groups-rings-fields

[next obsidian]

So in a monoid you have an identity

and this is unique I think

but just because x*y = y for some y

doesn't mean x is the identity

[pure crest]

ohh damn good point

[next obsidian]

So like define some element x as a "partial identity" if for some y

xy = y

then in group

partial identities are unique

becasuse you hit it with y^{-1} and you get x = e

but in a monoid

partial identities are not

[upbeat burrow]

Wait why wouldn’t x be the identity in that example?

It satisfies that property

[next obsidian]

in a monoid?

I only specified it for 1 single y

So take Z/2Z x Z/2Z under just multiplication

this is a monoid

but notice that (1,0)*(1,0) = (1,0)

so (1,0) is a partial identity

but so too is (0,1)

[upbeat burrow]

Ahhhhhhh

[next obsidian]

since (0,1)*(0,1) = (0,1)

so to get uniqueness of a partial identity requires inverses

(I mean maybe there's some weaker thing, but that's how you get it for groups)

[upbeat burrow]

Thank you!

[next obsidian]

Np

[pure crest]

i think if you have a principle ideal domain, you may be able to prove that all elements are unique

actually nvm

[upbeat burrow]

Are you thinking factorization?

In a UFD

[next obsidian]

All elements are unique

Like

if you say it like that you're saying for all x, that if y = x that y = x

but that's bruh

In terms of saying unique 0

you aren't framing it in terms of equality with something

[upbeat burrow]

It’s the property

[next obsidian]

you're saying "anything satisfying this property are equal"

[upbeat burrow]

I see what you’re saying

[next obsidian]

yeh

[pure crest]

right

but arent these the properties of identity elements

[next obsidian]

yeah but ai priori you don't know an identity element is one specific element

[pure crest]

but i think new nick is talking about a random element

[next obsidian]

I was adressing your point

which you said nvm about

[pure crest]

ohhh

but still

i dont understand how x = y, then y = x property holds for all elements

[next obsidian]

uh

because equality is symmetric

if x = y then y = x

Like if you talk about "elements are unique"

[pure crest]

hmm

i understand what you mean

[next obsidian]

Does going up hold for prime ideals in a polynomial ring?

Or more specifically, given k[x1,...,xn] if you have a prime ideal of k[x1,...,x_{n - 1}] can you extend this to a prime ideal in the larger ring in a way like going up?

[bleak abyss]

@next obsidian I would intuitively expect that using the same generating set would keep it prime even with a new variable

[next obsidian]

Hmm

So suppose p is prime

then for some field you have a morphism phi: k[x1,...,x_{n-1}] whose kernel is p correct?

[bleak abyss]

Yeah

[next obsidian]

so you can extend phi by just declaring x_n to go to something

[bleak abyss]

Field of fractions of the quotient

[next obsidian]

but then is the kernel still p, but this new fattened up version?

If phi necessarily were an algebra homomorphism I think I see how you could go about showing it

since it's determined by where the x_i go

Oh so an element of this

[bleak abyss]

Should be. My kinda low tech way of thinking about what I said was more like, let's say fg\in p

[next obsidian]

is a polynomial with coefficients in k[x_1,...,x_{n - 1}]

also sorry yeah

[bleak abyss]

And then kinda splitting stuff up. Like okay if you involve x_n you're just a polynomial whose coefficients are in the downstairs p, etc

[next obsidian]

hmmm

Idk, it seems kinda iffy

maybe you want to reduce to like

irreducible components

[bleak abyss]

I guess the problem would be that you take sums instead of straight products

Actually

Hold up

p is prime in k[x_1,...,x_{n-1}]

Let's say the quotient is A

[next obsidian]

yes

Oh

Is the quotient of the extra one just A[x]?

[bleak abyss]

Yeah I'm pretty sure we can just prove that

[next obsidian]

oh yeah that would be good

I need to think if that can solve my problem though

I think the only thing I need to worry about

is that two primes which are properly contained

somehow like umm

become the same when you fatten them up

but I think this should be obvious?

by like degree stuff in the different variables

[bleak abyss]

What was the OG problem?

[next obsidian]

To show that ht p + dim A/p = n for any prime in k[x_1,...,x_n]

it basically just means

that every maximal chain of primes has height n

so my idea

was to take a prime p

intersect with k[x_1,...,x_{n - 1}]

to get a length n-1 chain

do going up to get a chain of length n-1 in the original ring

then the maximal one I think you can get one more

by taking the old maximal one, and just throwing in x_n

Also Dami I think you're right about your thing

You know that for an ideal of R

call it I

that if you do R[x]/I that this is (R/I)[x]

So take R = k[x_1,...x_{n - 1}]

[bleak abyss]

Yeah I don't think it's possible if two primes are different downstairs they can become the same upstairs. Like write a generating set and just be like

[next obsidian]

well my idea was basically

if p_1 < p_2

take an f in only the x_1 through x_{n -1} in p_2 not in p_1

then you shouldnt be able to get this in the fattened up p_1

because you basically only add multiplication by stuff with x_n

[bleak abyss]

Yup

That's the general idea at least

[next obsidian]

Damn

If this works

It's a lot simpler than Matsumura's proof in the solution

[bleak abyss]

I could see there being something to think about if like

[next obsidian]

I glanced at it

since it's been hours

and it's whacky

[bleak abyss]

Magically x_n kills its own contribution

But I doubt that'll actually be a problem

[next obsidian]

hmmm

[bleak abyss]

Just like double check that it isn't and you're gg

[next obsidian]

I think it's this easy right

Actually I guess I see your point

I feel like if you consider a polynomial in k[x_1,...x_n] as an element of k[x_1,...,x_{n-1}][x_n]

that it should fall out by just knowing you need to have no x_n terms

But I guess it might end up getting messy

Perhaps you can like quotient out by p_1

and do stuff in the quotient and then adjoing the last variable

because then p_1 is 0

so it would fatten up to 0 still

but actually this doens't make sense nvm

this suggests to me that p_1 wouldn't fatten up at all when you adjoin x_n

Lol I actually almost had a proof

but I think it relies on a false result

if the fattened up versions were equal then

(A/p_1)[x] would be isomorphic to (A/p_2)[x]

but this doesn't mean A/p_1 is isomorphic to A/p_2

In fact maybe it's possible for those to be isomorphic without p_1 = p_2

gross

Wait this works I think

If p_1 = p_2 when viewed in A[x]

then A[x]/p_1 = A[x]/p_2 via identity

this translates to (A/p_1)[x] = (A/p_2)[x] via identity

restricting to scalars this would still be bijective

so that this gives rise to A/p_1 = A/p_2 via identity

so that the two are equal???

blech, this seems really weird idk what's going on. It feels bad to say there's an identity map between (A/p_1)[x] and (A/p_2)[x] I think I need to frame it in terms of like some categorical property or like commuting with some maps

I guess I do know the map

It all starts from like f(x) + p_1 maps to f(x) + p_2

so by tracing this out I think it descends to a + p_1 maps to a + p_2 for a in A

and from this I think you can get set equality of p_1 and p_2

by looking at where the identity coset goes

sweet

[next obsidian]

😔 even this isn't enough to give me what I needed god damn it

unless I prove another special things about how primes interact in polynomial rings over a field

Actually it just doesn't work period rip me

[next obsidian]

Okay, so I think I've figured it out. Even figuring out Matsumura's hint is a struggle

So, take $p$ a prime in $k[x_1,\dots,x_n] = A$, then consider $A/p = k[\alpha_1,\dots,\alpha_n]$ where $\alpha_i$ is the image of $x_i$. Then dim $A/p =$ tr. deg$_k \text{Frac}(A/p) = r$, so we can assume that $\alpha_1,\dots, \alpha_r$ form a transcendence basis of $\kappa(p)$ over $k$. Now set $K = k(x_1,\dots,x_r)$, note that for $\alpha_1,\dots,\alpha_r$ to form a transcendence basis this says that for any $f(x_1,\dots,x_r)$ that this is not in $p$. This means that $k[x_1,\dots,x_n]p = K[x{r + 1},\dots,x_n]P$ for some $P$, namely $P$ is the extension of $p$ into the ring $S^{-1}k[x_1,\dots,x_n] = K[x{r + 1},\dots,x_n]$ where $S = k[x_1,\dots,x_r]\setminus{0}$.

[cloud walrus]

UserFormerlyKnownAsMathemagician:

[next obsidian]

So the benefit here is that dim $k[x_1,\dots,x_n]p = \text{ht} p$, so what we want to do is show that dim $K[x{r + 1},\dots, x_n] = n - r$, but we can actually see that $P$ is a maximal ideal of $K[x_{r + 1},\dots, x_n]$, we can see this by looking the transcendence degree over $K$ of $\kappa(P)$, but notice that this actually is the exact same thing as $\kappa(p)$, the residue field of $p$ inside of our original polynomial ring. Noting however that $\alpha_1,\dots,\alpha_r$ formed the transcendence basis of $\kappa(p)$ over $k$, so it just suffices to show that $\alpha_1,\dots,\alpha_r$ are algebraic over $K$. This is however immediate, as $\alpha_i = x_i + p$, and $K = k(x_1,\dots,x_r)$, we can just take the polynomial $p(y) = y - x_i$, then plugging in $x_i + p$ due to how $k(x)$ acts this is $0$

So what we've done now is reframed this question into one we can answer when we have a prime ideal of a polynomial ring over a field, but that its MAXIMAL

So we just need to show then that its height is n

[night pier]

nerd

[tacit violet]

it always tru? $\cancel{o}\circ I=I\circ\cancel{o}$

[cloud walrus]

Desturu:

[scarlet estuary]

is I the identity function here? if so, yes

$I \circ \phi = \phi = \phi \circ I$

[cloud walrus]

Namington:

[scarlet estuary]

assuming I is the identity function

(and the domain/codomain makes sense)

[vestal snow]

Can someone verify if my solution for this problem is correct?

Let $u_i$ be the elements mapping to $e_i$, the canonical basis of $A^n$. Then every element of $M$ can be uniquely written as a sum $x + y$ where $x$ is in the module generated by the $u_i$ and $y$ is in the Kernel of $\phi$. Now we can define a surjective homomorphism $\tau$ from $M$ to Ker $(\phi)$ which simply "forgets" the $x$ in $x+y$. It is well defined and surjective, thus we get that Ker$(\phi)$ is isomorphic to $M$/Ker$(\tau)$, which is finitely generated because $M$ is.

[next obsidian]

There might actually me a lemma that implies this in 1 part

[cloud walrus]

Have a Banana:

[next obsidian]

I’ll check your proof in a second too but

Do you know what a module of finite presentation is?

[vestal snow]

No I don't

[next obsidian]

So it's stronger than finitely generated. A module M is of finite presentation if you have an exact sequence

$A^m\to A^n\to M\to 0$

[cloud walrus]

UserFormerlyKnownAsMathemagician:

[next obsidian]

finite generation gives you that like last part

it just says that the kernel is finitely generated too

So clearly $A^n$ is of finite presentation because you can just do
$$A^0\to A^n\to A^n\to 0$$

[cloud walrus]

UserFormerlyKnownAsMathemagician:

[next obsidian]

But you can show that if you have an exact sequence
$$0\to K\to N\to M\to 0$$

[cloud walrus]

UserFormerlyKnownAsMathemagician:

[next obsidian]

M is finitely presented (of finite presentation) and that if N is finitely generated

then so too is K

In this case you have
$0\to \ker\varphi\to M\to A^n\to 0$

[cloud walrus]

UserFormerlyKnownAsMathemagician:

[next obsidian]

So you can apply the lemma

[vestal snow]

Ah that's a slick proof

[next obsidian]

The proof of that is kinda annoying though, but it might end up actually being what you did basically

But like in a general setting

I'll look over it though

[vestal snow]

Thanks

I just feel that this proof is too easy

for a book like Atiyah MacDonald

So I just wanted to make sure

[next obsidian]

Oh huh that's pretty slick actually

It makes sense to me, I didn't bother to check that it's well defined

it's clearly surjective though

But bar that, it looks like it makes sense to me

But let me think about this hmm

Oh, okay

so the proof that it's well-defined and all

Uses something sort of deep I think

Has he introduced projective modules?

The fact that you can decompose it that way

is basically because M = ker phi oplus A^n in this case

and that's because A^n is projective

[vestal snow]

It's well defined because say $y+\sum k_iu_i = y' + \sum c_iu_i$. Then you get that $\sum (k_i-c_i)u_i$ is in the Kernel. This implies $\sum(k_i-c_i)e_i = 0$. This must imply $k_i = c_i$

[cloud walrus]

Have a Banana:

[next obsidian]

Yeah actually so this is really crucial

So yeah actually that's a right proof I think

[vestal snow]

Because there is no linear relationship between the elements e_i

[next obsidian]

basically what well-definedness means that any element m uniquely can be written as x + y

where x and y are in those repsective sets

[vestal snow]

Yes

[next obsidian]

This is really the fact that M = ker phi oplus A^n

So rather than it being "easy"

I think you just stumbled onto the like concrete realization of some really deep stuff hidden underneath this

and from that perspective

It's really really easy to see how to make your proof work in the abstract setting

you just stumbled upon the non-abstract implementation of it

if that makes sense

[vestal snow]

Yeah I think it does

[next obsidian]

Yeah, that's sick man

Look up the splitting lemma

[vestal snow]

Thanks!

[next obsidian]

It says for an exact sequence

0 -> A -> B -> C -> 0

[vestal snow]

I will once I'm done with other exercises

[next obsidian]

Cool

[vestal snow]

The next sequence of exercises introduces direct limits

[next obsidian]

Gotcha gotcha

If you haven't ever done them

or done category theory type stuff

[vestal snow]

Also, I just read your username

[next obsidian]

it'll be a little confusing I think haha

Yeah haha

Play on Prince's name

[vestal snow]

Prince?

[next obsidian]

The artist

the singer

[vestal snow]

Ahh gotcha

[next obsidian]

I think like something happend with his record label

so he changed his name to

The Artist Formerly Known As Prince

[vestal snow]

This gave me an idea for my new username

Does anyone know how to prove surjectivity directly in the proof that direct sum distributes over tensor products?

[next obsidian]

I think it follows by looking at simple tensors no?

[upbeat burrow]

@vestal snow I can kind of remember how.

[river fern]

is it true that if H is a subgroup of G, then (aH)(bH) = (ab)H for any a, b in G?

[solemn rain]

that's how you can define an operation on a set of cosets

if ur asking as in equality

first you have to make sense by what you mean by (aH)(bH)

[river fern]

I meant {xy: x \in aH, y \in bH}

maybe that is an invalid definition?

I guess coset multiplication is actually defined as (aH)(bH) = {(ab)h: h \in H}?

in which case the answer to my question follows directly from that

[solemn rain]

this isnt well defined

for all subgroups

[river fern]

oh it seems like it's well-defined for normal subgroups: https://math.stackexchange.com/questions/1752937/coset-multiplication-giving-a-well-defined-binary-operation/1753030

so i guess it's not well-defined in general

But I guess G / H is always a group?

just without that property being satisfied in general

[elder valley]

Normal subgroups are exactly the ones where that works. Normality is defined just for this reason

G/H is a group under operation aH bH = abH if and only if H is normal

[river fern]

Ok so I guess G / H is more of a space rather than a group in general

and you can have a group act on it under certain conditions (i.e. H having a left action on the left coset space G / H)

[elder valley]

'space' is kind of ambiguous. there are many kinds of spaces

in general G/H won't have any nice structure afaik. just when it's normal you get a group

[solemn rain]

G/H is just the set of cosets of H in G

it's a group iff H is normal

[chilly canyon]

If F is an irreducible polynomial over Q, is it true that F[X(X+1)] is also irreducible over Q ? If not, what would be conditions on F so that this statement holds ?
I have no clue on where to start. I'd like to try reduction mod p, but, how ? I also looked for "irreducibility of composite polynomials", and found an old paper [by [A. Brauer, R. Brauer, H. Hopf], but it seems way out of scope for what I'm looking for... Any clue/hint/whatever please ?

[elder valley]

there's an easy counter example 🙂

[pure crest]

can i ask questions about contractible numbers here, or is it too off topic

ahh

i see its being used

sorry

[chilly canyon]

Uhm F(X)=X is a counterexample for sure, but what would be a condition on F ?

Papers on the topic only show that there exists a certain amount of G s.t. F(G(X)) is (ir)reducible, it seems useless for me...

[pure crest]

one condition must be that F(x) has a constant

otherwise you can just factor out x(x+!)

but idk if that is enough

[elder valley]

you need F irreducible for sure. but that counter example shows that's not good enough

[pure crest]

you can just take x^2 - 2, it works

[pure crest]

hey guys

i am trying to solve this problem

so i did the basic thing so far

i wrote an equation for line y = mx+b and circle (x-c)^2 + (y - d)^2 = r^2

then substitued y into the circle equation

i was able to solve for m, c, d and r^2 in terms of coordinates of P,Q,R and S

but i am not sure how to solve for b

ahh i see

i can let x = P + |P-Q|

[elder valley]

take one of the points on the line, substitute into the line equation. you know x,y,m, so solve for b

[pure crest]

right

thats what i am trying to do

so i chose x to be on the circle

but i am not sure what to choose y to be

[elder valley]

you can choose R. x=r_1, y=r_2

[pure crest]

ohh right

because those are two points on the line

[elder valley]

yeah

also keep in mind that writing your line as y=mx+b assumes that r_1 != s_1. so you'll have to do that as a separate case

[pure crest]

yup you are right

so i would need a x = a line

i hate these computational questions

[elder valley]

yeah

alternatively you could write the line as ax+by=c and it should consider all cases

[pure crest]

ill try this out first

there seems to be less computation in each cases

[vestal snow]

How do I prove 15?

Here's what I've done

Suppose $(0, \cdots, x_i, \cdots 0) = \sum_k y_{i_k} - \mu_{i_kj_k}(y_{i_k})$.

Taking the projection onto the ith coordinate, we get $x_i = \sum_k y_{i_k}' - \mu_{i_kj_k}(y_{i_k})'$ where $y_{i_k}' = y_{i_k}$ if $i_k=i$ and $0$ otherwise. Similar definition for $\mu_{i_kj_k}(y_{i_k})'$.

[cloud walrus]

The formerly edible banana:

[vestal snow]

We get 0 in the cases where either both $i_k$ and $j_k$ are not equal to $i$ or both are equal to $i$

[cloud walrus]

The formerly edible banana:

[river fern]

Let U be a subring of R

Is it true that R / U (left cosets) is a ring iff U is an ideal?

[timid hull]

Yes

[next obsidian]

Do you require your rings to have unity?

Because if so that gets really really messy as 1 in R/U is going to be 0

[vestal snow]

I think that the universal property should also include the condition that $\mu_i = \mu_j \circ \mu_{ij}$

[cloud walrus]

The formerly edible banana:

[vestal snow]

Whenever $i \leq j$

[cloud walrus]

The formerly edible banana:

[vestal snow]

My question is whether the following results only hold for the specific construction in exercise 14 or to direct limits as defined by their universal property in general (as stated in exercise 16)

1. $\mu_i = \mu_j \circ \mu_{ij}$.

2. The results from exercise 15

[cloud walrus]

The formerly edible banana:

[vestal snow]

Let $(M, \mu_i)$ be the construction in exercise $14$ and let $(P, \tau_i)$ satisfy the property in exercise $16$.

One way to prove $1)$ and $2)$ hold for the universal property definition is to consider the isomorphism between the construction $M$ and $P$ (We would still need to assume that $\tau_i = \tau_j \circ \mu_{ij}$ to prove the existence of such isomorphism in exercise 16).

[cloud walrus]

The formerly edible banana:

[vestal snow]

Though I feel like there might be a slicker way to prove 1) and 2) using just the universal property

[chilly canyon]

Hello,
If I know some polynomial P(X) has α as a double root, is there an expression of P(X)/(X-α)² in terms of the coefficients of P ?

[woven delta]

That seems pretty hard

Probably you can write P(x)=(P(x)/(x-alpha)^2)(x-alpha)^2 and do some Algebra

Yeah there is

[chilly canyon]

Uhm yeah that's what I feared... Sadly, my P is given abstractly, I don't especially know its coefficients too, they're given using binomial coefficients and Bernoulli numbers

[woven delta]

Lol

One sec I think I have something

[chilly canyon]

Oh ! 😮

[woven delta]

Yeah I have an expression for the coefficients of P(x)/(x-alpha)^2 in terms of the coefficients of P(x) I think

Basically the main relation is if $a_i$ are the coefficients of P(x) and $b_j$ are the coefficients of $P(x)/(x-\alpha)^2$, then we can show $a_i = b_{i-2}-2\alpha b_{i-1}+ \alpha^2 b_i$ when i is greater then or equal to 2

[cloud walrus]

Liquid:

[woven delta]

And when i is 0 we get $a_0 = \alpha^2 b_0$

[cloud walrus]

Liquid:

[woven delta]

And when i is 1 we get $a_1 = -2\alpha b_0 + \alpha^2 b_1$

[cloud walrus]

Liquid:

[woven delta]

This should be enough information to figure out what you want

[chilly canyon]

Linear algebra with a tridiagonal matrix should work you mean ?

How did you derive those identities ?

[woven delta]

I looked at the identity $P(x) = (P(x)/(x-\alpha^2))(x-\alpha)^2$

[cloud walrus]

Liquid:

[chilly canyon]

Oh you developped it !

I get it !

Q(X)(X-α)²=P(X)

And you get Q

Thanks so much, that's clever !

[woven delta]

$a_i = b_{i-2}-2\alpha b_{i-1}+ \alpha^2 b_i$ so $ \alpha^2 b_i=a_i- b_{i-2}+2\alpha b_{i-1}$

[cloud walrus]

Liquid:

[woven delta]

So once you figure out $b_{i-2}$ and $b_{i-1}$ you've figured out $b_i$

[chilly canyon]

Yeah I assumed α isn't 0, otherwise the problem is... well, there's no problem in fact x)

[cloud walrus]

Liquid:

[woven delta]

And we have $b_0$ and $b_1$

[cloud walrus]

Liquid:

[woven delta]

So we have all the coefficients

That's the idea

[chilly canyon]

Thank you very much ! I'll try to apply this to my problem and see if the formula is not too ugly 🙂

[woven delta]

👍

[chilly canyon]

My polynomial is $T^{(2\delta+1)}(X)=\frac{1}{2^{2\delta}(\delta+1)X^2}\sum_{q=0}^\delta\binom{2\delta+2}{2q}(1-q)B_{2q}\left[(4X(X+1)+1)^{\delta+1-q}-1\right]$ with $\alpha=-1$... Wish me luck !

[cloud walrus]

Matplotlib:

[woven delta]

Lmao good luck

[ebon wyvern]

Where does that polynomial come from, out of interest?

[chilly canyon]

From my most recent question on MSE : https://math.stackexchange.com/questions/3770159/faulhabers-polynomials-and-irreducibility-in-the-sums-of-powers

I'm still working on it 😦

Oh wait I'm working with another expression for that polynomial

But I need to show its derivative cancels at -1, which requires me to evaluate a sum... May I ask for some clues ? I think Summation by Parts doesn't work

$\sum_{k=0}^{2\delta}\binom{2\delta+2}{k}(2\delta-k)B_k^+$

[cloud walrus]

Matplotlib:

[chilly canyon]

I equals $4\delta(\delta+1)-\sum_{k=0}^{2\delta}\binom{2\delta+2}{k}kB_k^+$ and I need to show it is zero...

[cloud walrus]

Matplotlib:

[ebon wyvern]

I'm afraid this is very much not my area

Good luck though

[chilly canyon]

Thanks !

[ebon wyvern]

Looks like a right pain in the arse that one

[shy bluff]

How does this work?

how can it be both irreducibelre and a product of primes?

[chilly canyon]

The product has one element only

[shy bluff]

Is'nt the definition of irreducible that if p is irreducible, p = ab, where a or b is a unit?

[chilly canyon]

p prime means that (p) is prime

It is p|ab => p|a or p|b

[shy bluff]

Do you have an example? I don't think I understand

[chilly canyon]

An example of what ? A prime element ?

[shy bluff]

Like prime vs irreducible?

[chilly canyon]

2 is not prime in Z[i] for instance, because 2=(1+i)(1-i) but 2 does not divide either 1+i or 1-i.

But it is irreducible, for if 2=ab, then 2=|a|×|b|, and thus |a|=1 or |b|=1, which means either a or b is invertible in Z[i]

(I'm assuming you're familiar with the ring Z[i])

[shy bluff]

that's the gaussian integers yes?

hrm

[chilly canyon]

So : Irreducible is "if you're a product, then it's a trivial product", whereas prime is "if you're dividing a product, then you must be diving one of the factors"

[shy bluff]

ok I think I see

[chilly canyon]

By "trivial product" I mean one of the factors is invertible

Which means you haven't done much

[shy bluff]

I see

[chilly canyon]

The "usual" definition for being a prime integer is misleading, because you're saying "p is prime iff its only divisors are 1 and p", which turns out to be equivalent to the prime definition in the ring Z

I.e. you're doing the converse : write p=ab, then a or b is 1

This is being irreducible

[shy bluff]

I see

[chilly canyon]

But in a PID, prime => irreducible

The generalization of "prime numbers" are irreducible numbers (see UFD's)

Wait no

Forget about that, it's prime, not irreducible in a UFD

So the terminology is fine (I remember I struggled when I first learnt about it too)

[olive mirage]

(1+sqrt(-5)) is irreducible, divides 2*3, but divides neither 2 nor 3, so is not prime.

(in, e.g. Z[\sqrt{-5}] )

[shy bluff]

I see

[smoky cypress]

How do you determine whether a number in Z[sqrt(-7)] or Z[sqrt(-5)] is irreducible?

[olive mirage]

The usual method is to use the norm, and the fact that it is multiplicative. so N(a+b sqrt(-5)) is defined to be a^2+5b^2. It's just the square of the absolute value, so it is automatically multiplicative

so the norm of 1+sqrt(-5) is 6

so the only way we oculd factor it is as a product of something with norm 2 and something with norm 3 (everything with norm 1 is a unit)

but we can see there aren't any such things.

[smoky cypress]

Oh yeah that makes sense

What about prime elements

[olive mirage]

I think you can sort that out with the norm. But the idea of prime elements turns out not to be very useful.

which is where prime ideals come from

[Not very useful outside a UFD I should say]

[smoky cypress]

Oh ok

Idk why but I just find primes to be pretty interesting

It would be cool if there's an easy criterion whether an element is prime in Z[sqrt(d)], where d is an integer

[knotty mason]

wouldn't the ring of integers for the first one be Z[(1+sqrt(-7))/2]

[olive mirage]

It is, yes

though the notions of prime and irreducible make just as much sense in an order

[woven delta]

@chilly canyon you can get an explicit expression for the coefficients of P(x)/(x-alpha)^2 btw, as opposed to the recursive one I did before

[chilly canyon]

Oh I'm still interested in that ! :)

[woven delta]

You can substitute u =x-alpha. Then x = u+ alpha, and you can expand that with respect to u. This just involves the binomial theorem and rearranging series. Then you can divide u out from that, and convert back to x

It's not very nice but in the end it seems like you can actually get the coefficients

It's bad because you have to deal with 3 series lol

But I think it's doable

I'm doing laundry rn so I'm fooling around with it while I wait

[woven delta]

$b_k = \sum_{j=k+2}^n \sum_{i=j}^n\binom{j-2}{k} \binom{i}{j}\alpha^{i+j-k-2}$

[cloud walrus]

Liquid:

[woven delta]

@chilly canyon

You can prove this works with the recursion I gave before

Or you can derive it

[elder valley]

doesn't just doing the regular polynomial division work?

[woven delta]

Yeah

[shy bluff]

GCDs exist wehther or not the ring is a PID right?

[olive mirage]

you can get by with GCDs in UFDs

[shy bluff]

wdym "Get by"

It's uh this, the last option that confuses me

hrm

Wait the last one is true right?

Ah no it's that if there is a principal ideal, not necessarily that they only exist in PIDs

[woven delta]

Why don't you look to see if 1, z, or xz+yz satisfy the definition you just posted

[shy bluff]

Well z divides both p and q

I'm just not sure (p, q) is part of a principal ideal?

[woven delta]

It's part of (z)

[olive mirage]

so I would say the gcd of x and y in Q[x,y] is 1. But you have to be careful what you mean by it.

[shy bluff]

Oh wait principal ideal is jusst any ideal that is made up of only one element?

[woven delta]

That can be generated by a single element

[shy bluff]

err yea

[woven delta]

It is all the elements of the form rt, for r in your ring and t fixed

[shy bluff]

Ok, then (p, q) is in (z), and z is a gcd of p and q? And 1 isnt' a gcd because p, q not comaximal?

And xz + yz is not a gcd because it doesn't divide q?

[woven delta]

Why is z a gcd?

[shy bluff]

Well (p, q) are a subset of (z), so then we fulfill the condition that a gcd can exist, and then we have that z | p, z| q, so it's a common divisor, and .... I'm not sure ?

[woven delta]

So suppose we have another principle ideal (m) with (p, q) inside (m)

[shy bluff]

ok

[woven delta]

q = z, so that tells us some relation between z and m

[shy bluff]

they're the same?

[woven delta]

Honestly you can just say (p, q) = (z) here

And that immediately gives you the second part of the definition

So I guess it's not such an interesting example

[shy bluff]

Ah, ok so z is indeed the gcd

and 1 isn't the gcd because p, q not comaxiaml?

[woven delta]

Yeah

p and q comaximal means (p, q) = (1)

But clearly that's not the case

[shy bluff]

Yea

ok epic I got like 3/4 of that

[shy bluff]

What is the difference between complete factorization and unique factorization?

like we have this

for complete ?

Is unique factorization that you can't factor the same element into 2 different... combinations of irreducible elements?

[stone fulcrum]

Note that this doesn't mention unique factorization anywhere

There are spaces that allow complete factorizations, but completely different such factorizations, for its elements

We say that two factorizations are "the same" if one can be changed into the other by multiplying units onto it

[shy bluff]

Yae we were supposed to have learnt about lunique factorizatios and UFDs but we're behind schedule, but these questions involve unique factorizations so

I don't really understand complete factorization

can you givem e an example of something that cant' be completely factored into irreducibles?

[stone fulcrum]

Haha hmm. That's a hard one lol

[shy bluff]

or like if I'm understanding it right then it' s saying that if R is a domain and r is not a unit, then r has a complete factorization if it can be written as a product of irreducibles

But where would you find an element that can't be written as a product of irreducibles?

Like if it's not a product of irreducibles... is'nt it just an irreducible?

[stone fulcrum]

Irreducible -> not a product of two non-units

[shy bluff]

uhhh can you explain further?

I feel like there's some subtlety that I'm not catching

A product of two non-units means that if p is irreducible, p = ab, then neither a nor b is invertible yes?

[stone fulcrum]

The exact opposite of that haha

That's a reducible element. One that can be written as ab for non-unit a and b

[shy bluff]

oh

[stone fulcrum]

An irreducible element is a non-reducible one

[shy bluff]

hrm

I think I need more examples :x

[stone fulcrum]

That is, it can't be written as ab for non-unit a and b

Example! Let's use Z:
8 = 2×4
2 and 4 are non-units.
So 8 is reducible

It can be reduced into 2 and 4 haha

5 can't though

[shy bluff]

5 = 1x5

And 1 is a unit?

[stone fulcrum]

Since 1 is a unit, the factorization doesn't count

We only want to consider non-units

Ooh, btw there's other way to factorize 8:
8 = (-2)(-4)
But this is the same factorization as 8 = (2)(4) because you can multiply it by -1 a few times to get the other

8 is uniquely factored in Z

[shy bluff]

Oh

Yea

[stone fulcrum]

And -1 is a unit. The other unit is 1 haha

[shy bluff]

Yea

Hrmmm so 5 is irreducible

Because we can only write it as a (1)(5), and 1 is a unit?

[stone fulcrum]

Because there's no way to write it as a product of non-units yeah

I can write 8 = (1)(8), but 8 is reducible because 8 = (2)(4) exists

[shy bluff]

oh ok I think I see

Ok, so complete factorization means that we can write it as a product of irreducibles?

[stone fulcrum]

Yus

[shy bluff]

So there's a complete factorization of 8

but there isn't one for 5?

[stone fulcrum]

I'm not sold yet that a domain exists that can't be written this way haha

Oh 5 is itself the "factorization into irreducibles"

It's already irreducible, no factorization needed haha

Z ends up being what we call a "unique factorization domain" or a UFD. Every element can be factored uniquely.

Non-UFDs are an object that we still have open questions about

[shy bluff]

oh

Ok so Z is completely and uniquely factorized?

[woven delta]

Z is also a PID

[stone fulcrum]

Every element in Z has a unique factorization! But that's the "regular prime" stuff you're used to

[shy bluff]

Yea

[woven delta]

Also what does "non UFD's are an object that we still have open questions about" mean

[stone fulcrum]

More interesting is considering other rings, like polynomials

[shy bluff]

What about like Z[x]

[stone fulcrum]

So for example
x² - 1 = (x - 1)(x + 1)
So x² - 1 can be factored

And is reducible

[shy bluff]

but x^2 + 1 is irreducible in Z[x]?

[stone fulcrum]

Exactly

[shy bluff]

But that's still a complete factorization of itself?

[stone fulcrum]

x² + 1 is written as a product of irreducibles

[shy bluff]

Yea

then doesnt every ring have a complete factorization?

[stone fulcrum]

I'm not great with the term "complete factorization" and I don't personally know of a ring that doesn't have that property. Maybe I'm ignoring one? Oop

[shy bluff]

Idk

I dont' understantd this

we're being asked this

I don't understand what the difference is?

[stone fulcrum]

Is there a ring in which an element exists that can't be factored into irreducibles? Anyone?

[shy bluff]

Yea like that's what I'm thinking

[stone fulcrum]

Interesting they're leading with Z[√-5] haha

[shy bluff]

yea iunno

[stone fulcrum]

That's a well known non-ufd

[shy bluff]

Like from teh examples taht you've given me and what they've said, can't every element of every ring be written as a product of irreducibles?

[woven delta]

Why?

Why would you think that

[shy bluff]

Well if p = ab only if a or b is a unit then it's irreducible right?

[woven delta]

Yes

That's the definition of irreducible

[shy bluff]

Then every time that you come across an element of a ring, it' seither reducible, in which case you can write it as a product of irreducibles, or it is irreducible, right?

[woven delta]

No

Why would you then be able to write it as a product of irreducibles

That's a bit of a leap in logic

[shy bluff]

Elements are either reducible or irreducibel right?

[woven delta]

Sure

But when do you stop reducing?

[shy bluff]

I have no idea

It has something to do with the ascending chain condition for ideals?

[woven delta]

Well okay let's think about that

So suppose we have an element a which is reducible

You're saying that by reducing you get an ascending chain of ideals which must stabilize if your ring is noetherian

I'm not sure I buy that, for example consider an idempotent element with a^2 = a

Then you can decompose a as a = (a)(a)

So that's a possible bit of trouble you can run into

[drifting plover]

@shy bluff @stone fulcrum I think a possible explicit example is this: take the ring C[X^{Q^+}]of polynomials over C but allow for rational exponents of positive powers in X. Then X is reducible but does not admit a complete factorisation into irreducibles (since a complete factorisation has to stop after finitely many steps, but any rational power of X can be factorised indefinitely).

[shy bluff]

Hrmmm that makes sense

[chilly ocean]

X is invertible inside your ring (with X^{-1}), hence it is not correct to say that it is "irreducible" or "reducible" element, but the general argument is correct I think. We can take Q[X^{1/2^∞}] instead

[drifting plover]

Ah right, I meant to say rational exponents of positive powers, but yeah that works too

(Updated my example above)

In general a ufd requires complete factorisation (which is a finitary process) and uniqueness of any factorisation (up to units of course)

[chilly ocean]

So all rings in the problem have complete factorization (because they are all noetherian, and noetherian easily implies complete factorization)

[woven delta]

@chilly ocean I don't think that's true because of idempotents

[chilly ocean]

Could you elaborate your point?

[woven delta]

Like take the ring Z/2 \times Z/2 and consider the element (0, 1). Then we have (0, 1)^2 = (0, 1)

Obviously this ring is finite so noetherian

But there isn't a factorization of (0, 1) into irreducibles

[chilly ocean]

yeah, i see, ok

you are right

[oblique river]

Z/2 x Z/2 has the unique factorization property in whatever sense a non domain can have it

1 is a unit

liquid what you wrote is akin to saying "1^2 = 1 so there's no factorization of 1 into irreducibles"

[woven delta]

Well we are talking about complete factorization

[chilly ocean]

I belive (0,1) is not a unit

[oblique river]

so then it must be a domain

[woven delta]

Oh you are right

The definition is for a domain

So we can't run into idempotents

[oblique river]

👍

[woven delta]

Okay so suppose you are working in a noetherian domain. Is it obvious that this implies that we have complete factorization?

I'm not so sure it does

[chilly canyon]

@woven delta thanks (again), the Binomial formula helped me ! I managed to extract a formula for my P(X)/(X+1)² !

[woven delta]

Awesome

[chilly canyon]

Sorry bout yesterday, I went to bed, I was exhausted !

[next obsidian]

Liquid

[chilly ocean]

@woven delta I mean, the height of the ideal generated by element x is the obvious upper bound on amount of multiplers in factorization of the element x

[next obsidian]

For a domain I believe being a UFD is equivalent to the rising chain condition for principal ideals

So yes

So here’s the actual equivalence

For an integral domain, A is a UFD if and only if every irreducible is prime and it satisfies the ascending chain condition for principal ideals

[chilly ocean]

No, almost all noetherian domains are not UFD

[next obsidian]

So whoops

I guess the irreducible => prime bit isn’t possible to get from just Noetherian

[woven delta]

Okay I found the proof of the complete factorization statement on the stacks project lol

[chilly ocean]

@next obsidian yeah, than it's true (GCD + ATOMIC <-> UFD)

[next obsidian]

Wtf is an atomic ring haha

[woven delta]

Okay I believe this now

[chilly ocean]

https://en.wikipedia.org/wiki/Atomic_domain

[next obsidian]

Ah, I’ve just called those factorization domains lol

Where you can factor everything but you don’t get uniqueness haha

[chilly ocean]

cool

[vestal snow]

How do I prove that S is maximal implies A-S is a minimal prime ideal?

I can't even prove that A-S is an ideal

Suppose $x,y \in A-I$ and assume $x-y$ is not. Then $x-y \in I$ which would imply that one of $x$ and $y$ is not $0$. WLOG, suppose $x$ is not $0$.

Consider the set $I' = $ ix^n: i \in I, n \geq 0$.

Then, $I'$ must contain $0$ by the maximality of $I$. Therefore, $ix^n=0$ for some $i \in I$ and $n > 0$

[next obsidian]

So I think it’s multiplicative because if a in A\S and ab not in A\S then S isn’t maximal. You can add in a, because if ac = 0 for some c in S then abc = 0 and ab,c in S so that S already contains zero divisors

[vestal snow]

Is it a problem if it S contains zero divisors?

[next obsidian]

It can’t

Because 0 isn’t in S

[cloud walrus]

The formerly edible banana:
Compile Error! Click the reaction for details. (You may edit your message)

[next obsidian]

Sorry so

By zero divisors I mean a pari a,b such that ab = 0

[vestal snow]

Ah yes

[next obsidian]

A localization at a set S is 0 iff 0 in S, iff there exists a,b in S such that ab = 0

So that’s why it specifies 0 not in S

[vestal snow]

That makes sense

Thanks

Though how would i prove that it's a subgroup under addition?

[next obsidian]

I’m not sure how to show it’s a subgroup additively though haha

[vestal snow]

Haha yeah

[next obsidian]

Also for showing it’s a minimal prime I guess you want to show the localization S^-1A is dim 0

[vestal snow]

I don't think so

[next obsidian]

Idk if that follows from S being maximal easily though

It would at least imply it

[vestal snow]

Because dim hadn't been introduced until chapter 6

[next obsidian]

Oh

[vestal snow]

and we're in chapter 3

I think minimality is easier

[next obsidian]

I see

Actually this is easy yeah

I think

If p < q

strictly

Then A\p > A\q

[vestal snow]

Yeah

[next obsidian]

Strictly

Yeah haha

[vestal snow]

and both are mult. closed

[next obsidian]

That’s way easier

Yup

Don’t contain 0

I think maybe you can extrapolate something from what you have so far??

Like umm

[vestal snow]

Hmm I don't think so

[next obsidian]

Hmmm

[vestal snow]

Because we have seen stuff about multiplaction

[next obsidian]

Actually yeah

[vestal snow]

and i don't see how one can get propeties about addition from those

One things for sure though

You need maximality to prove all 4 conditions

[next obsidian]

Okay so y ≠ 0

If it is

x in I

Then ix^n = 0

Contradiction right?

So also there exists some j such that jy^m = 0

By same proof

Take ij(x - y)^(m + n)

Or maybe larger exponent

This is 0

By binary expansion

You have a power of x >= n or y>= n

Then that bit dies to i or j

So every term dies

But i,j, (x - y) all in I

So 0 in I

There we go 😊

[vestal snow]

Nice dude

[next obsidian]

Damn lol

I feel like a G

[vestal snow]

I would never have thought to use the binomial expansion

[next obsidian]

Basically

[vestal snow]

Cause you are

[next obsidian]

Have you done the 3-fold equivalence of the umm

Presence of an idempotent

Spec disconnected

And stuff like that?

You did something similar with some binomial expansion thingy and this reminded me of that

[vestal snow]

Maybe my memory of this book is a bit hazy

[next obsidian]

It might be later

It’s 1.22

[vestal snow]

Oh yeah I did do that one

[next obsidian]

For uh, the implication Spec A disconnected implies orthogonal idempotents I think I did something like that

I remember all that happens is like the middle term died because of some nilpotence

And you let the idempotents be whatever’s left to the left or right respectively

Then they summed to 1

And multiply to 0

Which is all you needed

[vestal snow]

Very cool proof

The weird thing with this book is that the earlier exercises are often the most difficult

[next obsidian]

Yeah lmao

I didn’t do all of them in chapter 1

That shit was hard, like the stuff for his weird version of uh... Gauss’s lemma I think

[vestal snow]

When you switched to matsamura, did you start from the very beginning?

[next obsidian]

Yeah

But Matsumura has fewer exercises plus there’s a solution manual in the back basically

Almost every problem has a solution there

I’ve used in multiple times, and each time I wasn’t ashamed to

Those proofs were ludicrous

And right at the start the way Matsumura proves things is a lot different... idk how to describe it

It’s a method I’m pretty trash at so it’s nice, I feel like I’m improving some aspect of my math that I’ve sort of skirted around

But Matsumura has fewer exercises plus there’s a solution manual in the back basically
@next obsidian per section but there’s like 36 sections?

[vestal snow]

I might switch over to that

i always treated exercises as something to test my knowledge rather than build it

But 50% of the content in A-M is through the exercises

[vital quail]

if magician was lying he will not see this

[next obsidian]

I saw this

I might switch over to that
@vestal snow only thing to note is that A-M is way quicker. Matsumura covers a lot more stuff

It’s a middle ground between like A-M and Eisenbud I’d say. It’s nice if you know some comm alg already (which you do)

And want to know quite a bit

[vestal snow]

Yeah, I'll definitely check it out

[next obsidian]

I like it. There’s been times where i’ve been confused on some proofs he’s done, but I still like the style overall

Some of it was me being dumb tho lol

[shy bluff]

Why does xy = zw imply that R does noth ave a unique factorization?

[knotty mason]

the value xy can be factored as both xy and zw

[shy bluff]

oh I see

[next obsidian]

You have to show that x,y and z,w are actually irreducible tho

And don’t differ by units

This is pretty easy tho, the only issue is you’re working in a quotient so equality is a little messier

Oh lol that’s the “Exercise: (not so easy)” bit

[shy bluff]

yea

he said that he thinks it's a pain to do rip

[next obsidian]

I think you want to look at degrees

I forgot how I did it before

I remember it kind of being annoying though

[shy bluff]

Yea he says that if you know how to do it an easier way to email him lol

[supple marsh]

I'm sure Macaulay2 can verify it for you 🙂 .... I'll see myself out now.

[next obsidian]

I know a good proof that uses facts about graded rings, so if you professor wants a proof using that I can give him one lol

Also "good" just means you can use degrees easier

[charred pewter]

I know I have to prove it's homomorphic as one of the conditions, right? But I don't know what else.

[next obsidian]

$\mathbb{Z}[X,Y,Z,W]$ has a natural grading on it via total degree. The ideal $(XY - ZW)$ is a homogeneous ideal so that $\mathbb{Z}[X,Y,Z,W]/(XY - ZW)$ has a natural grading via total degree of the $x,y,z,w$ (which are the images of the $X,Y,Z,W$ in the quotient). This is basically the fact that you need. If you can express $x = fg$ then you require that one of the $a$ be degree 1 and the other degree 0, WLOG you can assume that $f = (ax + by + cz + dw)$ and that $a,b,c,d,g \in \mathbb{Z}$. This means that $(ea - 1)X + ebY + ecZ + edW \in (XY - ZW)$, but notice that the element is degree 1 while the ideal $(XY - ZW)$ has elements of at least degree $2$, so that the element must be $0$, implying that $ea = 1$ but then $e$ is a unit so that $x$ is irreducible

The main thing that you use graded rings for is just to get the grading inside of the quotient

[cloud walrus]

UserFormerlyKnownAsMathemagician:

[next obsidian]

Since ai priori you can have weird stuff happening, so if you bring this equality up out of the quotient all you know is that $x - fg = h(XY - ZW)$

[cloud walrus]

UserFormerlyKnownAsMathemagician:

[next obsidian]

From here you can't just assume that f is degree 1 and g is degree 0 or anything

@open torrent

[vestal snow]

@next obsidian I read through the first section. The exercises are a good mix of easy and difficult.

[next obsidian]

Yeah, I find the difficulty personally to be perfectly at the level I desire, and this is something I’ve repeatedly said, the fact that there’s less of them in a section motivates me to make progress

If you do A-M by reading a section, then doing every problem before moving on you spend huge chunks of time just sitting in the exercises section not ever turning a page, with Matsumura it feels like I make progress

Simply because I get to turn pages more often

Isn’t it?

Idk if it’s internal

[charred pewter]

This is a question my professor posted for homework.

[next obsidian]

But you can show it as a direct product really easily

The R>0 part tells you the norm

And then the |x| = 1 gives you the angle

Lol true

I totally skipped over that

Yeah that sounds right

[charred pewter]

What if I just assume it's the multiplicative group of C then

[next obsidian]

Multiplication is additive on angle and multiplicative on norm

And multiplication on the |z| = 1 is addition of the angle

[charred pewter]

I'm trying to find this stuff in the textbook because I honestly have no idea what's going on in the problem

The vids my professor made didn't really help

I found this definition in my book. Can I use this in this proof?

My professor posted this as a hint

problem 3: You have to check the properties. Is it the case that -1 * x = x * -1 for every positive real number x? What is the intersection of R_{>0} and {-1,1}?

uh, the arterisks aren't showing up

There

So the answer to the second one is that the intersection of R_>0 and {-1,1} =1=e, so the second condition checks out

And multiplication between a positive real number x and -1 is commutative such that -1 * x = x *-1 but I don't know if it's enough to just state that

[elder valley]

take H=U and K=R_>0 and show the 3 properties hold. it's pretty straight forward

use polar form of the complex number

[shy bluff]

Jesus magician... that was dense lol

Question: So (x) = (y) would mean that they have the same set of possible products right?

And that x and y not being associate, according to this definition of associate, means that there doesn't exist some unit u such that x = uy?

I think I'm going about this the wrong way because I tried to look for rings with (x) = (y) but x not equal to y?

Like I think that in Z[x, y]/(x - xy), we have that (x) = (xy) right?

But I'm not really sure how ot work with associates

[next obsidian]

wait what, I thought that if $(x) = (y)$ that x and y have to be associate... maybe this is only in a GCD domain or something

Or maybe you need domain

[shy bluff]

Idk

[next obsidian]

Like I think that in Z[x, y]/(x - xy), we have that (x) = (xy) right?
@shy bluff This is true but x = xy in that ring

[elder valley]

yeah in an integral domain

[shy bluff]

Yea integral domain

[elder valley]

i would look for a ring with a small set of units

[next obsidian]

Okay, so you need a ring with zero divisors

[shy bluff]

So I need to find a non integral domain

Yes

[next obsidian]

Ah okay

[shy bluff]

Z/6z has zero divisors right? 3 and 2?

[next obsidian]

Oh hold up

jeez this is mega weird

so if (x) = (y) then x = ay and y = bx for some a,b

so x = abx

if you combine the two

so I guess you want to find a ring where this is true, but a,b are not units

because then you can take y = bx

This won't let you be done I guess

Since you still have to show x and y aren't associate

[shy bluff]

So we want some ring where you can write x = abx with a, b not being units

[next obsidian]

but if you can find a ring with such elements my guess is that they'll end up not being associate

yeah this is necessary

but not sufficient

but I think if you can find that it'll just end up working out

is my guess

I think Auvera's suggestion is good

to consider rings with few units

[elder valley]

does your ring have to have 1? because if you find a non unital ring then you just have to find an ideal that has 2 different generators

[shy bluff]

what are some rings with few units? Because I'm not really understanding how to find those either

Yes, in this course we have to have 1 in our rings I believe

[next obsidian]

Lol Auvera that would be big brain

can't have units if there's no 1

[elder valley]

exactly xD

[shy bluff]

Oh I guess that's true

[next obsidian]

yeah this sounds mega grody

I don't see how this is helpful here

[shy bluff]

grody?

[next obsidian]

but when I look for non integral domains I've found that you can do weird stuff with nilpotence

like, gross

[shy bluff]

oh

[next obsidian]

I have no idea if this will work

but consider maybe k[x]/(x^2)?

x is nilpotent there

so all elements are of the form a + bx

where x is like infinitesimal

[shy bluff]

WHat does nilpotent mean?

[next obsidian]

x^n = 0 for some n

[shy bluff]

oh ok

[next obsidian]

so you can consider the b part as infinitesimal data

like it's a, but with this little bit

and when you mulitply two numbers

the product of the little infinitesimal data dies

I make no claims that this will work

but it might be a place to look since nilpotence does weird stuff

I think I found one

maybe

take x in k[x]/(x^2)

notice that x(1 + x)^2 = x(1 + 2x) = x

so look at x and x(1 + x)

I think these don't differ by a unit

but I don't see how to show it

wait

this is dumb

x(1 + x) = x

Rip

[shy bluff]

You've lost me lol

So I'm looking for a ring with zero divisors to set my a and my b to

[next obsidian]

Well you need zero divisors

I don't think you need a and b to be zero divisors????

Not sure though

[shy bluff]

hrm

So we want a, b to not be units

[next obsidian]

yeah

[shy bluff]

And that we need x = abx

[next obsidian]

yup

then hopefully from there x and ax don't differ by a unit

[shy bluff]

and because they're not units, a can't just be the inverse of b

[next obsidian]

yeah

I came up with another possible thing in k[x]/(x^2) but a ended up being b^{-1}

lol

[shy bluff]

well if x = 0 in the ring, and a, b are zero divisors

Then we'd have x = abx right?

At least that's kinda my thought process?

[next obsidian]

Well we can't have x be 0

because we wanted to set y = ax

or bx

does't matter

but if x = 0 then x = y

and we need them to be different

maybe k[x,y,z,w]/(xy - zw) is a good place to try

[shy bluff]

Oh

[next obsidian]

it doesn't even have unique factorization

wait no

[shy bluff]

That's a lotta variables

[next obsidian]

that's an ID

it won't even wokr

[shy bluff]

Ok so we need something that's not a domain yes?

[next obsidian]

😓

yeah

[shy bluff]

I'm very confused lol

what about uhh k[x, y, z]/(x - xyz)?

Woludn't we get that x = xyz

[next obsidian]

yes but how would that help?

[shy bluff]

and then we just need to show that y, z are not units?

[next obsidian]

inside that ring

that's literal equality

[shy bluff]

Oh

[next obsidian]

we want distinct elements

[shy bluff]

oh

[next obsidian]

I think Z/6Z might work lol

take 2 and 4

2 = 4(2)

and 4 = 2(2)

umm

The only units are umm

1 and 5??

and 2 is not 5 * 4

I think you suggested Z/6Z earlier too oof

[shy bluff]

Oh

[next obsidian]

Lmao

[shy bluff]

Ok so the units of Z/6z are 5 and 1

[next obsidian]

yeah

and (2) = (4)

[shy bluff]

2 * 5 = 10 = 4

[next obsidian]

because you can write both as products of each other

[shy bluff]

Right?

[next obsidian]

uhhh

wait

oh oof

[shy bluff]

In Z/6z

[next obsidian]

😔

I wonder if any cyclic group works then

[shy bluff]

As in "no cyclic group will work"?

[next obsidian]

Yeah

Maybe you can't get an example from Z/nZ for any n

but I'm gonna try a few more

[shy bluff]

hrm

What about some quotient ring of Q? That would'nt be cyclic... would it?

[next obsidian]

no quotients of Q exist

since it's a field

[shy bluff]

Oh

[next obsidian]

we also want to stay as far away from fields as possible I think

[shy bluff]

Oh

[next obsidian]

Damn

I've tried a few more Z/nZ

[shy bluff]

[next obsidian]

and I think there's some number theoretic reason that if (n) = (m)

that n and m differ by a unit or something

Something something, coprime idk

[shy bluff]

Ok so we want something that's not an integral domain

here are some examples on wikipedia?

We tried k[x]/(x^2 - n) and that did'nt work right

[next obsidian]

did we?

I just tried k[x]/(x^2)

because that introduces nilpotence

[oblique river]

Are you looking for (a) = (b) but a \neq ub for any unit u?

[next obsidian]

yeah

I'm struggling to come up with an example lmfao

I'm trying like Z/12Z[x]

[oblique river]

the example in the first answer is the one I always go to: https://math.stackexchange.com/questions/14270/are-associates-unit-multiples-in-a-commutative-ring-with-1?noredirect=1&lq=1

[next obsidian]

currentlyt

[oblique river]

it's in the ring of continuous functions on [0,3]

[shy bluff]

oh

[vestal snow]

Does the ring have to be commutative?

[shy bluff]

Yes

[next obsidian]

Buncho

that's such a ridiculous example

wtf haha

[shy bluff]

that

uh

[oblique river]

is it? I mean, the ring isn't "simple" but also like, the example is kinda just cooked up to be true

[shy bluff]

guess it's time to read that apper 😔

[oblique river]

no

why would you do that

just look at the example

lol

[shy bluff]

To look for other example sin the paper

[oblique river]

how many examples do you need?

[next obsidian]

I just never think about the rings of continuous functions

just 1

I think

I just never think about the rings of continuous functions
@next obsidian This is also really funny considering I do AG

[oblique river]

oh, then take that one, or modify it a little bit to find a new one

[shy bluff]

I want something that's ... we never talked about continous functions in this course?