groups-rings-fields

[quiet pelican]

Multiply both sides by f(1/n)

[crystal vale]

Yes, I am dumb

Got it, thank you

[cloud walrus]

Riku

[quiet pelican]

Consider x^2 + sqrt(2) x and x^3 - 2x

[boreal inlet]

Oh right

isn't the first element irreducible

[quiet pelican]

Yes
||But it’s not prime||

[boreal inlet]

I was already aware of this, there's already x for this

I was looking for a proof which does not use this fact

[quiet pelican]

Why?

[boreal inlet]

x is irreducible but not prime

[boreal inlet]

The simplest way I can say this is pedagogical reasons. I just want to give some examples of non-UFDs before I go into stating other properties

There's also Z[sqrt(-5)] which I'm already using

[rocky cloak]

So would like saying
2x^2 = 2x * x = (sqrt(2)x)^2
also be cheating? Or what kind of proof are you looking for?

[mental lake]

i dont rlly get how u can check if a function is irreducible in Q[x]

[rapid cave]

it usually depends on what polynomial are you working with

[rocky cloak]

What's the context here?

Do you have a specific polynomial in mind? There are lots of techniques for showing something is irreducible, but in general it's kinda hard

[mental lake]

f(x)=x^3+2x+1

[rapid cave]

so this is a degree 3 polynomial. Try to prove its irreducible if and only if it doesn't have a root in Q

and you have the rational root theorem to help understanding if it has a rational root

[rocky cloak]

Can even go iff no root in Z by gauss lemma

[mental lake]

oki ill try ty

[tardy hedge]

Kind of random question but I always thought of a map R->R/I being natural projection but it doesnt have to be ig . If we precompose the projection with idk an automorphism of R then I may not be in the kernel of R->R/I?

[south patrol]

Ye tho if they say "the" map R -> R/I or just "R -> R/I" it'll be natural proj

[copper kestrel]

i passed abstract algebra 1 :D

unfortunately did not get a B tho

[knotty badger]

Congrats!

[copper kestrel]

tytyty

i love thinking i did okay on the final but then got a 61

[uneven osprey]

quick wuestion: having a hard time understanding why the kernel of a group action is supposed to be g in G st g.x = x? when we talk about homomorphisms, the kernel values map the function to e, so why do they not do the same in group actions?

[velvet hull]

it does, but when you think about group actions as homomorphisms they are no longer functions from G to S

but instead G to S_X

and the identity in S_X is exactly the permutation on X that fixes everything

[uneven osprey]

so group actions are about permutations on S_x?

[vapid vale]

also there is no notion of an identity element of an arbitrary set

[velvet hull]

you can think about them that way

[uneven osprey]

and not necessarily talking about a function mapping like homomorphisms cuz you're mapping from one group to another in that context?

I see

[velvet hull]

there is a kind of duality going on with group actions, you can either think of them as functions G x X -> X with certain properties, or group homomorphisms G -> S_X

[uneven osprey]

I'm having some trouble internalizing the intuition this way, though it does make sense

this definitely helps, thanks

[vapid vale]

maybe some intuition is like, many groups you know are already defined in terms of group actions

the symmetric and dihedral group are defined by what they do to elements of a set and vertices of a polygon respectively

in general if a group acts on a set, it acts by permuting the elements of the set around – that's really the only thing you can do to a set. so abstractly, it should map to all the ways you can permute that set.

the definition of group action has some condition on group multiplication being respected by the action; homomorphisms are precisely those set maps which respect multiplication

[uneven osprey]

I understood that intuition far better now, I was using this to solve a practice problem and have a larger concern with proving part c: I understood that I can think of them like permutations so really finding the kernel of this group action is equivalent to finding the center of this group but don't know how to prove the center here

[velvet hull]

you know what G is

[uneven osprey]

yes it's defined in the question: the dihedral group D_8

[velvet hull]

so what is the center of the dihedreal group

[uneven osprey]

I was trying to exploit the relation yxy^3x = e but got stuck

[uneven osprey]

e, r^4? even powers of r?

[velvet hull]

almost

[uneven osprey]

except r^8 of course, although that itself is e

which is in the centere

[velvet hull]

it should just be e and r^4

[uneven osprey]

I did that but I double checked with AI (which could very well be wrong) but it told me that was incorrect: I read a rule about "not posting AI slop" but can i still paste the response here? It does make sense to me

[velvet hull]

sure

[thorn jay]

not posting ai slop is referring to that, slop

although i do urge you to just not use ai for this stuff

plenty of people online have had to answer the same questions as you

[uneven osprey]

it's true, I'm just a bit confused and have a test coming up with not much reference material to fal back on and found it as a quick source of clarification

but it can also often hallucinate

[velvet hull]

oh, I see the confusion

the group presentation given is not for the dihedreal group exactly, it's a different group

[thorn jay]

the AI is confusing the presentation commonly used for D_n with the presentation given lol

[thorn jay]

yup

and still the math the ai did is wrong lol not wrong but not written down correctly

yxy^3x = 1 => xyxy^3 = 1
so we have
xyx^-1 = xyx = y^-3
and from there the reasoning follows

if xy^k is in the center, then
xy^k = x(xy^k)x = y^kx
so k has to be even, and
xy^k = y(xy^k)y^-1 = y(y^kx)y^-1
=> xy^k+1 = y^k+1 x
which is a contradiction, so the center must indeed by y^k where k is even

ai got to the right answer but I wouldnt accept the "proof" it gave

[noble belfry]

Let C[x, y] be the ring of polynomials in 2 variables over the complex numbers
and let (xy) be the ideal generated by xy. Find all prime ideals of C[x, y]/(xy)

this is cooking me i've been studying too long lol

These are the prime ideals which contain (xy)

so they contain x or they contain y...

how does that work though since all maximal ideals are prime and the max ideals correspond to points

actually nvm that makes sense

so all of them are of the form (x, p(y)) where p(y) is an irreducible polynomial

or (p(x), y)

[noble nexus]

not sure about that

An ideal containing (x,p(y)) might not contain xy

oh actually no your right

but I would say (x, y-a) instead for some a

[noble belfry]

why

those are just the maximal ideals no

or

yeah

PID

so prime -> maximal iirc...

[noble nexus]

well the only irred polys are x-a since ur working over C

[noble belfry]

im losing it i've been studying too long

[noble nexus]

When I think about a problem like this I go first to geometry

idk if you've done any algebraic geometry though

[noble belfry]

yeah me too

[noble nexus]

but if you have the solution is immediately clear

[noble belfry]

i havent done enough algebraic geometry 🥀

[noble nexus]

since (xy) represents a union of the x-axis and y-axis

[noble belfry]

yeah

so the points are (x, y-a)

and (x-a, y)

[noble nexus]

so primes containing it include all the points (maximal ideals) on this curve

[noble belfry]

which correspond to maximal ideals

[noble nexus]

Together with the x and y axis themselves

aka (x) and (y)

[noble belfry]

yeah

this doesn't immediately tell us tho that all the prime ideals are maximal

[vapid vale]

it doesn't

(x) contains (x, y-b)

[noble belfry]

yeah

(x) and (y) are both prime ideals which contain (xy) i suppose

[vapid vale]

other way around lol

yeah

[noble belfry]

wdym

[vapid vale]

to my message

[noble belfry]

oh

yeah true

[noble belfry]

wait hold up lmfao

oh

other way around

cuz (x, y-b) is maximal

[vapid vale]

yeah

[noble belfry]

ok yeah

im really tired 💀 i think i'll take a nap or smtn and study some more

[vapid vale]

well youve classified the prime ideals

[tulip otter]

A is a commutative ring and G is a monoid written multiplicatively, how is the map φ_0 a monoid homomorphism? I mean φ_0(xy)=1.xy,
φ_0(x)φ_0(y)=(1.x)(1.y) but (1.xy)(xy)=1 while (1.y)(xy)=0 so that (1.x)(1.y)(xy) neq (1.xy)(xy)

[elfin wraith]

Where is all the stuff after the but coming from? I’m not sure I follow, is 1 not the identity here?

[tulip otter]

a.x is the map defined by a.x(x)=a and a.x(y)=0 if x neq y

[tulip otter]

where exactly are you finding a problem so that i can recheck my work

[tulip otter]

yes 1 is the identity of A

[crystal vale]

Do I need ufd here?

Because if it has more than deg(f) roots then that polynomial f will have deg > deg (f)

[crystal vale]

I see without ufd I can say there can be at most deg f different roots

But with ufd I can say there at most deg f roots

[quiet pelican]

You only need it to be a domain

The division algorithm works as long as the leading coefficient of whatever you’re trying to divide by is a unit, and 1 is a unit

[crystal vale]

No but i want to say that there can be at most deg f roots

[river heart]

You just need F to be an integral domain, but I think the text refers to F[X] being a UFD.

[quiet pelican]

?

[crystal vale]

I want to prove that f(x) have at most deg(f) roots

[river heart]

Use induction on deg(f).

[crystal vale]

No but how does it come from ufd ?

[river heart]

Basically a is a root of f iff (x-a) divides f, so maybe you can see that having restrictions on what can divide f would imply restrictions on the roots of f.

I'm putting it vaguely so that you can try to fill in some more details.

[crystal vale]

Okay thank you

I think I got it

[crystal vale]

If I take f = 2(x+1) and p = 3, then f is irreducible in F_3[X], right? But f is not irreducible in Z[x]

[quiet pelican]

The leading coefficient is 2, which is divisible by the prime 2

[crystal vale]

No I am taking p = 3

[quiet pelican]

f should be monic yeah

[crystal vale]

Is there any subring of R which is not a field but contains Q? Any hint?

[knotty badger]

I think yes

[quiet pelican]

Hint: Q[algebraic number] is always a field
Solution: ||Q[liouville’s number]
(You can put any transcendental there, I just put the number that’s easiest to prove it’s transcendental)||

[crystal vale]

Q[algebraic number] is a field because its degree is finite, right?

[quiet pelican]

[crystal vale]

Does Q[π] work here?

[quiet pelican]

[rapid cave]

you mean that its not a field

[crystal vale]

So is there any result that if [ E : F ] has degree infinite then we can always find such subring of E containing F but not a field ?

[crystal vale]

Yes

[quiet pelican]

No, let E be the algebraic closure of Q, and F be Q

There is such a subring if and only if E is not algebraic over F

If x \in E is transcendental over F, then ||F[x]|| is always a proper subring of E

[crystal vale]

Why?

Maybe it doesn't contain x^-1

[quiet pelican]

Yup

[crystal vale]

If it contains x^-1 then x will be algebraic number over F

Thank you @quiet pelican

[crystal vale]

What does it mean by E is finitely generated as a field over F ?

[quiet pelican]

There is a finite number of elements a_i such that E = F(a_1, …, a_n)

[crystal vale]

So if E\F is finite then it is easy to see E is finitely generated, right?

E\F denotes the degree of E over F

[quiet pelican]

Yes

[quiet pelican]

I’d usually denote that [E: F]

[crystal vale]

Is converse true?

[quiet pelican]

Q(x), so no

[lusty marlin]

Adjoining a transcendental element, say x, to a field in general provides a counterexample to the converse

[crystal vale]

I see

[sly crescent]

I think most people would interpret E\F as the set of elements in E but not F

[languid trellis]

I'm struggling with the following exercise from Morandi. Let $k$ be a field, $k(X)$ its field of fractions, $\sigma$ $tau$ automorphisms automorphisms of $k(X)$ given by $\sigma(f(X)/g(X)) = f(1-X)/g(1-X)$, $\tau(f(X)/g(X)) = f(1/X)/g(1/X)$. Determine the fixed field of ${ \sigma, \tau }$

[cloud walrus]

swifteeee

[languid trellis]

If the fixed field $F$ is such that $[k(X) : F]$ is finite, then $4 ; | ; [k(X):F]$, but that relies on $k(X)$ being a finite extension of $F$

[cloud walrus]

swifteeee

[languid trellis]

In fact I'm not even sure that the fixed field is bigger than k

I'd appreciate any hint in the right direction : )

[rocky cloak]

Something that is true that you might be able to prove is that if E is any field, and G is a group acting on E and F is the subfield fixed by G, then [E:F] <= |G|

[languid trellis]

oh

hm

no finite dimensionality hypothesis

is this not artins lemma

[rocky cloak]

It's called artins lemma yeah

[languid trellis]

So we have that k(X) is a finite dimensional extension of the fixed field

of degree no larger than 4

shouldn't [k(X) : F] be exactly 4?

[rocky cloak]

It will be exactly 4 yes.

And in this case I guess it's not that hard to see, as x has 4 conjugates, so you get a minimal polynomial of degree 4 (and of course F(X) = k(X))

[languid trellis]

haha i can explicitly calculate this using cayley hamilton xd

[rocky cloak]

Wait, does this group actually have order 4?

[languid trellis]

tau^2 = sigma^2 = 1 no?

[velvet hull]

it is "divisible" by 4

but when you compose tau and sigma together you get something with potentially infinite order

[rocky cloak]

Yeah, so then it won't be finite

[languid trellis]

hm

Maybe it'll be of some use to compute the fixed fields of just sigma, or just tau

as those will contain F

for instance aX+a/bX+b is fixed by sigma

Given S_1 \subset S_2 \subset Aut_K(L), denote the fixed field by F(S_1) \subset L. Then we have that F(S_2) \subset F(S_1)

if S = {sigma, tau}, S_1 = {sigma}, S_2 = {tau}, then F(S) \subset F(S_1) and F(S) \subset F(S_2)

so F(S) \subset F(S_1) \cap F(S_2) ?

unsure how useful this is

for one thing, the aX+a/bX+b is not fixed by tau

perhaps we can argue that F(S_i) is a finite extension of k?

because the group generated by sigma is just {1, sigma}

but this isnt really useful

I need food

[tardy hedge]

Can I show something like (a+b,c+d+e) is prime in k[a,b,c,d,e] by the map

k[a,b,c,d,e]/(a+b,c+d+e) -> k[b,d,e]

sending a to -b, b to b, c to -d, d to d, and e to e is an isomorphism?

[velvet hull]

are a,b,c,d,e indeterminates?

[tardy hedge]

yeah

[velvet hull]

then you dont need to do that

the quotient is isomorphic to k[a,d,e]

[tardy hedge]

I guess im trying to show k[a,b,c,d,e]/(a+b,c+d+e) is k[b,d,e]

[velvet hull]

the issue is that when you send a to -b and b to b, it is no longer an isomorphism

just show that the quotient is isomorphic to k[b,d,e] directly

[tardy hedge]

Why not?

[velvet hull]

a+b is in the kernel, for one

[tardy hedge]

I meant to define the map on the quotient already

so a+b is 0

[velvet hull]

oh, the quotient

thats not even an isomorphism then it's just equality

because a+b is identified with 0

[tardy hedge]

So should be fine right?

[velvet hull]

yeah sure, check that its a bijection

[tulip otter]

is lang here assuming that this chain contains all elements of S?

the way he writes it at first doesnt seem to imply that

but the conclusion later that both b and c admit factorizations into irreducible elements seem to imply that?

at least thats what I understood

I understood that b and c admit such factorizations since (a) is the union of all elements of S and (b),(c) neq (a)

[rocky cloak]

So I guess you're just trying to show the quotient is isomorphic to k[b, d, e]? Then this is a good approach, but you need to fix your map sightly.

Right not c+d+e does not map to 0.

[rocky cloak]

No, it's just an arbitrary chain in S

[tardy hedge]

Oh i meant to say c map to -d-e

Thank u

[rocky cloak]

(a) is not the union of everything in S, it's just a maximal element, like they say.

[tulip otter]

but then why does (b), (c) neq (a) imply that either b or c admits a factorization tho

[rocky cloak]

If a = bc then (a) is contained in (b), so (b) is not in S

[tulip otter]

ah its because (a) is a maximal element in S

i totally forgot that (a) satisfies this lol

tysm

[tardy hedge]

To show directly that ideal is not prime, you could do some degree argument? Thats what my supervisor was suggesting before

[white oxide]

Where can I find a proof of the first bullet point in the Remark? Or is it easy to prove?

[glass sinew]

new to group theory, would just appreciate feedback on my work. questions from Ash, ch 5.2. it's late so i'll be asleep for a while. C and N sub G of x denote the centralizer and normalizer of x in G resp

[tardy hedge]

bro's on da powerpoints now

[white oxide]

Fucking Atiyah and Macdonald referencing obscure algebraic facts that nobody knows

[tardy hedge]

i tend to forget its atiyah and also macdonald

not: atiyahmacdonald

[white oxide]

I want to say it has something to do with the minimal polynomial, assuming L is algebraic

[tidal schooner]

what definition of algebraic closure are you working with?

[white oxide]

Like take some element in l, find its minimal polynomial in k, and do something with coefficients

[white oxide]

I'm not sure tbh 💀 Probably just the algebraic extension of k which is algebraically closed

[tribal moss]

"If" is immediate. "Only if" requires a bit of footwork to describe the structure of an algebraic extension, but you might already have much of that in place.

[white oxide]

Sorry why is L/K algebraic => L embeds immediate 😭😭

[tribal moss]

It's the other direction that's immediate.

[white oxide]

Oh, but why is that immediate? I get that for every l in f(l) is algebraic over k, but how does that imply that l is algebraic?

[tribal moss]

That's the definition of "algebraic", isn't it?

Actually, after thinking a few minutes more, I think "L/K algebraic => L embeds" can be done without much of the footwork I had in mind, just by using Zorn's lemma on the set of embeddings of subfields of L.

[white oxide]

Well L doesn't have to be contained in \Omega right

[tribal moss]

If it embeds, then it's isomorphic to something that is a subfield of Omega.

[white oxide]

Might need to review some field theory lol

(Talking to myself)

[tribal moss]

Hmm, perhaps I'm too quick here. If L just embeds as a field, then there's no guarantee that the embedding preserves K.

Actually that seems to doom the entire claim.
For a counterexample let k be Q(x1,x2,x3,...), the field of rational functions in countably many unknowns, and let L be Q(x0,x1,x2,x3,...), the same thing but with a variable added.
Then L is isomorphic to k, so it clearly embeds in Omega, but L is not algebraic over k.

So the bullet point is only true if we rewrite "L embeds into Omega" to "L embeds into Omega via a map that preserves k".

@white oxide^

[white oxide]

Thanks

I've switched over to number theory rn lol but I'll take a look at it when I come back to it

[tribal moss]

And we should expect to need some kind of choice, because "L/K algebraic => L embeds preserving K" implies that an algebraic closure is unique up to isomorphism, which might not be true without choice.
(I've seen claims that we don't need full choice but just the ultrafilter lemma -- but if we're just interested in the algebra, that distinction doesn't need to concern us).

[rocky cloak]

For (2) I feel you should add in an argument for why
(nxn') s (nx'n') = s. As written now it's a little incomplete.

For (3) you've actually just shown that Stab(x)^g is a subset of Stab(gx). You also need the other inclusion.

[uneven osprey]

hi, I had a question on principal ideals: I understand that Ra, aR are ideals but I'm having trouble accepting the use of the 'generator' notation: isn't <a> = {a, a^2, …, a^n}?

[noble nexus]

for a group yes, but for rings generator brackets typically refer to ideals

in the integers, its the same because <a> = {na : n \in Z}

(which is a^n in additive notation)

[uneven osprey]

okay so the meaning of generators is just different when it comes to rings? this isn't some result of using the a^k notation besides the example you gave?

[velvet hull]

Could you provide the definition of a generator that you are using?

[tribal moss]

Generally, the intuition behind the notation <a,b,c> is that it denotes the smallest "thing" that contains all of a,b,c. But what kind of "thing" that means is context-dependent. When you're talking about groups, it's the smallest subgroup. But when you're talking about rings, it's the smallest ideal.

This convention is not completely systematic, but reflects the pragmatic fact that ideals are more important subsets of a ring than subrings are,

[glass sinew]

ah, thank you lots. i've appended a small verification that N is a subgroup, so nxn' is in the normalizer by subgroup closure. by hypothesis it should be safe to assume the elementwise conjugation works then?
for (3) i've just provided the converse inclusion. i appreciate that, didn't think of the hole there

[sullen salmon]

group
goup
goop

[copper kestrel]

i got a B in abstract algebra 1!!!!!!

i thought i got a C

[sullen salmon]

Celebrate that with jolly Christmas stuff.

[copper kestrel]

absolutely ‼️‼️

[tardy hedge]

Yayy good job 👏

[thorn jay]

super well done!

[copper kestrel]

ty yall :3333

[tulip otter]

$A$ and $A'$ are commutative unitary rings so there exists a maximal ideal of A', say $\mathfrak{m'}$. To show that $\mathfrak{m'}$ is unique, it is sufficient to show that if $\mathfrak{a'}$ is a maximal ideal of $A'$ then so is $\mathfrak{a}=f^{-1}(\mathfrak{a'})$. Indeed, if this holds then any such $\mathfrak{a}$ is equal to $\mathfrak{m}$ for otherwise $A$ would not be local.\\Now to prove that if $\mathfrak{a'}$ is a maximal ideal of $A'$ then so is $\mathfrak{a}=f^{-1}(\mathfrak{a'})$, assume that $\mathfrak{a}$ is not maximal so that there exists an ideal $\mathfrak{b}\subsetneq A$ such that $\mathfrak{a}\subsetneq\mathfrak{b}$. But then $\mathfrak{a'}=f(f^{-1}(\mathfrak{a}))\subsetneq f(\mathfrak{b})\subsetneq A'$ which contradicts the maximality of $\mathfrak{a'}$ and this completes the proof.

[cloud walrus]

ali yassine

[tulip otter]

is this correct?

[mighty spade]

having a “compilation error” here, and can’t quite figure out what the modules should be here

wait these are matrices

they’re the associated linear map, and you can recover the modules

[quiet pelican]

Yes

[mighty spade]

so i get the maps, but the modules aren’t clear to me

[elfin wraith]

Looks good to me, it could probably be written more clearly but it’s correct

[quiet pelican]

Another way to do it would be to prove that A’\f(m) are all units

[elfin wraith]

Ah mico beat me to it

[quiet pelican]

S is the polynomial ring C[x, y, z] (it’s pretty much the only thing it could be, given “free resolution over C[x, y, z]”)

[tulip otter]

ohhh, what do you mean by written more clearly, as in how would you more or less write the same argument but more clearly?

[mighty spade]

the matrices are again confusing me

[tulip otter]

how to do this? prove that given any element a' in A'\m', (a')=(1)? or is there an easier/faster way?

[quiet pelican]

Suppose c \in A’\f(m)
Choose b \in A \ m which maps to c
Then b is a unit, and f(b^-1)c = 1

[mighty spade]

are these presentation matrices?

[tulip otter]

ohh i see

[rocky cloak]

In general you represent maps between free modules by matrices.

A matrix like
[xy, yz]
takes the vector (a, b) to
xya + yzb

It works exactly the same as matrices in linear algebra, which you hopefully are familiar with

[mighty spade]

thank you, i figured this out while gathering the firewood for the day

for a while i was confused because i wasn’t sure the modules were free

maybe i thought of them over C for some reason

[tulip otter]

Let $K$ be the set of ideals described in the given and choose $x,y\in A$ in a way such that $xy\in\mathfrak p$ but $xy\notin\mathfrak a\ \ \forall\mathfrak a\in K$ (such $x,y$ exist since $\mathfrak p$ is maximal in $K$). Assume that $x\notin\mathfrak p$, i claim that $y\in\mathfrak p$. \Assume the contrary for the sake of a contradiction, then $xy\in\mathfrak p$ but $x,y\notin\mathfrak p$, it follows that $x,y\notin\mathfrak a\ \ \forall\mathfrak a\in K$ for otherwise if $\exists\mathfrak a\in K$ such that $y\in\mathfrak a$ then $xy\in\mathfrak a$ since $\mathfrak a$ is an ideal of $A$ which is a contradiction. Now since $x,y\notin\mathfrak a\ \ \forall\mathfrak a\in K$, it follows that $x,y\in S$ so that $xy\in S$ since $S$ is a multiplicative subset of $A$. But $xy\in\mathfrak p$ which is a contradiction. Hence $\mathfrak p$ is prime

[cloud walrus]

ali yassine

[tulip otter]

is this correct?

hmmmm the claim about the existence of such x,y in A might be wrong

[quiet pelican]

Yeah it’s not immediately clear that p is not a subset of the union of some of the a

[tulip otter]

i mean since no ideal of the union intersects S then this union is an element of K. Furthermore, none of the ideals in K contains 1 so this union neq A. Hence the union is a subset of \mathfrak p since \mathfrak p is maximal in K right?

[quiet pelican]

The union is not necessarily an ideal

[tulip otter]

ah right

[quiet pelican]

Also maximal in K doesn’t imply every element is less than it, only that no element is larger than it

[tulip otter]

but inclusion is a total order no?

[quiet pelican]

No

Consider the ideals (2) and (3) in Z

[tardy hedge]

I don't understand why you could not have two different maximal ideals in A' whose preimage is m.

[quiet pelican]

The proof proves that every maximal ideal is f(m) for the maximal ideal m of A

[tulip otter]

ah right i see, inclusion is a total order only if the set of ideals form an ascending chain ig?

[quiet pelican]

[tulip otter]

i see

[tulip otter]

from the way you phrased this it seems that its true, so ig i should try to prove it

[quiet pelican]

I don’t think that’s the nicest way to prove it, and it’s not necessarily true

[tulip otter]

ohhhh

[tulip otter]

so there is no way to fix this particular argument?

[quiet pelican]

I don’t think it’s easily fixable

[tulip otter]

i see

[tulip otter]

can you give me a hint towards the proof you have in mind

[quiet pelican]

Let x, y be st x, y not in p but xy in p
Consider (p + (x))(p + (y))

(More commonly called integral
But yeah that’s a very common proof strategy with “prove that [thing maximal wrt a property] is prime”)

[tulip otter]

yea, i was just using the terminology lang uses haha

he even tries to justify the use of this terminology lol

[quiet pelican]

No one outside lang really uses “entire” from what I’ve seen

tbh if you feel the need to write a whole paragraph to justify your use of terminology, there are probably very good reasons to use the terminology that you feel so motivated to argue against

[tulip otter]

yea thats a reasonable point ig

i also dont know if it really matters too much tbh, tho it might make a difference and i am just being ignorant

[tribal moss]

He's right that "integral" is rather overloaded, but "entire" seems to be as weird intutively as "integral".

[tulip otter]

of course as long as the terminology doesnt coincide with the name of another related concept

[quiet pelican]

It matters in that Lang is a reference that's using terminology you'd be hard-pressed to find elsewhere, and that makes looking stuff harder than it should be

[tulip otter]

ah yea i see your point, in this particular case its a bit troublesome

[tribal moss]

Both words seem to suggest an intutition about "there's nothing missing from the ring", but what would that have to do with "there are no zero divisors"?

[noble belfry]

guys

any tips on getting practice with module theory

[tardy hedge]

You could also show this pretty quickly with A/ker(f) = A’ i think

And with lattice isomorphism theorem stuff

[tulip otter]

idk if there is a good name that conveys this tho

[tardy hedge]

Dummit and foote is a good introduction

[thorn jay]

RAH

[tulip otter]

there is something that i found weird here which is why i deleted my comment on showing that A/p is not integral/entire which is (x) and (y). Shouldnt these be x and y instead or am i missing something

[vapid vale]

integral comes from integers, not the standard adjective definition of integral

as in rings of integers

[quiet pelican]

No, because p + (x) is an ideal, p + x is a "coset"
(I'm not actually sure if coset is used in rings, but it's the idea I'm trying to convey(

But showing the stuff with ideals p + (x) and p + (y) is equivalent to integrality

[thorn jay]

abelian heap with identity element x of course

[karmic moat]

funnily enough, "entire functions" in complex analysis are also called "integral functions" according to wikipedia

[tribal moss]

Yeah, but it still doesn't quite convince me, because "no zero divisors" is not a particular property of integers compared to rationals, reals, complex numbers, polynomials, ... it looks like we need to go all the way to matrices to find something that isn't.

[rocky cloak]

But integers does come from the adjective integral (whole numbers)

[vapid vale]

right but thats two steps removed

[tribal moss]

(Well, or integers modulo a composite number).

[rocky cloak]

Also I feel like commutative domain is right there

[tulip otter]

ah so you meant ideals, i thought you meant the cosets because it basically leads to the desired result too right

[vapid vale]

in any case it doesn't have to be completely convincing, im just commenting on the historical usage

[rocky cloak]

Why have integral domain

[tulip otter]

i mean (x+p)(y+p)=xy+p=p, but both x+y and y+p neq p

[tribal moss]

Wut?

[thorn jay]

what's the justification for the "domain" naming then?

[tulip otter]

why is working with these equivalent tho

[quiet pelican]

exercise!

[vapid vale]

iirc the original motivatino would have been rings of integers of number fields. so within such a field, this would be the "domain" where things were like integers

[thorn jay]

right I see, that makes sense

[vapid vale]

domain is not specifically motivated rather than being a word for a place and perhaps connoting some sort of safety or understanding

[tulip otter]

i thought about this at some point but then i dismissed it because ker f need not be equal to the unique maximal ideal of A

[thorn jay]

generalisation is that like any integral domain naturally lives in its fraction field

right?

[vapid vale]

ya

[tulip otter]

alright i will think about it and come back rq

[thorn jay]

:D

[tardy hedge]

Why would that matter?

[tulip otter]

whats this btw

[tribal moss]

If I were naming everything anew, I would consider something like "robust rings", signifying that the property of being nonzero is "robust" and won't be lost simply by multiplying two of them together.

[tardy hedge]

I dont actually think it needs it. A/ker(f) should still be a local ring and its isomorphic to A’

[tribal moss]

(Though if I were naming everything anew, I don't think I would come up with "ring" at all, so ...)

[tulip otter]

if ker f=m where m is the unique maximal ideal of A then A/ker f\cong A' implies A' is a field implies A' is local

[thorn jay]

correspondence theorem

[tulip otter]

that was the idea i had in mind at that time

[rocky cloak]

It being used in the noncommutative setting.

[tardy hedge]

ker f is contained in m at least

[thorn jay]

ideals of R/I naturally correspond to ideals of R containing I

closed immersion and allat

[tulip otter]

ohhh i see

tysm

[tulip otter]

it might be obvious and i am just being stupid rn, but i dont see this immediately

[vapid vale]

i havent really scrolled through this discussion so maybe this is redundant but a useful observation is that by maximality, if x \notin p then we have s = f + ax for s in S, f in P, and a in A.

now if you have x and y not in p, the property above can probably show xy not in p

[tardy hedge]

Well i think the correspondence theorem shows A/ker(f) is also a local ring

And then its isomorphic to A’

[white oxide]

As in f(a) = a for all a in k?

Where f: L -> \Omega

[tribal moss]

Yes

[white oxide]

Ah ok that makes sense then I think

Thank

s

[tulip otter]

(p+(x))(p+(y))=p+(xy) so if (p+(x))(p+(y))=p then since x and y in (x) and (y) respectively, it follows that (p+x)(p+y)=p too?

[hushed saffron]

group theory 3rd year exam tmrw, what do i study rn

[quiet pelican]

That last one should be a \subseteq, but yeah

[quiet pelican]

What’s the syllabus

[tulip otter]

ah right

tysm

[tulip otter]

ohhhh i see

tysm

[hushed saffron]

homomorphisms, quotients, group actions, symmetric groups, sylow's theorems, free groups and presentations

[tulip otter]

also tysm kiand

have a great day/night everyone

[thorn jay]

cya!

[karmic moat]

sounds like those are what u should study rn lowkey

[tulip otter]

cya!

(ig it will be in a few moments)

[thorn jay]

😭

study tip: go study!

[karmic moat]

ye subscribe to my patreon for more

[thorn jay]

buy my course

[karmic moat]

but yea idk i guess you can go find like old qualifying exams from some schools and work on those problems

[thorn jay]

big agree on the old exams part, always nice to see what kinda questions you can expect

[karmic moat]

lfg

[white oxide]

Probably a silly question, but if A is a commutative ring, and E is a proper subset of A, can we always write E as the intersection of prime ideals of A

[quiet pelican]

No
The intersections of prime ideals are exactly the radical ideals

[white oxide]

Ah damn okay

[velvet hull]

Arbitrary intersection of ideals is still an ideal

So that would imply any subset of a ring is an ideal which is unfortunately not true

[thorn jay]

it would be nice for the ideal lattice of a ring to always be a boolean algebra

[white oxide]

Could I have a hint for this problem? I am letting $E \subseteq B$ and am trying to show that $f^\bigl(V(E)\bigl) = V\bigl(f^{-1}(E)\bigl)$. Now $f^\bigl(V(E)\bigl) \subseteq V\bigl(f^{-1}(E)\bigl)$. On the other hand, let $\mathfrak{p} \in V(f^{-1}(E))$. Then we want to show that $\mathfrak{p} = f^{-1}(\mathfrak{q})$ for some prime ideal of $B$ with $E \subseteq \mathfrak{q}$. Proposition 5.10 shows that there is an ideal $\mathfrak{q}$ of $B$ with $\mathfrak{p} = f^{-1}(\mathfrak{q})$. However, I'm having trouble showing that $E \subseteq \mathfrak{q}$. We have the following diagram, which seems to resemble the ``Going-up theorem":

[
\begin{tikzcd}
A && B \
\
{\mathfrak{p}} && {\mathfrak{q}} \
\
{f^{-1}(E)} && {?}
\arrow[from=1-1, to=1-3]
\arrow[from=3-1, to=1-1]
\arrow[from=3-1, to=3-3]
\arrow[from=3-3, to=1-3]
\arrow[from=5-1, to=3-1]
\arrow[from=5-1, to=5-3]
\arrow[from=5-3, to=3-3]
\end{tikzcd}
]

Unfortunately, one of the criteria of the going-up theorem is that the chain of ideals in $A$ must all be prime. Since it is not necessarily the case that $f^{-1}(E)$ is prime, I thought about restricting my attention to closed sets of the form $V(\mathfrak{q})$ in $\text{Spec}(B)$. I'm wondering if it suffices to do this, or if some topological argument shows that every closed set can be written as an intersection of these closed sets, because then I can apply the going-up theorem (I can post for reference if needed)

[cloud walrus]

okeyokay

[thorn jay]

hint: take E to be a radical ideal

[white oxide]

Dude answered in less than 5 seconds

[thorn jay]

well you were asking if any subset could be written as an intersection of primes so I kinda figured :P

[thorn jay]

you can do this because V(E) = V(I) where I is the intersection of all primes in V(E)

[white oxide]

Let me ponder

[thorn jay]

that should be a fundamental property and is probably mentioned somewhere before in what you're reading

(not that it is a particularly deep fact - it's just lattice theory)

[thorn jay]

If you're lucky and B is Noetherian then every closed set can be written as a finite union of V(p) where p is prime, in which case you're in luck

[slender mulch]

whats the idea of subgroups, and whats the main use for them?

are they just small groups inside bigger groups

[thorn jay]

theyre useful because they allow you to relate certain symmetries to other, i suppose

theyre as useful as subsets are useful, just a ubiquitous concept

[thorn jay]

yeah

[slender mulch]

okay

[thorn jay]

for example, any group is essentially the same as some subgroup of Sym(G) where G is the underlying set of your group

[knotty badger]

Subgroups often arise by taking the collection of group elements that preserve some extra symmetry

[tulip otter]

let $\mathfrak a$ be the ideal of $A_{\mathfrak p}$ mentioned in the given. I claim that $A_{\mathfrak p}/\mathfrak a$ is a field. Indeed, $A_{\mathfrak p}/\mathfrak a$ is clearly a commutative ring since $\mathfrak a$ is an ideal of $A_{\mathfrak p}$ and $A_{\mathfrak p}$ is commutative. Furthermore, a non-zero element of $A_{\mathfrak p}/\mathfrak a$ is of the form $a/s+\mathfrak a$ with $a,s\notin\mathfrak p$, which means that $s/a+\mathfrak a$ is defined element of $A_{\mathfrak p}/\mathfrak a$ and in fact $(a/s+\mathfrak a)(s/a+\mathfrak a)=\frac{as}{sa}+\mathfrak a=1+\mathfrak a$ so that any element of $A_{\mathfrak p}/\mathfrak a$ is invertible, $A_{\mathfrak p}/\mathfrak a$ is a field and $\mathfrak a$ is maximal.\\Note that $a/s\notin\mathfrak p\implies a\notin\mathfrak p\implies\frac{as}{sa}=1\implies a/s\ \text{is a unit}$ so that the ideal $\mathfrak a$ is the unique maximal ideal of $A_{\mathfrak p}$

is this correct?

sorry for the too many edits lol

[cloud walrus]

ali yassine

[wooden rain]

I'm still learning the very basics of abstract algebra, but to put in my two cents, I like to think of it like this: imagine you're dealing with some specific instance of a group. It may happen that it's actually a subgroup of something well-known. Subgroups sometimes inherit some properties from their supergroups (e.g. being abelian), so if you know that, you can immediately say something about the group you're working with. That saves you having to repeat many reasonings again and again for similar objects.
Another "use case" I can think of is that because subgroups may inherit some properties of their supergroups, studying subgroups can help you understand the supergroup you want to learn something about. For instance, if you want to determine if a group is abelian, maybe it's easier to prove that some subgroup is not abelian. Then you know that the original isn't either. There are also "fancier" things you can do with them (e.g. taking a quotient) that lead to even more options. That is to say, subgroups can function as a tool when studying a given group

[karmic moat]

this ideal is commonly denoted pA_p btw

the point of this exercise is that localizing a ring at a prime ideal gives you a local ring (i.e., unique maximal ideal)

[tulip otter]

right, lang also mentions this in the section about localization of this chapter

tysm anamono and jagr, have a great day/night

[karmic moat]

u2

[next obsidian]

More like amanono

[thorn jay]

𝔪_x >>>>

[maiden crater]

Could someone explain the. motivation for a discrete valuations on a field please

this is what D&F has to day

[quiet pelican]

If you imagine the field of rational functions on the complex plane, the valuation of a function at a point is the order of its zero/pole at that point

[maiden crater]

I know basic group theory, and just started off with rings today

[maiden crater]

what

a variety?

[quiet pelican]

Slight edit

[maiden crater]

ah

I see

and the order of its zero is the number of zeros?

[quiet pelican]

Just imagine the complex plane (the example works just as well, but if you were doing it for AG stuff, the “reasonable” initial level of generality would be to do it on a “variety”)

[quiet pelican]

The order of a zero at a point is essentially “for which n does it ‘look like’ (x - p)^n near p ”

[maiden crater]

thanks

[thorn jay]

is this where valuation rings pop up in AG?

residue fields

[quiet pelican]

Yes

[iron arrow]

I was looking at a proof to show that every abelian finite group is isomorphic to a product of Z/(pi^ni) where pi is prime and it uses the torsions groups isomorphy to Z/(mi) and then by chinese residue theorem you can decompose these Z/(mi), but the theorem it uses is that every finitely generated abelian is isomorphic to Z^k × T(G), (where T(G) is the torsion group), is the argument to know Z^k is just the neutral element (k=0) that because G has finite dimension and for every k > 0 it would have infinite dimension?

[rocky cloak]

If k>0, then Z^k is infinite yes. So Z^k is only finite when k=0

[tulip otter]

any hint for this?

[next obsidian]

The fuck is a principal ring

But actually I get it

[thorn jay]

principal domain without the domain

[quiet pelican]

I assume “ring in which every ideal is principal”

[next obsidian]

Vro this is just about what ideals in a localization looks like

[quiet pelican]

And this follows from “every ideal in a localisation is extended”

[tulip otter]

PID

[thorn jay]

intuitively, youre only adding divisors, so that should only lessen the amount of ideals and not create any more

||i guess to see this, r/s is in some ideal I iff r is in I, so I is totally decided by I ∩ R||

[tulip otter]

what does that mean

[tulip otter]

hmmm i see

[thorn jay]

if f : R → S is a ring map and I an ideal of R, the extension of I is the ideal generated by the set f(I)

alternatively it is the kernel of the map from S to the pushout R/I ⊔_R S

perhaps not very helpful though

[iron arrow]

So the reasoning I used is fully correct? Because it isn't stated in the proof I assumed this is what the proof assumed

[tulip otter]

ohh ok

[tulip otter]

ohhhh i see.

tysm everyone

[tardy hedge]

Did you mean I cap S?

[tulip otter]

i dont think so. If r/s is in I, then r=sr/s in I. Conversely, if r is in I, r/s=(1/s)r in I right?

[thorn jay]

no, I ∩ R here is the preimage of I

youve got your localisation map R → S^-1R, and what i spoilered proves that the extension of I ∩ R, for some ideal I of S^-1R, is simply I itself

i.e. every ideal of S^-1R is an extension of some ideal of R, and then you basicslly immediately have what you want

[vagrant zinc]

Does anyone have the arxiv paper on how to prove that two groups are isomorphic?

[next obsidian]

Wut

[vagrant zinc]

If there was an article discussing techniques for proving that two groups are isomorphic, it was well written.

[barren sierra]

like a survey article?

[thorn jay]

holy shit they solved the word problem

[barren sierra]

I mean off the top of my head, Keith Conrad has some great notes on group theory (as well as other topics)

https://kconrad.math.uconn.edu/blurbs/

[barren sierra]

they did? Ok then list every word

[thorn jay]

{ a1 ... an | ai is a letter }

I do not care about comprehension

[tulip otter]

What if n< the number of all letters

[thorn jay]

n varies over N

[white oxide]

I'm a little bit confused - I know that if $E$ is any subset of $B$, then $V(E) = V\bigl(\bigcap_{E \subseteq \mathfrak{p}} \mathfrak{p}\bigl)$, but why is $\mathfrak{a} \coloneqq \bigcap_{E \subseteq \mathfrak{p}} \mathfrak{p}$ a radical ideal? Don't we need $E$ to be an ideal in order for $\mathfrak{a}$ to be an ideal in the first place?

[cloud walrus]

okeyokay

[next obsidian]

Why does that matter

a is defined to be an intersection of ideals so it’s an ideal

[thorn jay]

an intersection of ideals is an ideal

[next obsidian]

And besides, for E < I where I is an ideal it’s exactly the same as (E) < I

So you can just replace E with the ideal it generates anyway

[white oxide]

Yeah but the lemma I'm trying to apply requires that E is a prime ideal

[thorn jay]

it's a radical ideal because of the Nullstellensatz

[next obsidian]

Okay well this is completely unrelated to your question lol

At least the last part of it

[next obsidian]

I wouldn’t really say that, this is the radical of (E) by some commutative algebra, which is radical

I don’t think Nullstellensatz is needed here at all

[thorn jay]

oh I know the Nullstellensatz as the theorem that says the intersection of all primes above I is the radical of I

[next obsidian]

The hell

lol

I mean if that’s your Nullstellensatz then sure haha

[thorn jay]

lmao

[white oxide]

I'm confused though, it says that r(E) may not be an ideal all the time

So shouldn't we be worried about that

[south patrol]

bruh lol

[south patrol]

of a subset

[white oxide]

Oh I was taking E to be any subset of B tho

[thorn jay]

I believe they mean here r(E) = { r in R where r^n in E for some n } then

[next obsidian]

Wait but potato I’m confused isn’t this the same as the intersection of all primes containing (E)

[thorn jay]

yes it is

[next obsidian]

Ahhh

Okay yeah yeah yeah

[south patrol]

that meaning for nullstellensatz is cursed

[white oxide]

Oh so r(E) is not defined to be the intersection of prime ideals which contain E then?

If E is any subset?

[thorn jay]

idk that's the only logical explanation

[south patrol]

i think in atiyah macdonald they always define radical this way

[next obsidian]

I mean that also makes sense

[south patrol]

indeed otherwise proposition 1.14 is trivial

[next obsidian]

The radical is things whose powers fall in

The fact it’s an intersection of primes is a result

[thorn jay]

it's certainly not the definition of "radical of a subset" I would use but I've got my reasons lol

[next obsidian]

this is the law. It’s in the book of laws

[tall igloo]

um pretty sure the correct definition is that every T(n)-local E_\infty-ring admits a map into a height n Lubin-Tate theory

[south patrol]

Yeah

[next obsidian]

Man sybau

[south patrol]

Nullstellensatz is already cursed in that it means 25 different things

[thorn jay]

(look I work in a setting where lattice theory is the only thing I have okay I have little choice)

[south patrol]

fortunately doesn't matter outside idk exams lol

[white oxide]

Okay, so the intersection of all primes which contain E is not equal to r(E) (if we define r(E) to be { r in R where r^n in E for some n }) and is just an ideal period

[next obsidian]

If E = {1} then r(E) are all roots of unity

But the intersection of primes is an empty intersection and thus is all of A

[south patrol]

idk what menstruation has to do this with but sure

[next obsidian]

What

[thorn jay]

I mean the only actual interesting part of the nullstellensatz is that the intersection of all maximal ideals containing I is the radical of I (in a f.g. k-algebra)

[next obsidian]

No

[south patrol]

Disagree lol

[next obsidian]

I think Nullstellensatz in Matsumura is like

[south patrol]

What I usually think of is that I(V(J)) = sqrt(J)

for strong nullstellensatz

[white oxide]

The arbitrary intersection of prime ideals is not always prime right? So basically we can only reduce to the case where E is an ideal, not necessarily prime?

[next obsidian]

Finitely algebra generated extension is finite for fields

Or some shit lol

[south patrol]

you can also weaken slightly

[next obsidian]

Yes

[south patrol]

I guess you mean like

[next obsidian]

Yo the first part

[thorn jay]

the rest of the story can be boiled down to fairly simple lattice theory and UA

that's what I mean, "actual interesting" with "thing that is unique about this case"

[south patrol]

I mean agreed sure

[next obsidian]

Idk what you mean by “can only reduce” because that’s insanely vague

[white oxide]

Okay so we can't replace E with a radical ideal

Just a regular ideal

[next obsidian]

You can

But I mean it depends

[south patrol]

the only interesting bit of nullstellensatz is Noether normalisation cause after that you can just reduce easily

[next obsidian]

For what purpose?

[white oxide]

Bruh I'm so confused

[next obsidian]

If all you care about is the set then yes you can always replace E with sqrt(E)

[white oxide]

Aight imma just do it with a regular good old ideal

[next obsidian]

V(E) = V((E)) = V(sqrt(E))

Where sqrt is what you wrote as r I guess

[rocky cloak]

I guess the fact that all the residue fields are just k is also interesting. So you actually get a correspondence between points and maximal ideals

[thorn jay]

right that too

[next obsidian]

Nullstellensatz is any result that holds because of an algebraically closed field or whatever

[white oxide]

What is (E) here?

[south patrol]

or jacobson ig lol

[next obsidian]

The ideal generated by E

[south patrol]

eh

[next obsidian]

Nakayama is any result that says a module is zero

[south patrol]

Nakayama is inverse function theorem

[white oxide]

Ugh I would just really like E to be a prime ideal.. 😭

[south patrol]

Idk

[next obsidian]

Devissage is any result that says you can prove it just for curves

Or really any induction really

[rocky cloak]

Meh, IM = M implies exists i with im = m is also Nakayama.

[next obsidian]

Chow’s lemma says that anything true for projective is true for proper

[south patrol]

Devissage is also anything where you have short exact sequences

Idk

[next obsidian]

Spectral Sequence is any result that involves a computation you don’t want to do

[rocky cloak]

So I'd go for any result involving fg modules and ideals is Nakayama

[south patrol]

Proper schemes over Z admit ample line bundles

Sybau

[next obsidian]

[south patrol]

Hm

[next obsidian]

Stein factorization says all fibers are connected

[south patrol]

Tilting tells you everything is characteristic p

[next obsidian]

I thought it says everything is char 0

[south patrol]

Up to you

you know "Tilting" was named after Fortnite

[next obsidian]

Almost means you’re reading a paper by Bhatt

[south patrol]

I would say Scholze lol

[next obsidian]

I gotta rep my countrymen and brother

Damn I’m academic brothers with Bhatt. That means I’m kinda goated

[south patrol]

That is kinda goated

[white oxide]

Can I have another hint now that I'm letting V(J) \subseteq B where J is an ideal of B

[south patrol]

ig you mean subseteq Spec B

[white oxide]

Also sanity check: the ideal generated by a set E is just the set of all finite linear combinations with scalars in A right

[white oxide]

Well I want to show that any closed set maps to a closed set. So it suffices to show that f^*(V(E)) is a closed set since V(E) are the closed sets. But E is a subset of B right?

[south patrol]

I guess you choose what E is

Sure ye though can feel free to just choose an ideal

[white oxide]

But I thought an arbitrary closed set is of the form V(E) where E is anysubset of B

[south patrol]

V(E) = V((E))

[white oxide]

I'm just trying to see why we can replace E with the radical

Yeah

[south patrol]

If a prime contains an ideal then it also contains the radical

[white oxide]

I guess I have to workout why V((E)) = V(R(E)) lol

[south patrol]

yee it should be an easy exercise

[white oxide]

assuming that R(E) is not the intersection of all primes which contain E lol

[south patrol]

Oh lol idk why you are worrying about E not being an ideal though

Just assume it is

[white oxide]

No yeah I get why wecan assume that E is an ideal

I just want E to be a prime ideal

Or can you prove it without E being a prime ideal

[rocky cloak]

Wait, what kind of tilting are you thinking about?

[next obsidian]

https://tenor.com/view/cano-gif-2179805323852138377

[south patrol]

not the kind of tilting you are probably thinking of aha

I mean in the theory of perfectoid rings (and related stuff)

are you thinking of tilting modules and things

[white oxide]

If $E$ is a prime ideal, then we have $f^{-1}(E) \subseteq \mathfrak{p}$ a chain of prime ideals of $A$ and $\mathfrak{q}$ with $f^{-1}(\mathfrak{q}) = A \cap \mathfrak{q} = \mathfrak{p}$. Thus we can apply this theorem to obtain the desired prime ideal $\mathfrak{a}$ with $\mathfrak{a} \subseteq \mathfrak{q} \subseteq E$. And it shouldn't be that hard to show that since $f^{-1}(\mathfrak{a}) = f^{-1}(E)$ we have $\mathfrak{a} = E$. This all follows from $E$ being prime, and thus $f^{-1}(E)$ being prime

[cloud walrus]

okeyokay

[white oxide]

@south patrol

Oh sorry I guess I should be the Going up theorem

you get the idea

[south patrol]

Yeah so you can probably use this characterisation of the radical in terms of primes to make it work now

[white oxide]

I thought we said the radical wasn't an ideal

Ok I'm probably getting confused

Next exercise lol

[south patrol]

radical of an arbitrary subset (which isn't an ideal)

[white oxide]

Oh and we're working with a radical of an ideal now?

[south patrol]

ye

[next obsidian]

Bro you gotta just

Completely eliminate the notion of V(E) for a subset

You’re getting confused for no reason

Every closed set is always the V of an ideal because the ideals containing E and (E) are exactly the same

You always have an ideal always

[white oxide]

Yeah I get that we can just say V(E) for an ideal E

but I want to get is so that I can replace E to be prime for the above reasons

Or is that not feasible

[next obsidian]

No you can’t replace E with a prime because then every closed is irreducible

Being the V of a prime is special

But you can always assume the ideal is radical if you want

[white oxide]

Hmm yeah I don't see how that helps yet

[next obsidian]

But V(A\cap B) is just V(A) U V(B)

[white oxide]

Maybe I'm just zeroeing in on this Theorem

[next obsidian]

Well oky this isn’t a Noetherian ring

So you can’t do what I want

But like yes you need to use going up

I mean here’s a reduction you can do to start

The image lands in V(ker(f)) which is closed, so you can look at the map A/ker(f) -> B instead

So you can assume f is injective to begin with

So whats the issue with applying going up that you first run into?

[white oxide]

Lol I'm currently pondering why we can reduce to this

[next obsidian]

I mean the image lands in V(ker f) right?

[white oxide]

Yeah

I get that

And I know it has something to do with topological properties

[next obsidian]

And V(ker f) is the same as Spec A/ker f

And is closed

So if you can show that the image is closed in Spec A/ker f the image will be closed in V(ker f) which is closed in Spec A

So it’ll be closed in Spec A (closed in closed is closed)

[white oxide]

Ok wait just to make sure I'm on the same page we're talking about the image f*(Spec(B)) landing in V(ker(f)) right

[next obsidian]

Yeah

[white oxide]

Why is this true? If p is a prime ideal containing ker f, couldn't we have p = ker f? And then ker f is not in Spec A/ker f

ker f is prime I think right

[next obsidian]

No

That’s only if the image is an integral domain

[white oxide]

Oh wait

oh yeah

Is that iff?

[next obsidian]

Primes containing ker f are primes of A/ker f by the 4th iso

[white oxide]

Ah okay in that case makes sense

[next obsidian]

There’s a 1-1 correspondence and the topologies are the same because it’s all a lattice thing

Okay does this make sense though?

The reduction

[white oxide]

Yeah I think it does, I still need to verify the rest of the steps tho cuz I'm a noob

Thanks for the help

[next obsidian]

Verify what

There’s no other steps right?

If the image of f^* is of the form p_1,…, then the image of f’^* where f’: A/ker f -> B is p_1/ker f,…

So one of them is closed iff the other is closed

It’s just the difference of writing it as V(I) where I > ker f versus writing V(I/ker f)

[iron arrow]

I'm reading a proof for smith's normal form theorem and it goes like this, let A be a matrix from Mm,n(Z), and let fA : Z^n -> Z^m (x |-> Ax) be a group morphism, then Im(fA) = H is a subgroupz then it proceeds onto creating a morphism which sends the canonical vectors of Z^n into the diagonals we choose from using smiths algorithm, and I'm not quite sure what is the point of defining H in all of this, does it have to do with the group morphism that send to the diagonal elements be an actual morphism?

[rocky cloak]

I guess it depends on exactly what is meant by smiths algorithm. But the proof I've seen is just using row an column operations on the matrix A, so then thinking about the image wouldn't do much.

[knotty badger]

Hm maybe - similarly to matrices, you can interpret row operations as postcomposition and column operations as precomposition

So you’re just trying to find automorphisms on either side that reduce fA to a diagonal map

[rocky cloak]

That is the usual proof, but you're not using the image in particular for anything.

I guess it's hard to speculate without seeing the proof, but it could also be that they just use the image for something that isn't this proof. Like talking about Z^n/H to relate with classification of fg abelian groups.

[delicate wing]

Very elementary question, but, given a finitely generated group <a,b>, if the generators commute, is the group abelian?

[quiet pelican]

Yes

[delicate wing]

Alright, that tidies things up quite a bit, many thinks

[quiet pelican]

Everything can be written in the form a^n b^m, from which it’s easy to check

[delicate wing]

Yeah, that was my thought process, which then implies the groups generated by each of the generators is in the centre, and thought it should follow from there

But wanted to double check just in case

[iron arrow]

I'm pretty sure that now rereading it it's just stating it because this is the useful interpretation of smith's algorithm to classify finitely presented abelian groups

[twilit wraith]

yes because every element in the group is expressed as a word of those two elements

and a commutes with a and b

same with b

so you can move each part of any word around each other

[vagrant zinc]

The article was written for people who were learning and described the techniques and properties that could be used to prove that two groups are isomorphic. The paper is not what you think. wtf.

[twilit wraith]

they were for sure joking by the way

[thorn jay]

i couldnt have been less, yes

i figured it was clear since the word problem is famously unsolvable lol

[maiden crater]

How are poynomial rings commutative by defn

[velvet hull]

by definition of the multiplication described above

[maiden crater]

which implicitly assumes R is commutative right

[velvet hull]

The coefficient ring R above was assumed to be a commutative ring (...)

[maiden crater]

yes, so otherwise we just expand componentwise keeping the order in mind

I don't get this bit. How is it immediate that if R has no zero divisors, neither does R[x]

[velvet hull]

suppose that f(x), g(x) in R[x] such that f(x)g(x) = 0

suppose to the contrary that neither f,g are 0

then you can WLOG assume that f,g both have nonzero constant terms (why?) see below

hence R has a zero divisor

[maiden crater]

what

[velvet hull]

actually, even better, dont look at the constant terms look at the leading terms, which is exactly what your proof is doing

[maiden crater]

right, as we only care bout the coefficients

[maiden crater]

well, I want to show even when factorised there's no product such that f(x)g(x)=0 right

[velvet hull]

it is easier to prove the contrapositive

which is what your text does

[maiden crater]

Okay, I think I'm just confused now

Say we factor an elmeent of R[x]

as p(x)q(x)

now x belongs to the same field as its coefficients right

[velvet hull]

x is the indeterminate

x does not belong in R

[maiden crater]

Ah, that makes it clear

thanks

[velvet hull]

the idea of R[x] is that we are abstractly thinking about how an "extra" "unrelated" element can interact with R

[maiden crater]

cool

thanks

[next obsidian]

I think you belong, x. I think you belong

[maiden crater]

so group rings are basically the "dot produtc"

[velvet hull]

well the multiplication is more complicated

[maiden crater]

oh right

[maiden crater]

you basically perform multinomial multiplication, right

[velvet hull]

yes, but you respect the group multiplication as well

an alternative way of thinking about the group ring, is that R[G] is R[X] modulo the relations of G (written multiplicatively), where X is the base set of elements of G

[maiden crater]

noted

[velvet hull]

so you are multiplying and adding polynomials, there just so happens to be a little bit extra stuff going on with the indeterminates

[maiden crater]

mhm

thanks

[maiden crater]

Kind of confused here, isn't this true by definition( atleast if there's a b such that bp(x)=0...)

For the other direction, yes, it's not by defn

[mossy slate]

I am making a list of all the useful algebra rules I can think of, and today I was looking at the identity

$0 \notin {b, d}\rightarrow (a \div b) \cdot (c \div d) = (a \cdot c) \div (b \cdot d)$

and I wondered if you can transpose the additive operations in the same way. Turns out you can!

$(a - b) + (c - d) = (a + c) - (b + d)$

Kinda interesting... I don't remember being taught this anywhere, but it could be helpful.

[cloud walrus]

Unchaynd

[thorn jay]

this is way deeper than needed but this has to do with adding inverses to commutative monoids

so-called localisation

[mossy slate]

Interesting!

[thorn jay]

Basically if you have a commutative monoid (M, +, 0) (essentially an abelian group which does not necessarily have inverse elements, like the nonnegative integers), then you can "add all inverses" to the best of your ability: define G(M) to be the set of all pairs (a, b) subject to the equivalence relation:
(a, b) ~ (c, d) if there exists some t with a + d + t = b + c + t
and add the operation (a, b)/~ + (c, d)/~ = (a + c, b + d)/~. It turns out that this is well-defined and actually forms an abelian group!

Now, if you have an abelian group A, there is a natural isomorphism A -> G(A) given by a -> (a, 0). Then we have (x, y) ~ (a, 0) if and only if x - y = a. Now suppose that a, b, c, d are elements of A. Then, using our intuition of (a, b) being the element a - b in actuality, the addition in G(A) turns into the identity (a - b) + (c - d) = (a + c) - (b + d)

[tribal moss]

I think I vaguely remember something like you need the monoid to be cancellative for that to work well?

[thorn jay]

yes, iirc the natural map M -> G(M) is injective iff M has cancellative elements

[tribal moss]

Ah, I see you already handled that by including the +t.

[thorn jay]

I looked into localisation for a monoid structures on algebras such that multiplication distributes over over the operations a while back

pretty cool

localisation stays in your variety, for example

not exactly a super hard fact to prove but the notation gets super messy lol

[mossy slate]

I am screenshotting this, and hopefully one day I will know enough abstract algebra to where I can understand it all. I'm a freshman working toward a PhD in Mathematics, so hopefully it won't be too long. 😅

[tulip otter]

is G(M) the grothendieck group of M?

[elfin wraith]

Yes

Something something vector bundles K_0

[tulip otter]

i see

[tulip otter]

sadly i wont get that rn because idk what vector bundles are

[elfin wraith]

I don’t really get it either, K theory is scary, but yeah that’s where it first comes from afaik. You take like vector bundles over a manifold up to isomorphism and that forms a commutative monoid, then Grothendieck considered the abelian group from localising that and I’m pretty sure that’s what K_0 is

You can also make this a statement about projective modules over a ring and I think these statements are dual to each other

And that’s the easy K group to define!

[noble nexus]

not really dual they're basically the same thing

b/c vector bundles over a manifold are the same as projective modules over $C^\infty(M)$

[cloud walrus]

Blake

[noble nexus]

serre-swan

[elfin wraith]

Ahh yeah I see

[noble nexus]

super awesome theorem

[south patrol]

Should first come from algebraic varieties actually (though more or less the same definition)

Lol

[noble nexus]

works with C(X) for compact Hausdorff spaces too

[elfin wraith]

Yeah ofc ofc

I went to a very strange intro to K theory talk that went through this story a couple of weeks ago

99% of the room was 1st year students

[noble nexus]

Didnt K theory come from topology maybe I'm just misremembering my history

[south patrol]

Have you seen this funny like categorical proof lol

[elfin wraith]

He asked if anyone knew what a projective module was and I was the only person who put their hand up, great start

[noble nexus]

I didn't know you could prove it categorically, though of course serre swan has a clear categorical interpretation

[elfin wraith]

Didn’t it come from varieties in proving GRR?

[noble nexus]

(basically just making precise what "the same as" means)

[south patrol]

this is the history of i am aware of

[noble nexus]

ah ok

I think starting from topology is the fake history that makes it easier to understand

is probably what I'm remembering lol

although that might just be what I'm familiar w/ because it's the most relevant case for k theory of c*-algebras

[elfin wraith]

I really want to learn K theory nonsense

But like, shits hard

[tulip otter]

it turns out that everything interesting is hard 🥀

[south patrol]

Well it's slightly cheating but the point is that uh fix a smooth manifold M.
if you take the category of trivial vector bundles over M, then the "idempotent completion" is the category of all vector bundles. this idempotent completion basically means that you take anything that looks like a projection and create objects and maps which make it a projection lol

(here the point is like any vector bundle is a summand of a trivial one)

then you can do a similar thing with projective modules being the idempotent completion of free modules

And so you can formally reduce to the case that trivial vector bundles are the same as free modules over C^oo(M) lol

i guess somewhere here we should mention that like the K-theory you get from smooth vector bundles is the same as the K-theory from topological vector bundles for a smooth manifold. But that is not too hard to show

[noble nexus]

Oh that's cool

makes sense actually

If I remember that's like more or less how the non-category theory proof goes as well (start with trivial bundles and kinda reduce to that)

at least the one I've seen

[tulip otter]

let $a\in A$ and $s\in S$ and write $a=u\prod_{i=1}^n p_i, s=v\prod_{k=1}^m q_k$ where $u,v$ are units and $p_i,q_i$ are irreducible elements of $A$ for all $i,k$.\ WLOG assume that $(a,s)=1$ where $(a,s)$ denotes the gcd of $a$ and $s$, that $\frac as$ is not a unit and that $n<m$ then $\frac as=\frac{u\prod p_i}{v\prod q_k}=\frac uv\frac{\prod p_i}{\prod q_k}=\frac uv\prod_{i=1}^n\frac{p_i}{q_i}\prod_{k=n+1}^m\frac 1{q_k}$. Now itsclear that $\frac uv$ and $\frac 1{q_k}$ are units for all $k\in{n+1,\dots,m}$ (the inverses are $\frac{u^{-1}}{v^{-1}}$ and $q_k$), and that $\frac{p_i}{q_i}$ are irreducible for all $i\in{1,2,\dots,n}$. Hence $S^{-1}A$ is a UFD.\\Now assume that $(p)\cap S\neq\emptyset$, then $\exists x\in A$ such that $px\in S$ and $1=\frac{px}{px}=p\frac x{px}$ so that $p$ is a unit in $S^{-1}A$ and hence not irreducible. Conversely, assume that $(p)\cap S=\emptyset$ and write $p=\frac as\frac{a'}{s'}=\frac{aa'}{ss'}$. I have to prove that either $\frac as$ or $\frac{a'}{s'}$ is a unit, but i am not sure how to do that. \It might be something obvious but i am not seeing it. Any hints for this? and is the rest of the argument correct?

[cloud walrus]

ali yassine

[tardy hedge]

What book is all this from

[tulip otter]

lang's algebra

[quiet pelican]

So by definition if p = aa’/ss’, then pss’ = aa’ in A
||So p divides a or a’ in A
Now WLOG it’s a, and rewrite a = bp||

[tulip otter]

ohhhh i see. pss'=pba'\implies a'/s' b/s=1\implies a'/s' is a unit

[quiet pelican]

Yes

[tulip otter]

ohhh nice

what about the rest of the argument?

is it missing anything/can be written in a better way ?

[quiet pelican]

I feel I want this justified

The rest is fine I think

[tulip otter]

since $(a,s)=1, (p_i,q_i)=1$ for all $i\in{1,2,..,n}$, its possible to write $\frac{p_i}{q_i}=\frac {b_i}{k_i}\frac {b_i'}{k_i'}=\frac{b_ib_i'}{k_ik_i'}$ where $(b_ib_i',k_ik_i')=1$ so that $p_i=b_ib_i',q_i=k_ik_i'$ and then WLOG $\frac{b_i}{k_i}$ is a unit since both of $p_i$ and $q_i$ are irreducible

[cloud walrus]

ali yassine

[quiet pelican]

Justify the last sentence 🙂

(It’s right, but I don’t think you quite have a justification for it)

[tulip otter]

i mean b_i and k_i are units so they have inverses and then the inverse of b_i/k_i is b_i^{-1}/k_i^{-1} just like the way i justified that u/v is a unit here

[quiet pelican]

That’s how I thought you were trying to justify it, and it’s wrong
You can’t WLOG that, because what if you have b_i and k_i’ are your units?
k_i and k_i’ don’t play symmetric roles if you fix b_i

[tulip otter]

maybe i have to justify that k_i^{-1} and v^{-1} are element of S?

[quiet pelican]

You can’t assume that both b_i and k_i are units

[tulip otter]

but (b_i,k_i')=1 too because otherwise (p_i,q_i) neq 1

so then i can rewrite the fraction b_ib_i'/k_ik_i' as b_i/k_i' b_i'/k_i if b_i and k_i' are the units instead

[quiet pelican]

Yup but you either need to prove b_i/k_i is a unit, or b_i’/k_i’ is a unit

[tulip otter]

ah right lol

alright then, b_i/k_i is a unit whose inverse is k_ib_i^{-1}

this should be correct ig?

[quiet pelican]

Yes

And now you notice you didn’t actually need q_i to be prime

[tulip otter]

right, only that (p_i)\cap S=\emptyset as the other part of the question says

tysm, have a great day/night!

[tulip otter]

question 6. A is a UFD so by 5, A_{(p)} is also a UFD and the prime elements of A_{(p)} are the primes q of A such that (q)\cap A\setminus (p)=\empty set. But the only such prime is p. So p is the only prime element of A_{(p)} and A_{(p)} is principal

is this correct?

I feel like the answer is too short for this to be correct lol

[quiet pelican]

That works

[tulip otter]

i see, tysm

[next obsidian]

There’s another prime

It’s called (0)

Also, this works only if you know you can check if all primes are principal which it looks like you do

[tulip otter]

ah right since A (and hence A_{(p)}) is integral

[tulip otter]

are you talking about primes of A or A_{(p)}?

[next obsidian]

Doesn’t matter

[tulip otter]

but why is it necessary to check that

[next obsidian]

Your definition is that all ideals are principa but you only checked the prime ideals are principal

You need to know it’s sufficient to only check the primes

[tulip otter]

ah i see. What i had in mind was that A_{(p)} is a UFD so primes and irreducible elements are the same, thus if the only prime in A_{(p)} is p (other than 0) then any element in A_{(p)} is of the form pu where u is a unit

tysm Chmonkey

[balmy python]

Is the main point of rings to show how to construct finite fields?

what other uses are there

[noble nexus]

Rings show up everywhere

[wooden rain]

It's not possible to have a non-trivial group of subgroups, right? Or are there some constructions like that?

[noble nexus]

What do you mean by group of subgroups

[wooden rain]

We consider a subset of subgroups of a given group, and endow it with the structure of a group

The group need not be arbitrary

[noble nexus]

Well you could give it some random group structure but I don't think there's anything natural

I guess the ideal class group in ring theory is kinda like this

[wooden rain]

I'll look into that, thanks!

[mental lake]

Given that S is a Noether ring, is R also a Noether ring? With R in S rings.
Consider the example: R=\mathbb{R}[X,XY,XY²,...] in S = R[X,Y]. Now S is clearly a Noether ring, but R is not. This results in the chain: (X) in (X,XY) in (X,XY,XY²) in .... Explain why XY^k+1 is not an element of (X,XY,...,XY^k) and why this chain therefore does not stabilize.

can someone help me

[quiet pelican]

What are S and R

[mental lake]

It’s in the example

[thorn jay]

it has the structure of a lattice, not a group

[thorn jay]

for example: it turns out geometric spaces are best understood by looking at the (continuous or nice in some way) functions from open subsets of that space to some algebraic object (often the real or complex numbers). These have a natural commutative ring structure.

You can even go so far as to say that a geometric space is this weird structure of rings associated to open subsets which satisfy certain conditions (locally ringed spaces).

I would personally go a little further; to general algebraic structures; but that's just me and my silly universal algebra

[balmy python]

what's a geometric space?

like surfaces?

[thorn jay]

yeah, like manifolds

[balmy python]

topological spaces?

ahhh

[supple ice]

in R = R[X, XY, XY^2, …] set Ik = (X, XY, …, XY^k)
if XY^(k+1) were in Ik, we could write XY^(k+1) = f0·X + f1·(XY) + … + fk·(XY^k) with fi in R
every non-constant element of R contains at least one factor X (because R is generated by XY^m’s)
so if some fi is non-constant then fi·(XY^i) has at least X^2 but the left side has exactly one X so no X^2 terms can appear therefore all fi must be constants (real numbers)
then the equation becomes XY^(k+1) = c0·X + c1·(XY) + … + ck·(XY^k) divide by X in the bigger ring R[X,Y] and you get: Y^(k+1) = c0 + c1·Y + … + ck·Y^k which is impossible because the left side has degree k+1 and the right side has degree at most k
so XY^(k+1) is not in (X, XY, …, XY^k) meaning Ik is strictly contained in Ik+1 for every k so the chain never stabilizes and R is not Noetherian

[thorn jay]

like a topological space is not enough information to be properly "geometric"

[balmy python]

ah so you need something more like a metric

or something measurable

[thorn jay]

or a manifold structure

[balmy python]

ngl i haven't really come across manifold structures before

is it something locally homeomorphic to R^n?

[thorn jay]

this is especially emergent when you do algebraic geometry, that extra structure of those functions gives the Zariski topology (which is super coarse, basically never Hausdorff for example) "thickness" in some sense

[thorn jay]

yes, and transitioning from one "coordinate chart" to another has to be a homeomorphism too

or diffeomorphism depending on what you want to do

[balmy python]

ahhh

[balmy python]

ah so a zariski topology is never hausdorff

[thorn jay]

rarely

[balmy python]

ah

so you might not have properties of like limit uniqueness etc

i doubt that's a problem in algebra though

that's so cool though

ive never seen analysis tie in with algebra

usually i only see linear algebra link with groups and rings 😭

[elfin wraith]

Any examples off the top of your head where it is Hausdorff?

[thorn jay]

one point space

hehe

[elfin wraith]

I mean are there any non finite examples lol I wouldn’t think so but I’ve not like worked out the details before

[thorn jay]

no it doesnt seem probable

but yk edge cases

gotta cover my ass

[quiet pelican]

Iirc theres an exercise in Atiyah Macdonald that gives the exact criteria

[elfin wraith]

Topology is known for that

[thorn jay]

(with affine opens)

[supple ice]

finite product of fields: spectrum is a finite discret space

[quiet pelican]

It doesn’t seem like there is, but they do have one that shows it for rings where every element is idempotents

[thorn jay]

right cuz thats just Stone duality lol

[supple ice]

i think specA is hausdorff iff A is absolutely flat

[thorn jay]

flat for every R → A?

[mental lake]

but in principle, an X^2Y^i can appear in your right-hand side as long as it cancels out with other terms.
For example, if f0 = XY and f1 = -X, then f0X + f1XY = X^2Y - X^2Y = 0.
And I don't think you can just divide by X, but I'm not sure about that.

[supple ice]

okay u can reduce the whole equation modulo X²

[rocky cloak]

I think the spectrum is hausdorff iff it is zero-dimensional. So spec(F^N) for a field F for example

[elfin wraith]

That sounds believable enough yeah

[mental lake]

And for the first bit?

Or what do you mean I’m confused

[supple ice]

after reducing mod X^2 you get XY^(k+1) congruent to a real linear combination of X XY, …, XY^k and comparing the X-linear parts forces Y^(k+1) to be a real linear combination of 1, Y, …, Y^k impossible by degree

[supple ice]

i think we should assume that A is reduced

if it s not reduced if we take A=k[eps]/(eps²) then specA is hausdorff (single point) but A it s not absolutely flat

[thorn jay]

k[eps]/(eps^2) is a funky ring

[rocky cloak]

So say A is zero-dimensional, and m and n are two prime ideals.

Then the hausdorf condition can be formulated as there being closed sets V(I) and V(J) with m not in V(I), n not in V(J) and union of the two being everything.

Translated to algebra (let's say A is reduced) it's I not in m, J not in n, IJ = 0.

We have A/mn = A/m x A/n, let x be a lift of (1, 0). Then I = (x) is not in m. As A is zero-dimensional and reduced localizing at a prime p always gives a field. So the localization of (x)/(x^2) at any prime is 0, so (x) = (x^2). By Nakayamas lemma I = (x) is generated by an idempotent I = (e). Then we can pick J = (1-e). Boom boom.

Conversely if A is not zero dimensional any nonmaximal prime is a non-closed point.

[elfin wraith]

Very slick, nice!

[rocky cloak]

So if A is zero-dim reduced, the localization at any prime is a field. Being flat is a local property and fields are absolutely flat.

If A is not zero-dim, then we can localize to assume it's local, then A/m should not be flat, you can consider tensoring A/m with the inclusion I < A where I is a finitely generated ideal in m.

[supple ice]

so what I said is correct or not ?

we need that A should be reduced ?

[rocky cloak]

I'm at least saying spec(A) hausdorff iff A/nil is absolutely flat.

I'm not sure if A is never absolutely flat if A not reduced, but I would guess so

Hmm, okay if A is not reduced pick x with x^2 = 0. Then A/(x) tensor (x) -> A is 0

So yes you need A reduced

[supple ice]

yes

we are okay

[next obsidian]

This isn’t true, sorry. What about p^2?

There’s a lot of ideals in this ring

[tulip otter]

"this isnt true, sorry" sounded funny ngl

[tardy hedge]

Lol

[spare nymph]

What's the context

[tulip otter]

oh wait yea actually i missed the other powers of p you are right.

[next obsidian]

I guess this can directly show all ideals are of the form (p^n)

[tulip otter]

@spare nymph here

[tulip otter]

yea, probably the same argument i made but now corrected to include all powers of p?

[next obsidian]

Essentially

Find the min n such that p^nu is in I

And then you can show everything else is a multiple of that

[tulip otter]

I see

tysm

[tardy hedge]

Chmonkey W

[tulip otter]

real

let $I$ be an ideal of $\mathbb{Z}[i]$ and assume that its generated by $a_1+ib_1,\dots,a_n+ib_n\in\mathbb{Z}[i]$ and let $d=\gcd(a_1,\dots, a_n)$ and $d_1=\gcd(b_1,\dots, b_n)$. I claim that $(a_1+ib_1,\dots,a_n+ib_n)=(d+id_1)$.\Since $d\mid a_k$ and $d_1\mid b_k$ for all $k\in{1,\dots,n}$ then\$(a_1+ib_1,\dots,a_n+ib_n)\subset (d+id_1)$.\ Conversely, $\exists m_k\in\mathbb{Z}[i]$ such that $d+id_1=\sum_{k=1}^nm_k(a_k+ib_k)$ by applying bezout's lemma to $d$ and $d_1$ separately. Hence $d+id_1\in (a_1+ib_1,\dots,a_n+ib_n)$ and $(d+id_1)\subset (a_1+ib_1,\dots,a_n+ib_n)$ so that $I=(a_1+ib_1,\dots,a_n+ib_n)=(d+id_1)$ so that $\mathbb{Z}[i]$ is a PID and hence a UFD.\\The units of $\mathbb{Z}[i]$ are $\pm 1,\pm i$

is this correct?

[tardy hedge]

Shouldn't you start with an arbitrary ideal in Z[i] and then show its singly generated

[tulip otter]

ah i took for granted that every ideal in Z[i] is finitely generated

[tulip otter]

i probably worded it badly, i meant to start with an ideal and then let these a_k+ib_k be its generators

[cloud walrus]

ali yassine

[tulip otter]

i think its better now (hopefully) , thanks for the note

[tardy hedge]

But you are still assuming the ideals are finitely generated?

[tulip otter]

yes, idk if its true tho lol

[karmic moat]

it is

[tardy hedge]

every ring of integers is noetherian

[tulip otter]

i mean Z is finitely generated so Z[i] should probably be?

[tulip otter]

i see

nvm that was stupid. it turns out that noetherian means that every ideal is finitely generated by definition lol

[quiet pelican]

Actually, fun related exercise:
prove every ring with finitely generated additive group is Noetherian

[fading acorn]

while it's equivalent it's good to have in mind that noetherian (ring/module) means that any ascending chain of (ideals/submodules) has maximal element.

[tardy hedge]

ive been needing to review those equivalences

[tulip otter]

ohhh i see

i will keep that in mind

[fading acorn]

yeah this is what most textbooks use as definitions

and it's very much useful to prove stuffs