

Well if f = pq, then
f(g(x)) = p(g(x)) q(g(x))

So you can decompose f into irreducibles and argue from there

[opaque finch]

good question. It maybe or may not be.

[opaque finch]

yes seems so. But I think that it should be dimension of the hom space as @delicate orchid mentioned. Cause I clarified from my course instructor and that's what he mentioned

[mental silo]

do rings normally have multiplicative identity

cause my textbook is talking abt like "rings with identity" but i always thought it was part of the definition of ring

[opaque finch]

No.

But there is a construction by Dorroh that says every ring can be emdbedded as an ideal to a ring with identity

[tardy hedge]

For 11, why does it say consider a counterexample of “minimal degree”, as if the degree of F(a) over F can be different? Since f is irreducible, adjoining a root a to F should be the same degree no matter what root we adjoin right?

Like in x^2-2, adjoining the roots sqrt2 and -sqrt2 is same degree

[mental silo]

this makes sense i think

[mental silo]

is it like a pedagogical choice or is ring never defined with multiplicative identity

[opaque finch]

well by the construction of certain types of rings they do not contain identity.

[mental silo]

r there any famous examples of rings with no identity

[thorn jay]

Any (proper) ideal

[thorn jay]

Anti rng propaganda

[opaque finch]

no. If you try thinking as per hint, you will get maybe a counterexample of even smaller degree which will eventually be your contradiction

[opaque finch]

ring of even integers

[tardy hedge]

ok but what i was saying was right right, degree F(a) over F should be the same, but the point here is that this is not satisfied in this question ?

because we find something smaller or smth

[opaque finch]

and very nice fact it is an ideal in the ring of integers.

[mental silo]

o makes sense lmao

ooo yes ive heard of rngs

i thought that was additive identity for some reason

[thorn jay]

Without additive identity i would cry

[opaque finch]

no multiplicative. Else that wouldn't be a ring

[mental silo]

ok

[opaque finch]

wouldn't be an r even

[mental silo]

wait so mycroft what's an rng if rings don't need identities

if rings dont need multiplicative identity then what does the i stand for

[opaque finch]

i ?

[mental silo]

like isnt the joke u remove i from ring and get a different structure called an rng

[thorn jay]

It's the terminology for sane people who require rings to have multiplicative identity

[mental silo]

o ok looks like it's js a notational disagreement

there's lowk so many issues in math that the community is like half half on and it makes learning some stuff unnecessarily hard

[thorn jay]

It's always convention disagrement..

[mental silo]

like inclusion of 0 in N, how to write subset / strict subset, meaning of image / codomain / range

i mean image is pretty consistent

but still man someone should js drop the hammer and say these are the official opinions

[thorn jay]

As you can tell im on the opposite side as mycroft, i have my reasons

Im sure he does too

[mental silo]

hopefully john math patches this soon

[opaque finch]

wait wait

[thorn jay]

That will never work, I'll tell you

[opaque finch]

I totally got sidekicked from the discussion

[mental silo]

so enpeace would you say ideals aren't rings and are like rngs insteads

[thorn jay]

Im sorry 🥺

[thorn jay]

In my opinion yes

[opaque finch]

yes

so what's the problem in that ?

[mental silo]

well mycroft u said that rings didnt require multiplicative identity so by ur notation an ideal would js be a ring

ig this is js a notational disagreement with advantages on both sides

[thorn jay]

I do a lot of universal algebra, so requiring rings to have an identity just makes everything easier when talking about homomorphisms

[opaque finch]

yeah

[thorn jay]

It's just one extra term, and if you want to be extra general, if a rng is a ring, then the multiplicative identity is unique, making it a so-called completion, which gives you a lot of very nice extensions relating to homomorphisms and the retention of the unit

[opaque finch]

I mean it's like a disagreement

but that's like just how you consider

[thorn jay]

Yeah, but i do write in my papers that i consider rings with identity, precisely because of that reason

[opaque finch]

French people have arguments that 0 is a natural number.
We here don't consider 0 to be natural
Davenport on the otherhand, when discussing character theory didn't consider 1 to be natural

[thorn jay]

That last one is crazy 💀

[opaque finch]

Someday if @mental silo you write a book you can consider -1 to be natural (please don't do this though). As long as it doesn't effect majorly then I guess whatever convention you consider is fine

[opaque finch]

yes. Because for Davenport Natural Numbers = Primes union composite and 1 is neither

[thorn jay]

Considering -1 to he natural i mean, there is a successor function you can bestow upon it making it isomorphic to N as a model of the peano axioms

[thorn jay]

Mm, makes sense

[mental silo]

My book will have the most despicable notation of all time

[thorn jay]

You will be burned at the stake

[mental silo]

I will use question marks for inverse factorials

[thorn jay]

I mean, thats what happens every year right? Or did i accidentally join a cult again

[mental silo]

0 isn't a natural number but -1 is

[opaque finch]

exactly

[mental silo]

Are yall believers in including 0 in the natural numbers

Cause I'm lowk not

Peano stuff aside it's just inconvenient I like counting from 1

[thorn jay]

The induced semigroup would be weird
The grothendieck completion would then just be the group with operation
x • y = [x, -1, y] = x + 1 + y
Lmao

[thorn jay]

I do tbh

[mental silo]

But whenever I'm trying to actually write well I never say naturals I just say nonnegative integers or positive integers

That way there's no ambiguity

[opaque finch]

let's not overthink it

[thorn jay]

Read it and weep as I write a paper about diagrams of the shape 0 -> 1 -> 2 -> ... in a variety

[thorn jay]

You're asking a neurodivergent mathematician not to overthink something

[thorn jay]

Nah thats based though

[mental silo]

Here's what the survey says

[thorn jay]

The first blue slice is every commutative algebraist/geometer

[mental silo]

This sucks man now I just have to say unital or non unital ring all the time cause yall can't agree

[thorn jay]

:3

[thorn jay]

And definitely not rep theorist, they couldnt live without their G-modules

[wicked patio]

this was really funny when I studied math 319 where rings were unital

and then i took math 419 which apparently had the same prof as the 319 course where rings were not unital

[thorn jay]

Oh hai you're also here

[wicked patio]

and this wasn't even mentioned until i asked like wait what about the multiplicative identity

[thorn jay]

Prof must be constantly correcting himself

[wicked patio]

and these courses are literally just called like abstract algebra 1 and 2 or something lol

i think they were both out of dummit and foote as well but i could be wrong

[wicked patio]

i feel like all my good profs are constantly correcting themselves

the bad ones just make the mistakes and don't notice

[thorn jay]

My math teacher oml

[wicked patio]

my linalg teacher saying that any two distinct polynomials over any field are distinct as functions, even finite fields

[thorn jay]

Fermat's little theorem:

[wicked patio]

my measure theory prof saying that Z U {-infinity} is well ordered

[thorn jay]

Ah of course

Isn't Z itself not even well-ordered

[wicked patio]

and my analytic number theory prof messing up roots of unity

[wicked patio]

yes

but his logic was: it has a minimum, so it's well ordered

[thorn jay]

Impressive

Well i guess any closed set of the order topology has a minimum

[wicked patio]

mfw X is closed in X

[thorn jay]

Omg no way

You're telling me ø is open??

[wicked patio]

fr fr

[thorn jay]

Ah you see

The problem was that he isnt an algebraic number theory prof

[wicked patio]

so true

[thorn jay]

At least not at that momebt

[wicked patio]

all of these i pointed out to the profs which may be a mistake lol

well actually

these are stated in chronological order

the first one i pointed out and the prof was just like no you're wrong and i was like welp i tried i'm not gonna prove it to him

the second one caused the prof to despise me (he was completely nuts and everyone hated him but i didn't know that at the time)

so by the time we were at the third one i was like yeah maybe i should just let him be wrong lol

[thorn jay]

Hey at least you learn from your mistakes

[wicked patio]

someone else evidently tried to work the roots of unity thing out with the 3rd prof in office hours

he came back the next day with a correction but the correction was also wrong

[thorn jay]

Christ 💀

[opaque finch]

yes. you are correst

[thorn jay]

Oh yeah we were completely talking over him, whoops

[opaque finch]

people change colours

[wicked patio]

[thorn jay]

Insert the difficulty slider of the south park game

[clear ingot]

Yep, I was wondering about the general case tho

[tardy hedge]

thank you bro bro broski

[thorn jay]

Chad chadski ahh name

[south rain]

what is an example of a ring that where the jacoson radical is the whole ring?

[rocky cloak]

Any rng with all multiplications 0

[south rain]

We defined the Jacobson Radical as the intersection of all annihilators of simple $R$ modules. We defined a simple $R$ module as a module where the multiplication is non-trivial, and M has no non-trivial submodules. So in this case your example wouldn't work right?

[cloud walrus]

dellinger

[rocky cloak]

It would still work. The empty intersection is R.

But I don't see the point of requiring simple modules to have nontrivial multiplication. That just kinda messes up the correspondence between non-unital rings and unital rings with a map R -> Z

[south rain]

Thanks.

Well I can't really tell you why we defined it that way, sorry.

[rocky cloak]

I guess I'm not a big fan of non-unital rings in the first place

[thorn jay]

Same, it just makes a lot of definitions harder / more convoluted

And the only real use i have for them is with ideals, and else I'd call them rngs

[tardy hedge]

My prof is skipping so many proofs in my alg number theory class

[minor fulcrum]

Is there a subset of the permutations of N closed under composition which isn't a group

[storm kiln]

Don't think so, associativity still holds and repeated multiplication of an element with itself will give the identity permutation at some point, from there you can also deduce that inverses exist

I think this works for any finite group

[rocky cloak]

Sure consider the permutation
(1 2)(3 4)(5 6 7 8)(9 10 ... 2^4) ... ((2^n)+1 ... 2^n+1) ...
or just any permutation of infinite order

[storm kiln]

Ooh, capital N whoops, thought N would be some finite number lol

[rocky cloak]

Then closing the set under composition doesn't grant you inverses

[tardy hedge]

If g(x) over field F has odd degree, it should have an irreducible factor of odd degree right?

[rocky cloak]

A sum of even numbers is even, so yes

[tardy hedge]

Yes indeed hehe thank you

[barren sierra]

Throughout, R = k[x_1, ..., x_n] for some field k.

For the last part, I don't get why it says $t \ll 0$. We have that $R(-d_i)t = R{t - d_i} = 0$ whenever $t - d_i < 0$, meaning $t < d_i$.

[cloud walrus]

Spamakin🎷

[rocky cloak]

That's the proof right there yeah

[barren sierra]

t << 0 seems like a much stronger conclusion

[rocky cloak]

t << 0
means there exists some n<0 such that it holds for all t<n

[barren sierra]

Hm

I guess I'm thinking of all the d_i > 0

so you can take t = 0 even

[rocky cloak]

di can be negative

[barren sierra]

but all the d_i don't have to be greater than 0

yea

[glad osprey]

If F is a field and a, b is in F, then every non-zero element of F is a GCD of a and b, right?

[next obsidian]

Yeh

[coral spindle]

Unless they’re both 0 ig

[glad osprey]

does that mean that every polynomial over a field is primitive?

[coral spindle]

You mean that the gcd of its coefficients is 1? Yes, but iirc there is a different meaning to the word primitive in field theory

Just looked it up. Indeed sometimes this refers to a minimal polynomial for a Galois field

[rocky cloak]

Minimal polynomial of a primitive element*

[coral spindle]

Yup yup

[glad osprey]

I see this book writes under Eisenstein's criterion that "in particular, if p(x) is primitive, then it is irreducible" which is clearly not true for the definition I'm familiar with, so there's something I'm misunderstanding

[toxic zephyr]

in a UFD, if x is a product of n irreducibles, then the uniqueness guarantees all factorizations into irreducibles will have n irreducibles right (and they are equal up to associates, after reindexing)?

[rocky cloak]

Is there some context with more assumptions about p perhaps?

[glad osprey]

Not sure what they mean tbh, it seems like they're implying that any polynomial satisfying Eisenstein is primitive

[glad osprey]

Yep

[toxic zephyr]

ty!

[rocky cloak]

They're saying the theorem implies p is degreewise irreducible.

So in particular if you add the condition that p is primitive, then p will be irreducible.

So for example for R = Z, then the polynomial
p(x) = 2x^2 + 6
Then p is degreewise irreducible by Eisenstein at the prime 3, but p is not irreducible since it equals
2(x^2 + 3)

[glad osprey]

ah, of course thanks

[rocky cloak]

Also the formulation with a prime dividing the coefficients is kinda limiting. Would be better to formulate in terms of coefficients contained in a prime ideal

[glad osprey]

Hmm, I see I'm reading Field Theory by Steven Roman, not sure if it's any good. Do you have any suggestions for another book?

[rocky cloak]

I don't, but I'm sure there are people around with strong opinions on books

[tardy hedge]

Was naively going backwards by setting x to that number, but the polynomials i get dont seem to be irreducible by eisenstein

How else can i do this

It worked for a though

Nevermind 😂

I got eisenstein wrong again

[delicate bloom]

what'd you do for a, since it should work

or at least by inspection you can tell it's a sqrt(i) so it ought to be an 8th root of unity

[rocky cloak]

A different approach than Eisenstein could be to think about the extension the element generates.

For example Q(i + sqrt2) < Q(i, sqrt2) so the minimal polynomial must have degree 2 or 4. Once you eliminate 2 (by showing that the two extension are equal, showing nothing from the Galois group fixes it, or just brute force) you'll know it has degree 4. So then if you find a degree 4 polynomial with it as roots it will be the minimal.

[tardy hedge]

Yes cool i was thinking something like that too

[rocky cloak]

Also I don't think Eisenstein helps you with any of these anyway, unless you're doing some clever change of variables

[delicate bloom]

for the cyclotomic polynomials the eisenstein trick with change of variables works for it, which is what I had in mind for a) at least

I suppose that'd work for c) as well since that's already a shift by 2 on a 3rd root of unity so it ought to work find there as well

[rocky cloak]

I mean you don't really need Eisenstein to show that a degree 2 polynomial is irreducible

[thorn jay]

"Hey guys i have a novel proof that the n-th root of 2 is irrational for n > 2"

[rocky cloak]

Easy, just apply Fermat's last theorem

[thorn jay]

I cannot see any other way than this

[viscid pewter]

this has to be bait

[south patrol]

I like the "analytic" proof

[rocky cloak]

People get so hung up on these complicated techniques like "prime factorization" or "infinite descent".

Like bro, it's just FLT. Chillax

[thorn jay]

Oh no im 100.0001% serious, i believe we should teach wiles' proof of FLT before any of that elementary number theory or analysis nonsense

Because who needs that anyways

[rocky cloak]

I'm not sure students are really prepared to understand Euclid's lemma if they don't have a good grasp of the Taniyama–Shimura conjecture you know

[thorn jay]

Yeah no exactly

Some for something trivial like Stone Duality before the even begin to grasp general topological spaces

Hell, imo they should even get Isbell duality before topology

[worthy solar]

maybe a broad question but why does it seem the focus is on commutative rings and unity. Like I guess I dont know what the defintions of say a zero divisor or integral would be if you dropped communative part

Like some of these definitions will have baked in the criteria of communicative ring with unity. So i don't know if there is some defintion of say a... A "blank" is a ring with no zero divsors (but that prior defintion has communative already)

[thorn jay]

Because most theorems about integral domains require commutativity to work

[thorn jay]

You're thinking about a so-called domain

[worthy solar]

Yeah I was guessing that is the motivation for these definitions

[thorn jay]

https://en.m.wikipedia.org/wiki/Domain_(ring_theory)

[thorn jay]

Also, rngs (rings without unity) have little to no commonplace in places where modern ring theory is used (algebraic number theory, commutative algebra, geometry etc)

In my opinion, rings are standard equipped with a unit

[rocky cloak]

How can you tell whether you're doing analysis or algebra?

Well if you're rings have 1 you're doing algebra, if they don't you're doing analysis. If you don't have any rings I feel bad for you.

[thorn jay]

Analysis has infinitary monads and algebra has finitary monads

Wait no thats topology

Fuck

[rocky cloak]

Meh same thing

[thorn jay]

Nah topology is analysis where some category theory is accepted

[south patrol]

they do appear a lot in algebraic stuffs

[south patrol]

Yikes

[thorn jay]

Lmao

[thorn jay]

Oh, where?

[south patrol]

Well like as soon as you start to care about augmented stuff, that becomes equivalent to studying nonunital rings / (or more generally R-algebras)

[thorn jay]

Interesting

I've vaguely heard of augmentation

I suppose ive been doing too much commutative algebra when my R-algebras are unital

Lmao

[opaque finch]

Highly sensible answer.

[cloud walrus]

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

[serene dune]

having a hard time to comprehend this

so basically m \in I_n if n|m^r right

so its like m should have all the distinct prime that is in n ? the degeneracy doesnt matter cause we can always perform m^r to cover the degeneracy

[cloud walrus]

nastasya

[serene dune]

this sounds ?

[serene dune]

so i was wandering if symmetric matrices form a group under multiplication or not, but we have a problem with closure at the first place although the properties seem to give headache at this point.

if we add that structure of commuting do we get a group then, moreover is it recognised as anything important in the literature

i can see the group have some nice properties(having real characteristic values, diagonalisabe, sahring the same eigenspace[?])

although i might have done some blunder in my argument

[rocky cloak]

I mean you can't really "add the structure" of commutativity to symmetric matrices.

But there are groups consisting of symmetric matrices sure, for example diagonal (invertible) matrices

Being commutative is a property not a structure

[serene dune]

oh yeah I meant to say that(i feel), sorry for the language abuse

[rocky cloak]

Meant to say what?

[serene dune]

furthermore restriction, i.e. a nice subset of symmetric matrices

oh now i see the problem

everything has to commute with everything

[rocky cloak]

Yeah

[serene dune]

so scalar \times I_n

bruv

[rocky cloak]

That would be one example sure

[serene dune]

or is it like we can attach individual scalars to all the diagonal entries of I_n

[proper jolt]

what does non commutative mean in this case?

for End(M)

[velvet hull]

if M is a module of R, End(M) can also be viewed as a module over R

with pointwise addition and scalar multiplication

so viewing it as a module over R, End(M) may or may not be commutative

[proper jolt]

so like composition of the maps may not be commutative?

[velvet hull]

no, the composition of endomorphisms is not an operation within End(M) - if you choose to view it as an R-module

alternatively you can also view End(M) as a ring in and of itself, where multiplication is composition

and in that case the ring may or may not be commutative as well

[proper jolt]

as a module over R what is it non commutative over then?

[velvet hull]

what I said is kind of ambiguous, since I don't know if R is comm or not, but it's just a module over R

[proper jolt]

R is commutative

cuz

[velvet hull]

reading it now yeah they are viewing End(M) as a ring

instead of a module

[proper jolt]

can you explain to me

[velvet hull]

I don't see how that would imply R is comm

but you can certainly assume commutativity and the results still hold

[proper jolt]

cuz they say in the above screen shot that u_r is commutative

but compatible

means in this case

[velvet hull]

if all such mu_x's commute with each other then yes that would imply commutativity

I don't see that in the screenshot you sent

[proper jolt]

that R-> R' -> End(m) = R-> End(m)

right?

[proper jolt]

it's the last line, of the first screen shot

[rocky cloak]

End(M) is a ring. It being noncommutative just means there may be two homomorphisms whose composition is different if you compose them in the other order

[proper jolt]

when they say has a compatible R'-module structure

[velvet hull]

this

and if S->R is a ring homo, any R module can be made into an S module

[proper jolt]

does that mean that the scalar multiplication of R-module structure is the same as R'module structure?

[velvet hull]

through the induced ring homo

not necessarily, it just means you can made End(M) into a module over R'

in this case it is "the same" because R' is a quotient of R

[proper jolt]

then why this condition is necessary

from the first screenshot

[proper jolt]

if we only needed to do this then the above statement it seems like won't be necessary, because we can just change the scalar multiplication map of R

but if we apply the above condition then it seems like we are saying that the R-module should act exactly like the R' module

[rocky cloak]

An R-module structure is a homomorphism
R -> End(M)

if a is mapped to 0 this factors as
R -> R/a -> End(M)
Hence giving an R/a module structure

[proper jolt]

so compatibility means R -> End(M) = R -> R/a -> End(M) ?

also $u_R(R)$ is a ring right?

[cloud walrus]

thebirdsandbees

[rocky cloak]

\mu_R(R)? Yes

[regal zodiac]

Suppose that i have a group of order 2n for n >= 3
I showed that G is non abelian, that it contains a subgroup of order 2 which is not normal and a subgroup of order n isomorphic to Z/nZ, is it enough to conclude that my group is Dn ( the Dieadral group with 2n elements ) ?

[tardy hedge]

My prof wants to teach us galois theory from the category theory pov

Im lost asf rn ngl

[serene dune]

[tardy hedge]

Something with preorders

Does this ring a bell for ppl

[thorn jay]

Holy shit thats based

[thorn jay]

It's a strict ordering where incomparability yields an equivalence relation

[tardy hedge]

Dude right when he defined natural transformations is when i checked out

I need to review all of this

It feels like we covered a whole bucketload of shit in the past like 2-3 classes

[thorn jay]

Takes a while to get used to yeah

[velvet hull]

Holy based, please let me read your lecture notes

[tardy hedge]

Sure lol

[thorn jay]

Send them to me too 🥺

I gotta learn galois theory too

[chilly ocean]

not sure where this goes since it's a course in abstract algebra, but it's more of a linear algebra question:

Show that $T(\mathbf{x}) \overset{\text{def}}{=} Ax + b$ is a bijection if and only if $A$ is an invertible matrix. We call such $T$ an $\emph{invertible affine transformation}$.

I have already proven that if A is invertible then $T(\mathbf{x})$ is a bijection by showing that $T(\mathbf{x})$ is an injective and also a surjection. Not sure how to prove $(\implies)$ but I've already shown $(\impliedby)$

[cloud walrus]

s

[velvet hull]

T is invertible iff T-b is invertible. When is T-b invertible?

[glad osprey]

I'm studying galois theory this spring, could you send them to me too?

[chilly ocean]

if and only if?

ah

but why is that true

[velvet hull]

Well, you can quickly check that adding a vector is a bijective operation

[chilly ocean]

T-b is invertible iff Ax = 0 has only the trivial solution?