groups-rings-fields

[dense moss]

the follow up is considering $g \in G$ : $$g = \prod_{p | m , e_p \geq f_p} a_p \cdot \prod_{p | m, e_p < f_p} b_p$$ and since G is abelian the order of g , is the product of the order of $a_p, b_p$ these orders are powers of distinct primes (or trivial overlaps), so $|g| =m = lcm(r,s)$

[tardy hedge]

Ive been here for a lot longer i got kicked once lol

[cloud walrus]

Goëtia

[tardy hedge]

A shameful moment in my history indeed

[dense moss]

okee elder

[tardy hedge]

Yes yes ma’am or sir or something else

[glad lance]

something else?

😐

[dense moss]

🙂

[glad lance]

what are you?

[dense moss]

idk

im an average iwasawa theorist

[glad lance]

...

[tardy hedge]

Do u guys know each other

U both popped out of nowhere and are vibing

[proper jolt]

hello im trying to solve this problem

im not sure what a unique prime ideal

does that mean that R only has one prime ideal

and that ideal have no subset that is prime?

[dense moss]

me?

ion kno zoty

[rocky cloak]

It means R has exactly one prime ideal yes

[proper jolt]

and if an ideal is a proper subset of that prime ideal

it cannot be prime right?

[rocky cloak]

Well, then you would have two

[proper jolt]

yeh lol

ok thanks

[proper jolt]

im confused about part (ii) of this theorem

how come it says that all non units of R are contained in some ideal

M

wouldn't the nonunits of R just be a maximal ideal?

[mint seal]

that sounds right. And a ring is local if it has exactly one maximal ideal

[proper jolt]

and that maximal ideal would just be the non units of R right?

[mint seal]

actually, my bad. In general the set of non-units of a ring does not form an ideal

for example in Z, 3 + (-2) = 1

but as it turns out, if the non-units form an ideal, then that ideal is the unique maximal ideal of the ring (i.e. it's a local ring, and that ideal of non-units is maximal)

that's the content of (ii)

[proper jolt]

ok nice that was what im getting out

[mint seal]

wait, I just said the content of (iii) when I meant to be speaking of (ii)

but hopefully you see anyway

the non-units aren't generally contained in a (proper) ideal

afraid I don't know what transcendental set means here

[proper jolt]

but when they are contained in a proper ideal then it must be maximal right?

[mint seal]

yes, because (ii) is equivalent to (i) by the theorem

[proper jolt]

ok nice

thksa

[mint seal]

thinkin about it. My field theory is quite bad so I dunno if I can figure this out without doing some reading

I never even really learned galois theory despite seeing it as an undergrad and grad

just terrible

[jovial oyster]

Do not ping random users

[proper jolt]

can a local ring have more than one prime ideals?

why? it just says local rings have one maximal ideal not one prime ideal

oh wait maximal ideal

idk dimension theory

[rotund aurora]

why are rational functions defined on a variety "geometric"?

[proper jolt]

oh ok nice thks

[near gazelle]

Suppose H is a normal subgroup of G. Then obviously G/H forms a group. Let $x_1, ..., x_k$ denote representatives for the k distinct cosets. Then do ${x_i}_{i=1}^k$ form a subgroup in G?

[cloud walrus]

Ante0417

[near gazelle]

Actually no...

nvm

[south patrol]

Another easy example is any local domain (other than a field) since then the maximal ideal and 0 ideal are distinct prime ideals

E.g. Z_(p) or Z_p or F[[t]]

[coral spindle]

I imagine you mean just the set {x_1, …, x_k}. Sometimes you can choose representatives such that this holds (but you must choose them well). This is when G is a semidirect product of H and G/H. But not all groups are like this.

[proper jolt]

oh right yeh that's nice

[vivid kestrel]

So let $R$ be a factorial ring and $K = \mathbf{Quot} R$ and $f \in R[x]$ a monic polynomial. I'm trying to prove that if $a \in K$ with $f(a) = 0$ then $f \in R$:

$f$ factorizes as $f = (x - a)g$ over $K$ and we can write write $g = s \cdot h$ such that $h \in R[x]$ and $s \in K$ and then $f = (x - a)sh$ so $sh$ must also be a monic polynomial so since $h \in R[x]$, $s$ must be a unit in $R$ so $sh = g$ is a polynomial in $R[x]$ where the leading coefficient is a unit and we can perform the polynomial division $f / g$ to obtain a polynomial in $R[x]$ but $f / g = x - a$ so $a \in R$.

Is that correct?

[cloud walrus]

eggman

[rocky cloak]

So it doesn't seem to me that you're using that R is factorial (which is necessary).

Also I don't see how you're concluding that s is in R. For example
(1/2)(2x + 1) = x + 1/2
is a monic polynomial, but 1/2 is not an integer.

[vivid kestrel]

ah i see where i went wrong, thanks!

[crystal vale]

To prove that $\mathbb{Z}/n\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} = 0$ if $m$ and $n$ are coprime, we have the following:

Consider the relations $a x \otimes y = x \otimes a y$ and $x \otimes y = 0$. Since $m$ and $n$ are coprime, we can choose $a \in \mathbb{Z}$ such that $a \equiv 1 \pmod{n}$ and $a \equiv 0 \pmod{m}$.

This gives us $x \otimes y = x \otimes 0$. Since the right-hand side is equal to 0, we conclude that $x \otimes y = 0$.
is it correct?

[cloud walrus]

Notknow🙇

[next obsidian]

Yeah

[crystal vale]

thanks

In the construction of extension of scalars, we start with a ring homomorphism ( f: A \to B ). Given an ( A )-module, we want to extend its scalars.

\textbf{Step 1:} Since we can view ( B ) as an ( A )-module, the module action of ( A ) on ( B ) is given by:
[
a b = f(a)b \quad \text{for all} \quad a \in A, , b \in B.
]

\textbf{Step 2:} We construct the ( M_B = B \otimes_A M ), where ( M ) is an ( A )-module. Since we are tensoring with ( B ) over ( A ), this gives a new module over ( B ), and it naturally inherits an ( A )-module structure.

\textbf{Step 3:} Now, we define the ( B )-module structure on ( M_B ) by the following action:
[
b \cdot (b' \otimes x) = b b' \otimes x \quad \text{for all} \quad b, b' \in B \text{ and } x \in M.
]

This defines the scalar extension of the module structure to ( A)-module

[cloud walrus]

Notknow🙇

[next obsidian]

But you should prove that R/I (x) R/J = R/(I+J)

[crystal vale]

i wrote in my language and then use ai to latex it, so it is not copy from ai, is i got it correct?

[next obsidian]

To get another proof of this

[crystal vale]

i know from chinese - remainder theorem, I and J are co-maximal ideal

right?

[next obsidian]

Yes and that shows I + J will be (1)

[crystal vale]

yes

[next obsidian]

But what I wrote is true always

The (x) is a tensor product

Not a direct product

[crystal vale]

i see

okay i will try to prove this

[next obsidian]

Maybe first try to show that M (x) R/I = M/IM

And from there this is a corollary

[crystal vale]

equal, you mean isomorphic?

[bitter rover]

tfw no isomorphism key on keyboard

[next obsidian]

Yeah

[crystal vale]

yeah

[bitter rover]

yeah

[next obsidian]

I mean really that there’s a canonical isomorphism

It’s just “multiplication”

[crystal vale]

so first we define the mapping $M\times R/I \rightarrow M/IM$ such that $(m, r+ I)\mapsto rm + IM$ it is bi-linear so it induce the mapping $M\otimes R/I \rightarrow M/IM$ such that $m\otimes (r+I) \mapsto rm + IM$, right?

[cloud walrus]

Notknow🙇

[crystal vale]

then we construct its inverse mapping

[next obsidian]

No

[crystal vale]

why?

[next obsidian]

Show it’s bijective

[crystal vale]

so idea is correct?

i mean the way

[next obsidian]

Yeah

[glad osprey]

If k <= F is a splitting extension, then every embedding of F in k-bar has the same image, right? While for example Q(cbrt(2)) has 3 embeddings with 3 different images in Q-bar (the canonical one, plus 2 others corresponding to the other roots of x^3 - 2)?

[next obsidian]

What is a splitting extension?

[glad osprey]

Splitting field (I prefer the term splitting extension, since it's a property of an extension, not of a field itself)

[crystal vale]

since $(m, r+I)\mapsto rm + IM$ is onto so induce map also onto, but i am not sure about one-one

[cloud walrus]

Notknow🙇

[next obsidian]

then no

http://sporadic.stanford.edu/Math121/SeparableDegree.pdf

The number of embeddings into k-bar (I am assuming embeddings which fix k) is equal to the separability degree

[next obsidian]

https://stacks.math.columbia.edu/tag/09GZ

Okay the above doesn’t really define separable degree in the right way

[glad osprey]

Hmm what assumptions do we need for the image to be the same? Splitting and separable?

[next obsidian]

Image is same if and only if tbe extension is purely inseparable

https://stacks.math.columbia.edu/tag/09HD

Sorry, this is the actual page you want to look at

Look at Lemma 9.14.8 in particular

[next obsidian]

Think for a while

[glad osprey]

Hmm, are you sure we are talking about the same thing? Like, Q[x]/(x^2 - 2) has 2 distinct embeddings in Q-bar, but they have the same image, since Q(sqrt(2)) = Q(-sqrt(2)). The same is not true for Q[x]/(x^3 - 2) for example

[next obsidian]

Same image

Sorry

[glad osprey]

Sorry, what's the proper term? 🙃

[next obsidian]

That is the right term

I just

I don’t know

It seems like a funny thing so I just immediately thought about distinct embeddings

[glad osprey]

Ah, I see I've just started learning Galois theory, so purely inseparable extension and stuff is still a bit above my head

But my understanding is that in Galois theory we want to look at splitting fields because then there is a correspondence between the Galois group and the set of embeddings in the algebraic closure, is that correct?

[rocky cloak]

Yes, the image of F into k-bar is the same for all embeddings iff F is normal iff F is the splitting field for some family of polynomials.

[crystal vale]

So I am going for elementary proof, let rm in IM, then rm = im_1 for some i in I and m in M, and now I have to prove m \otimes r +I = 0.
I think this approach is not good
Any hint?

[unkempt stream]

Does anyone mind providing some insight why I should assume that M is primary. I know that the module will decompose into primary components, and so will the subgroups, so it makes the quotient have a nice structure.

Should I try to incorperate a similar technique that jacobson uses for the case of trying to prove the invariance of the decomposition of the module, by considering the descending module chain p^n M and taking the quotients to consider them as vector spaces over D/(p)?

I just don't think it would behave as well due to the quotienting by p^kM not p^k N

I'm still trying to wrap my head around this sorta stuff tbh, since p^kN is the intersection of N and p^kM

[rocky cloak]

If
M = D/p^n1 (+) D/p^n2 ...
With n1 >= n2 ...
then any quotient of M will be if the form
D/p^m1 (+) D/p^m2 ...
with
m1 >= m2 ...
and mi <= ni.

This you can see by just ordering the mi by size and looking at the dimension of p^(ni-1) M / p^ni M.

Once you have the result for the primary parts you just use that if X is p-primary with invariant factors x1, x2, ... and Y is q-primary with invariant factors y1, y2,...
Then X(+)Y will have invariant factors x1*y1, x2*y2, ...

[unkempt stream]

Damn I thought about considering p^(ni)M and trying to quotient by the next power but that wouldn't necessarily be a vector space. i didn't think about just removing one lol

thank you very much

[vivid kestrel]

So let $G$ be a (finite?) group and $g, h \in G$ with $hgh^{-1} \in <g>$ and let $U = <g, h>$. Can we then conclude already that $<g> $ is normal in U and $U = <g><h>$?

[cloud walrus]

eggman

[crystal vale]

Which textbook are you doing?

[vivid kestrel]

none at all

[crystal vale]

Source of this question?

[vivid kestrel]

doing exam prep and there is one question where if that were true i could save a lot of labourious calculations

[serene dune]

How is <g> defined ?

[rocky cloak]

So hgh^-1 in <g> ensures that
h<g>h^-1 is a subset of <g>.

For <g> to be normal in U you need that h<g>h^-1 equals <g>.

If <g> is finite this is true, because they have the same size. But for infinite G it could be a proper subset

[vivid kestrel]

sorry the typesetting is probably wonky, just the cyclic group generated by g

[serene dune]

Lol, alr, very sleepy, also jagr is here

[rocky cloak]

To take an example, let G be the group of functions of the form
f(x) = ax + b
for a and b rational (a non-zero). This is a group under composition.

Let g be g(x) = x+1. Then <g> consists of functions of the form x+k for k an integer.

Let h(x) = 2x.

Then h g h^-1 = x + 2 is in <g>. But h^-1 g h = x + 1/2 is not in <g>

[vivid kestrel]

ah okay, interesting counterexample for the infinite case, thanks

[bitter rover]

Hmm, counter-examples when G is infinite would be an interesting way to introduce the semi-direct product.

[crystal vale]

what is $(\mathbb{Z}/n\mathbb{Z})/(m\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z})$?

[cloud walrus]

Notknow🙇

[crystal vale]

any hint?

i mean i have result M\otimes R/I \cong M/IM, so i let M = Z/nZ and R = Z and I = mZ

[candid patrol]

GUYS !!!

Is Aut(G) always a non-trivial subgroup of S(G) for |G| > 3?

I'm struggling to know if it is true :/

[frigid epoch]

If |G| > 1, then G contains a nonneutral element g. Then you can simply write down a map sending e to g in S(G)

[rocky cloak]

For G non-abelian G/ZG is a nontrivial subgroup of Aut(G). For G abelian, inversion is an automorphism. Inversion is the identity iff all elements have order 2. That would make G an F2-vector space, so you would get a nontrivial automorphism by permuting basis vectors.

That leaves the G = C2 or the trivial group as the only cases where Aut(G) is trivial.

[candid patrol]

Aut(G) is trivial with G = C2 or the trivial group yes

But I was asking about Aut(G) != S(G)

But it's okay I got it

[rocky cloak]

I see, that's usually called a proper subgroup

[candid patrol]

f(e) != e, so f € S(G) \ Aut(G)

[candid patrol]

Ok ok didn't know it in english haha

[toxic zephyr]

suppose $m\in\bZ$ and $\bar N=a^2-mb^2$ is prime in $\bZ$. can $\bar N\mid b$? if so, under what conditions will that be impossible?

(if $m<0$ them definitely not, but what about $m>0$?)

[cloud walrus]

eigentaylor (STfFGMOaPID)

[toxic zephyr]

(this is in service to prove that if N bar is prime then a+bsqrt(m) is prime)

[rocky cloak]

If N divides b, then N also divides
N + mb^2 = a^2
since N is prime N divides a.

But that means N^2 divides a^2 - mb^2 = N

Which is impossible

[toxic zephyr]

wait why does N^2 divide that? because N^2 divides a^2 and b^2?

[rocky cloak]

Yup

[toxic zephyr]

ahhhh genius!

thank you so much!

okay so this seems to work even if m is a perfect square (i.e. were in Z[m]). because then the norm function never yields a prime so it's vacuously true.
Norm prime implies prime might work in general...
this might be super obvious but I'm trying to figure out and prove it on my own.

[clear ingot]

i think this is not well defined

[south patrol]

Yeah lol

Presumably it means (Z/nZ)/(mZ/nZ)

[white oxide]

How can we tell that y - x^2 is irreducible over k[x, y]

[south patrol]

View it as a polyomial over k[x]

Or work out what k[x,y]/(y - x^2) is

[white oxide]

Oh, so then it's a polynomial of degree one over k[x] and hence irreducible right

[south patrol]

Yeah

More precisely any factorisation provides one over k[x] (of course)

[white oxide]

Cool thanks

[clear ingot]

then by one of the isomorphism theorem its just Z/mZ

[south patrol]

Indeed

[tough raven]

If A → B is an injective integral morphism of commutative rings, then Spec B → Spec A is continuous, closed and surjective, hence a quotient map. This means that the map I ↦ sqrt(I^e) from radical ideals of A to B should be injective, which is true iff for all radical ideals I of A, I = sqrt(I^e)^c = sqrt(I^ec) iff I = I^ec, where ^e means extension of ideals and ^c means contraction of ideals. Is there a direct algebraic argument to see this?

i.e., I want to show that if A → B is injective and integral, I is a radical ideal of A and a ∈ A, then a^n = b_1 i_1 + ... + b_k i_k for n ≥ 0, b_j ∈ B, i_j ∈ I implies a ∈ I.

[tough raven]

OK, I found a proof in a book: replacing B by A[b1, ..., bk], we can assume that it is finite; it is also a faithful A-module (because a ⋅ 1 = a). x^n B ⊆ IB, so picking a spanning set, multiplication by x^n can be given by some matrix with coefficients in I, so its characteristic polynomial p has coefficients except the leading coefficient in I. By the "determinant trick" + faithfulness, p(x^n) = 0. Modulo I, this means x^n is nilpotent. Thus some power of x is in I so x is in I.

If this can be turned into an application of Nakayama's lemma, I'm not sure how (maybe consider the fg A-module B/x^nB?).

[elfin wraith]

Could anyone explain how to compute character tables, or point me to somewhere that does? I need to find the characters of C_6 and prove theyre all distinct, but as far as I can tell all my notes say is that its easy to compute the characters of a cyclic group, without saying how

[proper jolt]

10.4 of artin has characters and some examples

[crystal vale]

okay now, i have to show R/(I+J) \cong (R/I)(J (R/I) )

[next obsidian]

Is it not clear that J(R/I) = (J+I)/I?

[crystal vale]

no

i mean how?

J(R/I) = {jr + I, j in J, r in R}

oh i can send jr + I \mapsto jr + I, right?

it is onto and also one-one

right? @next obsidian

sorry to ping

[crystal vale]

If E is sequence then what is the meaning of sequence Hom(E, Hom(N,P) )?

[tough raven]

Hom(0, Hom(N, P)) → Hom(M'', Hom(N, P)) → Hom(M, Hom(N, P)) → Hom(M', Hom(N, P)).

[uneven bobcat]

You don't need character tables for that. (but they're cool, so learn them anyways)
Let C_n be the cyclic group of order and let g be a generator. A character is a homomorphism f: C_n --> C* to the non-zero complex numbers. Every element of C_n is of the form g^m, and f is a homomorphism, so f(g^m) = f(g)^m. In other words, each character f is completely determined by the value of f(g).

Now, using the same sort of trick I just used above, you can determine the exact possibilities for what complex values f(g) can take on.

[crystal vale]

How do I get this?

[brave remnant]

can we apply regular "algebra" on cosets for example if aH = Hb then we can do a^-1 on both sides to obtain H=a^-1 Hb

[mighty kiln]

Yes

[velvet hull]

as long as it's cosets of normal subgroups (assuming group theory here)

but yes, that is kind of the point of taking cosets

[brave remnant]

why if H is not normal it fails?

I used this in the technique of proving H is normal, is this not allowed?

[velvet hull]

well, if aH = Hb and we take a^-1 on both sides, we get that aHa^-1 = Hba^-1

but now we can no longer assume that aHa^-1 = H

[velvet hull]

that sounds like a circular argument to me

because algeba with cosets is defined if an only if you are doing it with a normal subgroup

that's the entire point of normal subgroups

[brave remnant]

you took a^-1 on the right side then the left side?

[velvet hull]

aHa^-1 = Hba^-1 is still true, so you can do ""algebra"" like that

but using that, you cannot then conclude that H = Hba^-1

because aHa^-1 is no longer necessarily equal to H

[brave remnant]

that is true but im confused why you applied a^-1 to the right of aH and then to the left of Hb

why don't u do left and left consistent and right and right consistent

[velvet hull]

that works, it's just slight notation abuse

what do you mean

[velvet hull]

but my point is, you essentially assumed that H is normal to prove that H is normal, you see the problem?

[brave remnant]

suppose 4x=8 then u can apply the inverse of 4 on the left side with 1/4 times 4x = 1/4 times 8

see how i applied 1/4 on the left side of both sides of the equation

why did you apply a^-1 to the right of aH and left of Hb

[velvet hull]

oh, you're right, it's a typo I meant to say Hba^-1

but the point still stands

[brave remnant]

okay yes that makes sense I used a^-1 aH = H = a^-1Hb to show H=a^-1 Hb but did not conclude that means H is normal

[velvet hull]

hmm, so what exactly is your proof here?

[brave remnant]

I later used that a is in aH and since aH=Hb a is in Hb and since a is in Hb e is in a^-1Hb then since e is in a^-1Hb that is equal to H

[velvet hull]

why is H equal to a^-1Hb

[brave remnant]

because Hb contains a since Hb=aH

[velvet hull]

so is that a given?

[brave remnant]

trivial

e is in H so a is in aH

since a*e=a

[velvet hull]

no no no, why is Hb = aH

[brave remnant]

that one is given in the problem

[velvet hull]

okay, so that is given

is it true for all a,b in G

[brave remnant]

yes

[velvet hull]

okay, now could you repeat your proof for me please

[brave remnant]

well actually given a exists a b

[velvet hull]

next time please tell people what your givens are

[brave remnant]

yes I just wanted to know if you can perform algebra on cosets

but the problem i think i figurd it out by myself

[velvet hull]

depends on what you mean by algebra

[brave remnant]

like applying things to both sides of an equivalence relation

[velvet hull]

because the aH notation always makes sense for any subgroup, and things like a*bH = abH are still true

but if you want any meaninful way of doing algebra between the cosets, you need your subgroup to be normal

[brave remnant]

youre saying as in commuting elements

like if H is normal then c^-1Hc=Hc^-1c=H

[velvet hull]

thats one way to call it

[brave remnant]

but in general supose you have two subgroups H and K

then take some element a,b in G containing H and K

then is it true that if aH=bK I can do stuff like a^-1 aH=H=a^-1bK

[velvet hull]

that always works regardless if they are normal or not

you just can't "push" things past the cosets

[brave remnant]

right right that wasn't what my intentions were okay that makes sense

can you also help me on another problem involving intersections

[velvet hull]

sure, just ask

[brave remnant]

H and K are normal

then the intersection is normal

is this not just inherited since the intersection i have proven before is a subgroup

like the way we inherit associativiy

[velvet hull]

it's a nontrivial thing

for example if only H is normal, then the intersection is no longer guaranteed to be normal

normality is a slightly more subtle property than you think

[brave remnant]

right but we are given that both H and K is normal

just like both H and K have associative elements

then like we can say H $\cap$ K is associative we can also say H $\cap$ K is normal?

[cloud walrus]

redoftwored

[velvet hull]

then it is inherited, sure, but I don't like that word

because again, normality is more subtle than that

for instance, if A is normal in B and B is normal in C, that does not guarantee that A is normal in C

in fact it's "usually" not true

[brave remnant]

i guess the implications of the proof would mean that you can say the property is inherited

but im not sure if I can just say that for the proof

if I want to prove that H $\cap$ K is normal how would I do that?

[cloud walrus]

redoftwored

[velvet hull]

well, usually the simplest way to prove things in beginner abstract algebra is to just go back to definitions

show that gxg^-1 is in HnK, for all x in HnK

[brave remnant]

H is normal then cxc^-1 is in H for x in H and K is normal then dyd^-1 is in K for y in K is this true?

[velvet hull]

that is the definition, yes

[brave remnant]

oh I see wait does not necessarily need to equal H

just needs to be in H

then we can have the possibilty that cxc^-1 for x in H forms something that is not all of H but must be in H

[velvet hull]

..what? cxc^-1 is just some element in H

I think you may be overthinking it a little

[brave remnant]

I meant cxc^-1 for all x in H forms something that is not all of H but must be subgroup of H***

this is possible^

[velvet hull]

well, it has to be all of H actually

(think about why)

but that's not needed for this problem

it's just some random element of H, that's all you need

[brave remnant]

wait so youre saying that cHc^-1 = H no exceptions?

when c is in G

[velvet hull]

if you fix c, yes

[brave remnant]

isn't this only true if c in H?

[velvet hull]

nope

[brave remnant]

why does my textbook give a definition that H is normal when cHc^-1 contained in H then

[velvet hull]

because it is true

[brave remnant]

if this case never happens

[velvet hull]

it just so happens to also be equal

[brave remnant]

no they give both

one contains one equal

[velvet hull]

what?

[brave remnant]

one contains or equal one equal

cHc^-1 contained in H

cHc^-1 = to H

[velvet hull]

which textbook are you using, which page, let me see

[brave remnant]

hungerford thm 8.11

sorry are u assuming H is normal?

because that is true if H is normal

but not necessarily true if H is nort normal

[crystal vale]

How it is exact?

[velvet hull]

I think I see the thm you're referring to

they're equivalent conditions, is just what your book is saying

if cHc^-1 is in H iff cHc^-1 = H

[brave remnant]

oh i see

its basically if it contains it must be equal to

so proving it contains is equivalent to proving equality

[velvet hull]

think about why, that's a good exercise as well

but, the most convenient definition for normal subgroups (in my opinion at least) is just that if n is in N, g is any element in G, then gng^-1 is in N

that's it

[brave remnant]

closed under conjugation right

[velvet hull]

yes, that is essentially what my definition says

[brave remnant]

the book defines it as Na=aN then N is normal I don't really like this definition

[velvet hull]

they're equivalent, but that is a sucky definition

also, having a subsection on category theory right after introducting the symmetric group is kinda crazy

maybe skip subsection 7, and get back to it later (the section titled "categories: products, coproducts and free objects")

wait is this a grad textbook?

okay that explains it

[brave remnant]

just a weird book not even entirely sure

they do rings before groups too

[velvet hull]

oh boy, this book looks dense

is this the one your prof chose for an introductory AA series?

I hope they know what they're doing

[brave remnant]

thomas hungerford right?

[candid patrol]

How do you say Int(G) in english ?

[brave remnant]

the content is somewhat overwhelming

[candid patrol]

Interns morphisms ?

[velvet hull]

do you mean Aut(G)?

automorphisms?

[brave remnant]

inner auts?

[candid patrol]

No I mean Int(G) ~ G/Z(G)

[velvet hull]

yeah I think it is Inn(G)

inner automorphism group

[lusty marlin]

Yeah that's Inn(G), the inner automorphism group of G

[candid patrol]

Oh ok ok thanks !

[brave remnant]

I don't really like group theory i like rings more for NT

[velvet hull]

yes, it looks like a grad textbook, i hope whoever's teaching you know what they are doing

[candid patrol]

And how do you write Aut(G)/Inn(G) ?

[brave remnant]

im forced to learn this lol

[velvet hull]

Out(G)

[candid patrol]

Ok ok same

[velvet hull]

if you ever get to study galois theory, you'll see group theory and ring theory come together beautifully

[brave remnant]

yea for solvability by radicals for degree 5 polynomials cry

[velvet hull]

but uh groups are really cool, just stick with it you'll see

[velvet hull]

if you like ring theory, a lot of the knowledge from ring theory carries over into group theory

normal subgroups are just ideals for groups, that's it

[brave remnant]

okay i see

is that how kernel is a normal subgroup in group theory but kernels are ideals in ring theory

[velvet hull]

yes

[brave remnant]

then u can always do a quotient group with the kernels

and do modular stuff

okay i see

[velvet hull]

yes

in groups, since you only have one operation you only need conjugation for taking the quotient to work

[brave remnant]

my intuition for groups is terrible im worried for the tests I will get stuck a lot

[velvet hull]

you actually need conjugation for ideals as well, it just so happens that commutativity makes the conjugation invisible

[brave remnant]

I se

e

[velvet hull]

if you have a solid foundation in ring theory, you should not do too bad in group theory

in a lot of ways they are extremely similar

[brave remnant]

so what advice would you give regarding the problems

i seem to get stuck a lot more often

and it takes a lot longer to get unstuck

since the holes i fall into seem harder to get out of metaphorically

[velvet hull]

I mean, since you're already familiar with ring theory just think about their ring theory equivalents

it's really not that different

[brave remnant]

alternating group symmetric group etc the structures are lot bendier idk

like D_4 is crazy to me

[velvet hull]

okay yeah those are harder to find equivalents for

true

[brave remnant]

like I had a problem that was prove Aut(z_2 x z_2) iso to S_3

and that was a very intersting problem

[velvet hull]

the interesting thing you'll soon realise, is that one big difference between group and ring theory, is that commutative rings are still very interesting

but commutative groups are actually really boring

[brave remnant]

I see that's probably why I havent done a lot of proofs where the abelian nature is given

[velvet hull]

in fact we have a complete classification of noetherian abelian groups, they are known fully up to isomorphism

see "fundamental theorem of finitely generated abelian groups"

[velvet hull]

fun fact, this falls out as a special case of the fundamental theorem of fintiely generated modules over a PID

because every abelian group can be viewed instead as a module over the ring Z, if you know what that means

[brave remnant]

I haven't learned modules yet

[velvet hull]

yeah don't worry about it then

[brave remnant]

are you an algebraist

[velvet hull]

I... would like to be one 😂

still very much a student

[brave remnant]

not a researcher?

are you graduate?

[velvet hull]

no, just a humble undergrad

[brave remnant]

nice

but regardless youve already done the basic modern algebra sequences

[velvet hull]

yes

[brave remnant]

do you know much about AG ?

[velvet hull]

I know enough algebra to know that I know nothing about AG

and that AG is fucking massive

but am currently trying to study some

[brave remnant]

I need to finish learning basic algebra yikes

[proper jolt]

hello does anyone know if the following fact is true? If R is a commutative ring with a unit and all its non units are zero divisors then all of its elements are nilpotents. I've been struggling to show this.

[rocky cloak]

It will be hard to show, because it's not true

[proper jolt]

damn that sucks, but is there a counter example

[rocky cloak]

Consider products of rings

[proper jolt]

wait doesn't this question imply it then?

like if i try to sho

[rocky cloak]

The nonunits all being contained in a common ideal is a very strong property, called being a local ring

[proper jolt]

im confused as to why (iii) imply (ii) doesn't just show what i said above

[rocky cloak]

Because (iii) is much stronger than what you wrote above

It has the added assumption that all nonunits are contained in a common ideal

[proper jolt]

so all non units form an ideal?

i mean it forms a maximal ideal?

[rocky cloak]

In (iii) yes, it says they form a prime ideal

[proper jolt]

but the common ideal is just the ideal of all non units right?

[rocky cloak]

Yes, and that is usually not an ideal

Consider Z, then 2 and 3 are nonunits but 3-2 = 1

[proper jolt]

ok i think i see what you mean now

[vivid kestrel]

So we know that for a polynomial $ f \in \mathbb{F}_p[x]$ where $p$ is prime and $\alpha$ from some larger field containing $\mathbb{F}_p$, then $f(\alpha^p) = f(\alpha)^p$. Does that still hold when $p$ is a prime power?

[cloud walrus]

eggman

[rocky cloak]

Notice that
x^p^2 = (x^p)^p

Or, you're asking for Fp with p a prime power maybe?

Yes, it is true

[coral spindle]

You are really asking if in a field of characteristic $p$ (i.e., a field containing $\mathbb F_p$) the map $x \mapsto x^p$ is a homomorphism of fields. Indeed this is the case. The proof is in fact completely the same as the proof you have likely seen for the field $\mathbb F_p$.

[cloud walrus]

$\mathbf{Boytjie}$

[tough raven]

You just have to do the verification. Hom(0, whatever) = 0. Exactness at M'' is because M → M'' is epi. Exactness at M is because of the universal property of cokernels.

[crystal vale]

yes i got it, thank you

[glad osprey]

In this proof, we find that [k(a) : k(b)]_s [k(b) : k]_s = [k(a) : k(b)] [k(b) : k], but why does this imply that [k(b) : k]_s = [k(b) : k]?

[next obsidian]

Look at the two terms in the <=

In the left, both things in the product are <= the corresponding thing in the product of the right term

So for these to be equal, you need both of those things in the product on the left to be equal to the corresponding thing in the product on the right

[glad osprey]

Ah, so we have ab = cd and also a <= c and b <= d, which gives b = d?

[next obsidian]

And a = c

But in particular, b = d is what they highlight

[glad osprey]

I see, thanks

[next obsidian]

[crystal vale]

In Hungerford, he defined cardnial number of a set A, is the equivalence class of A under the equivalence relation of equipollence, now my doubt is there are many equivalence class or it means number of equivalence classes ?

[rocky cloak]

The equivalence class itself is what is the cardinality of A.

If I understand what you're asking

[crystal vale]

But there are many equivalence classes, right?

[rocky cloak]

I'm not sure what you mean exactly.

A is only contained in one equivalence class.

But if you forget about A and are just asking in general, then there are many cardinal numbers yes

[crystal vale]

Oh I see you mean the equivalence class which contain A

Is it correct?

[rocky cloak]

Yes

[crystal vale]

I got it

Thank you ❤️

[misty cosmos]

i'm just double checking, is the $\phi(2\cdot2k) = \phi(2k + 2k)$ step valid in both scenarios

[cloud walrus]

sleepi

[misty cosmos]

i'm sure it is but it feels like i'm not adding in the ring because i think about stuff like $\phi(2k) \neq 2\phi(k)$ here because $\phi(k)$ isn't defined for odd $k$.

[cloud walrus]

sleepi

[coral spindle]

2.x = x + x in any ring, yes

Oh god wait you're looking at rungs, not rings

It is true that 2.2k = 2k + 2k in 2Z

[misty cosmos]

no

rings

wait

yeah rungs

these definitions are weird imo

gallian just weird

groups so cool for having a standardized definition

[frigid epoch]

functional analysts are annoying af for calling monoids "semigroups"

[bitter rover]

roings

[potent condor]

why is Z -> O_X -> F zero? It takes n to e^{2\pi i n}, which is one, not zero

[rocky cloak]

By the zero-map here, one means the constant map mapping everything to the identity.

So in this case mapping everything to 1 would be the "zero map"

[clear ingot]

I believe that for it to be isomorphism 2 need to be mapped to 3

Use that to show the map is not surjective

[potent condor]

😐

[vivid kestrel]

So let $z$ be an n-th root of unity, then we can represent $C_n \coloneqq \langle a \coloneqq \mathbf{diag}(z, z^{-1})\rangle \le GL_2(\mathbb{C})$. Set $R \coloneqq \mathbb{C}[x, y]$. Then $C_n$ acts on $R$ by $a \cdot p(x, y) = p(zx, z^{-1}y)$. By looking at monomials I can see that the invariant ring $R^{C_n}$ of polynomials in $R$ fixed under the operation of $C_n$ is equal to $\mathbb{C}[x^n, xy, y^n]$. How does one follow from that, that
$R^{C_n} = \mathbb{C}[xy, x^n + y^n] \oplus (x^n - y^n)\mathbb{C}[xy, x^n + y^n] $? Sleeping now, please ping if you reply

[proper jolt]

hello im confused about the wording here

lets say p is the minimal prime ideal in question.

Is p a minimal prime ideal, which happens to contain all the zero divisors

Or is p the smallest ideal which contains all the zero divisors, and that p can have a prime ideal as subset

im trying to show that this if and only if R has a unique prime ideal

[cloud walrus]

eggman

[proper jolt]

i think its the former, because i can't prove assuming the latter lol

[south patrol]

Hm in functional analysis these are probably honest semigroups right

As in nonunital monoids

[tough raven]

Yes, the former is equivalent to R having a unique prime ideal.

I think so; I've tried reading some Banach algebra theory and they usually reduce non-unital to unital by adding a unit (if A is a Banach algebra, ℂ (+) A is a unital Banach algebra, with the new ℂ being "span(1)", containing A as the closed maximal ideal 0 (+) A).

[somber bluff]

im trying to prove properties of automorphisms of reals.

ive already deduced that for positive r. f(r) must be real because f(r) = f(sqrt(r))^2 > 0
but how do i show it preserves order

[mint seal]

suppose a < b

i.e. b - a > 0

apply the automorphism and the result you just mentioned about sending positives to positives

[proper jolt]

when they say bijection between Hom(F(A), A') -> Hom(A, F'(A')) do they mean that

some arbitary rule or map between the elements that are injective and surjective?

[mint seal]

not arbitrary, natural

[proper jolt]

but by bijection is it just a rule that is injective and surjective. The reason im confused is because they define maps as follows

[mint seal]

Hom_C'(F(A),A') and Hom_C(A,F'(A')) are sets, for any A in C, A' in C' (assuming C and C' are locally small)

the bijection between them for a given A, A' is an actual function between sets

but the way the bijections vary with A and A' has a naturality condition

[proper jolt]

but that naturality condition

what does locally small mean?

[mint seal]

that all the collections of morphisms between two objects are in fact sets

the Hom things

[proper jolt]

oh ok so just to make it clear that natural bijection is a mapping between the Hom sets that satisfies the naturality condition and is a bijection?

[mint seal]

the "natural bijection" they speak of isn't really one bijection between two sets

but a collection of bijections between all possible pairs of hom sets of that type

with the naturality condition

the naturality amounts to that compositions of morphisms go through the adjunction in an, uh, natural kind of way

[proper jolt]

ok i think i get what you mean now thanks

[coral spindle]

Why is this in #groups-rings-fields, are you going to ask a question about specific adjoint functors on the category of groups maybe? Perhaps #category-theory would be a better fit if not – people there are used to explaining this stuff

[proper jolt]

idk this is in a commutative algebra textbook

[mint seal]

"Adjoints are everywhere." - S. Mac Lane

but yeah without more specific context this is just category theory

[proper jolt]

yeh maybe i should ask in cat theory then

[tardy hedge]

How do we rephrase this in terms of natural isomorphism of functors

You can show adjunction by natural isomorphism of some Hom functors right

[mint seal]

yeah, hom(F(-),-) is naturally isomorphic to hom(-,G(-))

[tender wharf]

If F is a char 0 field is the ideal (x-y, x+y-4) prime in F[x,y,z]? I feel like it's probably true because the quotient ring formed has x=y and 2x=4 so x is just 2; it's really just F[z].

because x and y are both identified with units

[coral spindle]

That ideal is equal to (x-2, y-2) which is maximal

& in fact the quotient is F!

[tender wharf]

this part I get

[tender wharf]

but not this

[thorn jay]

It's more of a surprise when a functor which arises naturally isnt an adjoint tbh

[elfin wraith]

An ideal I of a (commutative) ring R is maximal if and only if R/I is a field

So if you see why the ideal is maximal, you have that it is the ground field

[tender wharf]

i see that it is equal to (x-2,y-2) but it's not obvious why that ideal is maximal

like

(x-2,y-2,z) is surely an ideal containing it properly

[elfin wraith]

Yeah I was about to say, without working it out im not conviced it is maximal, but thats just vibes based ive not actually sat down to work it out

Actually, assuming F is algebraicly closed I dont think it is maximal. For K algebraicly closed, maximal ideals of K[x_1,...,x_n] have the form (x_1-a_1,...x_n-a_n)

[tender wharf]

it's not maximal, and F[x,y,z] is a ufd so it cannot be prime

wait

[elfin wraith]

It is prime no? F[x,y,z]/(x-2,y-2) \cong F[z] which is a PID

[tender wharf]

that doesnt work

my bad

[tender wharf]

i accidentally assumed it was a pid

[tender wharf]

this makes sense to me yeah

[elfin wraith]

I believe the argument is to show that (x-y,x+y+4) = (x-2,y-2), then take the quotient, thats a PID which is in particular an ID, so then the ideal must be prime

I could be mistaken, Boytjie is far better at this than I am, but I think he may have missed the z

[tender wharf]

I defined a homomorphism on the generators of F[x,y,z] by identity on F, x and y to 2 and z to itself

so the kernel is seen to be (x-2,y-2)

it's surjective onto F[z]

seems easier than quotienting

[elfin wraith]

Same difference

[tender wharf]

true

anyway thanks guys

[coral spindle]

Think of the map F[x, y] -> F sending both x and y to 2, and compute the kernel.

Then use first isomorphism theorem. It's always first iso!!!!

Oh wait.

Darn, sorry I did indeed misread. I thought it was just F[x, y] not F[x, y, z]. That's my bad

[coral spindle]

@chilly radish A while ago you asked about G x H^op-sets. It turns out there is a relatively well-known terminology for these things: bisets! In fact these bisets are crucial for the concept of biset functors, which are (as a friend of mine put it) a sort of 'global Mackey functor'.

One thing that may interest you is that there is a well-defined tensor product of bisets. If X is a (G, H)-biset and Y an (H, K)-biset, then write x (x) y for the H-orbit of (x, y) in X x Y, where h(x, y) = (xh^-1, hy). Then note that xh (x) y = x (x) hy. Write X (x) Y for the set of H-orbits here, and you can easily verify that this is a (G, K)-biset in a natural way.

This is helpful because certain bisets encode certain operations on (for example) representations. If H is a subgroup of G, then G is an (H, G)-biset and tensoring by this biset gives you induction of H-representations to G-representations, for example. This is where the idea of biset functors comes from.

Forgot to mention: Serge Bouc has a book on biset functors where he classifies all the simple biset functors. Pretty neat!

[tardy hedge]

I replied in category theory

[vivid kestrel]

so i see that the p adic integers contain the integers and also some fractions, what im wondering, are there p adic integers that have no corresponding representation in Z or in Q?

[coral spindle]

Yes

You are aware of the p-adic expansion I hope?

You can show that a rational has a repeating p-adic expansion, using the same argument that you use to show it has a repeating decimal expansion. So simply choose an expansion that has no period.

[vivid kestrel]

yes

[coral spindle]

But also, you can use size arguments. It's not too hard to argue that there are uncountably many p-adic integers – this is just Cantor's diagonal argument (without the fluff that the reals add in!). So obviously they can't all be rational.

And indeed this shows there are transcendental p-adic integers too.

[vivid kestrel]

oh, i see, makes sense

[delicate orchid]

am I the friend

I'm a fiend for the biset category

the biset DOUBLE category if one were to be so bold

[coral spindle]

Alas you are not the person in mind... afaik

Unless you moved to the states

[delicate orchid]

grrrrrrrrrrrr

[hidden wind]

thank you, finally getting around to trying this noe

on another note, tfw Aut C

[toxic zephyr]

we can consider vectors in $F^n$ as functions from ${1,2,\ldots,n}$ to $F$. so does that mean like... $F^n$ is the free left $F$-module on the set ${1,2,\ldots,n}$?

[cloud walrus]

eigentaylor (STfFGMOaPID)

[coral spindle]

Or right, yeah. It's the free F-module.

[toxic zephyr]

since it's a field, it doesn't matter right?

[coral spindle]

Yes

[toxic zephyr]

because of commutativity

that's really cool thank you

[coral spindle]

Np

[tardy hedge]

What part doesnt matter?

[toxic zephyr]

if it's left or right

[tardy hedge]

Ohh ok yea

Thx

[hidden wind]

ok i don’t understand your idea; but i found another which i’m pretty sure works except that taking the radical of an infinite intersection doesn’t necessarily distribute, which i’m not sure how to deal with

[glad osprey]

Btw, in what way are covectors dual to vectors? Like, I know the definition, but which arrows are actually reversed?

[thorn jay]

There is a sense in which you can see vectors as functions from the base field to the vector space

As in, given the vector space V as abstract object in the category K-vect, then the vectors "in" V can be seen as the elements of Hom_K(K, V)

[glad osprey]

Oh, I see is it because K is a terminal object in K-vect, so it plays the same role as the singleton set in Set?

Wait, K isn't a terminal object

[thorn jay]

K-Vect has a zero object hehe

K is the free vector space on a singleton though

So Hom_K(K, V) is naturally isomorphic to Hom_Set(*, V)

By the free functor adjunction

(Similar thing holds for any variety of algebras)

[tough raven]

V → V* is a contravariant functor.

[tough raven]

Yes, it doesn't distribute. For example consider the intersection of (X^n) over all n in ℂ[X]. All the radicals are (X), but the intersection is (0) with radical (0).

However, an arbitrary intersection of radical ideals is radical, so in that case the radical does respect the intersection.

I can't really say anything else without knowing if there's anything specific about my hint you're finding confusing, or what your idea is.

[potent condor]

My prof. showed that if F/E is a finite normal extension and K is the maximal subfield of F which is purely inseparable over E, then F is separable over K. His argument was: if a in F has minimal polynomial g(X) in E, then write g(X) = h(X^{p^n}) for a separable irreducible h with coefficients h_i. Let l_i the the p^n th root of h_i in the algebraic closure of F and let l(X) be the polynomial with these coefficients. Then he showed that the l_i are in K and claimed l has no repeated roots, so since l(a) = 0, a is separable.

How does he know l has no repeated roots?

[rocky cloak]

The roots of l are the p^n-th roots of the roots of h, and since h is seperable they're distinct.

[vivid kestrel]

so about hensels lemma, is the converse trivially true, as in, if we have an equation in the p adic integers does that equation also hold mod p^n for any n?

[rocky cloak]

Yes, you can just reduce the equation modulo p^n

[hidden wind]

my idea is to take some prime J, and note that the intersection of the maximal ideals containing it, is the same thing as the intersection of the maximal ideals corresponding to the points of Z(J), so when we take the zero set we get an expression for Z(J) in terms of this intersection, taking the ideal then by the nullstellensatz we get that the radical of the intersection of maximal ideals is equal to J

[tough raven]

Right, so maximal ideals are prime hence radical and the intersection of radical ideals is radical. So in your last sentence phrase, you don't have to take the radical of the intersection.

[hidden wind]

oh

i’m not sure if i’m following actually

[tough raven]

You can also take J radical instead of prime (if you want), since you don't use the fact that Z(J) is irreducible anywhere.

[hidden wind]

does radicals distribute over arbitrary intersections if the ideals are all radical ?

[tough raven]

Yep, because taking the radical does nothing then.

[arctic trail]

Z mod p^k is is just Z_p mod p^k

[hidden wind]

right, se we also need that the intersection of radical ideals is radical

[thorn jay]

That is true :3

In fact, the set of radicals of R is the intersection closure of Spec R, but that aside

[hidden wind]

lovely

[thorn jay]

Using the fact that rad I is the intersection of all primes containing I, the proof becomes a fairly routine order theory exercise

[hidden wind]

guh plz don’t tell me i need to use zorn

[thorn jay]

No dw lmaoo

[hidden wind]

i’ve had three too many exposures to that today

[thorn jay]

Well i suppose it's more lattice theory

It's surprising how much the study of primes can be done using lattice theory (if the desired connections between the algebra and the lattice theory are established)

But big math does not want you to know this

:(

[coral spindle]

I'd like to hear more about this if you're willing to take the time to explain

[arctic trail]

can you give a reference?

[thorn jay]

Well in typical UA fashion i saw how many different types of ideals there were, and saw that congruences only really were classified into "minimal, nonproper, maximal", and i wanted to classify them better in some way.

That begs the question: "what is a classification?"
And the answer i came up with (taking properties from primes, radicals, etc) is as follows:

a classification of congruences \phi in a variety V is, for any algebra A in V, a set of congruences Spec^\phi A such that for any homomorphism h : A -> B we have the preimage h^-1 : S_B -> Spec^\phi A, and if R in Spec^\phi A, and S \subseteq R, then R/S in Spec^\phi A/S

then you can define (for a classification \phi) notions of the \phi-radical, which satisfies a lot of the same identities w.r.t. preimage and quotients and all.

The central theorem really is that such classification is equivalent to a class of algebras contained in V closed under subalgebras and isomorphisms.

Then you can define a special subset of those classifications which are defined by some set of model-theoretic sentences in some way (in actuality the corresponding class of algebras is the class of algebras satisfying said sentences), but I'm still looking into that.

I suppose it's not really the study of primes themselves? But a lot of basic facts one uses plenty are simply general facts about these classifications of congruences.

[thorn jay]

I dont know the paper but I remember seeing somewhere that the notion of a prime ideal and prime element of a lattice coincide

:P

[thorn jay]

Oh yeah! Under certain conditions (namely that Spec^\phi A consists of prime elements), you can define the Zariski topology too

[hidden wind]

[hidden wind]

so real

[thorn jay]

Oh, fun fact! The class of these classifications (let's call them coherent systems) can be complete lattice ordered giving a poset category with arbitrary small products and coproducts.

You can define a radicalisation closure operator (in the form of a monad) which takes a coherent system to it's radicalisation (like it takes the coherent system of primes to that of radicals), and in the corresponding classes of algebras that is actually reflected in taking closure under products! So this is a general version of the fact that a commutative ring is reduced iff it can be embedded into a product of fields

And as any prevariety (that is a class closed under products subalgebras and isomorphisms) is defined by a particular collection of sentences, so is automatically such a special coherent system

I know i havent given a lot of details but i wouldnt want to clog this chat more than I've done already

Its also not really groups/rings/fields :P

[arctic trail]

and how does lattice theory help?

[thorn jay]

in congruence-modular varieties (where the congruence lattices are modular) you have a well-defined notion of the commutator of congruences, which for rings is I•J, from which you can define a specific type of coherent system, and is studied in a paper called "The spectrum of a universal algebra", ill look for it, hold on

[thorn jay]

Im not sure if it explicitly helps there, i havent read that paper

[thorn jay]

https://link.springer.com/article/10.1007/BF01195383
this was the article

It is my life goal to get published in algebra universalis one day 🔥

[hidden wind]

godspeed

[vivid kestrel]

so i want to prove there is a non abelian group of order 55, let $N = C_{11}$, $U = C_5$, then obviously there is a non trivial map $\varphi : U \xrightarrow{} \mathbf{Aut}N = C_{10}$, then is it already guaranteed that the direct product of $N$ and $U$ with $U$ acting by $\varphi$ is non abelian?

[cloud walrus]

eggman

[arctic trail]

semidirect product

and yes

In the semidirect product N x_phi U we have (1,u).(n,1) = (phi_u(n), u) while (n,1).(1,u) = (n,u)

take n in N that's not in the kernel of some u in U

[vivid kestrel]

oh yeah mb im tired

i see, thanks!

[arctic trail]

is this even nilpotent*?

I don't think it is

[arctic trail]

U won't be normal but it's a Sylow 5-subgroup, so it's not nilpotent!

[vivid kestrel]

Okay, so trying to prove that $f = 1 + x + x^2 + x^3 + x^4$ is irreducible over $\mathbb{Z}$ by showing that it's irreducible over $K = \mathbb{F}_2$, since if it obviously has no root in $K$ so if it were reducible it would have to factor into 2 polynomials of degree 2 but if that were the case let $g$ be a quadratic factor of $f$ in $K[x]$ and then $g$ would have a root $\alpha \in K[x]/(g) = \mathbb{F}_4 = L$ and that root would have multipicative order $5$ in $L^*$ which obviously isn't possible. Is that correct?

[arctic trail]

why didn't you just write 1 + ... + x^4

[vivid kestrel]

no good reason, ill edit

[cloud walrus]

eggman

[arctic trail]

Oh I really just meant

1+ ... + x^4

like

1 + \cdots + x^4

because to me it just takes up less time to write and looked cleaner than the \sum_[...]

[vivid kestrel]

oh yeah i forgot mhm

[arctic trail]

it has multiplicative order 5 since f(x-1) = x^5-1

so all the roots are 5th roots of 1

[vivid kestrel]

that therei s only one irreducible degree 2 polynomial

[arctic trail]

nice argument

the proof I knew is just saying f is cyclotomic

and proving cyclotomic polynomials are irreducible

[void cosmos]

is hard 😄

[velvet hull]

pretty sure it only holds for cyclotomic poly's of degree p-1 for p prime

let me check

[vivid kestrel]

oh idk never saw a proof for that, is it simple

[arctic trail]

nah cause one of the possible definitions is as the minimal polynomial of a primitive nth root

[velvet hull]

yeah, you're right, the proof for nonprime is a bit more involved

but it's true

[arctic trail]

I meant for primes p

Q[w] has degree p-1, where w is a primitive pth root

[rocky cloak]

You could also use change of variables + Eisenstein for this

Notice f(x+1) = x^4 + 5x^3 + 10x^2 + 10x + 5

This also works in general to show that the pth cyclotomic polynomial is irreducible

[vivid kestrel]

oh I forgot about change of variables

But if I'm not missing anything that seems like a bit of a hassle to calculate? Or is there a reason it's easily seen

[rocky cloak]

f(x) = (x^5 - 1) / x - 1
f(x+1) = ((x+1)^5 - 1) / x
= x^4 + 5C1 x^3 + 5C2 x^2 + 5C3 x + 5C4

And you don't actually have to compute 5Ck to finish the argument

[vivid kestrel]

oh that's real neat, thanks

[velvet hull]

there is no calculation, you just consider when pCk is divisible by p (and p^2 when k=0)

[south patrol]

ye

once you know (a+b)^p = a^p + b^p mod p by that you're good

[velvet hull]

btw, it's not just change of variables, this specifically works because f(x) = x+1 is a ring automorphism of Z[x]

so you'll have to prove that as well (not too hard)

[vivid kestrel]

yeah I already had seen that that holds previously, thanks though

[chilly radish]

Makes sense

I didn't know the terminology, thanks for sharing!

[misty cosmos]

hi chat

[solemn garden]

How can statements like [ \big(\mathbb{Q}[x,y,z]/(xy-z^2)\big)/(x,z) \cong \mathbb{Q}[x,y,z]/(x,z,xy-z^2) \cong \mathbb{Q}[y]] be justified in general? What theorems are used in stating these isomorphisms?

[cloud walrus]

soup_norm

[south patrol]

A combination of the usual isomorphism theorems will give you this

[solemn garden]

oh right, for this example $(xy-z^2)$ is contained in $(x,z)$ so the first isomorphism follows by the third isomorphism theorem. How about [ \big(\mathbb{Q}[x,y,z,t]/(xy-z^2)\big)/(x,t) \cong \mathbb{Q}[x,y,z,t]/(x,t,xy-z^2) ]?

[cloud walrus]

soup_norm

[rocky cloak]

In the quotient ring xy - z^2 = 0, so (x, t) = (x, t, xy - z^2)

[solemn garden]

I don’t understand, (xy-z^2) isn’t contained in (x,t), so the third isomorphism theorem can’t be used here, right?

so it can’t directly be applied using $(xy-z^2), (x,t)$, and $\mathbb Q[x,y,z,t]$ to show [ \big(\mathbb{Q}[x,y,z,t]/(xy-z^2)\big)/(x,t) \cong \mathbb{Q}[x,y,z,t]/(x,t,xy-z^2) ]

[cloud walrus]

soup_norm

[rocky cloak]

So (x, t) means the ideal generated by x and t.

But we still need to consider which ring we're working in.

Since we wrote Q[x,y,z,t]/(xy-z^2) / (x, t)

The ideal (x, t) is in the ring Q[x,y,z,t]/(xy-z^2). And there it does contain xy-z^2 because that's 0.

More generally ideals in Q[x,y,z,t]/(xy-z^2) correspond to ideals in Q[x,y,z,t] that contain (xy, z^2)

[solemn garden]

Oh, I see, so the statement actually hides some notation, and it should be written [ \big(\mathbb{Q}[x,y,z,t]/(xy-z^2)\big)/(x,t) = \big(\mathbb{Q}[x,y,z,t]/(xy-z^2)\big)/\big((x,t,xy-z^2)/(xy-z^2)\big) ] \ [\cong \mathbb{Q}[x,y,z,t]/(x,t,xy-z^2) ] and here the third isomorphism theorem can be applied to the ideals $(xy-z^2), (x,t,xy-z^2)$ in the ring $\mathbb Q[x,y,z,t]$. Thanks!

[cloud walrus]

soup_norm

[unkempt stream]

doom

[vivid kestrel]

so let $M = \mathbf{Mon}(K[x_1, \ldots, x_n])$ the set of all monomials in $n$ indeterminates. according to lecture notes the (formal) sum of all elements of $M$ is given by $$ \frac{1}{(1 - x_1)(1 - x_2) \cdots (1 - x_n)} ,$$ i fail to see why

[cloud walrus]

eggman

[unkempt stream]

Consider it inductively when you factor out x_n^k from all the monomials with an x_n^k

What is left over?

[vivid kestrel]

hmm

[rocky cloak]

My first thought for this one would be to set up a sequence like

[\begin{tikzcd}
\bullet & \bullet & \bullet \
\bullet & \bullet & 0
\arrow["B", from=1-1, to=1-2]
\arrow["AB"{description}, from=1-1, to=2-1]
\arrow[from=1-2, to=1-3]
\arrow["A"{description}, from=1-2, to=2-2]
\arrow[from=1-3, to=2-3]
\arrow[from=2-1, to=2-2, equal]
\arrow[from=2-2, to=2-3]
\end{tikzcd}]

apply the snake lemma and reason using the resulting exact sequence.

[cloud walrus]

jagr2808

[unkempt stream]

I haven't learned the snake lemma in this textbook

I don't expect jacobson to expect you to use a homological lemma he doesn't introduce until midway through the second book

[vivid kestrel]

ah i see how it follows inductively, thanks!

[rocky cloak]

Nah, you can probably do it by fideling with matrices or something, but I like this approach

[unkempt stream]

plus I don't see why this is helpful.

The problem is that you can't just multiply the diagonal matricies because your original matricies are only similar, not conjugate (which is not a congruence relation)

[rocky cloak]

Well, you would go from the language of matrices to the language of modules. Then you have a lot of tools, and you can reduce to the case of p-primary modules and suddenly is starting to get easier

[unkempt stream]

actually that might do it.

I just need to go from the characterization in forms of matricies to splitting it up into a sum of p-primary modules

and how that goes down

[unkempt stream]

But we're only talking about D^n here no?

not some quotient or anything

Unless you decompose A

into prime diagonal matricies

[rocky cloak]

Well it would be about
D^n / AD^n

[unkempt stream]

where the p-valuations are on the diagonal

[rocky cloak]

And D^n / BD^n and for AB

[unkempt stream]

I might try to prove it with just matricies first and then come back to your idea and do it twice

But I think you can do it via the prime decomposition of the diagonals into like a "prime decomposition" of the whole diagonal matrix and see what I can do from there

[unkempt stream]

Since you can break D^n / BD^n down into a direct sum of prime components, do you think there is a way to reflect that as a matrix?

[tropic obsidian]

I don't understand how the last line makes sense semantically

the left hand side has a B module structure and the right side has an A module structure, where does the isomorhpism take place?

[coral spindle]

In Z-modules, i.e, Abelian groups

[vivid kestrel]

so hensels lemma applies to valuation rings of complete fields with a discrete valuation, so for example Q_p with the valuation ring Z_p if i understand correctly, and since Z_(p) is contained in Z_p, can i use hensels lemma to show that Z_(p) is not complete?

[tropic obsidian]

Hi, I am trying to do this problem and I don't see how it immediately follows from (3.5)

(3.5) just says $$ S^{-1} M = S^{-1} A \otimes_A M$$

[cloud walrus]

Musa

[hidden wind]

recall, how is S^-1 M defined?

oop i’m speaking to myself

[tropic obsidian]

The suggested solution avoids that by passing to tensor products

but I do not see how you can use the usual tensor product isomorphisms to get the other side

[rocky cloak]

rewrite S^-1 N as S^-1 A (x)_A N

use the fact that M (x)_R R = M

[vivid kestrel]

(for a fixed p not for all p)

[tropic obsidian]

not sure how this works in this situation, since N is not originally a $S^{-1} A$ module, but $S^{-1} A \otimes N)$ is

[cloud walrus]

Musa

[rocky cloak]

I'm not entirely sure what your confusion is, but you start with the expression

$$S^{-1}M \otimes_{S^{-1}A} S^{-1} N$$

Then rewrite it to

$$S^{-1}M \otimes_{S^{-1}A} S^{-1} A \otimes_A N$$

[cloud walrus]

jagr2808

[rocky cloak]

Now I'm saying you want to use the fact that for any R-module X
X(x)_R R = X

Do you see how you can use that to simplify the expression further?

[tropic obsidian]

honestly, I don't 😭

[rocky cloak]

Well do you see something that looks like $\otimes_R R$ anywhere in the bottom expression for some value of R?

[cloud walrus]

jagr2808

[tropic obsidian]

yeah like the $\otimes_{S^{-1} A} S^{-1} A$?

[cloud walrus]

Musa

[rocky cloak]

Right, so then we can cancel that part

Then we could also rewrite S^-1M

And then I claim we're done

[tropic obsidian]

right, I am a bit worried in how you would justify the associativity of the tensor product in this case

[rocky cloak]

I see, you haven't proven that tensor product is associative

[tropic obsidian]

oh no I have

[rocky cloak]

Then I'm not sure I see the issue

[tropic obsidian]

just so I understand what you are saying, is the claim that $S^{-1} M \otimes_{S^{-1} A} (S^{-1} A \otimes_A N) = (S^{-1} M \otimes_{S^{-1} A} S^{-1} A) \otimes_A N$? and then use the cancellation thing you were talking about on the first term?

[cloud walrus]

Musa

[tropic obsidian]

if so, I don't see how you are putting a S^{-1} A structure on N

or wait I guess you dont need that

but I don't think this follows from regular associativity

but the iso is correct anyhow

[wraith cargo]

but this isn't happening at any point

you never say N is a S^-1A module

[rocky cloak]

I'm not sure what regular associativity is, but

(A (x)_R B) (x)_S C = A (x)_R (B (x)_S C)

is all that's happening

I guess something that maybe isn't stated explicitly is that the right A-module structure of
S^-1 M and S^-1 M (x) S^-1 A is the same.

But that is the case

It's just (m/s)a = ma/s in both cases

[tropic obsidian]

are A, B, C bi modules?

by regular associativity I just meant the usual $(A \otimes_R B) \otimes_R C = A \otimes_R(B \otimes_R C)$

[cloud walrus]

Musa

[rocky cloak]

Yes

[potent condor]

B,C are galois extensions of k contained in E. Is this hint completely unnecessary? can i not just say

G(E/k) = G(E/B cap C) = G(E/B) v G(E/C) = G(E/B)G(E/C)?

[mellow crescent]

if P ∈ Rn[X] and Q ∈ Rn[X]
is possible deg(P+Q) <= max{deg(P),deg(Q)}?

[rocky cloak]

Think about for example if P = -Q

[rustic rapids]

any tips on how to approach these problems? I keep getting the trivial ring but im pretty sure thats not true. For example, in a i have x^2 = 3 and 2x = -4, but im not sure I can do these because it may have a zero divisor. Well, it would be trivial if it could, so how should I go about it? for b i multiplied by conjugate and that worked, but I dont see how to work with the other ideals

[velvet hull]

For a), notice that Z[x] is a ufd, so it can’t have any zero divisors

And you can check that any quotient ring of an integral domain must also at least be an integral domain lmao.

[potent condor]

Z is an integral domain, but Z/4Z isnt

[velvet hull]

update: I got skill issued

[south patrol]

Now for which rings R is it possible to write it as a quotient of an integral domain

||All||

[rocky cloak]

So you can use the trick that
R/(f, g) = (R/(f))/(g)
That at least simplifies it down quite a bit.

And if you have guessef for what the rings should be you can construct some homomorphisms and use the first isomorphism theorem

Another useful thing might be to try to find some integers in your ideal, then you can at least reduce it to some quotient of (Z/n)[x]. Then you just have something finite to deal with

[rustic rapids]

I see, i will try that

[tough raven]

commutative*

[south patrol]

True, tbh I hadn't even considered the notion of a noncommutative ring when I said that lol

[potent condor]

neither did anyone else tbh

[south patrol]

Actually true

Group rings get a pass too

[rocky cloak]

I mean, just change "integral domain" to "domain", problem solved

[south patrol]

Sure

[uneven bobcat]

But if we're meeting on convolution algebras of finite groups surely we want to let in convolution algebras for infinite groups.

[rocky cloak]

I propose, all rings are commutative, but not all algebras are.

Then we have Z-algebras in case we need them.

And we get to keep our group algebras, and path algebras and endomorphism algebras

[hidden wind]

[south patrol]

Lol

To be fair I can get behind that too

[thorn jay]

What if i want the addition in my ring to be not necessarily associative

[coral spindle]

Honestly a pretty good separation in terminology

[rocky cloak]

What if I want the square root of fish to be a flying monkey

[south patrol]

If my grandma had wheels she would be a bike

[thorn jay]

Exactly, those are the real questions

Why arent we mathematicians answering those

[rocky cloak]

To be fair, ham in carbanana is pretty good

[thorn jay]

I would like this to cease

And keep it... ceased

[rocky cloak]

Something ceased forever. We might call it
"The ceased"

[thorn jay]

Damn

That would be a 1 2 2 50

[hidden wind]

i am very confused by this, how would an algebra be defined if not as a ring over another?

[rocky cloak]

Just as you would any algebraic structure I guess.

You have a multiplication and an addition, and it satisfies some axioms

[thorn jay]

There are many ways one can think of certain structures.

In particular, here an R-algebra isnt just a ring homomorphism from R anymore (as that would make the algebra commutative), but defined as it would be in universal algebra, as abstact operations and axioms

[sly crescent]

As far as I’m concerned, a ring is an associative unital algebra

[rocky cloak]

As far as I'm concerned, an algebra is an associative unital algebra

[hidden wind]

is the word algebra used for a structure that is not over another? i’ve really only come across algebra over a ring

[rocky cloak]

The word algebra is used in quite a lot of contexts, but it usually is defined over a (commutative) ring yes

[thorn jay]

The only actual correct definition of "algebra" is ours, tyvm

[rocky cloak]

But you also have sigma-algebras and F-algebras

[thorn jay]

Your structure should be called monoid-enriched R-module or smt

[rocky cloak]

Probably other uses aswell

[hidden wind]

guh i forgot sigma-algebras have the word algebra in them

[rocky cloak]

Tbf sigma-algebras are just certain algebras over F2

[hidden wind]

i kinda just don’t want to check that

[thorn jay]

Yeah qwq there is at least one

[hidden wind]

i am not looking forward to doing measure theory someday

[thorn jay]

I mean, you can choose not to

[tardy hedge]

i chose not to

I dont like it

[thorn jay]

[rocky cloak]

I think measure theory is kinda fun. If it wasn't for all those pesky integrals

[hidden wind]

i just dont have the analysts capacity to name magic sets

[tardy hedge]

same i find it really hard to think about

not that im good at algebra but i find it muuch more comfortable to think about

and reason through

[rocky cloak]

I feel like algebra is all about building something beautiful, and analysis is about showing you how ugly the world really is

[thorn jay]

Just do algebra like you normally would except drop the assumption that your algebraic theory is locally small

[tardy hedge]

Hahaha

[hidden wind]

so real

[tardy hedge]

Ya jagr actually spit some heat

[rocky cloak]

There's a reason all those horrible AoC paradoxes were proven by analysists

[hidden wind]

there’s some ugly AoC algebra things as well though

[sly crescent]

I’ve seen that quote before

[tardy hedge]

Yeah i guess this is why i prefer algebra

Its more satisfying to learn because of this

[thorn jay]

Universal algebraists solve this by assuming it

[hidden wind]

like, the cardinality of the automorphism group of the complex numbers…

[tardy hedge]

Right

[rocky cloak]

Cool, who is it due to?

I just pulled it out of my hat

[sly crescent]

Probably you

[tardy hedge]

Hahahaha

[thorn jay]

Every ideal is embedded into a maximal one

or any maximal congruence

[hidden wind]

that’s not all that wild

[rocky cloak]

There's also this thing about how you eat your corn on the cob. Don't remember how that one goes

[thorn jay]

No but its slightly annoying that a should-be-trivial fact is equivalent to AoC

[glad osprey]

I think the claim is that analysts eat corn on the cob in a kind of spiral, while algebraists eat row by row

[rocky cloak]

Hmm, and is there an armchair psychology that goes with it?

[hidden wind]

there are only two continuous automorphisms, identity and conjugation, but $|\mathrm{Aut}(\mathbb C)| = 2^{2^{\aleph_0}}$

[cloud walrus]

rødbet-jens

[glad osprey]

No idea about the psychology, I only know that eating corn in a spiral is for psychopaths

[thorn jay]

That analysts like uzumaki?

[sly crescent]

Does complex conjugation generate a normal subgroup of Aut(C)?

[rocky cloak]

Nope

[hidden wind]

the kernel is zero, as is the kernel of any hom out of a field

oh of Aut C

[sly crescent]

Very cursed

[rocky cloak]

But iirc complex conjugation is the only automorphism of finite order up to conjugation

This is for the algebraic closure of Q. C has others

[sly crescent]

Again, very cursed

[south patrol]

I believe there are models of ZF where Aut C is generated by complex conjugation, but with choice you can get rather bizarre automorphisms by e.g. extending autos from Q bar I guess

[tough raven]

Makes sense, any two real-closed fields of the same transcendence degree over ℚ (or as a special case, uncountable cardinality) ought to be isomorphic.

[unborn helm]

While this is eventually going to get used for things like for #numerical-analysis I'm trying to figure out how to algebraically characterise things I'm thinking of.

1. There are sequences of vector spaces and/or algebras, often polynomial-centric.
2. The basis REALLY matters, so some way to say that whatever these things are, a different basis means it's a completely different thing could help.
3. Maybe some good examples might be orthogonal polynomials, interpolatory Hermite polynomials & Bernstein polynomials for the varieties of things that need to be taken into account. (Note these «Hermite polynomials» are not the sequence of orthogonal polynomials, but rather, the basis polynomials for the interpolation problem that includes derivative values beyond just the function values, where the basis sets for each degree are distinct.)
4. There needs to be some sort of way to have an idea that a nontrivial projection can happen from a higher to lower grade space, e.g. that things expressing degree constraints on e.g. Bernstein polynomials are dealing with a completely different set of basis functions, not just a subset of the ones already there, and that it matters that the basis is different.
5. It would be super good if there were some way to throw things like condition numbers or some sort of description of numerical behaviour into the various kinds of mappings between the spaces.

What kinds of algebraic structures am I looking for?

Is there some sort of representation-theoretic thing I'm looking for?

It's mostly a terminological question, to be clear.

[lusty marlin]

By Aut(ℂ) here do you mean the group of group automorphisms of (ℂ,+)?

And by continuity you mean with respect to the euclidean topology on ℂ, right?

[lusty marlin]

This should come from some sort of Hamel basis construction if I'm not wrong

[tough raven]

As a field, and the Euclidean topology.

[lusty marlin]

Ah ok

[lusty marlin]

So the Hamel basis part is correct right? Viewing the field ℂ as a vector space over ℚ

And then characterising automorphisms using that

Or is something else going on here

[tough raven]

No, because it wouldn't give you field automorphisms.

[lusty marlin]

Oh right

My bad lol

[tough raven]

You use a transcendence basis instead.

(And a lemma about extending automorphisms to algebraic closures.)

[lusty marlin]

Hmm ok

Thanks!

[tough raven]

Going to take a stab at this although I know almost nothing about the examples you've mentioned: (i) Are you trying to represent vector spaces with a specified basis (and maybe linear maps between them), or elements of such vector spaces? (ii) Do you need to do anything special for 4. beyond defining a new vector-space-with-basis and a linear map from it to the old one which happens to be injective?

[unborn helm]

Vector spaces (well, mostly algebras) with those specified bases.

[tough raven]

I don't think there's any term more specific for those than "vector space with a (specified/privileged) basis", if that's what you're searching for.

[unborn helm]

I guess the point of 4 is that the grading scheme's interaction with concrete instances of the bases isn't necessarily a nesting set inclusion tower.

[tough raven]

At least in algebra. I suppose it's possible in theory that another subfield like numerical computing might have invented its own terms? But the point of "basis" is to capture this.

[tough raven]

I'm sorry, but I have no idea what a grading scheme is (unless it's about how to score tests...), or "concrete instances of the bases".

You do mean that you might want to consider a subspace with a basis that is not a subset of the original basis, right?

[unborn helm]

By the grading scheme, I mean how the algebra is graded.

[tough raven]

What I was saying is that you could define a new vector space with its own basis and then a linear map from the second to the first which is injective, representing the inclusion.

[unborn helm]

Yes, lower-degree Bernstein polynomials are the premier example. Maybe a different one would be interpolatory Hermite bases that forget a knot or tie derivatives' values to adjacent intervals via implicit differentiation with the denominators cleared.

[tough raven]

It seems to me there's nothing you can do here except declare each vector-space-with-basis (so "polynomials with the 1, X, X^2, ... monomial basis" and "polynomials with the 1, X, X(X-1), ... falling factorial basis" are two different vector spaces) and represent linear maps between them by how they act on components with respect to the respective bases (i.e., by matrices with respect to the chosen bases). For example, the identity map from "polynomials with the monomial basis" to "polynomials with the falling factorial basis" would be represented by the change-of-basis matrix between those two bases - which makes sense, because the operation is doing something computationally non-trivial by converting from components wrt one basis to another.

And I'm not sure what else you need that you don't get by doing this.

But I suppose your question was about terminology, not implementation. And regarding that, I can only say I haven't heard a term other than "vector space with a (specified/privileged) basis".

Maybe someone else will know more.

[unborn helm]

That'll have to do, then. I'm surprised there's nothing already out there for it, but if that's the answer, that's the answer. Thanks!

[tough raven]

I mean, I suspect there is, just that no-one bothered to name it anything more specific.

[unborn helm]

If I knew what to Google for, I might be able to survey what theorems exist that I can use.

[hidden wind]

girls i need a sanity check, am i writing nonsense here?: $\mathbb Z[x,\sqrt x] \cong \mathbb Z[\sqrt x] \cong \mathbb Z[x]$

[cloud walrus]

rødbet-jens

[rotund aurora]

No, this is fine

[coral spindle]

Note the first isomorphism you list is in fact just equality, since $\sqrt x$ generates $\bZ[x, \sqrt x]$.

[cloud walrus]

$\mathbf{Boytjie}$

[hidden wind]

sure equality between the sets, but the map from formal polynomials in x and sqrt x to polynomials in sqrt x requires some shifting coefficients around so i prefer writing it like this

it’s possibly stupid of me to be thinking about polynomials in x and sqrt x though

[coral spindle]

No they are literally the same ring if you take the right perspective

Remember that $\bZ[k]$ can mean a subring generated by $\bZ$ and $k$. So in particular $\bZ[\sqrt x]$ is the subring of $\bZ[x, \sqrt x]$ generated by $\sqrt x$. So they are literally equal

[cloud walrus]

$\mathbf{Boytjie}$

[coral spindle]

Now if you want to construct these rings in different ways, that's fair enough

[hidden wind]

yeah my problem is that i’m separating these i shouldn’t be doing that

[coral spindle]

I have no idea how you would justify that without first seeing it the way I described

[thorn jay]

Magic

[tardy hedge]

Is the fixed field corresponding to all of Aut(K/F) always F?

[glad osprey]

No, the extension needs to be Galois

[tardy hedge]

Oh ok thanks

[glad osprey]

For example, Aut(Q(∛2)/Q) is trivial, so the fixed field is all of Q(∛2). And Q <= Q(∛2) is not Galois

[unborn helm]

Coordinate frame or affine frame is at least a piece of it.

[unborn helm]

@tough raven That idea of a frame describes the basis as an attribute of an object involving a vector space.

[tough raven]

Um, reading Wikipedia, an affine frame would essentially amount to choosing a basis of the form (monomials in p1, ..., pn), where p1, ..., pn are linear polynomials with exactly one common zero (equivalently, for some scalars a1, ..., an; p1, ..., pn forms a basis of span {X1 - a1, ..., Xn - an}). That doesn't sound nearly general enough for what you want to do.

[unborn helm]

The frame of the framed space doesn't seem to have the constraints you're describing. Why would they all need to be linear?

[unborn helm]

Maybe the neologism of «framed space» could make sense.

[tough raven]

I'm just going off what I read at https://en.m.wikipedia.org/wiki/Affine_space#Affine_coordinates.

[tardy hedge]

In 1, when it says A is possibly noncommutative, it means the multiplication in A as a ring could not be commutative?

[violet spade]

Hi, I need to find an example of a cubic field extension of Q (i.e. [Q(alpha):Q]=3) such that alpha cannot be written as the cube root of some number

I know that this exists. I was thinking about complex extensions maybe, but i can

can't really find anything

[arctic trail]

yes

[tardy hedge]

ty

[violet spade]

i was thinking about maybe trying something simple, like 1+i or something.

but the minimum polynomial of this is of degree 2

[arctic trail]

that's just

finding a polynomial of degree 3

with a root which is not a cube root

[violet spade]

Yes

Well we always have Q(1+cbrt(2)) =Q(cbrt(2)) right

So like it doesnt really hold?

But yes that's essentially where i'm stuck @arctic trail

[arctic trail]

here's an example

[velvet hull]

an affine function isn't necessarily linear, it's linear plus a constant

so, similar but that's where the confusion may be coming from

[tough raven]

Sure - that matches with what I said originally. My point is that I doubt you can get interesting polynomial bases like anything similar to Hermite polynomials with just products of powers of n given affine-linear polynomials.

[rocky cloak]

One thing that comes to mind would be, say
E = Q(a) for a cuberoot a. Then the splitting field of E contains a 3rd root of unity, hence must have degree 6.

So if you can find an extension whose splitting field has degree 3, then that proves it doesn't have any cube roots.

You can determine the splitting field from the discriminant.

For example ||x^3 - 3x + 1|| should work

Or possibly easier, if the splitting field is real, then it can't contain any roots of unity. Splitting field is real if the discriminant is positive

[hidden wind]

i’m not used to asking this kind of question but i’m curious if anyone have some insight to this: under what conditions is there a constructive proof that every ideal is contained in some maximal ideal?

obviously if the is finite there will be such a proof (just compare all ideals), but surely there are more general conditions?

[rotund aurora]

if it's noetherian?

[mighty kiln]

Does the argument use countable choice

[rocky cloak]

Seems finitely generated algebras would work

https://mathoverflow.net/a/7062/157483

[rocky cloak]

Depends on your definition of Noetherian

[hidden wind]

i’m not sure if i understand this idea of “constructively noetherian,” i guess the idea that whenever you have an ideal you must have a constructive way to check whether or not there is a larger proper ideal?

if so i don’t see how to apply the nullstellensatz

[tough raven]

https://arxiv.org/abs/1605.04832 is a good reference for doing commutative algebra constructively, though I can't point you to a specific result of the form you ask. I suspect there aren't many: constructive reasoning usually avoids using prime ideals entirely (proving something for every prime ideal of a commutative ring A can often be replaced by a "finite cover" of A by rings like A[a^{-1}, ...]/(b, ...) and a local-global principle).

[tough raven]

Constructively there are several ways you could define Noetherian (any infinite ascending chain stabilises, any infinite ascending chain has two equal consecutive terms, no infinite strictly ascending chain exists, every ideal is finitely generated) and it's not always clear which one to use.

[tough raven]

And this book says (although I haven't read enough to see it backed up) that it seems better to replace "Noetherian" and "finitely generated" hypotheses with "coherent" and "finitely presented".

[hidden wind]

amazibg, thank you, i’m not intimidated by the length at all

[tough raven]

Fair . I find a section on something specific and read it when I get interested in that.

[hidden wind]

i’ll see what i can get out of it mwahaha

[rocky cloak]

Say R = k[x1, ..., xn]/I

The Nullstellensatz says that if I is not (1), then there is a point in k^n where I vanishes. This then gives you a map R -> k.

[tough raven]

Let V be a vector space over a field K and k a positive integer.
For w: V^{k-1} → K alternating multilinear, define
i_w: ∧^k V → V: v1 ∧ ... ∧ vk ↦ w(v1, ..., v(k-1)) vk - ... + (-1)^(k-1) w(v2, ..., vk) v1.

For any subspace W of V, if h ∈ ∧^k W then i_w h ∈ W for any w.
Is the converse true: for
W = {i_w h: w alternating (k-1)-multilinear},
is h ∈ ∧^k W?

[tardy hedge]

Why is it that a simple tensor m x n is 0 iff all R bilinear maps take (m,n) to 0?

I dont think i really ever understood tensor products that well

[south patrol]

Okay consider a vector space $V$ and we have an element $x \in V$. Note that $x$ is zero iff all maps $V \to k$ send $x$ to $0$. $\Rightarrow$ is trivial, and $\Leftarrow$ follows because, for example, if $x \ne 0$ then you can extend it to a basis of $V $ and find a map $V \to k$ sending $x \mapsto 1$

[cloud walrus]

Prismatic Potato

[south patrol]

Now you want to apply this to the case where V is a tensor product and use the universal property of tensor products (in terms of bilinear maps)

[rocky cloak]

The map that takes (m,n) to m(x)n is bilinear.

So if all bilinear maps take it to 0, then m(x)n is 0.

Conversely if f is a bilinear map, then the universal property says there is a linear map f' with f'(m(x)n) = f(m, n). If f(m, n) is nonzero then m(x)n must be as well