groups-rings-fields

[wild sapphire]

if there's that kind of relationship, then I'm sure there's something to say with it

[next obsidian]

Yeah it’s just 1/x

Note that the term that used to be degree d

Turns into degree n-d

Since you have x^n times 1/x^d

So it does indeed flip all the coefficients

[wild sapphire]

so, what would that procedure do the the leading coefficient? It feels like it would be inflating it to larger degree, maybe I'm not understanding

[next obsidian]

a_nx^n goes to x^na_n(1/x^n)

= a_n

So it becomes the constant

And I think I have a proof idea

If the reversed polynomial factors as g•h

Then I think you can maybe try to reverse g and h and that factors f

Since like we know that

x^nf(1/x) = g(x)h(x)

Then say deg g = k, deg h = r, then k + r = n

Then if we plug in 1/x to this

We get (1/x)^nf(x) = g(1/x)h(1/x)

Multiply by x^n

f(x) = x^ng(1/x)h(1/x)

Break this up like f(x) = x^kg(1/x)•x^rh(1/x)

So f(x) is factored via reversing g and h and by assumption those were degree > 0

So indeed this shows f wasn’t irreducible to begin with

[wild sapphire]

Damn, that's some good stuff

I think I understand

[next obsidian]

Schmoney

I did a problem which I think involved doing a 1/x thing before

So I remembered the idea

This also doesn’t require Q at all lmao

This just works over any field

[wild sapphire]

Ah, make sense. So it would work in like Z_5 all the same?

[next obsidian]

Pretty sure

The algebra I did didn’t use any specific elements

Just x

[wild sapphire]

yeah, just inverse and multiplicative cancellation stuff, all field properties

[next obsidian]

Yup

The one thing that needs a field and not a ring

Is that if x^nf(1/x) factored as gh

That degree g and degree h > 0

Since that’s used to show when we reverse g and h that this actually shows f wasn’t irreducible

But it might work out even over any ring

Or maybe an integral domain

Idk, I don’t feel like trying it out lol

Actually I’m pretty sure it does

Kekw

But yeah, you can try to find the most general statement if you’d like

[wild sapphire]

yeah, I think it all makes sense. Thanks for the help !

[next obsidian]

[next obsidian]

Thanks everyone who helped me do commutative algebra the past few months. I just finished the last exercise in Matsumura! Big thanks to @gritty sparrow

[robust pollen]

Is the following true:
An $R$-module $M$ is \emph{faithfully flat} iff the functor $M \otimes -$ is exact and faithful.

[cloud walrus]

expectTheUnexpected

[robust pollen]

So of course I'm only asking "if $M$ is flat, then it is faithfully flat iff $M \otimes -$ is faithful."

I guess in one direction it's: If $M \otimes f \colon M \otimes A \to M \otimes B$ is zero, then we have an exact sequence $0 \to M \otimes A \xrightarrow{M \otimes f} M \otimes B \to 0$, and since $M$ is faithfully flat, $0 \to A \xrightarrow{f} B \to 0$ is exact again, whence $f = 0$, establishing faithfulness of the functor.

[cloud walrus]

expectTheUnexpected

[robust pollen]

Ok, so apparently, faithful additive functors between abelian categories reflect SES's, so the other direction follows as well.

[next obsidian]

That’s the definition is it not?

Oh you mean faithful in the general categorical sense

[robust pollen]

yeah, sorry if I was imprecise

[next obsidian]

I think my “definition” of faithful as a functor which reflects SESes might be one I just made up lol

[robust pollen]

like what, "an additive functor between abelian cats is called Ab-faithful if it reflects SES". But then I think that F is Ab-faithful iff it is faithful, basically the same proofs as above for both directions. So your notion seems to work, right?

[next obsidian]

Maybe lol

¯_(ツ)_/¯

I didn’t think about it haha

[cloud walrus]

mns

[simple mulch]

Where S is a subset of an arbitrary Vector space V

[cloud walrus]

mns

mns

mns

[robust pollen]

Your lemma is certainly wrong (since S is just a subset, but <S> is a subspace).
If you take an arbitary element in <<S>>, how would you write it down? Can you massage the expression to get something which looks like an element in <S>?

[simple mulch]

it would be a vector from V which belongs to <S>

[robust pollen]

So, every element in <S> can be written as a (finite) sum of elements in S with coefficients, right?

[simple mulch]

Well.. these lectures notes I am using didn't introduced linear combinations yet (?)

just a sec, I will show you the author argument

[robust pollen]

Wait, how do you define <S> then?

Just as smallest subspace containing S?

[simple mulch]

yes

this was his argument, but I felt something didn't clicked

[robust pollen]

Does <<S>> contain <S>?

[simple mulch]

yes it does

It was the sentence "But then <S> is a subspace of V containing <S> itself".

that I didn't understand

[robust pollen]

Well, <S> is a subspace, so it certainly contains itself, right?

[simple mulch]

well yeah for any space V, we have $V \leq V$

[cloud walrus]

mns

[robust pollen]

But by definition, <<S>> is the smallest subspace containing <S>. So <<S>> \subset <S> \subset <<S>>

[simple mulch]

but why is it the smallest?

oh nvm

The smallest subspace containing <S>, is the subspace of V generated by <S>

Which corresponds to <<S>> such that <S> is contained in <<S>>

so <S> subset <<S>>

[robust pollen]

yes

and since it's the smallest, the reverse inclusion follows

[simple mulch]

So we're claiming <<S>> is the smallest subspace produced by <S> ?

[robust pollen]

No, that's the definition

[simple mulch]

Basically we're showing that the smallest of the smallest subspace generated by S is equal to the smallest subspace generated by S?

[robust pollen]

Yep, that's the assertion $\langle \langle S \rangle \rangle = \langle S \rangle$

[cloud walrus]

expectTheUnexpected

[simple mulch]

Ok, it started to make sense xd

Thank you!

[golden grail]

Okay, this is probably a fairly basic question but I'm trying to prove something for a course in commutative algebra and I'm floundering about. I have a ring A with an ideal I, and I'm looking at their quotient. I've gotten to the point where for $a \in A$ I know that $a + I = 0$. I want to say that then $a = 0$ or $a \in I$, is that true?

[cloud walrus]

TheKikko

[chilly ocean]

just a in I, because 0 is always in I

[golden grail]

Oh, alright. Perfect! Thank you 😄

[dusty tangle]

When can we write $Z[x]/I$ as $Z[\alpha ]$? (My guess is not always as if $I=(p,x)$ then we have $Z/pZ$ which is not $Z[\alpha]$ for any alpha ). In particular does $Z[\alpha]$ always contain $Z$? I am confused about what $Z[\alpha ]$ means in general.

[cloud walrus]

Ashutosh Jangle

[hidden haven]

Is alpha a complex number?

[dusty tangle]

need not be

[hidden haven]

is it given to be an element satisfying certain polynomial equations?

[dusty tangle]

$\alpha $ comes from quotienting by $I$, so if $I=(x^2+1)$ then we have that $\alpha =i$

[cloud walrus]

Ashutosh Jangle

[hidden haven]

So it seems alpha is the equivalence class of x by that definition

[dusty tangle]

yea

[hidden haven]

So Z[alpha] won't necessarily contain Z

because I could be, for example the ideal generated by 2 in Z[x]

Then the quotient is (Z/2Z)[x]

There is a map from Z to Z[alpha] but it may not be injective

the map is
Z → Z[x] → Z[x]/I = Z[alpha]

The composition of these

[dusty tangle]

So for some alpha Z/pZ=Z[alpha] , that seems weird

[hidden haven]

If I is maximal then yeah I guess

Idk what your exact definitions are

[hot lake]

Z[alpha] is either a polynomial ring or there is a ring R containing Z and alpha and then Z[alpha] is the smallest ring containing Z and alpha

[hidden haven]

There is also an abstract adjoining of elements satisfying some relations, like if you say that alpha is an abstract element satisfying some set of relations R, then you define Z[alpha] = Z[x]/(R) and say alpha = [x]

[slate summit]

is it correct to say that in a strict UFD, every element can be decomposed into prime factors, or is it better to say irreducible factors?

[rustic crown]

what's a strict ufd? like one that is not a pid?
don't think that matters because in UFDs irreducibles and primes are the same thing.

[slate summit]

i'm trying to find or come up with a definition of a PID that doesn't mention ideals

like trying to word it without using the ideal concept

[rustic crown]

i think dedekind hasse valuation might be useful...

[slate summit]

hmm just might be

like UFDs have the notion of unique factorization, and Euclidean domains have Euclidean division

so what basic notion would be intrinsic to PIDs?

[rustic crown]

PID's have a form of "euclidean division" in a little weaker sense

so say you have an integral domain D with a valuation say d : D\{0} --> N
and say that a, b are elements such that b is not 0.
a normal euclidean division would say that you can find q such that a-bq is either 0 or an element with smaller valuation
but what if you're also allowed to have a factor with a? that's exactly what a dedekind hasse valuation is, you can find p and q such that ap-bq has smaller valuation or b just divides a.

you can practically use the same proof of Euclidean => PID to show that if D has a dedekind hasse valuation then D is a PID.
start with an ideal I, take the nonzero b with smallest valuation. then for any a in I, ap-bq can't have a smaller valuation as b was the minimum, is b is forced to divide a showing I = (b)

[slate summit]

so like O_k of Q(√-19) has a dedekind hasse valuation but p=/=1?

[rustic crown]

yea, if you could assert that p = 1, then it would just be a euclidean domain, but it's quite well known that O_k has no universal side divisors so can't be euclidean

i remember that for dedekind domains the notions of UFD and PIDs coincide...

[slate summit]

basically Q(√-19) is not eucidean, so there does not exist a q such that d(a-bq)<d(b) for any choice of valuation d?

[rustic crown]

yea

and to show that, one way is to first get some sort of a criterion which is independent of the valuation

universal side divisor is one I know of. if D is a euclidean domain, then look at the non-zero non-unit with smallest valuation, call it u. Then for any non-zero element of a of D, either u divides a or u divides a + some unit

for k = Q(√-19), it's easy to find all units in O_k, so not hard to see that it has no universal side divisors, hence it can't be euclidean

[slate summit]

to sum it all up

UFD - unique factorization

[slate summit]

PID -

And Euclidean domian - p can be set to 1

[rustic crown]

yee

try to find a nice dedekind hasse valuation for a PID

[slate summit]

alright thanks m8

[rustic crown]

[dire summit]

i have this question here on my homework:

Let $G$ be a finite group and $H\leq G$. For $a\in G$, let $f(a)$ be the smallest positive integer $d$ such that $a^d\in H$. Prove that $f(a)\mid|a|$.

[Hint: Consider the cyclic group $\langle a\rangle$].

[cloud walrus]

Snodlop

[dire summit]

i think i worked out the case regarding cyclic groups using the fundamental theorem of cyclic groups, but i'm not sure how to extend that to non-cyclic groups

i asked this question in a help channel here as well, and while typing it up i briefly ruminated on the notion of applying lagrange's theorem (but again, i'm not sure how to accurately apply it)

is there some way to find f(a) with the information that |H| divides |G|?

[viscid pewter]

let a^d be h, and let the order of h be e; is the order of a just de?

[wooden ember]

am i correct for addition in saying that $(b_0 + b_1p + ...) + (b_0'+b_1'p+...) = (b_0 + b_0') + (b_1+b_1')p+...$ where each addition $b_i + b_i'$ is done mod $p^{i+1}$?

[cloud walrus]

𝓛’ai Eric Stove ✓

[wooden ember]

im getting a bit confused frankly

[rustic crown]

nah that's not true

you have to carry over

[wooden ember]

yeah i expected that

but im getting really confused with patching that together with the addition as seen in the inverse limit

cause in the inverse limit you just add component wise

[rustic crown]

right but you're adding different things

so what's an element of the inverse limit?

[wooden ember]

an element of the inverse limit would be a sequence $(a_0,a_1,...)$ where each $a_i$ is a least residue mod $p^i$ and $a_j \equiv a_i \operatorname{mod} p^i$ for $j\geq i$

[cloud walrus]

𝓛’ai Eric Stove ✓

[rustic crown]

right

[wooden ember]

and then i took the hint and wrote each least residue as a base p expansion

[rustic crown]

nice nice

[wooden ember]

so that the morphisms $\mu_{ij}$ have the effect of truncating powers of $p$ that are greater or equal to $i$

[cloud walrus]

𝓛’ai Eric Stove ✓

[rustic crown]

right

[wooden ember]

so then if I write $a_1 = b_0, a_2 = b_0 + b_1p, a_3 = b_0 + b_1p + b_2p^2$ and similarly for the $a_i'$, dont i get that $(a_i){i\in \mathbb{Z}^+} + (a_i'){i\in \mathbb{Z}^+} = (a_i + a_i'){i\in \mathbb{Z}^+} = (\alpha_i){i\in \mathbb{Z}^+}$ so that each component has base p expansion $\alpha_1 = b_0+b_0' \operatorname{mod} p, \alpha_2 = b_0+b_0' + (b_1 + b_1')p \operatorname{mod} p^2, ...$?

[cloud walrus]

𝓛’ai Eric Stove ✓

[rustic crown]

but when you add two base p number you have to carry over the stuff, else that extra part is lost

maybe think it like this...

you have a power series b0 + b1 p + ...

[wooden ember]

but since the b_0 + b_0' for example is done mod p doesnt that carry over get lost?

like i get that you should have a carry over but i dont see where that comes in by arguing purely from the inverse limit construction

[rustic crown]

right, lets go over that

[rustic crown]

like you said, the corresponding element of the inverse limit is just the collection of all it's partial sums...

think about how you write a real number... you just give an expansion in base 10... with as much precision as you want

so think of 'p' as being very smol

so yea... if you have two power series (b_i) and (c_i)... how should be add them?

looking at a precision of p^n, that is looking this modulo p^n we know what the numbers (b_i) and (c_i) correspond to in Z/p^n

so do you now see that all the lower powers of p should have a carry over?

but since we're mod p^n all the things that cross it are kinda lost

but they'll be just recovered again if you started with a higher precision

[wooden ember]

right

so when i was adding the a_1 components i was losing information since i was adding mod p but that's recovered when i add a_2 and a_2' since im adding mod p^2?

[rustic crown]

right

[wooden ember]

right i think i see

[rustic crown]

so if you define the sum with carry over and say you got a power series (a_i)...
then going mod p^n we will definitely have
(b0 + b1 p + ... + b_{n-1}p^{n-1}) + (c0 + c1 p + ... + c_{n-1}p^{n-1}) = (a0 + a1 p + ... + a_{n-1}p^{n-1})
so this has to be the right multiplication

[wooden ember]

im not sure i get what you're trying to say

[rustic crown]

okei so let's take an example

lets say p = 5

and i wanna see what -1 looks like

mod 5, it's 4

mod 25, its 24 = 4 + 4 * 5

mod 125, it's 124 = 4 + 4 * 5 + 4 * 5^2

so -1 in the inverse limit is represented by the power series 4 + 4 * 5 + 4 * 5^2 + ...

[wooden ember]

and adding that to 1 gives you 0 because of the carry over i can see that

[rustic crown]

yee!

it might seem that you're getting higher and higher powers of p... but like we think of p as smol so that's not an issue

[wooden ember]

how do you properly "define" carry over addition for an infinite series though?

[rustic crown]

i think you can define some topology on it where p actually becomes smol

[wooden ember]

yeah the p adic metric does it i believe

[rustic crown]

like do you want to explicitly write an algorithm?

it's not very different from the usual base 10 thing... but in "reverse" direction

[wooden ember]

well im just not sure how formally the exercise wants me to describe addition

and i cant really write an algorithm can I? Since the series are infinite so the process would never end

ah but in the inverse limit each component has a finite expansion

so would i just say we get regular finite carry over addition in the components which translates to infinite carry over globally?

[rustic crown]

by algorithm i meant like how to get a_i if you've computed everything till a_{i-1}

just define it formally, and say it works?

[wooden ember]

ah right

[rustic crown]

so definition of ai would be something like (b_i + c_i + previous carry over) % p and next carry over = floor((b_i + c_i + previous carry over)/p)

[wooden ember]

yeah really not bothered to write that

[rustic crown]

yea lol

[wooden ember]

ill stick to a verbal description

[rustic crown]

arithmetic becomes weird when you have bases lol

a simpler situation occurs when you instead look at the ring k[x] and the prime x

if you do the same construction, you'll see that you recover k[[x]]

[wooden ember]

fair

ugh multiplication is going to be a pain

[rustic crown]

yea lol

but if you think about it... addition and multiplication of real numbers is also very painful

we just don't need to do it lol

[wooden ember]

true

[thin geyser]

is it possible to deduce the addition table for F4 given no other information aside from the defn of a field

[rustic crown]

i think so

it's a field, so it's characteristic must be a prime number...

so you know adding anything to itself in F4 should give 0

and the underlying abelian subgroup has order 4, which forces it to be Z/2Z * Z/2Z

[thin geyser]

wait how does prime characteristic -> x + x = 0, for x in F

was struggling to make that link earlier

[rustic crown]

oh because F4 has size 4... characteristic is the additive order of the element 1, so it divides 4

characteristic is either 2 or 4

but it can't be 4

the characteristic can't be composite follows directly from the definition of a field that every non-zero thing has an inverse

like if characteristic was m*n then (1 + ... +1) m times and (1 + ... +1) n times would multiply to give 0

[thin geyser]

why is it that the characteristic must divide 4? i get everything else

[rustic crown]

have you seen some group theory?

if yes, i'm just using lagrange's theorem with the underlying additive group of F4

order 1 divides the order of the group = 4

if not, then consider this
if you add 1 to every element of F4, you should get a permutation of F4. if the sum of elements of F4 was s, then it should also be 4 + s showing 4 = 0 in F4

[thin geyser]

studied some group theory a year ago but im currently limited to results shown in class, not including group theory

but yeahi see

tysm !!

[rustic crown]

the exact same thing shows that additive subgroup of any F_{p^n} is (F_p)^n

char has to divide p^n, so must be p. Then only abelian subgroups of order p^n such that multiplying stuff by p gives 0 is (F_p)^n

[shell brook]

If G is not a cyclic group then there exists at least one subgroup H ≤ G that is not cyclic.
Is this true in general if we restrict it to proper subgroups that aren't the identity element?

I am sure it is not but I don't know that many examples of groups

lol

[wooden ember]

No it's not true in general

every subgroup of Z_2 x Z_2 is cyclic for example

[shell brook]

ah yes

thank u

[urban acorn]

narwhal gave you a specific counterexample, but here's a cool argument:

if this were true, since H is not cyclic, the proposition applies to it again, and so you can keep getting non cyclic groups as small as you want

but there aren't non cyclic groups as small as you want

[languid walrus]

could this be true if G were an infinite noncyclic group?

nope, https://danaernst.com/an-infinite-non-cyclic-group-whose-proper-subgroups-are-cyclic/

[cloud walrus]

██▓▒░⡷⠂Braham Gödel⠐⢾░▒▓██

[last cargo]

I know it's not reflexive, cuz (1,1) not in omega.

it's symmetric, cuz for each (a,b) in omega, there's a (b,a) in omega.

But how do I prove if it's transitivity?

[hidden haven]

You can interpret the relation as "if x is large, then y must be small" (obviously not very descriptive) and vice versa

With this can you come up with a guess?

[last cargo]

Can I use 3 pairs like a, b, c? Then I sum a with b, and so on. So I prove that it's not transitive?

[hidden haven]

If you are trying to show that it's not transitive, you have to give a counterexample

Here's a hint ||make a and c large, b small||

[hidden haven]

This is pretty easy with model theory. We have groups of arbitrarily large finite size with that property (Zp x Zp) and this property can be expressed in first order logic, so by the compactness theorem there is an infinite group with this property

In fact this gives you any infinite cardinality

First order expressibility is only there if you restrict to abelian groups (this doesn't affect the result, you get infinite abelian examples)

[last cargo]

thanks, moldi

[wooden ember]

nice one

[kind temple]

audi?

[languid walrus]

yoo whats good

[kind temple]

yoo lol

[chilly ocean]

If you have a chain complex (C,d) How is 0 —> Hom(H_n(C);G) —> H^n(C;G) —> Ext(H_n-1(C);G) —> 0 exact

I am reading hatcher and it is a lot to keep track of, Does anyone know of any good explanations online or MSE answers?

[languid walrus]

thanks that's really neat!

[tropic spade]

Sort of a dumb question, but if somebody says "K is a root field over F" they just mean K is the root field of some polynomial over F right?

My book defines the root field of a polynomial a(x) over F but not that other wording.

(I'm not sure if root field is common terminology but I think other books call it a splitting field instead.)

[rustic crown]

Haven't seen that terminology, but yea seems right if you replace it with splitting fields.
There's actually an equivalent description of a splitting field which doesn't require mentioning the polynomial. It's (finite) normal.
Or more generally an extension is normal iff there is a collection of polynomials such that the extension is the smallest field where all those polynomials split completely.

[tropic spade]

That makes sense.

[weary terrace]

Let $V$ be a representation of some group G. We know that $V^{\otimes 2} = S^2 V \oplus \wedge^2 V$.
For which $k$ does $V^{\otimes k}$ has $\wedge^n V$ as a direct summand?

[cloud walrus]

Gromov

[weary terrace]

Put differently, I'm trying to figure out for which $k$ does $V^{\otimes k}$ contains the sign representation as a subrepresentation.

[cloud walrus]

Gromov

[stark sigil]

What if $\wedge^n V$ is a different 1-dimensional representation of $G$? Or is this not considered?

[cloud walrus]

Icy001

[stark sigil]

I can only think of using the character table and orthogonality relations for any given example, but not any criterion in general

[weary terrace]

If I'm not mistaken, since the sign function on a matrix is calculated by a determinant, the sign representation is obtained by $\wedge^n V$ on an n-dimensional space.

[cloud walrus]

Gromov

[weary terrace]

How would you get the result in this way?

[stark sigil]

The determinant of the trivial $n$-dimensional representation would be the trivial 1-dimensional representation, no?

[cloud walrus]

Icy001

[stark sigil]

We just find a formula for <character of V^n, sign representation> in terms of n

[weary terrace]

Sorry, I meant for a nontrivial V. Let us even make it irreducible nontrivial

[stark sigil]

Okay, let G be Z/nZ. Irreducible nontrivial representations have determinant equal to themselves!

[weary terrace]

So what you are saying is that for each V we have k such that $V^\otimes k$ contains a one dimensional representation

[cloud walrus]

Gromov

[stark sigil]

Nope, even simpler

Every irreducible representation of Z/nZ is 1 dimensional

[weary terrace]

No, I mean for a general group.

[stark sigil]

I was just showing you that the determinant of a nontrivial representation isn't always the sign representation

[weary terrace]

Yeah, I get it, I'm just trying to generalize my first question: prove that for each irrep V of some group G have k such that $V^\otimes k$ contains a one dimensional representation

[cloud walrus]

Gromov

[stark sigil]

Okay that sounds better

[weary terrace]

I thought about doing it through the sign representation

[stark sigil]

what if there isn't a sign representation 👀

such as G = Z/nZ for n odd

[weary terrace]

In general I still think we should use the det

thus we have just 1 as a result for the det

[stark sigil]

Ah yes

[weary terrace]

more precisely a constant

[stark sigil]

Maybe it's true that $V^{\otimes n}$ always contains $\wedge^n V$ as a subrepresentation?

[cloud walrus]

Icy001

[stark sigil]

I believe there's a known decomposition of tensor powers into symmetric and alternating powers

and if you just care that there's a 1 dimensional subrepresentation

[weary terrace]

there is the Schur-weyl duality

[stark sigil]

Well, $\wedge^{\dim V} V$ is always 1-dimensional

[cloud walrus]

Icy001

[weary terrace]

but it's an overkill

[stark sigil]

Maybe just provide a G-injection $\wedge^n V\to V^{\otimes n}$

[weary terrace]

correct. So we're left to find that tensor power of V

[cloud walrus]

Icy001

[stark sigil]

Just use the dimension of V 👀

[weary terrace]

How?

[stark sigil]

Didn't you use n as the dimension of V?

[weary terrace]

yeah

[stark sigil]

I was saying there's a decomposition of $V^{\otimes n}$ with one of its factors being $\wedge^n V$ which solves your problem

[cloud walrus]

Icy001

[weary terrace]

I agree.. I mean how do I construct this injecton

i

[stark sigil]

and if you didn't want to use some overkill theorem, I think it should just be possible to show that $\wedge^n V$ G-equivariantly injects into $V^{\otimes n}$

[cloud walrus]

Icy001

[stark sigil]

Maybe try the obvious map

[weary terrace]

I think it's a good time to admit I'm terrible at tensor calculations. Do we know for sure that $V^{\otimes n}$ includes $\wedge^n V$?

[cloud walrus]

Gromov

[stark sigil]

I have high confidence in it but I don't know it for sure

[weary terrace]

What about combinatorial considerations? Can you think of a way to generalize the $V^{\otimes 2} = S^2 V \oplus \wedge^2 V$ case?

[cloud walrus]

Gromov

[stark sigil]

The map $S^2 V\to V^{\otimes 2}$ should go something like $vw\mapsto v\otimes w+w\otimes v$ right?

[cloud walrus]

Icy001

[stark sigil]

and $\wedge^2 V\to V^{\otimes 2}$ is $v\wedge w\mapsto v\otimes w-w\otimes v$

[cloud walrus]

Icy001

[stark sigil]

That sounds pretty generalizable to higher powers

[weary terrace]

I tend to agree.. I mean it is rather obvious by the definition of wedge power that it is contained in the tensor power

[stark sigil]

ye, that's one of the constructions of the exterior power

You can take the tensor power and look only at elements on which $S_n$ permutating the factors acts by sign

[cloud walrus]

Icy001

[stark sigil]

The other construction is to take the tensor power and quotient by the relation $v\otimes w+w\otimes v=0$

[cloud walrus]

Icy001

[stark sigil]

Equivalent constructions in characteristic 0

[weary terrace]

Great! this was very insightful

Thanks

[stark sigil]

Glad to have helped!

[robust pollen]

What is the name for the following algebraic structure:
$S$ is a set with a binary operation $(a,b) \mapsto ab$, and $X \subset S$ is such that $ab, b \in X$ implies $a \in X$.

[cloud walrus]

expectTheUnexpected

[robust pollen]

Actually, in my specific case, $X$ is even a submagma, but $ab, a \in X$ does not imply $b \in X$.

[cloud walrus]

expectTheUnexpected

[rustic crown]

just check if sums, products and additive inverses of symmetric polynomials are symmetric or not.

[wooden ember]

how do we know that 3 does not divide 2-sqrt(-5)?

and similarly that 2+sqrt(-5) is not a multiple of 3

cause they just say "3 doesnt divide the coefficients" but i dont see why that should mean it doesnt divide the quadratic integer

[rustic crown]

a multiple of 3 looks like 3(m + sqrt(-5)n) = 3m + sqrt(-5)(3n)

[wooden ember]

oh right of course we cant get any cancelling since there's no sqrt(-5) term in 3

silly me

[rustic crown]

yee

[wooden ember]

thnx

is there a way in which we can quotient by units to get equality of the gcds?

in general really is there a way to quotient by units for a ring

so of course this wouldnt be a regular quotient of a ring by an ideal since we'd just get the trivial ring

but maybe there's some other equivalence relation such that this notion is useful?

maybe quotienting the additive group by the normal closure of the group of units (viewed additively) 🤔

[rustic crown]

just use ideals lol

ideals are called ideals because they are idealized elements

[wooden ember]

i know that we dont need the gcd's to be equal but im just wondering

[rustic crown]

so the notion you're looking is called "associates"

a ~ b iff a = bu

[wooden ember]

yeah im trying to work with it on paper right now

see if it inherits any structure

[rustic crown]

but if you quotient by this equivalence relation, you'll obviously lose the additive structure

[wooden ember]

yeah

[rustic crown]

1 and -1 are same now

but we still have multiplicative structure

[wooden ember]

you keep the multiplicative structure if the ring is commutative yeah

[rustic crown]

a lot of people like to phrase unique prime factorization theorem modulo this relation

[wooden ember]

does it have any proper uses though?

[rustic crown]

not any serious ones ig? because ideals just do the job far better?

[wooden ember]

cause all you get in the end is a monoid...

[wooden ember]

yeah fair enough

[wooden ember]

we dont have a guarantee that (a,b) is a unit in the more general case??

how can we just divide

[cloud walrus]

mns

[wooden ember]

oop sorry

[simple mulch]

sorry me

[wooden ember]

but no it is not true @simple mulch

[simple mulch]

exactly!

[wooden ember]

however $x\in S \implies x\in \left<S\right >$

[cloud walrus]

𝓛’ai Eric Stove ✓

[simple mulch]

So what can we say when we take x in <S> ?

[rustic crown]

by definition gcd(a, b) divides both a and b. so that makes sense

[wooden ember]

that $x$ is expressible as a finite product of elements of $S$

[cloud walrus]

𝓛’ai Eric Stove ✓

[wooden ember]

fair enough

i must be tired today ugh

[simple mulch]

so, now suppose we want to prove <S> = the set of all linear combinations of elements in S, we can't assume that (?)

[wooden ember]

oh are you working with modules

i thought you meant in the context of groups

but yeah in the context of modules $\left <S\right > =$the set of linear combinations of elements of $S$

[simple mulch]

I am sorry, with modules? This is from lecture notes in linear algebra 😐

[wooden ember]

why is there an emoji

[cloud walrus]

𝓛’ai Eric Stove ✓

[wooden ember]

ah okay so a vector space

it is also true in vector spaces yes

the way to show it is to show that the set of all linear combinations is a subset of $\left < S\right >$, which isnt too hard to show, and that any subspace containing $S$ must contain all linear combinations of elements of $S$

[cloud walrus]

𝓛’ai Eric Stove ✓

[wooden ember]

so you get equality by double inclusion

wait sorry

lemme rewrite that

nvm it's correct

lmao i really am tired

[simple mulch]

Yeah I was thinking about taking v in <S> and show v in W

and the converse

take v in W and show it is in <S>

[wooden ember]

what is your definition of W here?

the set of linear combinations?

[simple mulch]

(W is the set of all linear combinations of elements in S)

[wooden ember]

yeah so the converse is easier

instead of handling elements v of <S> individually it's better to work straight with the definition of <S>

[simple mulch]

my definition of <S> is that it is the smallest subspace of V (the vector space) containing S

[wooden ember]

Namely $\left <S\right > = \bigcap_{S\subset U, U\subset V}U$ where $V$ is the ambient space and $U$ is a subspace of $V$

[cloud walrus]

𝓛’ai Eric Stove ✓

[wooden ember]

so if you show every $U$ contains $W$ you're all set

[cloud walrus]

𝓛’ai Eric Stove ✓

[simple mulch]

uh

[rustic crown]

just say directly? it's clear that W is a subspace containing S. And if a subspace U contains S, then since U is closed under addition and scalar multiplication, it contains all finite linear combinations of elements of S, so it contains W. no need to first show that <S> exists by taking intersections.

but yea it's almost the same lol

[wooden ember]

this is the same lmao

just not writing it out with the intersection

[rustic crown]

yea ignore lol

[simple mulch]

interesting, I wasn't thinking about W as being a subspace

[wooden ember]

well you dont know that it is yet

but you're showing that it is <S>, which is a subspace

[rustic crown]

(nitpick)also it's nicer to write finite instead of all

[simple mulch]

yeah indeed

[wooden ember]

actually nvm you can see that W is a subspace straight away as det said

but i mean the fact that it is a subspace isnt particularly important for the proof, only that it contains <S>

[rustic crown]

wait what?

[wooden ember]

you get that W contains <S> by showing that every subspace U containing S contains W

and the reverse inclusion is simple

i guess it's all a matter of style really you could also say W is a subspace containing S so it must contain <S> but you first have to show W is a subspace

nvm im an absolute idiot

listen to det

stop listening to me

idk if it's meds but im not thinking straight obviously

[rustic crown]

take care uwu

[simple mulch]

I am showing $\langle S \rangle \subseteq W$. Note that $<S>$ is the smallest subspace of V containing S. By definition of subspace it contains finite linear combinations of S. But I can't assume straight away that it means that it is a subset of W (?)

[cloud walrus]

mns

[simple mulch]

Oh yeah, take care eric!

[wooden ember]

lmao eric

call me narwhal this is a joke name

[simple mulch]

👀 ok narwhal xd

[wooden ember]

there back to normal

[rustic crown]

i thought that eric was your real name for a second

[wooden ember]

nope

it's just the other day autocorrect corrected derivative to eric stove

[simple mulch]

Can you check my answer above?

[rustic crown]

yea anyway, so to show that <S> is a subset/subspace of W, all we need to do is show that W is a subspace containing S. Since <S> is the smallest such thing, we'll get <S> is contained in W.

[simple mulch]

Oh I see, W is a subspace of V because for all v,w in W and k,f in F(ield) we have kv+fw in W

we showed closure of addition and scalar multiplication

[rustic crown]

oh we need a little stronger... v and w should be allowed to come from all of W, and not just S

basically, "sum of two finite linear combinations is a finite linear combination" same for scalar multiplication

[rustic crown]

yee now it works

[simple mulch]

and by taking v,w in W we're using those linear combinations from S

[rustic crown]

yep

[simple mulch]

I see, so W is a subspace containing S. since <S> is the smallest subspace containing S, we deduce <S> subset W ?

For the other inclusion, we shall prove $W \subseteq \langle S \rangle$. Note $\langle S \rangle$ is a subspace, thus it is closed under addition and scalar multiplication. Since $S \subseteq \langle S \rangle$, then the finite linar combinations of S are elements in $\langle S \rangle$. But then $W \subseteq \langle S \rangle$.

[cloud walrus]

mns

[rustic crown]

yep

[simple mulch]

this was very instructive. Thank you very much and @wooden ember too

[rustic crown]

try multiplying s_1 and s_2

s_2 = sum of T_iT_j for i < j

yee

it would sum of T_iT_jT_k where i, j, k vary over 1, 2, ..., n with i < j

notice this equals (the required sum) + 3 * s_3

to get a particular T_a T_b T_c for a<b<c there are 3 ways to get it, (i, j, k) can be (a, b, c) or (a, c, b) or (b, c, a)

[bronze jay]

If I wanted to show a subgroup N is normal in G, i have to show that gNg^{-1}=N for all g in G. But someone said that it is enough to show that it holds for the generating set of G and N, so let x be in the generating set of G and n be in the generating set of N, then show that xnx^{-1} \in N.

Why is this possible?

[rustic crown]

first notice that it's enough to show that gNg^-1 is contained in N for all g.

this is because you can use it for g^-1 to say g^-1Ng is contained in N which is equivalent to N is contained in gNg^-1 giving us the reverse inclusion

so it's enough to check that for each g in G and n in N, gng' is in N

i'll use ' for inverse cuz it's easier to type

now notice it's enough to show that gng' is in N where n belongs to a generating set of N

this is because any element of n then can be written as a product of elements or their inverses from the generating set

similarly just reduce it to generating set for G

[simple mulch]

Another question. Assume V is a vector space over an infinite field F. Assume the "initial vectors" (if that means anything (?)) in V are a and b. R is an infinite field, so let V be a vector space over R. Can't we build two proper subspaces, say A and B, such that A contains all those vectors with a? and B all those vectors with b? But then we could create a finite union of its proper subspaces?

This is because I am supposed to prove: Letting V be a vector space over a infinite field F. Show that V cannot be written as a finite union of its proper subspaces

[bronze jay]

Is it because if you have n1 n2 in the generating set of N, then you have gn1n2g'=gn1g'gn2g' in N?

you can always just insert g'g between the elements

[rustic crown]

yep, and also with inverses, you can pull out gn'g' = (gng')'

[rustic crown]

what do you mean "contain all vectors with a", which one will contain a + b? what about a + 2b?

if you say both are in the "thing of a" then their difference should also be in "the thing of a" but that's b...

try to show this... if V is a union of n proper subspaces, then it's also a union of n-1 proper subspaces.

[long obsidian]

Can anyone give me an idea how to approach this problem?

[median pawn]

Hey

So I've been reading about free abelian groups, and just want to make sure I understand things right

[cloud walrus]

Hausdorff

Hausdorff

[median pawn]

Is this the right way of thinking about it? That is, multiplication by (-1) in the formal sum is the multiplicative inverse? and + is "multiplication"

[rustic crown]

yep... but just a smol thing about writing it, it's usually nicer to write sum (n_i b_i) where n_i are in Z, and all but finitely many are 0. saves you some boring cases to check.

[median pawn]

yes okay! thank you

[rustic crown]

say f is the inner automorphism which we get from the element alpha in Aut(N). so alpha : N --> N.
just write out what the homomorphism condition should be, and work your way back.

spoiler : ||use the map given by : (n, h) maps to (alpha(n), h)||

[long obsidian]

Thank you! I really appreciate it

[plucky flicker]

Hi guys! How can one prove that in an arbitrary ring R the intersection of all maximal left ideals is a two-sided ideal?

I mean, if R is a ring with identity it's obvious

cuz then it is the Jacobson radical, which is a two-sided ideal

but what if we don't have an identity?

it's obviously a left-ideal, but why is it a right-ideal?

[rustic crown]

does the usual proof fail?

not sure... i haven't worked with Rngs at all

but feels like it should be true

[plucky flicker]

what do u mean?

on usual proof

[rustic crown]

if R is a ring, then J(R) can be defined as the intersection of annihilators of all simple R-modules

simple R-modules are just R/(maximal left ideal)

so that's same as intersection of all maximal left ideals

now if an element x is in J, then it kills every simple V. but then clearly for any a, b in R, axb kills V

like take v in V, x will kill bv in V. hence axb will kill any simple R-module.

so axb should lie in J

[plucky flicker]

but this is the case when 1 is in R

right?

[rustic crown]

yea i'm trying to see if we ever used it

[rustic crown]

this step looks a little fishy

okie this might be wrong... but i think we can just avoid that

[plucky flicker]

well, we can embed our rng into a ring with identity but idk if this helps

[robust pollen]

https://math.stackexchange.com/questions/1516990/category-of-modules-over-a-ring-without-unity
here it's said that modules over a Rng are isomorphic to modules over the Dorroh extension

[rustic crown]

don't have any experience with Rngs... so can't say if anything works

[plucky flicker]

yeah, i'm thinking bout the Dorroh extension as well

[plucky flicker]

@rustic crown are you here? i have the solution, let me know if you are interested in

[rustic crown]

yee i'm interested

[clever falcon]

Guys, a little bit off-topic question: where should I ask a question about an optimization problem? I cannot find any channel related to optimization theory

[plucky flicker]

maybe in applied computational mathematics

[clever falcon]

Okay, thanks 🙂

[plucky flicker]

@rustic crown

[rustic crown]

,rotate

[rustic crown]

,rotate

[rustic crown]

oooh pretty neat

[robust pollen]

But isn't this exactly like the proof for unital rings?

[plucky flicker]

Proving that the intersection of all maximal left ideals is a two-sided ideal is the same for unital rings, so we don't have to distinguish the proof

We don't have to complicate our life with the Jacobson radical

When is saw the problem I thought about the Jacobson because that's precisely the intersection of all maximal left ideals, but this is true only for rings with identity

[robust pollen]

Right. But since I dont know about rngs, I guess I'm worried about whether the expression "intersection of maximal ideals" makes perfect sense, or if "ideal" means something else.

[plucky flicker]

In fact in the case of rng-s, the intersection of all maximal left (resp. right) ideals is contained in the Jacobson radical of the rng

[robust pollen]

e.g. wikipedia says rngs don't need to have maximal ideals (then of course the statement that the intersection of maximal left ideals is a right ideal is true lel)

[plucky flicker]

Well we don't really need an identity to define ideals

[robust pollen]

True

[plucky flicker]

maximal ideals have the extra property that they are the biggest

in the sense of inclusion

[robust pollen]

Right, so what you wanted to prove was just " intersection of max left ideals is right ideal, nothing else?

[plucky flicker]

yes

I just complicated with the Jacobson stuff, just forget that 😄

[elfin patrol]

I’m given that for o(x)=n for x in a group and n< infinity. Also know d is a positive division of n and I’m trying to prove o(x^d)=n/d. Not sure if my proof makes sense though. First said o(x^d) implies (x^d)^y=e and choose y=n/d then x^n=e as given. To show n/d is minimal I did contradiction and assumed there was c<n/d wheee o(x^d)=c. Then o^(cd)=e contradicting the given x^n=e

I feel like I messed up somewhere though since I never used fact that n/d is integer

[robust pollen]

"First said o(x^d) implies (x^d)^y=e and choose y=n/d then x^n=e as given"
whats the first "implies" supposed to mean?

[elfin patrol]

Wasn’t sure how to phrase it but finding o(x^d)=c means (x^d)^c=e and just chose c

[plucky flicker]

actually u did

when you chose y to be n/d

because the order can't be anything else than an integer

[elfin patrol]

Oh yeah ig u right. I’ll mention that explicitly in write up, but other than that does proof seem valid?

[plucky flicker]

yes

I guess it's okay

[elfin patrol]

How could I make it better

[plucky flicker]

in the last row x^(cd) = e u wrote o^(cd) = e

[plucky flicker]

what do you mean?

[elfin patrol]

U said I guess it’s okay does that mean it could be written better or something. I feel like it could be better so open to criticism

[plucky flicker]

Ohh, I meant it's totally fine 😄

[elfin patrol]

Oh lol thanks 🙌

[robust pollen]

So, the "then just choose c" is maybe not so good, because you don't choose anything there

[elfin patrol]

How could I phrase it better to show (x^d)^(n/d)=e?

[prisma ibex]

Is there a particular reason you are thinking about rings without identity?

[plucky flicker]

yes, I'm a masochist lol jk jk

[plucky flicker]

I'm a theoretical mathematician master student, and I have a course where we study ring and module theory

[robust pollen]

maybe make the logic a bit more clear, e.g. "we want to show that o(x^d) = n/d. So we first show that (x^d)^(n/d) = e, and then we show that n/d is the minimal integer...." yadayada
but I'm also just nitpicking 😬

[prisma ibex]

I mean is the course specifically studying rings without identity? that seems really nonstandard

[plucky flicker]

It's enough if you say that (x^d)^(n/d) = e following from the calculation rules in the group

[plucky flicker]

nope, it was just an exercise, usually we assume that our ring has an identity

[prisma ibex]

right yea I haven't seen a single important use for rings without identity in modern mathematics so it's always kinda funny to see textbooks that reference these

[plucky flicker]

yea, sure

not really useful, but interesting

... to see how a single 1 can change the story 😄

[wooden ember]

apparently rng's are useful in functional analysis from what i hear

[prisma ibex]

yea practically the only interesting examples of rngs I've seen appear "in nature" are things like rngs of compactly supported functions

[next obsidian]

https://tenor.com/view/baby-crying-baby-crying-gif-5943733

When your ring doesn’t have 1

[prisma ibex]

though in the situations where these things come up, it's not like module theory and the like is used in any kind of serious way

[robust pollen]

But the point of maths is not to be applicable, so

[next obsidian]

nG means applicable to math itself

[long obsidian]

Do you think my question reads as "an inner automorphism of the automorphism group" or "an inner automorphism inside the automorphism group"?

[rustic crown]

of

f o phi should be a map from H --> Aut(N)

[robust pollen]

imagine dealing with Lie algebra objects in Set

[long obsidian]

Thank youuu 😢

[rustic crown]

aww why sad?

[bronze jay]

Does the image of a homomorphism form a subgroup?

[chilly ocean]

yes

inb4 first iso thm

[plucky flicker]

yes, the image and the kernel are two "classical" subgroups

[bronze jay]

ah

thanks

[robust pollen]

Try to prove it 🙂

[bronze jay]

Ok, let G and H be groups, and f a homomorphism from G to H. Define K = {f(g) | g \in G}, we show that K is a subgroup of H. We see that K is included in H trivially per definition. K is nonempty since f(1) is in K, for any two elements a,b in K, we can write a = f(x), b=f(y) for some x,y in G and we have that ab^{-1}=f(x)f(y)^{-1}=f(xy^{-1}) \in K, therefore K is a subgroup.

[waxen hedge]

Try to prove that the inverse image is a subgroup is a subgroup ! This give a short proof of the kernel being a subgroup

[bronze jay]

that the image of f^{-1} is a subgroup?

[waxen hedge]

Yes

But it works for any subgroup of H

[waxen hedge]

In this context it's called the inverse image (of {1} )

[robust pollen]

does this work generally in the universal algebra context?

[waxen hedge]

What do you mean by universal algebra ?

[plucky flicker]

well

[robust pollen]

https://en.wikipedia.org/wiki/Universal_algebra but I dont know anything about that

other than the name

[plucky flicker]

by category theory it does, but that's like shooting a mosquito with a bazooka

[bronze jay]

Maybe something like this?

Consider f^{-1}(H), we will now prove that it is a subgroup. We know that f^{-1}(H) is nonempty, since we have that f(1) = 1 \in H and therefore 1 \in f^{-1}(1) \subset f^{-1}(H). Furthermore for any two elements x, y \in f^{-1} we can write a=f(x), b=f(y) and observe that ab^{-1} = f(x)f(y)^{-1}=f(xy^{-1}) and therefore we have that xy^{-1} \in f^{-1}(ab^{-1}) \subset f^{-1}(H). We conclude that f^{-1}(H) is a group

[waxen hedge]

Perfect

[robust pollen]

what do you mean? What is the "inverse image" of a morphism in a generic category?

[waxen hedge]

Well in a general category you don't have a notion of a subobject

[waxen hedge]

It works with the usual algebraic structures yes (groups, rings, vector spaces, modules, algebra over a ring, ..)

[robust pollen]

don't you? A subobject is just a monomorphism (or rather an equivalence class of monomorphisms), no?

[waxen hedge]

Oh right maybe

[robust pollen]

sorry, not equivalence class though, isomorphism class says nLab

[waxen hedge]

In that case with this definition I don't know what's left to prove

[robust pollen]

uhhh... lemme jus uh. Confused. prove what?

[waxen hedge]

That the image of a subobject is a subobject

Because there is not a notion of "image" anymore

[robust pollen]

Oh, I mean't that the preimage (whenever this notion makes sense) of a subobject is a subobject. I guess like this:
\begin{tikzcd}
A \arrow[r, "f"] & B \ & I \arrow[u, hookrightarrow]
\end{tikzcd}

[cloud walrus]

expectTheUnexpected

[robust pollen]

And so you pullback the inclusion I -> B along f

to get a monomorphism into A

So f is your homomorphism in that picture, and I is a subobject of B. Would the pullback then be the "preimage of I"?

[waxen hedge]

Sound like a reasonable definition but I don't know more !

If I remember well in that case you'll get a monomorphism from your pullback to A right ?

[robust pollen]

Yes, monomorphisms are preserved by pullbacks in that sense

And actually, if you do all this in Grp for example, then the pullback will actually be (isomorphic) to f^{-1}(j(I))

I just computed it and I find it really fascinating

[waxen hedge]

That's funny

[robust pollen]

This is my proof that the kernel of a homomorphism f is a subgroup, because it is just the pullback of the trivial subgroup * along f

[long obsidian]

Any idea how I might show $S_{4} \cong \hbox{aut}(S_{4}) $?
A proposed isomorphism is given by conjugation $f(g)(\tau)=g\tau g^{-1}$.
I'm not even sure how to show this is injective lol

[cloud walrus]

fajitas

[long obsidian]

I'm a bimbo and algebra is hard

[waxen hedge]

Try to show that the kernel is the center of S_4

[long obsidian]

This is equivalent to injectivity?

[waxen hedge]

Yes

For any group G, you have a morphism from G to Aut(G) given by the conjugation

And the kernel of this morphism is exactly the center of G

[waxen hedge]

Well not exactly, because you'll have to show that the center of S_4 is trivial

[long obsidian]

Ahhh that makes sense actually

[uncut girder]

What does Sp(4,Z) look like?

[past temple]

what is the multiplicative identity in the group ring ZG ?

[robust pollen]

1

The group ring is a free Z-module with basis G.
Or, if you want, every element in ZG can be written as a finite Z-linear combination of elements of G

then multiplication is defined component-wise, so the identity of the group is the identity of the group ring

[prisma ibex]

What do you mean

[uncut girder]

what kind of matricies are they

[prisma ibex]

4x4 integer matrices

Namely those which preserve a standard symplectic pairing on Z^4

[uncut girder]

Is there a standard generating set

[prisma ibex]

There are some standard generators you can write down sure, depending on how you present the group

[wise igloo]

hey, sorry to interrupt but I saw a video that said we cant go beyond the complex numbers because there are no algebraic solutions to polynomials that are complex... sooooo what about quaternions?

[chilly ocean]

I think you are talking about algebraic extensions?

[wise igloo]

field extensions I believe

[prisma ibex]

The quaterions don’t form a field and there’s not really an analogue of Galois theory in that setting

[past temple]

how are there no algebraic solutions to polynomials that are complex?

[wise igloo]

there are

[past temple]

there are always complex solutions to polynomials that are complex lol

[prisma ibex]

In particular talking about solutions to polynomial equations isn’t as well behaved

[past temple]

thats the fundamental theorem of algebra

[chilly ocean]

Complexes are algebraically closed so every extension is itself is what you are referring to I think.

[wise igloo]

oh I meant beyond complex numbers sorry

[past temple]

? that still doesnt make any sense

polynomials with complex polynomials always have complex roots

[viscid pewter]

all the roots of complex polynomials are complex

[wise igloo]

I know that

but not like quaternion solutions

that's what I meant to say

[wise igloo]

but anyway this is all I needed, thanks

[chilly ocean]

What are you thinking for the generating set? Are you computing generators? I am not sure how to go about that, so what are you doing?

[past temple]

u mean polynomials with quarternion coefficients?

[wise igloo]

nvm

[past temple]

like if u take ZQ_8[x]

the group ring

or RQ_8

[wise igloo]

nevermind

[past temple]

lol ok

[past temple]

how are these two definitions different?

[chilly ocean]

When I looked up Sp(4,Z)and definition for symplectic group it looks like special linear group

[past temple]

if its a free Z-module with basis G

doesnt that mean that every element can be written as a Z-linear combination

of elements of G?

[robust pollen]

they're not different, I was just being more explicit

So, you have that a in ZG is written as sum_{g\in G} a_g e_g where e_g are the generators and a_g are in Z , finitely many non-zero.
Multiplication is defined on generators e_g e_h = e_{gh}, which is enough by standard linear algebra.
And so you see that e_1 acts as the multiplicative identity of your group ring

(which is actually even a Hopf algebra in Z-mod)

[past temple]

ah okay

[prisma ibex]

@uncut girder if you want a particular presentation you can read

BENDER, P. “Presentation of Symplectic Group Sp(4,Z) with 2 Generatrices and 8 Definitive Relations.” Journal of Algebra 65, no. 2 (1980): 328-331.

But having these presentations honestly isn’t very useful unlike for SL_2(Z)

[chilly ocean]

How did you find that? Google?

[prisma ibex]

Yes lol

The more useful thing is being able to list the conjugacy classes of torsion elements and their characteristic polynomials

That’s also a huge pain to do but that at least gives you more useful information

Idk that there is much you can do with the actual presentation that is very useful

[barren sierra]

Can someone confirm this?
Say we have F = Z/2Z. Then F(x) = Frac(F[x]) is a countable infinite field with characteristic 2? But then F((x)) = Frac(F[[x]]) is uncountably infinite by a diagonalization and inclusion argument from F[[x]]?

[waxen hedge]

Seems good for me

[barren sierra]

Let's go

[next obsidian]

I want to point something out that is incredibly cursed

But it may save you some trouble some day

[barren sierra]

[next obsidian]

F((x)) = Frac(F[[x]]) is indeed true where the left hand side is formal Laurent series

So power series with negative exponents, but only finitely many negative exponents have non-zero coefficient

HOWEVER

If A is an integral domain

It is not true that Frac(A[[x]]) = Frac(A)((x))

[barren sierra]

Oh in my class we defined F((x)) as Frac(F[[x]])

[next obsidian]

Okay

[barren sierra]

wdym

[next obsidian]

The field of fractions

Of formal power series over A

Is not Laurent series over Frac(A)

[barren sierra]

Ah

[next obsidian]

It embeds into this

But it is not always everything

[barren sierra]

I haven't learned what Laurent series are lmao

[next obsidian]

Literally just

[barren sierra]

But I shall look into this

[next obsidian]

Sum_-infinity^infinity a_ix^i

[barren sierra]

Oh

[next obsidian]

Except you require there to be only finitely many nonzero a_i for i < 0

Else when you multiply…

[barren sierra]

Got it

Yea

[next obsidian]

Uh oh infinite sums

[barren sierra]

My favorite

[next obsidian]

These show up in complex analysis

Anyway, I just want to warn everyone about this now

Because I spent like 8 hours fucking

[barren sierra]

O no

[next obsidian]

Screwing around with these things

[barren sierra]

I feel that

[next obsidian]

Put this in that like

“Will remember something about it if I ever deal with this” bin

I know we all have it

“Wait… I swear I remember hearing something about this…”

[barren sierra]

I shall start that list and add it

[barren sierra]

how in god's name does the hint pertain to the problem

or well

[hidden haven]

Lol it seems like they want you to give Zp as the example in the previous problem

[barren sierra]

the hint about using problem 5

to solve 6

[hidden haven]

Or something like that 😵‍💫

[barren sierra]

this is talking about using 6 to solve 5 I guess? But that's not even what it says

like

[hidden haven]

No it doesn't even with lol

[barren sierra]

how does "yea some rings have polynomials who have a derivative equal to 0" relate to "oh yea Euclid's Lemma"

_>

[hidden haven]

Work*

[barren sierra]

also yea

you need a ring that isn't a field

cause you need zero-divisors

[hidden haven]

Ye

[barren sierra]

for 5

[rustic crown]

you can use fields as well

[barren sierra]

I used 2x^3 in Z_6[x]

[barren sierra]

sounds harder lol

anyways

[rustic crown]

x^5 - a in Z_5[x]

[barren sierra]

oh lol

yea

anyways

how the fuck does that have anything to do with 6

like ??????????

[rustic crown]

btw are you sure by previous they mean "immediately previous"?

[barren sierra]

here are the other previous

but even still

[rustic crown]

yea i think they mean 3

[barren sierra]

yea

[rustic crown]

you know Zp is an integral domain, so Zp[x] is too.

[barren sierra]

yea

I mean divisibility isn't well defined for non-integral domains

[rustic crown]

but is it needed?

[bronze jay]

if you have a subgroup N in G, and you show that x^{-1}yx \in N for x in G and y in N, can you necessarily conclude that xyx^{-1} \in N?

[next obsidian]

Doesn’t sound true to me

But I don’t have a counterexample off the top of my head

[bronze jay]

what if x^{-1}yx=y^{-1} ?

[next obsidian]

Idk, still doesn’t feel true to me

But I might be wrong

[bronze jay]

Ok let x' denote x^{-1}, we have that x'yx=y', but then
xyx'=x(y')'x'=x(x'yx)'x'=x(x'y'x)x'=(xx')y'(xx')=y' \in N

I think it's true

[hidden haven]

Yeah it's true, since inversion is a bijection

Assuming you mean "for all y in N, x in G"

Rather than some particular ones

In your proof i didn't get how
x(y')'x' = x(x'yx)'x'

[rustic crown]

i think you might need the group to be finite for that

[bronze jay]

we have assumed x'yx=y' to begin with

so i just replaced y' with x'yx in the parentheses

[hidden haven]

What do you mean by this? Do some notions of divisibility clash?

[hidden haven]

Where

[bronze jay]

Here

I just said we assume that x'yx=y'

[hidden haven]

oh lol I see

But why do we have that

[bronze jay]

it's a part of the problem lol

[hidden haven]

oh epic lol

[rustic crown]

as in if you know for all y and a particular x, then
x'yx in N, then (x'^2)yx^2 in N... and so on, if order of x is finite, we can write x' = x^k, so it'll work out

[hidden haven]

Who are you responding to, det

[rustic crown]

this 😶 , without the assumption

[hidden haven]

oh ok

[old zenith]

Is there actually any difference between $\mathbb F[x]$ and $\mathbb F(x)$? I've seen both used to denote field extensions and was wondering if there was anything actually different about them that I hadn't realised.

[cloud walrus]

A Fellow Human

[upper pivot]

so F[x] is the ring of polynomials over F

while F(x) are all the rational functions

so 1/x would not be in F[x] but it would be in F(x)

for instance

[old zenith]

oh

wait that doesnt matter if we're talking about field extensions of nth roots of rationals to Q right

[upper pivot]

yeah in that case it would be the same

if x is algebraic these two are the same things

[old zenith]

thx

[wooden ember]

wasnt bothered to do gaussian integers gcd computations so i wrote a program for it

what a waste of my time in retrospect

[bronze jay]

if a homormorphism is injective between two groups of the same order, can we necessarily conclude that it is also surjective, hence an isomorphism?

[chilly ocean]

ye

same finite order ofc

[bronze jay]

what is the proof?

[chilly ocean]

obvious

[south patrol]

An injection between finite sets of the same cardinality is always a bijection

Informally if the domain of the function has n elements then by injectivity its image has n elemets

And is thus surjective

[robust pollen]

And a bijective group homomorphism is automatically an isomorphism, meaning that its inverse is also a homomorphism (prove this if you've never seen it)

[bronze jay]

a bijective homomorphism is per definition an isomorphism

[wooden ember]

that's a definition you see in group theory yeah

more generally an isomorphism is a homomorphism that admits a homomorphism as an inverse

those are equivalent for groups though

it's a cat theory thing really: in that language bijections are isomorphisms of sets

but just stick to isomorphism = bijective homomorphism for now

[waxen hedge]

What do we get then if the take Frac(A[[X]]) ?

[hot lake]

something smaller

for example, for Frac(Z[[x]]), I think you can have only a finite number of primes in the denominators

while in Q((x)) you can have 1/2 + 1/3 x + 1/5 x² + 1/7 x³ + ...

[wraith obsidian]

Weird curiosity, but does anybody have an idea how we could prove associativity of (ℕ, +) without using induction by referencing something more high-level?
Addition here is defined inductively via 0 + n := n, s(k)+n := s(k+n)
My idea was to use the left regular representation somehow + then use the fact that function composition is associative.
If λₖ is left addition with k, then the inductive definition of addition precisely ensures that λ₀ = id and λₛ₍ₖ₎ = s◌λₖ

So… doesn't that mean that addition from the left is in some sense a homomorphism from the universal (nil, succ)-algebras (ℕ, 0, s) → (End(ℕ), id, s_∗), where s_∗ is postcomposition with s: ℕ→ℕ?

Ah, that would just be the unique homomorphism due to (ℕ, 0, s) being the iniitial (nil, succ)-algebra by definition

Now the question would be why $λ_k \circ λ_n = λ_{λ_k(n)}$

[cloud walrus]

lux

[waxen hedge]

That's weird

[wooden ember]

so im really struggling with this one. Ive drawn the elements of O in C as hinted to get a lattice generated by 1 and (1+sqrt(D))/2. So now the idea is to perform euclidian division on gaussian integers, we perform regular division in Q(sqrt(D)) (say a/b) then get the distance d to the closest point on the lattice so that we have a/b = x+y(1+sqrt(D))/2 + wd where w is some complex number of norm 1 and multiply everything by b to get a = (x+y(1+sqrt(D))/2)b + r with r=bwd and if we can get d<1, then |r|<|b| and so everything works out. My issue is in getting this distance less than 1: the hint says the distance should have (1+|D|)^2/(16|D|) as an upper bound but i fail to get this expression anywhere: geometrically ive taken the maximal distance to a point on the lattice to be the maximum distance of an incenter of the iscoceles triangles that make up the lattice to its vertices, and i get max(sqrt(|D|)/3, sqrt(9+|D|)/6) for that

originally i wasnt even thinking in terms of the lattice and just had a/b = x+y(1+sqrt(D))/2 for some x, y in Q and then rounded x and y to the nearest integer but that gave a bound on d of (9+|D|)/16 which only works for D=-3

cause that was more like the proof in the text which did that for Z[i]

it's a really nice exercise and i feel im close but i dont see what im missing

[hot lake]

I find the same thing as them by drawing some triangles

I'm not sure what the incenter has to do with anything

[next obsidian]

A subfield of Laurent series. It’s something about coefficients

There’s some MSE thing about it

[trim grove]

How can i show that a $x^3$ an isomorphism from group G to G ,then G is abelain? Any hint .

[cloud walrus]

Algebra

[scarlet estuary]

that isnt true in general

[hot lake]

that's one of the "write a lot of things until magic happens" exercises

[scarlet estuary]

do you have any other facts about G?

like 3 doesnt divide |G|?

oh wait

isomorphism

shit

LMAO

ignore me, im being dumb

yeah just try playing around with stuff

[final oasis]

x^e e a power of 3

[trim grove]

Can you please just give a starting hint ,so that i can proceed further

[final oasis]

if continue applying the iso do we get back to G

[scarlet estuary]

it suffices to prove that squares commute with elements in G, i.e. ab² = b²a for all a, b. why?

||(ab)³ = a³b³ since cubing is an isomorphism, so ababab = aaabbb, and therefore (ba)² = baba = aabb = a²b², and if squares commute we can then write (ba)² = a²b² = b²a² and cancel this to just ab = ba||

so that tasks you with proving arbitrary squares commute

to do this, take b²a and write a as the image of some element under your isomorphism

im not sure of the details from here on but that probably works

if you fiddle with it enough

[trim grove]

So i have to prove that square commute?

[final oasis]

rambling (bit of a tangent),
if G is finite then Aut(G) finite , let f be the iso taking x to x^3 , then f to some power is the identity map id , that is x^e = x , e = 3^o(f), then, o(x) | e-1.
so is G iso to a multiplication group of a finite field?
Is this wrong ?

[chilly ocean]

What are automorphisms from Z/4Z to Aut(N) where N is normal subgroup of order 7 contained in group of order 28?

I remember some fact about homomorphisms from Z/nZ to G having images of a subgroup in G with order n, but this might be for G abelian.

[hidden haven]

Homomorphisms not automorphisms

N being contained in some order 28 group is irrelevant. There is only one group of order 7 upto isomorphism

Once you know what that group is, things become easy

[chilly ocean]

Sorry I meant what are homomorphisms between Z/4Z and Aut(N)

I think i figured some parts out

For homomorphism f, f(1) needs to have order 4 in Aut(N)

[hidden haven]

This is also not true in the way you've said it. Homomorphisms from Z/nZ to G correspond to order d elements of G with d | n

[hidden haven]

Not necessarily 4

It just needs to divide 4

[chilly ocean]

But doesnt f(0)=4f(1)=0 in Aut(N)?

[hidden haven]

Yes, that only gives you that it divides 4

Order could be smaller

Order 2 elements also become identity when you put an exponent of 4

[chilly ocean]

Oh I see, thanks so much!

[wooden ember]

I think it’s me being bad at geometry: wouldn’t the max distance of the incenter to the triangle vertices give the maximal possible distance?

How did you go about the geometry?

[past temple]

somehwat basic linear algebra question

that im struggling to understand

given vector spaces U, V, and a linear map A: U -> V, how do i construct a linear map A**: U** -> V**

[hot lake]

well I think I'd look at the centre of the circumscribing circle ?

but maybe they're the same for isosceles triangle idk

[hidden haven]

Suppose P is in U**, ie P takes some f in U* and gives a constant. A(P) needs to take a map g in V* and give a constant. Do you see any ways of taking this g in V*, turning it into an f in U*, then applying P?

Discord and asterisks 🤧

[golden pasture]

easier jus from A: U\to V, you have A^*: V^*\to U^*c which is the "transpose" and then taking dual again you get A^**: U^**\to V^**

[hidden haven]

Funky 😌

[golden pasture]

in matrix-language this is just transpose of the transpose

[past temple]

hmm

im guessing P maps f in U* to a constant by evaluating it somewhere?

[robust pollen]

$\infty \times \infty$-matrices yuss

[cloud walrus]

expectTheUnexpected

[past temple]

but how do i determine where to evaluate that f

[hidden haven]

What do you mean by that?

[golden pasture]

well

[golden pasture]

what is A^* here

like explicitly

[past temple]

it like

sends an f: U -> F

to f compose A

i think

[golden pasture]

a bit

flipped

[past temple]

oh wait

f: V -> F

sry

[golden pasture]

A^* maps V^* to U^*

yup

so lets say you have a f: V\to F, what is A^*(f)

[past temple]

its a mapping from U to F

given by f compose A

[golden pasture]

yup so A^*(f) = f\circ A

[past temple]

mhm

[golden pasture]

ok nice so now we have
A^*: V^*\to U^*

so we want

A^**: U^**\to V^**

now lets say you have some \phi: U^*\to F

what would A^**(\phi) be

[past temple]

hmm

[golden pasture]

oops mb

[past temple]

wait

shouldnt the domain of phi be U**?

[golden pasture]

this is your elements of U^**

[past temple]

oh wait yeah

hmm

[golden pasture]

yee

[past temple]

so we want to send it to some map

[golden pasture]

so A^**(\phi) becomes?

[golden pasture]

(rmb it should be similar)

am element of V^** is simply a map from V^* to F

[past temple]

do we map it to like

phi \circ A*

?

fuc im so confused lol

oh wait yeah

[golden pasture]

yup

so A^* maps V^* to U^*

and \phi maps U^* to F

so \phi\circ A^* is yesh

maybe like if you're uncomfortable with it, we can look at how it acts on the elements

[wooden ember]

Do you think once the current q is done you could roughly indicate how you calculated their value?

[golden pasture]

so
A^*: V^*\to U^*
for any
f: V\to F
u in U
A^*(F)(u)=F(A(u))

[chilly ocean]

How are you so good at algebra?

[hidden haven]

By being old

[robust pollen]

Moldi, I bet you're younger than me

and I'm not as gud, fugg

[hidden haven]

I was more referring to the high schoolers on the server who know pretty advanced stuff

Hello Ari

[chilly ocean]

Old like fourth year phd or prof?

[hidden haven]

Lol younger than both of those

I'm doing MSc

[golden pasture]

A^**: U^**\to V^**
for any
\phi: U^*\to F
f: V\to F
A^**(\phi)(f) = \phi(A^*(f)) which is in F, probably as good as it gets i think

just got to

be old enuf

to start finding a house and settling down

[past temple]

yeah i think this double dual stuff is making a lot more sense now

[golden pasture]

it's kinda tricky to like work with but yea

[robust pollen]

fug, I'll quit. This server is not good for self esteem lel

[golden pasture]

nonononono

[past temple]

im trying to prove that the functor sending vector spaces to their double duals is isomorphic to the identity functor on the category of finite-dimensional vector spaces

[hidden haven]

[hidden haven]

Is this actually serious