groups-rings-fields

[wooden ember]

connection is saying no

[rustic crown]

aww

[trim grove]

can someone help me with How many maximal ideals does $\frac{\mathbb{C}[x,y]}{\langle y^2-x \rangle}$ has?

[cloud walrus]

honey99

[gritty sparrow]

Uncountably many

[trim grove]

how to approch?

can you give me a hint?

[gritty sparrow]

Here is a hint: (x,y) contains (y^2-x). So does (x-2,y-4)

Do you get the pattern

[chilly ocean]

question regarding this: where can I learn about such stuff? is this some intro AG?

or comm

[gritty sparrow]

You’ll learn atleast some of this stuff in an abstract algebra class that does a lot of stuff with rings in it

[trim grove]

about my question?

[chilly ocean]

i didnt:(

[gritty sparrow]

Ah, then it will be in an intro to commutative algebra class probably

Might also be there in an intro to ag class as well

[chilly ocean]

is there some way to visualize this quotient?

[gritty sparrow]

Sorry this is a typo i meant (x-4,y-2)

[gritty sparrow]

Yes, in a way it “corresponds” to the curve y^2=x in the complex plane.(and that is another hint to help you find maximal ideals)

[trim grove]

i think this is related to algebraic geometry

[gritty sparrow]

Yep exactly

In any case, did you find any more examples of maximal ideals containing this ideal?

[trim grove]

okk so here $<x-a,y-b>$ are maximal ideals of $\mathhb{C}[x,y]$?

[cloud walrus]

honey99
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[gritty sparrow]

Yes, can you see why?

[trim grove]

please don't say because of Hilbert's Nullstellensatz, i am afraid of that term

[gritty sparrow]

No

This isn’t by nullstellensatz

Firstly do you see why (x,y) is maximal as an ideal?

[trim grove]

because $\frac{\mathbb{C}[x,y]}{<x,y>}$ is isomprhic to $\mathbb{C}$

[cloud walrus]

honey99

[gritty sparrow]

Yep, and you can use similar logic to show that ideals of the form (x-a,x-b) are all maximal for any complex a and b

[trim grove]

okk , i will try to solve further

[gritty sparrow]

Cool, so once you have that all you have to do is select suitable a and b that make it so that (x-a,y-b) contains (y^2-x). (There will be uncountably many)

[trim grove]

thankyou @gritty sparrow , firslty i will try to show how <x-a,y-b> is maximal ideal .

[trim grove]

can you help me with this , i am trying ,but not getting here

[gritty sparrow]

Think very deeply about the isomorphism between C[x,y]/(x,y) and C. What is the isomorphism doing, and how can you generalize it to (x-a,y-b)

Ok maybe i’ll be more explicit, we can construct a map from C[x,y] to C that send the polynomials to their constant coefficient. So xy+1 goes to 1 and so on

[chilly ocean]

yeah but liek this curve corrseponds to 0 right

[trim grove]

hmm , now i have homomorphism , i can try form here

[gritty sparrow]

And the kernel of this map is (x,y), so by the first isomorphism thm C[x,y]/(x,y) is C

[gritty sparrow]

No it corresponds to a parabola, so (0,0) is in the curve, so it (4,2) and (9,3) and so on

Here the brackets mean coordinates, not ideals or anything

[chilly ocean]

ok so I thought about it rather like you have a plane and this parabola is like annihilated or sth, thats how I understand quotient

[trim grove]

now i understand this

[gritty sparrow]

Ok so here is a glimpse at the correspondance, let f in be a polynomial in n variables, we will correspond the set of all zeroes in C^n of f with the ring C[x1,..xn]/(f).

And here is something nice, if (a1,..,an) is a zero of f, then the ideal (x1-a1,x2-a2…,xn-an) is a maximal ideal in C[x1,..,xn] that contains (f) so it is in correspondance with a maximal ideal of C[x1,..,xn]/(f)

[chilly ocean]

hmm this is some nullstellensatz or sth right

I think I've seen it only in one variable case

[gritty sparrow]

Well so far i haven’t done anything that needs fancy stuff, the nullstellensatz is the reverse, that all maximal ideals containing (f) are of this form

[delicate hull]

Let, $G$ be a finite subgroup of $GL_n(\mathbb K)$ where $\mathbb K$ is an algebraically closed field then there exists a basis of $\mathbb K^n$ with respect to which every element of $G$ is diagonal matrix.
$\textbf{Statement is false} $ . But I can't get any counter example. Can anyone give me any idea that how to proceed?

[cloud walrus]

Aritra

[delicate hull]

If $char(\mathbb K)>0$ then I can prove that there exists a group $|G|>1$ such that $G$ is contains the upper triangular matrices

[cloud walrus]

Aritra

[delicate hull]

So, at least I can say that if characteristic of the field is greater than 0 then we can say there does not exists such G

[chilly ocean]

Can't you just find a G that is noncommutative?

Or is that hard?

[thorny knoll]

If you found a finite subgroup of, say, GL_2(C), which contains a non-diagonalizable matrix you'd disprove the statement by counterexample, wouldn't you?

[delicate hull]

Can we find such G? Here the field is C, characteristic 0. I am not sure.

I mean I think this result is true for characteristic 0 fields

[thorny knoll]

Hmm fair I can't come up with an example of the top of my head either

[gritty sparrow]

I think the matrix {{1,i},{0,i}} should work

It is non diagonalisable and I think its fourth power is identity, so it generates a finite group

(But do check if it really works)

[delicate hull]

Ah! Thanks I got it. We can take ${(0,-1), (-1,0)}$ as a basis. It will work and $2nd$ power is identity.

[gritty sparrow]

Oh nevermind, the matrix is diagonalizable

[cloud walrus]

Aritra

[delicate hull]

No no it's correct. Means we got a basis such that the elements of G is non diagonal

[gritty sparrow]

We need to find a group such that no basis makes all elements of the group diagonal, we can’t pick any basis we like

[delicate hull]

Hmm.

[gritty sparrow]

Anyway, the group can’t be generated by one matrix as if it were then it would satisfy x^n-1 which is a separable polynomial, so it would be diagonalisable. I suppose you would have to take two noncommuting such matrices. That will work because then they won’t be simultaneously diagonalizable, the hard part would be showing the group generated will be finite

[delicate hull]

Can we proceed in this way: If $\exists$ a basis in $\mathbb K^n$ such that every elements of $G$ is diagonalizable so $\exists P\in GL_n(\mathbb K)$ such that $A=P^{-1}DP$ where $\forall A\in G$ and $D$ diagonal matrix?

[cloud walrus]

Aritra

[delicate hull]

Then $A^{|G|}=P^{-1}D^{|G|}P\implies Id_n=P^{-1}D^{|G|}P$. So $D^{|G|}\sim Id_n$

[cloud walrus]

Aritra

[delicate hull]

Then can we arrive at any contradiction?

[gritty sparrow]

Why are you trying to arrive at a generalised contradiction? Just find one counter example. In any case, you can’t find a contradiction for all G because there are infact groups that are simultaneously diagonalizable (the identity group) so this seems like it won’t help

[delicate hull]

Yes I am trying to find it but not getting

Yeah we need to take |G|>1 otherwise it make no sense. Right?

[gritty sparrow]

Ok here is an example:
Consider the group generated by {{i,0},{0,i}} and {{0,1},{-1,0}} this is finite and non abelian

Finally lol

[delicate hull]

Yes, thanks!

[delicate hull]

it's {{0,1},{-1,0}} right?

[gritty sparrow]

Yes that’s right

Lemme edit that now

[delicate hull]

Yes!

[unique juniper]

Why is the dimension n-1?

[crimson falcon]

Doesn't the equation add a constraint reducing the dimension by 1?

While we are at it so people might see it a have a question of my own. I'm told to show that, given a group, the subgroup generated by S is the following.

I "did" it by construction. i.e. i start from one s1 element, its generated subgroup i need his inverse and every element of the from s1* s1* s1* s1* ..., add s2 and say the subgroup is, by construction now, includes that inverse, and every element of the form s1* s2 too, and so on so forth. As in each step i add only the minimum of elements necessary to be a subgroup, then it's the subgroup generated by it.

But the answer given goes for a more complicated process of doble inclusion and other stuff. So ...what am i missing?

[unique juniper]

i dont really know

[crimson falcon]

what exatly are A and X there?

[unique juniper]

oh im asking about the next part, V is the vanishing set of 1 polynomial equation so the roots, x's are just indeterminates, the dimension of V is the transcendence degree of k(V) over k, but i dont understand why a set of n-1 elements in k(V) will only satisfy the 0 polynomial

[crimson falcon]

ahh you're asking about "the converse is also true" ?

[unique juniper]

no

like

i dont follow that example

why is there no non zero polynomial in k[x1, .. ,xn-1] that has a1,..,an-1 as roots

when a's are roots of that initial polynomial

[crimson falcon]

You're parting from A^n, in which there is no non zero polynomial, right?

If V had such non zero polynomial, then you would have the same polynomial in A^n, no? (just get 0 for the X^n)

[unique juniper]

hmm

[crimson falcon]

If you have X, X2...Xn algebraically independent, you add a constraint/equation you basically make one of them dependent of the others (this is your non zero polynomial that gives you V) but the other one are still independent

[unique juniper]

right

in the first part, k(A^n) has a dimension of n, thats saying that for a set of n things, the only polynomial it can satisfy is the 0 polynomial?

[crimson falcon]

I'm assuming that dimension here is the same dimension i've seen before aka the cardinality of the basis

[unique juniper]

yes

[crimson falcon]

assuming you're not pikcing stuff lke, n things that are just scalar products of the same thing (idk what A^n is there). Mind you, i'm just seeing this as somethig totally analogous to vector spaces with the exception that instead of linear independance we ask algebraic independence, so take it more like "bouncing ideas" rather than "getting answers" lol

[unique juniper]

hm

for the first part

im guessin the a's are not in F, so in our example, it would the x's right?

then f(x1, .. xn) cant be 0 unless f was 0, if we only take coefficients from F

is this it?

[crimson falcon]

I think yeah, if i understood you correctly

https://en.wikipedia.org/wiki/Algebraic_independence the first example is pretty clear i think

[trim grove]

not getting this, can you please elaborte?

[unique juniper]

ah yeah that makes sense

[delicate hull]

Can not understand how to proceed.

[wooden ember]

im feeling a bit stupid... In the hint it says G/N acts as automorphisms of N but I fail to see how

i dont want a solution to the exercise Im just wondering how G/N acts on N as automorphisms

[cloud walrus]

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

[wooden ember]

not looking for a solution

[delicate hull]

Just wait

please

[wooden ember]

alright

[delicate hull]

I will let you know when it is complete

[wooden ember]

oh i figured it out it's because $N_G(N)/C_G(N)\cong H\leq Aut(N)$, and we have $N_G(N) = G$, $C_G(N)=N$

[cloud walrus]

Little Narwhal ✓

[wooden ember]

wait no that's not quite right, i cant justify $C_G(N)=N$ sorry

[cloud walrus]

Little Narwhal ✓

[delicate hull]

@wooden ember please have a look

[wooden ember]

I would not know how to show the very final part

(and i dont see what it has to do with my question)

[delicate hull]

You can give an isomorphism $f:G \to \mathbb Z_p_{1} \times \cdots \times \mathbb Z_p_{r}$

[cloud walrus]

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

[delicate hull]

by kernel is identity

[wooden ember]

im sorry i appreciate that youve taken time to type this but i simply dont see how it helps with my question, even if i took the conclusion to be true as that would anyhow solve the problem in the first place

[delicate hull]

Your question is to show G is cyclic right?

[wooden ember]

.

[wooden ember]

.

[delicate hull]

Oops sorry sorry

[wooden ember]

it's aight

[delicate hull]

I just have seen the problem that G is cyclic

extremely sorry

[wooden ember]

haha it's fine i tried to tell you to save you the typing

frankly even if I suppose I understand, and take it for granted G/N acts as automorphisms of N, right now i dont quite see why N<=Z(G) or why when we get the final conclusion that G is abelian we can then conclude G is cyclic

I need to spend more time on this one...

[delicate hull]

Actually I also can not understand the sentence G/N acts as automorphisms of N

[wooden ember]

ah so im not the only one at least

[delicate hull]

yes yes

but this problem can be solved in some easy way also

[wooden ember]

yeah but id like to follow the given hint

[delicate hull]

as i did...

[wooden ember]

maybe they made a typo and meant to say G/C_G(N)

then it makes more sense

and then i can see how maybe by divisibility considerations we must have C_G(N) = G

[delicate hull]

yes, I am trying to prove in your approach

[delicate hull]

yes yes I also think so

[wooden ember]

ah right i think i know: The quotient G/C_G(N) is of order $p_1p_2...p_s$ (renumbering the prime divisors of G as necessary), and N is of order $p_{s+1}...p_r$ so that the automorphism group of N is of order $(p_{s+1} -1)...(p_r -1)$ and since we have the divisibility condition on the primes we must have that G/C_G(N) is trivial

ah right i think i know: The quotient $G/C_G(N)$ is of order $p_1p_2...p_s$ (renumbering the prime divisors of G as necessary), and N is of order $p_{s+1}...p_r$ so that the automorphism group of N is of order $(p_{s+1} -1)...(p_r -1)$ and since we have the divisibility condition on the primes we must have that $G/C_G(N)$ is trivial

bruh latex why

[cloud walrus]

Little Narwhal ✓

[wooden ember]

there

yeah this is right

[delicate hull]

now fine

[wooden ember]

cause the automorphism group of N has order phi(|N|) where phi is the euler totient function

and since we had p_i does not divide p_j-1 the quotient is forced to have order 1 by lagrange's theorem

sound good?

alright with this the next steps follow easily towards showing that G is abelian

all that's left for me is to show G is cyclic, thanks for the help @delicate hull

right the last step is easy

[urban ice]

I don't understand how $\sigma^{-1}\overline{\sigma}\sigma(x)=x$, is there some cancellation or property of $x$ that I'm missing?

[cloud walrus]

Syst3ms

[delicate hull]

yes it's fine

[rustic crown]

first apply sigma^-1 to get y then apply sigma bar to get y again, lastly apply sigma to get x

[unreal portal]

Can I get a quick sanity check? The groebner basis of a principal ideal is just that one generator $g$ right? Because we can arbitrarily write any polynomial $f$ as $f=q\cdot g + r$ where no term in $r$ is divisible by the leading term of $g$, and $f$ is in $\langle g\rangle$ if and only if $r=0$

[cloud walrus]

cgodfrey

[unique juniper]

yes

[rustic crown]

idk about gobner basis

[unique juniper]

(i think)

[rustic crown]

but you can't always do division with remainder

take Z[x] and divide x^2 + 1 with 2x

[unreal portal]

right

but I'm in Q[x,y], so I can do division

[unique juniper]

if f was inside the ideal then you will

have no remainder

[rustic crown]

idts?

Z and k[x] are a lot similar

so Z[x] and k[x, y] are very similar

try to divide x^2 + 1 with xy

[unreal portal]

Dummit and Foote defines polynomial division on F[x_1, ..., x_n] with F a field

your example is okay because the no term in the remainder (x^2 + 1) divides the xy

[rustic crown]

no but usually the point of division with remainder is that the remainder is in some sense "smaller"

[unreal portal]

ooh

well I don't care about that part of division

[rustic crown]

you can ofc do f = 0*g + f

[unreal portal]

what I care about is that if I have a principal ideal $I$ with generator $g$ in a ring $R$, given a polynomial $f$, $r$ is a unique representative for the coset of $f$ in $R/I$

[cloud walrus]

cgodfrey

[unreal portal]

(specifically if the generator of the principal ideal also serves as a groebner basis for the ideal)

[unique juniper]

if its a grober basis then id say yes

[unreal portal]

I'm basing it off this, the special case where the ideal is principal with generator g

particularly (3)

[unique juniper]

principle ideals are already grobner basis

[unreal portal]

okay yes, that's what I wanted the sanity check for

thanks

[unique juniper]

radicals of an ideal is always prime right?

[unreal portal]

yeah I think so

[unique juniper]

i havnt done this yet

but why is it asking to show that (rad I) is not prime?

[unreal portal]

oh wait, that's the other way around

the radical of a prime ideal is just that same ideal

[unique juniper]

but the radical of an ideal I will be the intersection of all prime ideals containing I yes?

[unreal portal]

that part is right, yeah

[unique juniper]

i might be missing something

like if rad I is prime how will we show that rad I is not prime

[maiden ocean]

rad I doesnt have to be prime

[unique juniper]

if we intersected prime ideals, would we not get a prime ideal?

[maiden ocean]

Uh no

[unique juniper]

oh lol

[urban ice]

oh right, "left to right", what bullshit

[hidden haven]

2Z ∩ 3Z 😌

[maiden ocean]

In a PID (2) cap (3) = (6)

Which is not prime

[hidden haven]

[unique juniper]

lol

[maiden ocean]

fuck you moldi

[unique juniper]

my bad lol

[rustic crown]

yea lol

[thorny flame]

What is Bazo's Lammet?

[mild laurel]

maybe you're looking for Bezout's Lemma?

[thorny flame]

I think so

[upper inlet]

Bozo's Lament

[south patrol]

Bezout's lemma states that if gcd(x,y)=d, there exist integers a,b such that ax+by = d

importantly, if x,y are coprime then you can write any integer as a linear combination of x and y

[urban ice]

it's like "oh yeah, let me revert the order of composition that literally everyone uses everywhere for a sec"

[south patrol]

how common is that order for permutations? lol

[urban ice]

i'd hope not

[south patrol]

my uni for some reason said 'at this uni everyone uses this order' [where they act from the right lol]

[urban ice]

worst thing is, sometimes he uses the other order, and says "read from right to left"

[south patrol]

oof

[urban ice]

anyway, that does clear things up

[delicate orchid]

just out of curiosity does anyone recognise this group presentation and is it isomorphic to anything funny
<a, b : a^8=b^4=1, a^4=b^2, b^-1ab = a^-1>

it's kinda like a dihedral group but with a four-cycle reflection thing which is quite nifty

[kindred mist]

So I notice that Hungerford has a little bit about cardinal arithmetic, does anyone have any recommendations for further reading about cardinal arithmetic, or is what's in Hungerford probably enough for commutative algebra & algebraic geometry? (which is what I plan to study)

[fossil shuttle]

cardinal arithmetic is not a huge deal but it's also like, fairly basic set theory. any textbook on set theory will have stuff on cardinal arithmetic. for example, Enderton's Elements of Set Theory.

Ordinals and cardinals are important in algebra for doing induction arguments

like the proof that every submodule of a free module over a PID is itself free relies on transfinite induction

you might also encounter cardinals when you're studying field extensions of infinite transcendence degree, but that's kind of a stretch. Really you should be able to go a long way with just a good working knowledge of Zorn's lemma (which is used in lots of places in algebraic geometry, i.e. proving that the radical of an ideal is the intersection of all primes containing it, proving that flasque sheaves have vanishing higher Cech cohomology, etc)

[chilly ocean]

Whenever you don't know how to prove a statement, just use Zorn's lemma. It works about half the time.

[vestal snow]

Can someone explain why p_X(0) equals the characteristic of X here?

p_X(m) is defined as h^0(X,O_X(m))

[gritty sparrow]

p_X(m) is defined to be X(X,O_X(m)) right?

[vestal snow]

If you look at definition 18.6.2, you'll see that it's defined as the 0th cohomology dimension of the m^th line bundle over F

[gritty sparrow]

So what he’s doing is defining p_F(m) as X(X,X(m)) and showing that this is a polynomials in m. Then for m>>0 all the higher cohomologies will vanish, so h^0(X,F(m)) is eventually equal to p_F(m) hence it is eventually polynomial

[vestal snow]

Oh my bad

[vestal snow]

Then what is so surprising about this result?

Like it's literally the definition

[gritty sparrow]

I have no clue

[vestal snow]

Thanks for the help!

[gritty sparrow]

Sure thing

[thorny flame]

Can you use euclidean algorithm to check the gcd of 3 numbers?
(15,37,21)

[gritty sparrow]

Sort of, given 3 numbers a,b,c to find the gcd what you can do is find the gcd of a and b, call it d. Then find the gcd of d and c, call that e. e will then be the gcd of the 3 numbers

[thorny flame]

so you need to do it 3 times

[gritty sparrow]

No, 2 times

Once for a and b to get d. Then again for d and c to get e

[thorny flame]

ohh. I see

thanks

[gritty sparrow]

Np

[next obsidian]

Abstract algebra moment

[wooden ember]

so ive solved exercise 6 here and im more or less confident in my method but id love some feedback on the formulation: i tend to write things very vaguely in constructions... Here it is:
Fix some $\sigma \in A$ that moves a finite subset $N_1$ of $\Omega$. $\sigma \in A_{N_1}$ and $A_{N_1}$ is a finite simple subgroup of $A$. Suppose we have picked $A_{N_1},...,A_{N_k}$ for $k<m$. Pick a $\tau \in A$ such that the subset in moves $N_m$ is a superset $N_1\subset ...\subset N_k\subset N_m$, and append $A_{N_m}$ to our list of finite simple subgroups. Then we get an inclusion chain $A_{N_1}\leq A_{N_2}\leq ...\leq A$ with $A=\bigcup_{i=1}^{\infty}A_{N_i}$ so that by the previous exercise $A$ is an infinite simple subgroup of $D$.

[cloud walrus]

Little Narwhal ✓

[wooden ember]

whenever i have to deal with infinite constructions by induction like this i always feel my formulation to be very awkward

[hidden haven]

The only problem I see is that you didn't prove that the union is equal to A, whatever you said only tells you that it is included in A

[wooden ember]

yeah that's my main issue it feels obvious to me but i dont know how to say it properly

[hidden haven]

Ah are you taking N_m to be a strict superset in each step

[wooden ember]

yes

so like eventually the set that any tau of A moves is included in some N_m if you catch my drift

[hidden haven]

Yeah then it should work because any permutation's non fixed points should be in some N_i eventually

[wooden ember]

but i dont know how to say that correctly

[hidden haven]

Yeah

Lol

That is fine I'd say. What part of it do you think is not formal enough?

[wooden ember]

the fact that i cant write it out formally. If you catch my drift isnt exactly a phrase i should use in a proof lol

[hidden haven]

Ah lol ok

So you have to show that A ⊂ that union

[wooden ember]

yep

[hidden haven]

oof I see why

There's an error

[wooden ember]

ouch

[hidden haven]

The N_m, even if they grow each step, need not equal the whole set eventually

[wooden ember]

oh i see what you mean

[hidden haven]

Is Ω an arbitrary set or countable?

[wooden ember]

arbitrary

[hidden haven]

[wooden ember]

which makes everything harder for me or i couldve just imposed an indexing

[hidden haven]

Then this won't work, because finite sets can't countable union to uncountable

[wooden ember]

fair enough...

[hot lake]

maybe exercise 5 also works with uncountable chains

is the thing in brackets supposed to be a hint ?

[wooden ember]

yeah but i couldnt see how to apply it

[hot lake]

it's very unclear what they mean

but maybe it's "forall x,y in G prove that y is in the smallest normal subgroup of G containing x" ?

ah the group is named A and not G

[wooden ember]

yeah but for some reason the hint tells you to work with D

[hot lake]

yeah idk they're weird

[wooden ember]

i mean the hints arent always flawless i spent hours on an exercise trying to parse the hint yesterday because of a typo

so after a quick search online it seems we do not need to require that omega is countable

right okay the proof is far simpler....

[hot lake]

did they do it with the weird hint

[wooden ember]

sort of??

lemme send a screenshot

they keep saying G but mean A

but yeah they're considering pairs of elements of A not of D

and even then it's not exactly the same as the hint

[hot lake]

looks like what I was imagining

[wooden ember]

i see

i always get lost in arguments about infinite sets tbf cause i stop myself from visualizing since i know if I do ill make a mistake from "intuition"

as i did earlier

[hidden haven]

Based finitist

[rustic crown]

henlo Moldi

[hidden haven]

Henlo det

[rustic crown]

didn't you tell me that all sets are finite 😆

oh oops, i think that was countalbe

not finite

[hidden haven]

Did I

[wooden ember]

any set too big to imagine doesnt exist

[chilly ocean]

[hidden haven]

Are you sure I wasn't joking

[rustic crown]

it was in the context of logic and foundations, so idts 😛

[hidden haven]

Wait did I really say that

[chilly ocean]

[delicate orchid]

That’s gonna be a ban I’m afraid

[wooden ember]

maybe it was one of the imposter moldis

[hidden haven]

Oh was I talking about skolem's paradox

Which gives a countable model of ZFC and makes all sets meta countable 😌

[rustic crown]

our sem 1 problem proved that R is countable

😆

[hidden haven]

prof* right

[rustic crown]

oh oops

[wooden ember]

how so?

[rustic crown]

yea

so she defined R using Q and Q^c
and defined Q^c using R

[hidden haven]

Didn't seem to know the definition of countability

[wooden ember]

circular reasoning is where it's at

[hidden haven]

Why is the second statement getting sullied that one's fine

The first one's the issue

[wooden ember]

im sullying the message as a whole

[rustic crown]

lol

[trim grove]

here he is asking for trace of permuation matrix so total martrices are 10! , and posability of trace are 1,2,3,4,5,6,7,8,9,10 but answer given is something else.

[dusty river]

Expected value weighted average of possible values

Weighted by their respective probabilities

[trim grove]

so i have to find each type permutation and multiply with each its sprobabilty

?

[dusty river]

Yeah, for each n from 0 to 10, you find the probability of a random permutation matrix having that trace

[hot lake]

I assume the matrix is picked uniformly at random

[dusty river]

Ye randomly usually mean uniform distri

[foggy merlin]

does there exist a subgroup of order 14 in s7? I kind of struggle with that question, according to the sylow theorem it needs to have an element of order 7 (which must be a 7-cycle) and an element of order 2. but we cannot pick a 2-cycle as our element of order 2 since both of those elements would generate s7 by themselves, so I would have to pick another element of order 2, (if it even is possible).

[trim grove]

but find each kind of permuation is like finding partition of 10, that will be a lenghty process

[dusty river]

You can find an expression for number of matrices with trace = n

[trim grove]

okk like for trace 0 it will be of from (12345678910) permuation, and for trace 2 it will be of form (1)(2)(345678910) form , am i rigth?

[dusty river]

I don't think so, but I'm not sure if I understood what you said

The trace = number of fixed points of the permutations

[trim grove]

i edited

[dusty river]

Ye something like that but the remaining things need not be in a single cycle

[trim grove]

let me try

[foggy merlin]

nvm I found sth

[hot lake]

that sounds way too complicated

[trim grove]

hmm, for trace 1 it comes out 10!/9 , for 2 = 10!/2! 8 and so on..

[hot lake]

how many matrices are you saying have trace 0 ?

[trim grove]

(12345678910) total 10!/10 =9!

[dusty river]

It shouldn't be that simple yeah

You're counting derangements

[hot lake]

you counted the number of 10-cycles

but there are many many more permutations that have no fixed points

counting derangements is difficult

[dusty river]

Isn't there an alternating sum formula

[trim grove]

isnt they will be conjugate to it?

[dusty river]

From inclusion exclusion

But yeah if you haven't seen that

[hot lake]

no ... the set of 10 cycles is already a conjugacy class ...

[trim grove]

like?

[trim grove]

okk like (12345)(678910)?

[hot lake]

and many more

counting the number of derangements is much harder than figuring out the average trace

[trim grove]

i am using this fomrula for expected value = $\sum P(X_{i} )\times X_{i}$

[cloud walrus]

honey99

[hot lake]

well that one isn't going to be very helpful

[trim grove]

can you please tell me another way?

[hot lake]

count the number of permutation matrices that a 1 in the upper left corner (the entry indexed by (1,1))

then count the number that have a 1 in the entry indexed by (2,2)

and so on for each entry of the diagonal

[trim grove]

so that means i have to fix each daigonal entry like (1)(2345678910), (2)(1,34567910) so on ...?

[hot lake]

yeah you go through each square of the diagonal

and count permutation matrices that have a 1 on them

[trim grove]

like fixing 1 on counting others ,hmmmm, so that means i have to fix 1 and rotate the elemnts in (2345678910) okk so in (2345678910) there are total nine numbers and so ways should be 9!?

wait , so for each permuation the ways are 9! , so total marices are 10(9!)= 10! and out of 10! so 10!/10! =1 ?

[hot lake]

are you saying that there are 9! 9-cycles made of 2,3,4,5,...,10 or am I misunderstanding your notation

[trim grove]

sorry for my english(this is not my first language), yes i mean to say that there are total 9! matrices which have 1 at (1,1)

[hot lake]

then yes

[trim grove]

so total will be 9!+9!+9!+...... 10 times = 10(9!) =10!

[hot lake]

yep

[trim grove]

okk =, now what to do for expected value

[hot lake]

you divide by the total number of matrices

and you get 1

[trim grove]

yes

beacue each matrix have probabilty of selection 1/10! (because each matrixx is unique in itself)

[delicate hull]

For every positive integer n and for every m with 1 ≤ m ≤ n, S_n has a cyclic subgroup of order m. How can I prove it? Could you please give me any hints?

[dusty river]

Can you think of a permutation of n elements that has order m?

[delicate hull]

Yes but permute m elements and other (n-m) fixed

So that become cyclic

[trim grove]

like we want to make a cyclic group of order 3 then we can make form (123) , it has order three and can generate cyclic group of oder 3 ,?

[delicate hull]

Yeah that is true, but I am talking about if m<n then

Just permute m elements and other are fixed

Is it the process?

[dusty river]

You should specify how you're permuting them

Many ways to permute m elements, and not all are order m

[delicate hull]

No no, Like f(i)=i+1

and f(m)=1

[dusty river]

f(m) = 1 right?

[delicate hull]

(123...m) then other are fixed

[dusty river]

Yeah

[delicate hull]

yes yes

[dusty river]

Cool so the powers of that element form a subgroup

[delicate hull]

Yeah I got it. Thank you

[trim grove]

i have not written but that (123) is of type (123)(4)(5)(6)....(m)

[dusty river]

Aritra are you from India btw?

[trim grove]

never heard this name in Indian name list 😅

[dusty river]

I've seen it once I think it's a Bengali name

[delicate hull]

Yes I am from India

You?

[dusty river]

Which year

Ye india, in CMI

[paper flint]

Flex

[dusty river]

[trim grove]

this is dream institute for lots of Indian students

[paper flint]

I'm one of them

[trim grove]

same here

$f(x,y)=x+y$ maps open to open and closed to closed? for open part i used open maping theorem and that is true , but for closed one what method should i use?, any hint

[cloud walrus]

honey99

[dusty river]

Closed iff sequentially closed for R^n

Might help

[chilly ocean]

#groups-rings-fields

[dusty river]

Don't you see the + terra

[chilly ocean]

[dusty river]

Are you sure that this is true

I feel like graph of -x + (1/x) for positive x is a counterexample because this is closed but its image is (0, ∞)

@trim grove

[delicate hull]

One can think of $\mathbb Z\times \mathbb Z\sqrt2$

[cloud walrus]

Aritra

[trim grove]

but here function is in two variable, is it legal to use -x+1/x?

[dusty river]

Ah that one is good

[dusty river]

I mean the set of points (x, -x+1/x) for positive x

[delicate hull]

is colsed in $\mathbb R^2$ but image is dense in $\mathbb R$

[cloud walrus]

Aritra

[delicate hull]

So, image may not be colsed

[dusty river]

Right, countable and dense

[delicate hull]

Yes

[trim grove]

okkk , so image is dense , but for closed set it should be same as that of set , but here its clouser is whole R?

[dusty river]

Ye

[cloud walrus]

Little Narwhal ✓

[wooden ember]

because since D is generated by transpositions you can view the generating set as being in bijection with $\Omega \cross \Omega$

[cloud walrus]

Little Narwhal ✓

[wooden ember]

so then generating D from a set of cardinality equal to $|\Omega \cross \Omega| = |\Omega|$ it would be very practical to conclude $|D|=|\Omega|$

[cloud walrus]

Little Narwhal ✓

[rustic crown]

are these problems continuing from some previous ones you asked here?

[wooden ember]

yes though for the particular question im asking i dont think it matters

im more interested in knowing if in general, a group generated by a set of infinite cardinality n also has cardinality n

[rustic crown]

so the free group?

or generated in some other sense?

i see S_A is that permutation group?

[wooden ember]

again i havent seen free groups yet but generated in the sense G is generated by A iff G is the closure of A

[rustic crown]

there is this thing equivalent to axiom of choice that if you have an infinite set then |A x A| = |A| and by induction |A^n| = |A|

[wooden ember]

this is assumed in the hints yeah

[rustic crown]

so anything in G could be written as a finite product of elements in A

cardinality of G can't be more than cardinality of the disjoint of A^n for n >= 0

[wooden ember]

but could it be less?

oh no i see

[rustic crown]

nope!

it contains A

[wooden ember]

since it's lower bounded by the generator yeah

right thanks

[rustic crown]

more formally, you have injective maps A --> G --> disjoint union of A^n

[wooden ember]

i had the same idea somewhat but couldnt quite make it formal

[wooden ember]

right

[rustic crown]

oh oops

one more thing

[wooden ember]

and then we just use cardinality facts on injective functions

[rustic crown]

make that, you have injective maps A --> G --> disjoint union of (2A)^n

we can also use inverse of generators, i always forget about them

[wooden ember]

fair enough

[rustic crown]

yup

[wooden ember]

i dont quite understand why you mention disjoint unions?

[rustic crown]

the reason is more in the lines of abstract non-sense

but it really doesn't matter

[dusty river]

Do you mean why not union instead of disjoint union here?

[wooden ember]

no just why there is a union at all

[dusty river]

Because you want to take the set of all words in 2|A| many letters

Each word has some length n

(2A)^n is set of all tuples of length n

[wooden ember]

ah cause we cant say there is some maximal such n

[dusty river]

Yeah, we need all finite lengths

[wooden ember]

i see

so why disjoint then

[rustic crown]

they way i was thinking was, we have a map (2A)^n --> G which intuitively is like looking that word and evaluating in G

so abstract non-sense gives us a map from disjoint union of (2A)^n --> G, and by the assumption that A is a generating set, this map would be surjective

[dusty river]

Disjoint here is redundant yeah

Because they'll be all discount anyway

[wooden ember]

wait is abstract non-sense an actual thing or do you mean it's more complex than that

[rustic crown]

discount

[wooden ember]

discount on our sets today folks

[dusty river]

[rustic crown]

abstract non-sense is a cute name given to category theory

[wooden ember]

haha okay

[rustic crown]

[dusty river]

As opposed to cat theory which is a horrifying name

[rustic crown]

cat theory is also cute

[dusty river]

[rustic crown]

[prime sundial]

when did this turn into #cats

i can participate now

[rustic crown]

[sharp dirge]

how to do this?

the second part i mean

dimV = dimKer + dimImage

but if image is isomorfic to V/ker

does kernel containing image make dimker smaller

[chilly ocean]

other way around

[sharp dirge]

i can only see that dim image is less than kernel

yes

wait thats it

[sharp dirge]

[kindred mist]

trying to solve this for a while now. I have "most" of it figured out I guess. More precisely: letting M be a f.d. extension of F, I think I know what to do once I can show that M is normal over F (assuming that M is contained in K bar (1), which I guess we can assume). Thus, we can make a few assumptions by way of contradiction (to rule out all of the cases where M is automatically normal over F), namely: (i) M is not normal over F, (ii) F is not finite, and (iii) every extension E properly intermediate to the extension F\subset M is normal over F, and thus, M = F(\beta), where the minimal polynomial g of \beta over F is separable (this comes from (1) and the fact that K bar is Galois over F by Hungerford's definition of Galois), and not every root of g lies in M. By (iii), the primitive element theorem gives (by separability of M over F) that every proper intermediate field E is of the form F(\alpha) and contains a splitting field over F of the minimal polynomial of \alpha over F (since it is normal, by our assumption on M). --- This is as far as I have gotten, I just don't see how to get a contradiction from this stuff, it has to do with the fact that F is the fixed field of the subgroup of Aut_K K bar generated by \sigma, I just don't see how exactly (I do see how to use this fact once we can prove M is normal over F, just not prior to)--- tldr: M is a f.d. extension of F, I need to show M is normal over F and I know what to do from there I think (yes I notice that F is the fixed field of <\sigma > btw)

(maybe there is some way to do this without assuming that M is contained in K bar, but then I have no idea what to do)

[dusty river]

F is the fixed field of the cyclic subgroup generated by σ. Now you can use Galois correspondence (any finite extension of F will be contained in F closure = K closure, so will correspond to some subgroup of Gal(K clos/F). All subgroups here are normal)

[trim grove]

is group free ?can i ask question?

how can i find number of niloptent elemnts in $\mathbb{Z}_{n}$? ( cyclic group of order n)

[cloud walrus]

honey99

[kindred mist]

I understand everything except the part about the subgroups of Gal(K bar/ F) being normal, why are they normal?

[dusty river]

It's an abelian group, because it's cyclic

Isomorphic to Z

[kindred mist]

Wait it's <\sigma> right?

[dusty river]

Yeah, sorry but isomorphic to Z

But it's cyclic

[kindred mist]

is it possible that \sigma has finite order though?

[dusty river]

All cyclic groups are abelian

Yeah then it will be some Z/nZ

[kindred mist]

right ok, now it all makes sense to me

thanks

[dusty river]

[kindred mist]

Idk why I didn't think to just use this fact

[dusty river]

😌

[dusty river]

An element is nilpotent in this iff a power of it is divisible by n

Prime factorize n

[gritty sparrow]

This is a bit of a weird q because he’s asking for a nilpotent in a group, which technically would just be every element right? (Because here the group product is addition). I think they maybe meant to ask about nilpotents as a ring

[dusty river]

Ye I was assuming ring

[trim grove]

Let me check once again

Yes they are asking for ring Z/nz

[trim grove]

Why so?

[gritty sparrow]

By definition, pretty much. If a^m=0 mod n, that is just saying that some power of a is divisible by n

[trim grove]

Is this because of characteristic?

[dusty river]

This is how you define a quotient

a = 0 in R/I iff a ∈ I

For a ring R and ideal I

[trim grove]

Got it.

[trim grove]

elements of Q/nZ will br of form p/q +nZ , form , so for first part it is true , but in second part , there are countable many elements like for n=5 , we have 5/2 ,15/2,25/2..... so on elements of order 2

[gritty sparrow]

5/2=15/2 (can you see why?)

[trim grove]

because there remainder is same for 5Z?

like 5/2 +5z, 2.5+5z and for 15/2 + 5z = 7.5+5z = 2.5+5z?

[gritty sparrow]

Yes pretty much

[trim grove]

can you make that "pretty much " to yes definitely?

[gritty sparrow]

Yes definitely

[trim grove]

no no means that , if i made some mistake , can you please fix that?

[gritty sparrow]

No mistake

[trim grove]

but how to prove that ? any hint

[gritty sparrow]

Well, I guess the hint is, given some order t, try to find out what elements of order t look like. Basically they will be elements of Q, q, such that tq is an integer and divisible by n. So what sort of rationals are possible, and what are do the equivalence classes look like? Here is another hint: all elements are equivalent to some element less than n and greater than 0.

[trim grove]

like for rational of form p/q we say it has order t if t*p/q = mn, but how to find p/q elements?

[gritty sparrow]

I feel like I maybe shouldn’t give any more hints, try to work this out in the case n=1 atleast

[trim grove]

ok

[trim grove]

i'm trying this but not getting this ( trying for n=1 ,i.e integers) , like for element of p/q , i want to find the elements that have order 2 , i.e 2p/q is a integer , but here p an q are co prime , but how can i find q such that this should be coprime of p and also divide 2p?

[dusty river]

If 2p/q is an integer, the denominator must have cancelled with something in the numerator

[gritty sparrow]

Ok, one more hint, consider the rationals that have t in the denominator, do you see that every element of order t is equivalent to some fraction of this form, and I can even take this fraction to be a positive less than n? (By fractions with denominator t i don’t necessarily mean reduced fractions). Now of course there is the possibility that not all such fractions are of order t, but we don’t really care (if you want you can figure out which ones have order t) in any case there are only finitely many such fractions less than n, and that will give us an upper bound on the order t elements.

[carmine fossil]

I have a feeling continuity is important here

Like I think you will conclude f is a constant function via some epsilon delta argument

[rustic crown]

so if f is in the family, it would be bounded on [1, 2] as its continuous and [1, 2] is compact. for any other positive real you can scale by 2s to get in the interval [1, 2]

to see that, consider 2^n where n varies over Z. for any positive real you can find 2^n <= x <= 2^(n+1)
so 1 <= 2^-n * x <= 2

[trim grove]

does same work for (0,infty)?

[delicate orchid]

topology in abstract alg

[trim grove]

every time i make this mistake ,sorry for this

[rustic crown]

let's move to #point-set-topology

[trim grove]

okk

[rustic crown]

wait i'm confused, should we move to analysis-pde?

[trim grove]

i think i should stop asking hints, , should ask for a complete solution

[knotty mason]

2p/q = n

2p = nq

p | n since p not | q

q | 2 since q not | p

so q = 2 or 1

p can be anything

[rustic crown]

p,q positive primes?

[dusty river]

Probably answering honey99's question 😌

[maiden ocean]

Is there a short proof that if A is noetherian and M is an A-module then for x in M, r(Ann(x)) prime implies Ann(x) is prime

i know that letting p = r(Ann(x)) we have p^n subseteq Ann(x) subseteq p for some n, so for a in p we have a^nx = 0

i guess if we let n be the minimal such n that this is true a^n-1 x is nonzero so a(a^n-1 x) = 0 and then if A is an integral domain we're done

But without that i dont see what to do

actually no im 4headed integral domain does nothing because x is an element of a module

[gritty sparrow]

Is this really true? I think with A=Z and M=Z/p^nZ viewed as an A module, the annihilator of 1 is p^nZ, whose radical is pZ, but p^nZ is not prime

[maiden ocean]

actually

ok wait yeah you're right

but it should be true that r(Ann(x)) = Ann(y) for some y in M

associated prime of the 0 module

[gritty sparrow]

Yeah that should be a zorn’s lemma argument

(I think)

[maiden ocean]

my idea was to replicate the proof in the case of ideals

but it doesnt really translate

maybe im supposed to use ideal quotients or something

actually let me try that

[gritty sparrow]

Hmm so i thought that you could just choose a maximal ann(y) that will contain ann(x) or something along those lines, and ann(y) will automatically be prime

But I haven’t checked this very properly, just to be clear: basically if r(ann(x)) is a prime, you need some y st ann(y)=r(ann(x))?

[maiden ocean]

Yeah

contextually this is a primary decomposition in modules thing

Im not sure how you'd conclude that a maximal ann(y) would be prime here

[next obsidian]

Maximal annihilators are prime is just an element argument

You just grab ab in ann(y), and if neither are in ann(y) then ann(ay) strictly contains ann(y)

[maiden ocean]

but i dont think its for all x in M

[next obsidian]

Then what does maximal ann(y) mean here?

[maiden ocean]

we have a primary decomposition of 0 into N_1 cap ... cap N_m so if r(N_i) = p then its for the x contained in the N_j, j neq i

like its the ann(x) whose radical is p

[next obsidian]

Wait so what are you trying to do?

[maiden ocean]

uh basically prove that p belongs to 0 in M implies that p is an associated prime of 0

[next obsidian]

Belongs to 0 in the sense that there’s a minimal primary decomposition with p as the radical of something in that?

[maiden ocean]

Yeah

[gritty sparrow]

Isn’t p=r(ann(M/N)? I don’t get what you meant by r in the above thing

[maiden ocean]

equivalent defn i think

[gritty sparrow]

No, i mean what is your r here

[maiden ocean]

r(N_i) = {x | x^n M is in N_i } = r(Ann(M/N))

radical

[gritty sparrow]

I see, yeah it is equivalent

[maiden ocean]

yeah i dont think the proof in AM for the case of ideals can be adapted

Agh

[gritty sparrow]

I think i have a proof of the statement about the primary decomposition of 0 having only the associated primes of 0, but I think your original q is a little stronger.
If 0=int(Ni) is a minimal decomposition, then picking x in the intersection of N2..Nk, as you observed, P^nx/=0 and P^(n+1)x=0. Picking a y/=0 in the former, we see that y is in the intersection of N2..Nk st Py=0. Also as ann(M/N1) is a subset of P, we see that if t is st ty=0 (which is in N1) then t is in P. Hence ann(y)=P. This required using the decomposition of 0 nontrivially, but I think this is very close to the argument for ideals

[maiden ocean]

i see

I think this works

Oh i see right

if y = a^n x then y = a^n x p subset x p^n+1 = 0

so p is in Ann(y)

for y in N

[gritty sparrow]

Yeah

Oh and to be more explicit about the converse, as ty=0 and y is not in N1, as N1 is primary, t is in r(N1)=P

[maiden ocean]

[gritty sparrow]

Nice

[wooden ember]

alright ive got an intuition issue here that i cant seem to solve. In this exercise I originally thought that $\phi_{\sigma}(\phi_{\tau}((g_1,...,g_n))=\phi_{\sigma}((g_{\tau^{-1}(1)},...,g_{\tau^{-1}(n)}))=(g_{\sigma^{-1}(\tau^{-1}(1))},...,g_{\sigma^{-1}(\tau^{-1}(n))})$ and it's an argument that still makes sense to me, but then $\phi$ would not be a homomorphism.

[cloud walrus]

Little Narwhal ✓

[wooden ember]

By contrast here is the correct argument which also makes sense to me (i subscript denotes ith coordinate)

why is my "order of execution" wrong in the first case?

and it seems the authors knew it could cause confusion and so gave an explanation but even if that explanation makes sense to me i cant work out what's wrong about my argument

[gritty sparrow]

Your argument fails because phi_sigma applies on elements of the form (a,b,..f) where a belongs to G1 b belongs to G2 and so on. So once you applied phi_tau, notice your elements are not of this form. g_pi^-1(1) does not belong to G1

[wooden ember]

hmmm i see

[gritty sparrow]

So in a way you are composing functions where the target of one is not the source of the other

[wooden ember]

but in the case of exercise 8 G1= G2 =... = Gn

[wooden ember]

that makes a lot of sense yeah

[wooden ember]

plus without this for the same reason the second argument would fail too would it not?

since $\phi_{\pi_2(g)}$ isn't necessarily an element of $G_1 \cross G_2 \cross ... \cross G_n$ without the condition

[cloud walrus]

Little Narwhal ✓

[gritty sparrow]

Ah yes, in the case where all the Gi’s are the same, when you apply phi_pi, it takes whatever is in the pi^-1(1) coordinate and puts it in the first coordinate. Now what is in the pi^-1(1) coordinate? Well phi_tau takes whatever is in the tau^-1(pi-1(1)) coordinate and puts it in the pi-1(1) coordinate. Hence what ends up in the first coordinate is whatever was in the tau^-1(pi^-1(1)) coordinate

Also sorry about the earlier comment, I didn’t look at q8 when I said that, ofc when all the G’s are the same, we can compose these functions as the targets and the sources are the same

[wooden ember]

so see that argument makes sense to me but i still dont see at which point the first argument goes wrong

which of the equalities is false

[gritty sparrow]

The second one

[wooden ember]

so how am i misinterpreting the definition there?

[primal pier]

Anyone know the etymology of graded in graded algebras?

Naively I'm thinking it's referring to the physical grading, creating notches that correspond to each direct subspace.

[gritty sparrow]

In the second thing, imagine we are doing phi_sigma(a1,..an) where ai=g_tau^-1(i). In the first coordinate we will have a_sigma^-1(1). So what is that? By what i just said, it will be g_tau^-1(sigma-1(1))

[wooden ember]

aha yes

it's such a subtle thing

it feels like you would just kind of apply whatever to the index

[gritty sparrow]

Yeah, this threw me for a loop too

[wooden ember]

like if i came across something similar in the future i feel id make the same mistake

i understand it but it doesnt jump out at me

[gritty sparrow]

Yeah true

[wooden ember]

i still dont feel like ive nailed the fundamental misunderstanding

but ill ponder over it again tomorrow, thanks

this already helps

[gritty sparrow]

Well, what you are doing will taking the element g_sigma^-1(tau-1(1)) to the first position, now which position is this element in? This element is actually in the tau(sigma^-1(tau^-1(1))) position. Because the map phi_tau takes the element gi and sticks it in the g_tau(i) position. So I suppose that is exactly where your map falls apart, I guess the fundamental misunderstanding is assuming that g_sigma^-1(tau-1(1)) is at the position sigma^-1(1) after applying phi_tau.

I hope this is a more satisfactory way of describing the mistake made

[vestal snow]

What does it mean when we say that L has g sections?

Does it mean that the global sections of L form a dimension g vector space over k?

[ivory dust]

Is there formal reason why E = Q(sqrt2, sqrt3) = Q(sqrt3, sqrt2) and the degree of E over Q is 4

like Q(sqrt2):Q = 2, Q(sqrt3):Q = 2, but how do we know that Q(sqrt2, sqrt3):Q(sqrt2) = 2 and Q(sqrt2, sqrt3):Q(sqrt3) = 2

like these are intuitively obvious to me

Or is actually much harder to formally show it

than it seems trivial to me

[hidden haven]

The smallest subfield of C that contains Q[sqrt 2] and sqrt 3, is the smallest subfield of C that contains Q, sqrt2 and sqrt3

Because Q[sqrt2] is the smallest subfield of C that contains Q and sqrt2

So that gives the equality Q(sqrt2, sqrt3) = Q(sqrt2)(sqrt3)

To show that [Q(sqrt2, sqrt3): Q(sqrt2)] =2, you need to show that sqrt3 is not contained in Q(sqrt2)

[ivory dust]

Ohh which is actually easy

[hidden haven]

yep

[ivory dust]

ive done something similar before

cuz its quadratic

[hidden haven]

Yeah

[ivory dust]

sqrt3 = a+bsqrt2

a,b e Q

[hidden haven]

And squaring both sides you get that sqrt2 is rational

[ivory dust]

so min polyn is x^2 - 3 in Q(sqrt2)[x]

[hidden haven]

Yeah

[ivory dust]

ok period tysm

Also quick question comparing questions

https://puu.sh/I1SeI/39cd4a71ef.png

ignore the x^2-5 in the 2nd one

does the fact that the irreducible factors are squared, or cubed matter in the first compared to the second?

[hidden haven]

No, the second f (ignoring the last factor) splits in an extension iff the first f does

So they will have the same splitting fields

[ivory dust]

so f(x) in the first question

[hidden haven]

A polynomial splits iff all of its irreducible factors split, their powers don't matter

[ivory dust]

can be "reduced" to (x^2-7)(x^2-3)

[hidden haven]

Yeah

[ivory dust]

So i can say splitting field of f(x) (reducible) over Q in first question is splitting field of g(x) = (x^2-7)(x^2-3) (irreducible version of f) over Q

[hidden haven]

Yeah, except g(x) isn't irreducible

[ivory dust]

wait what rly

[hidden haven]

You have written a factorization of g(x)

[ivory dust]

isnt it unique

[hidden haven]

what is?

[ivory dust]

the factorization of g

[hidden haven]

Yeah

How does uniqueness of factorization matter?

[ivory dust]

like i cant redistribute and try to factor another way

[hidden haven]

Factorization is unique for all polynomials, reducible or not

[ivory dust]

o

wait im an idiot

its legit 2 factors

[hidden haven]

[ivory dust]

i meant its factors are irreducible*

[hidden haven]

Yeah

[ivory dust]

blonde moment

[hidden haven]

I mean of course its irreducible factors would be irreducible

You can say that it is square free

ie no perfect square divides it

[ivory dust]

ok so it splits over E which is Q(s7, s3)

[hidden haven]

Yes

[ivory dust]

as (x-a1)(x-a2)..(x-a5)

where a1=a2=a3 = sqrt7, a4=a5=sqrt3

[hidden haven]

That is not how (x^2-7) splits in Q(s7)

[ivory dust]

So splitting field E of F = Q is actually F(a1,a2...a5) = Q(a1,a2,..a5) =

o

[ivory dust]

i meant (x^2-7)^3

over Q

err Q(s7)

[hidden haven]

Nope

first see how x^2-7 splits

Then you'll see how its cube splits

[ivory dust]

Im reading this to try to replicate it formally

https://puu.sh/I1SlK/b6bcd1a13d.png

isnt it (x+s7)(x-s7)

[hidden haven]

Yes

[ivory dust]

wait omg im so fried rn

should be f(x) = (x-a1)(x-a2)..(x-a10)

a1=...=a6=+-s7

[hidden haven]

yeah

[ivory dust]

a7=..=a10 = +-s3

[hidden haven]

3 +s and 3-s

[ivory dust]

So E is field ext i.e. splitting field of f(x) over Q as

Q(a1...a10)

=Q(s3,s7)

due to redundancy

and since +- doesnt matter

But i guess thats actually given in the statement...

😐

Do these

higher powers even have an effect on Gal(E/Q)

[hidden haven]

No

[ivory dust]

no since they do not matter right, since E is determined afterwards

okk

also would

Gal(E/Q) be a subgroup of S4, with 4 elements hence iso to Z2 x Z2

[hidden haven]

It is, but you have to eliminate the possibility of cyclic on 4 elements

[ivory dust]

like id map: s7 s7, s3 s3
phi1: s7 -s7, s3 s3
phi2: s7 s7, s3 -s3
phi3: s7 -s7 s3 -s3

[hidden haven]

Yes

[ivory dust]

Cuz roots map to ea other

and if phi(s7) = s3

then

phi^2(s7) = phi(7) = 7 =/= s3^3 = 3

so not possible

[hidden haven]

ye

[ivory dust]

okk period

ok sorry one last question

https://puu.sh/I1SpJ/f4ce3ff824.png

Does the F being defined as Q(s2) matter

since s3, s5 arent contained in Q just as much as they arent contained in Q(s2), and their degrees over Q(s2) are both 2

so F may as well be defined as

Q

[hidden haven]

Well you have to check that

That Q(s2, s3) doesn't contain s7

[ivory dust]

wait why does s7 matter

o u mean s5?

[hidden haven]

yeah s5

[ivory dust]

Hm

ok well intuition tells me it doesnt , but i see i should show that

im assuming

Q(s(p1), s(p2), ...)

for all primes

distinct

dont contain one another?

[hidden haven]

Yeah

[ivory dust]

in subfields of one another

ok

tyty

[hidden haven]

[crimson falcon]

Easy peasy question. anyone could explain to me what are R-algebras in this context? As always wikipedia is not great to learn new concepts 😒

[unique juniper]

i think its like a polynomial ring

like R[x_1, .. x_n] would be an R-algebra (i think)

[crimson falcon]

Yeah, but what is an R-algebra 😦

[crimson falcon]

If i understand that correctly. I have R on the background, I take A (Don't all rings have identity?), an homomorphism from R in my background to A such that f(r)y=yf(r) for all y in A (it's image it's on the center of A), I bundled them together and call that an R-algebra, is that correct?

[unique juniper]

yeap

[crimson falcon]

Alright, thanks, wikipedia sent me looking for what an R-module is before even starting

[clever mountain]

I attempted to show every maximal ideal is a prime ideal. Is this correct?

R is a commutative ring with unit 1. Let M be a maximal ideal.
Say ab is in M but a is not. Want to show b is in M.
The ideal (a) must be R, since M is maximal.
So the unit 1 is in a. Thus 1 = qa for some q in R.
This implies b = b·1 = b·qa = q·ab, and q·ab is in M since ab is.
This shows b is in M and we are done.☐

[chilly ocean]

ye

[dusty river]

I think you get (a) + M is R

(a) itself need not be R

[clever mountain]

a is not in M. If (a) is a subset of M then a is in M, which is contradictory.
Therefore (a) must contain M, ergo be the full ring R.
Is this wrong?

[gritty sparrow]

{3} is neither contained in nor contains {1,2}

[clever mountain]

Oh right, maximal ideal doesn't mean there is only one maximal ideal. There can be several

[clever mountain]

Is this a valid argument for showing Q[x]/(x^2-2) is not isomorphic as a ring to Q[x]/(x^2-3)?

Consider Q[sqrt2] and Q[sqrt3].
Any isomorphism between these would simply relabel sqrt2 to sqrt3 and vice versa.
If we consider the element 1+sqrt2, it would correspond to 1+sqrt3.
But observe:
(1+sqrt2)^2 = 1 + 2 + 2sqrt2 = 3 + 2sqrt2.
This should equal 3 + 2sqrt3. Yet we find
(1+sqrt3)^2 = 1 + 3 + 2sqrt3 = 4 + 2sqrt3.

Therefore they can't be isomorphic as rings.

[hot lake]

why would the morphism send sqrt2 to sqrt3 and not to -15+32sqrt3 ?

[clever mountain]

True, I had some misgivings about stating that.
Trying to think of other ways to attack this hmm..

[oblique river]

you can just say "suppose that sqrt(2) goes to a + b*sqrt(3)"

and then repeat your same argument

[wooden ember]

did part a) but ive got to say im having trouble gaining intuition on the central product, so I dont really see how to approach b except writing out every element explicitly and then constructing the first isomorphism that jumps out at me

oop sorry ill let you finish the previous q

[oblique river]

you should find that there aren't rational numbers a and b for which (a + b*sqrt(3))^2 = 2

@clever mountain

[clever mountain]

@oblique river thank you for the assistance.
How can we know there must exist an a+bsqrt3 in Q[sqrt3] whose square is 2?
I understand Q[sqrt3] contains elements of kind a+bsqrt3, and certainly contains 2
since 2 is rational. But I don't see why 2 must have a square root in Q[sqrt3]

[oblique river]

I'm following your exact argument

if Q[sqrt3] is isomorphic to Q[sqrt2]

[clever mountain]

Oh right sorry

[oblique river]

then sqrt(2) has to go to some element of Q[sqrt3]

[clever mountain]

Yes true true!

Thanks!

[oblique river]

np

[clever mountain]

@oblique river Hm, were you tacitly assuming the isomorphism phi takes 2 to 2?
phi(sqrt2) = a +bsqrt3
phi(2) = (a+bsqrt3)^2

but I would need to assert phi(2) = 2 here I feel

[oblique river]

yes but it's not really much to assert

isomorphisms will always preserve the integers

because they have to send 1 to 1

and therefore they send 1 + 1 to 1 + 1

isomorphisms will also preserve all rational numbers

[clever mountain]

Interesting, the definition I'm working with does not assume 1 goes to 1.
Only that it preserves addition and multiplication

[oblique river]

in that case

1^2 - 1 = 0

which means phi(1^2 - 1) = phi(0) = 0

(have you proven that phi(0) = 0 for an isomorphism?)

[clever mountain]

yes, follows from group homomorphisms

[oblique river]

so phi(1)^2 - phi(1) = 0, so phi(1) is a root of x^2 - x = 0

there are only two roots of this, x = 1 and x = 0.

so phi(1)^2 - phi(1) = 0, so phi(1) is a root of x^2 - x = 0

sorry those messages just got switched

but basically phi(1) cant be 0

cuz then not an iso

therefore phi(1) = 1

alternatively, phi is a group isomorphism of the multiplicative group Q \ {0} to itself

[clever mountain]

same phi now?

[oblique river]

yeah

just remove 0 and forget about addition

now you have a group under multiplication

and this map must preserve the multiplicative identity

for the same reason that phi preserves the additive identity 0

[clever mountain]

interesting, so if phi is an isomorphism of integral domains, then phi(1) = 1?

Since x(x-1) = 0 forces either x=0 or x=1.
phi(0) = 0, so 0 must be the only one mapped to 0, if phi is an isomorphism

[oblique river]

that's right

you can actually prove something more

if phi(1) = 0

then phi must be the 0 map

phi(r) = 0 for all r in R

so therefore any nonzero homomorphism of integral domains must send 1 to 1

[clever mountain]

phi(r)=phi(1*r)=phi(1)*phi(r) = 0, for each r in R

[oblique river]

yeah

that's right

[clever mountain]

Hm interesting, that's a key takeaway. A ring homomorphism (between integral domains), has property that phi(1) is in {0,1}

[oblique river]

i think many sources just require phi(1) = 1 as an axiom

regardless of whether or not your rings are domains or not

[clever mountain]

I have seen that, but e.g. D&F don't

[oblique river]

personally, I prefer that definition

yeah D&F tries to do things in the "most generality" but imo like, definitions should reflect what is actually useful

[clever mountain]

Fair

thank you very much @oblique river !

[oblique river]

np and gl!

[wooden ember]

this doesnt satisfy $\alpha((1,i))^2=\alpha((1,j))^2$ though so why do they say that it is a homomorphism

[cloud walrus]

Little Narwhal ✓

[rustic crown]

yep, doesn't look right

[wooden ember]

aight thanks, thought so

this question is driving me nuts

[wooden ember]

putting it up again in case anyone didnt see before but part b is really driving me nuts any help would be appreciated

i think ima just end up solving it by solving the next exercise first which is to give a presentation for the two groups

[maiden ocean]

does length classify modules over a PID up to isomorphism or at least up to being contained in K_0(A)?

i guess this can be boiled down to: is there some exact sequence relating A/(p) and A/(q) where p, q are primes in a PID that makes them equal in K_0(A)

[sturdy marsh]

K_0 of a PID is Z

projective modules are classified by rank

[maiden ocean]

thats what im trying to prove

[oblique river]

I dont think A/(p) would even be an element of K_0(A)

it's not going to be projective over A

[sturdy marsh]

this is just the structure theorem for modules over PID

[maiden ocean]

oh A is Noetherian here

[oblique river]

all PIDs are noetherian

[maiden ocean]

oh right

anyway the way i defined K_0(A) was by taking the free abelian group with basis isomorphism classes of finitely generated modules over A

and then quoting out by the SES relation thing

[oblique river]

i thought the usual definition was finitely generated projective modules

[maiden ocean]

i dunno im just going off of what AM does

[oblique river]

all right, here you go then

A/(p) is trivial in K_0

because of the SES 0 --> A --> A --> A/p --> 0

first map is multiplication by p

so it doesnt matter if you exclude nonprojective things as a definition or not cuz you can just show they're all 0 that way

[maiden ocean]

but dont the A/(p) generate K_0(A)

[oblique river]

no

[maiden ocean]

like that was the past exercise

that A/p generates K_0(A) for p prime

[oblique river]

if A is a PID then K_0(A) is Z

generated by A itself

[maiden ocean]

oh well 0 is prime

so it works i guess

[oblique river]

what exercise is this?

[maiden ocean]

AM chapter 7 number 26

[oblique river]

oh, yeah the previous part is for any noetherian domain A

or i guess doesnt even need to be a domain

[maiden ocean]

Here most of the generators end up being 0

[oblique river]

when A is a PID, yes

[maiden ocean]

ah hmm so length isnt the isomorphism here then

[oblique river]

I think you should just reference the structure theorem for modules over a PID

and combine that with the observation that A/(p) is trivial in K_0 for nonzero p

[maiden ocean]

that is kind of because length works for showing it for fields right

its equivalent to dimension

[oblique river]

I mean, it is just length once you remove all of the A/(p)

if AM had defined it just using projective modules, then it would just be classified by length

but like "all modules over a field are projective"

[maiden ocean]

Uh hm so length is an additive function from the class of modules to Z right

so there needs to be an induced map K_0(A) -> Z making this commute

[oblique river]

yes

it is length

or like sorry how about this

[maiden ocean]

but then doesnt this send zero elements of K_0(A) to non-zero elements in Z

[oblique river]

tensor with the fraction field of A

and then it's length

[maiden ocean]

like A/(p^n) has length n but its 0 in K_0(A)

[oblique river]

yeah

also wait a minute

[maiden ocean]

[oblique river]

just look at

[maiden ocean]

what the heck

[oblique river]

0 -> Z -> Z -> Z/2Z -> 0

first map is multiplicaiton by 2

all of those modules have length 1

so length isnt additive?

i mean like this is exactly the issue. Z/2Z is trivial in K0

[maiden ocean]

Huh

But theres a proposition in AM that length is additive

on modules of finite length

[oblique river]

also wait a minute

Z doesnt have length 1

[maiden ocean]

yeah

lmfao

[oblique river]

lol

it's infinite length

[maiden ocean]

Right

[oblique river]

so yeah, length is not an additive map on finitely-generated modules

that's the issue here

finitely generated modules need not have finite length over a PID

[maiden ocean]

i see

[oblique river]

so you just can't even talk about length in this context

[maiden ocean]

that makes sense

Right

but it works in the field case

[oblique river]

yes

[maiden ocean]

because being finitely generated and having finite length are exactly the same thing

[oblique river]

that's right

[maiden ocean]

ok this makes more sense