#groups-rings-fields

406252 messages · Page 634 of 407

next obsidian
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Where both are < p

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You get some relation like

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k^nh^a = k^mh^b

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Now like a≠b

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Else you can undo and you get k^n = k^m

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So multiply by h^-a

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Then multiply on the left by k^-m

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You get like

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k^{n-m} = h^{b-a}

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Note that n-m and b-a are both like, non-zero

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And < p

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So this says some non-identity power of k is in H

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But like, from this you can show k is in H by like

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Taking a suitable power of k^{n-m}

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Basically cuz of Bezout’s lemma, and p being prime

barren sierra
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How do I show that that's what I'm lost on

next obsidian
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Okay

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So let N = n - m

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0 < N < p

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Now gcd(N,p) = 1

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So by Bezout there is some a,b so that aN + bp = 1

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So do

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(k^N)^a = k^{aN} = k^(1-bp) = k•(k^p)^-b = k•e^-b= k

unique berry
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Well I learned some reasons why I failed my exam

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In question 1 I cited a result we proved in the homework; we were supposed to prove it again (I had forgotten the proof so fair enough).

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Question 2 I'm pretty sure I got points taken off for initial scratch work but idrk.
Question 3 I got correct
Quedtion 4 I didn't prove a fact about non-trivial automorphsms which was crucial to the overall proof

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In all cases I got half points taken off which means I got 5/8 of a score which is just below failing

past temple
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how does multiplication in graded algebras work

next obsidian
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By multiplying

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Do you just have a graded ring?

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Or an a graded algebra over a graded ring?

past temple
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like

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a graded ring

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that is an algebra

next obsidian
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Over…

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What

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Just a normal ring?

past temple
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itself...?

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yeah

next obsidian
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Wut

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I mean

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Every ring is an algebra over itself

past temple
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right yeah

next obsidian
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I mean in this case

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If it isn’t over a graded ring it just like

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Works

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Like I guess the two are separate kinds

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Like when you wanna multiply by the “scalars”

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It totally fucks ur grading up

past temple
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like somehow the graded ring that i have is an algebra

next obsidian
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Because there’s nothing ensuring it works right unless like the map goes into the degree 0 part

past temple
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ok ig my question is

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like

next obsidian
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This happens if like

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A -> B -> B[x1,…,xn]

past temple
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hmmm

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ok so like

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since our ring is graded

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we can express elements like

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(a_i)_i

next obsidian
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Yeh

past temple
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so then like

next obsidian
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Just like index by N

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For ease of writing

past temple
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ok

next obsidian
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Or rather I don’t wanna think about arbitrary gradings lol

past temple
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u mean like

next obsidian
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You can grade wrt any ordered abelian group

past temple
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finite gradings?

next obsidian
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No like

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I want to grade by either N or Z

past temple
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oh sure

next obsidian
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Not like your stupid fucked up ordered group

past temple
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yeah just imagine the i

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ranges from 0 to infinity

next obsidian
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Yeh

past temple
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so then if i have another element (b_i)_i

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like

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how do i multiply them

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and get another element in the form

next obsidian
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In the ring?

past temple
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(c_i)_i

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yeah

next obsidian
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You do it like

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Okay so

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It’s a direct sum

past temple
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ye

next obsidian
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So these are like finite

past temple
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yeah

next obsidian
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So you want to write it out like

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(a1 + … + an)(b1 + … + bm)

past temple
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ohhhhhhhhhhhh

next obsidian
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You can like truncate cuz

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Finite

past temple
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and then distribute out

next obsidian
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Then just like

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Distribute

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Yeh

past temple
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ok ok

next obsidian
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So my advice

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Pretend ur ring is always

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K[x1,…,xn]

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And think “how do I do it here”

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This will work often times

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And it gives you intuition of like

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When does it suffice to prove for just the graded parts?

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And shit like that

past temple
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righttt

next obsidian
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Like here’s another example is

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Suppose you had

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(ai)(bi) turns into

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Like umm

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A homogeneous element

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Of degree n

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Then by some magics in the graded parts besides the n-th degree

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We can like…

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Okay whoops what I’m about to say is false but

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I guess we can always assume that ai and bi

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Have nothing above degree n

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Cuz whatever was there has to cance

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This is like saying yes we might be able to take some

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(a + bx)(c + dx) and end up with an ex^2

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Cuz of cancellation

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But we can just assume the two things have nothing in degree >= 3

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Cuz like they can only contribute stuff in degree >= 3

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And those all have to become 0 anyway

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This is bad explanation

past temple
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wait

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u mean they can only contribute stuff in degree <=3?

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im a little confsued

next obsidian
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In degree greater than or equal to 3

past temple
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how can they have nothing in degree >=3 and also contribute only in degree >=3

next obsidian
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The terms that would look like

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No no sorry

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I meant like

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Okay a priori

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Maybe our shit could look like

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(a0 + a1x + a2x^2 + … + akx^k)

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And similarly for b

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But because the result is just x^2

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Usually you can just cut out all the ai and bi where i > 2

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Cuz they can only show up in the parts of degree >=3 which have to be 0 anyway

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So like they all have to cancel out in the end

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This is useful cuz like let’s say we had a homogeneous ideal I

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And we have that (ai)(bi) = x^2 where (ai),(bi) in I

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Then we know that aix^i are all in I

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So when we subtract away all the aix^i and bix^i for i>=3

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The result is something like
(a0 + a1x + a2x^2)(b0 + b1x + b2x^2) = x^2

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Where both things on the left are still in I

past temple
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ah okay

next obsidian
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Anyway me point is

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This is like pretty obvious for a poly ring

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But the intuition holds true in general

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I’ve had to do something like this a number of times

past temple
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yeah i think it makes sense for the most part

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looking at it like a polynomial ring helps a lot

next obsidian
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Yah

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And at some point I thought I came up with a reason why that’s okay

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Like umm

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Oh yeah

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Most graded rings R are generated over R_0

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By R_1

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Let’s say for right now it’s finitely generated

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In particular if R is Noeth this is true (wait this isn’t true, Noeth gives fg over R_0 but not in degree 1 necessarily)

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Okay so now take generators a1,…,an

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Then we know R = R_0[a1,…,an]

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This is surjected onto by R_0[x1,…,xn] an actual poly ring

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And this is like a graded surjection

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Cuz xi map to ai

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Sends degree 1 to degree 1

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Then the kernel should be graded or some shit

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So we can write R as a quotient of R_0[x1,…,xn] by a graded ideal which means the quotient has a grading by like, lifting

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My point is as a graded ring, so even considering the grading

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R looks like a quotient of a poly ring over R_0

past temple
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i see

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ohhh i think thats what i was meaning to say

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yeah graded rings are algebras over A_0

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generated over a finite collection of homogenous elements

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which gives it the same structure as A_0[x_1,...xn]

next obsidian
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Well the latter one is an assumption

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You need to assume it’s fg

past temple
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oh

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i see

next obsidian
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Yeah like

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Consider

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K[x1,x2,…]

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This isn’t finitely generated

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Cuz infinite

past temple
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ah ok

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another quick question

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so like

next obsidian
past temple
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we have this way of defining multiplication

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by just like

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considering the elements

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as literal sums

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but then like

next obsidian
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Yeh

past temple
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hmm

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im kind of like

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trying to do it in a general way

next obsidian
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You have to just do it this way

past temple
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using something like this

next obsidian
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Like

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Yeah I mean

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That’s the same thing

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You just haven’t written it out as a sum

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On the left

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I highly recommend against this tho like

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Anytime you deal with a graded ring

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You really won’t be able to do shit unless you write out like a in R

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a = a_0 + … + a_n

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Then manipulate it like that

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I guess for arbitrary gradings it doesn’t work well

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But if you don’t do this it’ll be really hard to know wtf is going on I think

past temple
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yeahh im kinda in the spot where idk wtf is going on

next obsidian
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Just write stuff as

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a1 + … + an

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Always

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Like a, b in R

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Write them out as graded parts

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Then do shit

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This is what we do for polynomial rings

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:^)

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And you’ll get more used to it

past temple
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gotcha

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ty for the help

next obsidian
shell brook
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if i have |G| = m, |A| = n, A abelian, then wbhat is |Hom(G, A)|? pls do not tell me but give hint or something

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i have no clue how to start

next obsidian
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You can’t say in general I think

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I think using classification of finite abelian groups you can give the set of values it can take on

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Like this isn’t a function of m and n I think

shell brook
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this is a problem on my pset so I hope i can say something

next obsidian
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I mean, this seems mega fake like

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Idk, maybe there's something silly I'm overlooking but

shell brook
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Let me make sure I have the full context

next obsidian
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If you take G = Z/p^2Z

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and Z/pZ x Z/pZ

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I don't see why there number of maps from here to an abelian group have to be the same

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but maybe I'm being stupid

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Maybe this case actually works out cuz it's all about divisibility criterions

shell brook
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its supposed to be. |A| = m

unknown.png
next obsidian
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Hurbed

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Also omegalol

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|H| = m

shell brook
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is it easier

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cuz of context

next obsidian
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Not really, I guess this is just true...

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wtf

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Like

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if G is abelian too

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I can maybe see how to do it

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Do you have the classification of finite abelian groups?

shell brook
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absolutely not

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we just learned about actions & orbits today

next obsidian
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This isn't determined by size

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There are counterexamples where G is abelian too

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find one

shell brook
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wdym

next obsidian
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You can find

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G, G', A, A'

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|G| = |G'|

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|A| = |A'|

shell brook
next obsidian
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but |Hom(G,A)| is not equal to |Hom(G',A')|

shell brook
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LMAO

next obsidian
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I knew this problem sounded like bullshit

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Hint:

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You don't need to think about complicated groups to find a counterexample

shell brook
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idk what is complicated to chmonkey algebrain

next obsidian
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You don't have to go past size 4

shell brook
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nice ok

latent anvil
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I knew it

next obsidian
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;)

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I thought it was bullshit Sham

latent anvil
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Yeah haha

next obsidian
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But I just didn't compoot

latent anvil
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I just figured it came from here

next obsidian
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I was thinking Z/p^2Z and (Z/pZ)^2

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but I was like

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"wait maybe these are the same"

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I should've plugged a number in for p lol

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Oh actually

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I was fixing A

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that's why

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I was pretty sure by classification that like

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the number of things of order dividing p^2 is the same as (order dividing p)^2

shell brook
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did you ask shamrock

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im coming up with a counterexample to own my prof on piazza rn

next obsidian
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I asked in Chmonkeymath

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and someone else came up with the counterexample

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the legendary boy who went to Berkeley

shell brook
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how do i just enumearte all the homomorphisms

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this seems hard to write down every map

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i guess there are very little elements

next obsidian
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Maps from a cyclic group

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are easy to describe

shell brook
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oh yeah

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i was trying to use dihedral groups r smth

next obsidian
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wtf

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I mean

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one of them is dihedral of order 4 lol

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but there's 2 groups of order 4

shell brook
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yeah

barren sierra
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is any notion of gcd for polynomials or just normal elements only well defined for integral domains?

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like do we require no zero divisors

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for gcd to make sense?

past temple
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is there a way to construct a surjection M (x) N -> M

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where (x) is the tensor product and M, N are A-modules

hidden haven
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Tensoring can make modules smaller

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M ⊗ A/a ≈ M/aM

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So take a finite module to get a counterexample

chilly ocean
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Z/2Z (x) Z/3Z is trivial, good luck coming up with a surjection to Z/2Z

past temple
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shoot

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okay right now

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im trying to construct a bijection

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so M is an A-module

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and B is an A-algebra

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and i want to construct a bijection from

chilly ocean
hidden haven
past temple
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$Hom(\oplus_{i=0}^{\infty} M^{\otimes i}, B) \to Hom(M,B)$

cloud walrusBOT
past temple
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where the Hom on the left

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are algebra homomorphisms

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and the Hom on the right

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are module homomorphisms

hidden haven
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Do you have the universal property of the tensor algebra?

past temple
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uhh

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im not sure

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i dont think so

hidden haven
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Any progress you've made so far?

past temple
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not much, but i was thinking like

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if we have a mapping from like

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the tensor algebra to B

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i wanted to use some kind of projection

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from the tensor algebra to M

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and like

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get a mapping from M to B

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that way

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like a universal property kinda

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idk

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thats why i was asking

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if it was always possible to construct

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a surjection from a tensor product of modules to the factors

hidden haven
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So you have a module map from M to TM (TM being the tensor algebra) which is the natural map as m ↦ m in the degree 1 component. If you have an algebra map from TM to B, you can precompose that (viewing it as a module map) with M to TM to get a module map M to B. This should be the bijection

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To go the other way, if you have a module map from M to B, then you get a k-linear map from M^k to B by multiplying the images. This gives a map from the k-fold tensor to B for each k, which gives a map from the direct sum over all k, to B

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You'd have to show that these 2 are inverse operations (or that either of these is a bijection)

past temple
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ahhh okay

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yeah i had just realized how to construct the map from M to TM

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phi -> [m -> phi(0,m,0,...)]

hidden haven
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Shouldn't it just be m ↦ (0,m,0,0,...)

past temple
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sorry

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from M to B

hidden haven
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Or are you talking about the map between hom sets

past temple
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yeah between

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the hom sets

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the image

hidden haven
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Right ye

past temple
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is a mapping from M to B

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here

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yeah

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so thats the same thing as like

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precomposition by inclusion onto the first factor

hidden haven
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Yes

past temple
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wait im a little confused

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by ur construction of the inverse though

hidden haven
past temple
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wait by M^k

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do u mean like

hidden haven
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Which part

past temple
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M^(x)k

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?

hidden haven
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Ordinary product

past temple
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oh

hidden haven
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M^⊗k is the k-fold tensor

past temple
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ok so ur getting the map from the k-fold tensor

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via the universal property

hidden haven
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Yep

past temple
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but then how r u extending that

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to the direct sum

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just like

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concatenating them?

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adding them?

hidden haven
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If you have a family of maps f_i from M_i

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They combine to give a map from ⊕M_i

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Universal prop of direct sum

past temple
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ahh okay

hidden haven
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It happens by mapping the tuple (m_i) ↦ summation f_i(m_i)

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Only finitely many m_i are non zero so sum makes sense

past temple
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ahh okay

regal carbon
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Proof require 10000 hours of practicing?

rustic crown
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,w 10000 hours

cloud walrusBOT
hidden haven
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🤡

trim grove
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The total numbers of maximal ideals of $\mathbb{Z}_n \times \mathbb{Z}_m$ are

I know that maximal ideals in Zn are prime divisers of n , but what about in product ? Any hint

cloud walrusBOT
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Algebra

863703908576002061.png
next obsidian
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Prove that maximal ideals of R x S look like m x S or R x n for m a maximal ideal of R, or n a maximal ideal of S

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Bonus: show this statement is true for prime ideals

rustic crown
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Bonus: show that ideals of R x S look like I x J. for ideals I of R and J of S.

trim grove
rustic crown
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nope

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look at chmonkey 's statement

trim grove
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like in case of Z10 X Z15

Z10 X <[5]>
Z10 X<[3]>
<[5]> X Z15
<[2]>XZ15?

rustic crown
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check the third one

trim grove
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My bad

rustic crown
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but yea, notice

cloud walrusBOT
chilly ocean
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I am going through my exercises, could someone tell me if the following linear transformation of R-vector spaces which maps a continuously differentiable function to its derivative. (Hint: FTC) is epi/mono/iso morphism? Thanks

unknown.png
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I proved that its a linear transformation by now

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I guess it is bcs it is continuous

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no

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both are 3

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its not 1-1

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so not isomorphism

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so it is epimorpshim

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since it cannot be monomorphism since they can match to the same derivative

cloud walrusBOT
chilly ocean
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yeah I have been thinking it is surjective as there cannot be a derivative in the set with no preimage

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oops

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yeah that makes sense, thank you a lot

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now I have concluded it is injective so i think iam right

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its C^1[0,1] -> C[0,1]

hidden haven
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Continuously differentiable and continuous

chilly ocean
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surjective

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no

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yeah but why is that related if it is surjection or injection

regal carbon
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Here i can just consider "R:R×R mapping to R" just as real number right?

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Example 1

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Sir did you understood the above example?

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I am new to proof not understanding anything

cloud walrusBOT
regal carbon
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And also a and b belongs to the group of real numbers right?

regal carbon
cloud walrusBOT
regal carbon
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Yes... I have to recall those.

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Yeah, I'm looking at it. I concentrated on proofs of contradiction, Direct, contrapositive and those sets functions and even relations got partially erased.

chilly ocean
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Indeed, all continuous functions have antiderivatives. But noncontinuous functions don't.

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I have not understood why we care if it is continuous I just want to see if it is injective or surjective

barren sierra
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if we want to have gcd well defined for a ring

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do we require that the ring is a PID?

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or just integral domain?

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nvm answered my own question cause gcd is well defined for polynomials but <2, x> can never be principal in Z[x]

rustic crown
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just integral domain isn't enough

barren sierra
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wut

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what else do we need?

rustic crown
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okie so consider the ring Z[sqrt(-5)] what is the gcd of 2(1+sqrt(-5)) and 6?

barren sierra
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1?

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yea

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no

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2

rustic crown
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2 and (1+sqrt(-5)) are common divisors

barren sierra
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I didn't see the 2 at the front

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gcd 2

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wait what?

rustic crown
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6 = (1+sqrt(-5))(1-sqrt(-5))

barren sierra
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oh

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ok hm

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but you can't say one is greater than the other

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so we need a notion of order in addition to integral domain?

rustic crown
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nah, that's not how gcd is defined

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gcd of a and b is an element d such that d | a and d | b, so d is a common divisor. And for every common divisor c of a and b, we have c | d.

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the partial order | does the job for us.

barren sierra
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ahhhhh I forgot | is a partial order and that's how we define greatest

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ok so wikipedia says GCD domains are a subset of integral domains

rustic crown
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yee

barren sierra
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but all it says is "GCD domain is an integral domain where gcd is defined"

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which like

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thanks but that means nothing

rustic crown
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lol

barren sierra
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cause "where gcd is defined" is really my question

rustic crown
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there are also domains which are called bezout domains where bezout's theorem holds

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so you can write gcd as linear combination

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but you're right, unless you study them specifically the name doesn't help lol

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in general you can define gcd for UFDs

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if a and b are two non-zero elements, factorize both into primes and take minimum power corresponding to each prime

barren sierra
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hm

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ok

lilac trench
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someone explain why isomorphisms only send elements to those of the same order

rustic crown
# rustic crown 6 = (1+sqrt(-5))(1-sqrt(-5))

so if you wanna continue this proof, say d was a gcd of 2(1+sqrt(-5)) and 6
then notice that norm is a thing in Z[sqrt(-5)]
N(2(1+sqrt(-5)) = 24 and N(6) = 36
this means N(d) divides both 24 and 36, so it divides 12
but since 2 and (1+sqrt(-5)) are common divisors, they need to divide d, which means 4 and 6 divide N(d). All this forces N(d) = 12
but there are no elements of norm 12...
indeed, 12 isn't a perfect square and 12 - 5 = 7 also isn't a perfect square and 12 - 4 * 5 already gets negative

delicate bloom
lilac trench
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f(x)^n right?

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ohhhhhh

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✅

delicate bloom
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be careful, it takes a little more thought to be sure that this works, how do you know that if x has order n that f(x) doesn't end up having order d, some number that divides n?

lilac trench
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hmmm

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can I say that the inverse function of f is also an isomorphism and hence the same rule must apply?

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nvm the same problem arises

delicate bloom
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you're on the right track

lilac trench
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all I could think of is that it would be a contradiction since the inverse doesnt map back to x

delicate bloom
#

sounds good 👍

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might help to write this all out a bit more formally if you haven't already done so

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but that's really it

lilac trench
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thx

barren sierra
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contradiction is the way I would go too

lilac trench
#

discussion is fun

barren sierra
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wait for cosets
suppose G = finite group
and H subgroup of G
we have |H| divides |G|
but we also have |gH| = |Hg| = |H| for all g in G right?
does that mean all cosets have the same cardinality
so for all g1, g2 in G, |g1H| = |g2H| = |Hg1| = |Hg2|?????

chilly ocean
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yes

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this is Lagrange's theorem

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cosets partition the group

barren sierra
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I knew the partition part but I didn't realize that all the partitions were equal in size

next obsidian
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You can just

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Express a bijection between the different cosets

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They all have size |H|

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You go from H to gH by sending h to gh

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And the inverse is h -> g^-1h

barren sierra
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oh true

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cause cancellation proves uniqueness of such mapping

next obsidian
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I don’t know what that means

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They’re inverses

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So the maps are bijections

barren sierra
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I mean like proving that this map you described is indeed both injective and surjective

shell brook
#

Chmonkey does Hom(Z/4Z, Z/4Z) and Hom(K4, K4) work for that question I posted yesterday

#

I think thats 4, and 16 respectively

#

maybe i cant count tho

sturdy marsh
#

yeah that looks alright @shell brook

shell brook
#

do you knwo what i referring to brofib xd

unknown.png
#

i think is counter example

sturdy marsh
#

yeah

#

it is

shell brook
#

okay

sturdy marsh
#

Hom(K4, K4) = Hom(Z/2, K4)^2 = (Hom(Z/2, Z/2))^4

#

so it has 16 elements

shell brook
#

K4 is klein four for you right

sturdy marsh
#

yup

shell brook
#

i dont know what its supposed to mean

#

okay dope

sturdy marsh
#

Z/2 \oplus Z/2

next obsidian
#

I wrote it down!

#

gH -> H by h -> g^-1h

#

that’s what I was saying

#

You don’t need no silly cancellation to show injectivity or whatever it just has an obvious inverse

shell brook
#

Chmonkey is this the example ur firend came up with

next obsidian
#

Yes

shell brook
#

Okay thanks

barren sierra
#

ahhhh

#

ok

shell brook
#

or however you call what he did here

sturdy marsh
#

it's the universal property of (co)products

shell brook
#

lmao

sturdy marsh
#

Hom(A \oplus B, C) = Hom(A,C) x Hom(B,C)

shell brook
#

k i asked on Piazza

#

maybe im actually stupid and its obvious cuz i misread something

#

fuck this professor!

languid moss
#

can anyone help?

#

Suppose R is a PID, and I ⊂ R is a nonzero ideal. Prove that I is a prime ideal if and only if I is generated by a prime element of R.

next obsidian
#

Show that the statement is true assuming I is a principal ideal for any ring R

#

This is… basically how a prime element is defined

#

Now every ideal in a PID is principal… that’s how they’re defined

#

Then you GG win EZclap

languid moss
#

@next obsidian SO contrapositive proof?

next obsidian
#

Umm

#

No

#

I mean to prove th (p) is prime iff p is

#

I guess one direction is maybe easier with contrapositive

#

But honestly it’s kinda definitional

chilly ocean
lavish mantle
#

No, it's not definitional

#

A prime ideal $I$ is an ideal in which ALL elements $c \in I$ satisfy $ab = c \implies a \in I$ or $b \in I$

cloud walrusBOT
#

polikuj2

278476727708155904.png
lavish mantle
#

The statement above is wrong in a not-PID ring

#

(the minimal hypothesis being factorial, I think)

chilly ocean
#

at least one direction is definitional hmmCat

#

just change ur definitions so it's trivially true

next obsidian
#

I said that (p) is prime iff p is

#

A principal ideal is prime iff its generator is, this is definitional

#

This proved the result in PIDs because all ideals are principal

languid moss
#

@lavish mantle what?

chilly ocean
#

Hello, I am attempting on working on this question but have no idea where to start… Why does it suffice to show that <s> (s is symmetric reflections on the n-gon) is not normal in Dn?

IMG_0223.png
chilly ocean
#

<@&286206848099549185>

scarlet estuary
#

hint 1: products of abelian groups are abelian
hint 2: ||what do you know about subgroups of abelian groups?||
hint 3: ||theyre normal!||

#

this is weird though

#

there should be a far easier proof

#

(like, hint 1 should give it automatically?)

chilly ocean
#

I see!

chilly ocean
#

Cyclic groups are abelian, and products of abelian groups are abelian. If Dn were isomorphic to Cn x C2, to the contrary, then Dn would be Abelian. And all subgroups subgroups are normal in abelian groups. So if we show that the subgroup <s> of Dn is not normal, we are done.

ivory dust
#

Is this enough to say $H_{n+1}(C^kX) = \bigoplus_{i=1}^{k=1} H_n(SX)$?

unknown.png
cloud walrusBOT
#

Namington + mniip 2024

381174875215364109.png
ivory dust
#

This is from alg top, but this is really more of a pure algebra question

#

and might just be first iso theorem idk

#

like if you have groups G,A such that $(\bigoplus_{i=1}^{k+1} G)/A = G$, can you say $A= \bigoplus_{i=1}^k G$

cloud walrusBOT
#

Namington + mniip 2024

381174875215364109.png
ivory dust
#

or maybe this is rank-nullity on Z-modules?

next obsidian
#

No

#

At least not obviously

#

Rank nullity won’t work it doesn’t exist

ivory dust
#

hm?

#

rank nullity does exist for Z-modules though right?

next obsidian
#

no

#

rank nullity only works over a field

#

you can get some sort of shitty statement maybe from PID-ness of Z

ivory dust
#

bruh

#

that's literally just false though

#

I explicitly remember my professor using rank-nullity over a Z-module

thorn delta
ivory dust
#

maybe it only works in special cases?

next obsidian
#

If you’re free you get what’s basically rank nullity

#

But there’s plenty of non-free abelian groups

robust pollen
#

If the localization of a commutative ring at a prime is a field, what can I say about the prime?
E.g. would it imply p=0?

next obsidian
#

No

#

It’s a minimal prime

#

And the localization is reduced (these two combined are equivalent to A_p a field. If A is an integral domain so that 0 is prime then this forces p = 0)

#

Im not sure if you can turn the latter into a statement about p

#

But this happens if eg your ring was reduced to begin with

#

As a related thing, you can prove the following theorem is true

#

Let A be a reduced Noetherian ring, and p1,…,pn its minimal primes

#

The total ring of fractions of A is isomorphic to A_p1 x … x A_pn, so in particular it is a finite product of fields

robust pollen
#

Oh, that's cool!

#

Thank 🪑

next obsidian
#

This fact I said

#

Is unexpectedly useful

#

Lmfao

robust pollen
#

And when does a ring have a (unique) minimal prime ideal?

next obsidian
#

It’s used to prove Serre’s criterion for normality

next obsidian
#

It’s equivalent to its Spec being irreducible lol

#

It’s also equivalent to having the nilradical be prime

#

Which is interesting I think

#

two things which aren’t nilpotent can’t go have a baby which is nilpotent (have a baby means multiply)

robust pollen
#

What am I getting myself into 🙈

next obsidian
#

The chmonkey zone

robust pollen
#

Why is algebra(icish geometry) so interesting? :( It's making real life hard

next obsidian
robust pollen
#

What about local rings with unique minimal prime?

next obsidian
#

Doesn’t say shit I think (I bet you can find something you can say but nothing comes to mind that isn’t mega esoteric)

#

I mean the lattice of primes looks cool

#

But that’s kind of on the wrong side of things

robust pollen
#

But assuming that a ring as a unique min prime, is localization at it a field?

next obsidian
#

No

#

There could be nilpotence

#

So the result has just 1 prime

#

So that prime is exactly the nilpotents

#

If your ring was reduced to begin with any localization also is reduced

#

So the only prime of the ring has to be 0

#

Aka it’s a field

robust pollen
#

I'm asking because I was thinking: forget from (field) to (commutative rings + conditions). When does forget functor have which adjoints?
For example, if conditions are "local pid = DVR", then the left adjoint should be localizing at (0), and the right adjoint should be quotienting out the maximal ideal

#

Is that right?

next obsidian
#

I don’t see why that’s a right adjoint

robust pollen
#

This particular example btw feels to me like the discrete-forget-indiscrete adjunctions for topological spaces, and there is probably some deeper relationship that I don't know since I don't know grotehendieck stuff

next obsidian
#

Like

robust pollen
next obsidian
#

I don’t see why we can’t get K -> R/m

#

To be given by

#

Two different maps K -> R

#

Like quotients are a mapping out thing

#

So it should be on the left I feel

#

Also

#

Forgetful functors shouldn’t have a right adjoint I think

#

Just like

#

Morally

#

Or rather you shouldn’t expect them to

#

I feel like if you’re gonna call something a forgetful functor it needs to have a left adjoint haha

#

If not you shouldn’t call it a forgetful functor lol

robust pollen
#

Yes they should, if for example you forget from the category of comodules over a comonads, then you'll get a right adjoint, the "cofree functor"

next obsidian
#

Sure

#

Lol

#

But like, I don’t think they ought to in general

robust pollen
#

I mean, "forgetful functor" is not a rigidly defined thing, it can mean what you want it to mean pacman

next obsidian
#

I mean all the examples I think of are right adjoints

hidden haven
robust pollen
next obsidian
#

Without being a left adjoint, or at least an obvious one

#

Yeah I mean this isn’t always true

#

Yes

hidden haven
next obsidian
#

My point tho is I don’t think field-> DVR

#

Ought to have one

#

Like there’s no natural way to grab a field from a DVR which has a mapping into property

#

At least one that I feel is natural

hidden haven
#

Theorem: no functor with domain or codomain in Field has a left or a right adjoint

next obsidian
#

Anyway I need to go drive

hidden haven
robust pollen
hidden haven
#

Obviously not true

#

I think category of domains with injective maps is a reflexive subcategory of Field

next obsidian
#

All maps factor through FOF

hidden haven
next obsidian
#

🚗🚗🚗🚗

hidden haven
#

You got caught in a 4 car bumper to bumper crash?!!

#

Shouldn't have been texing while driving

robust pollen
#

uh... what? how does it fix no faces

unknown.png
#

if I rotate about the blue line, the action on the faces is (1,2,3,4,5,6) -> (6,5,3,4,2,1) [= (1,2,4,3,5,6) morally]. Am I being kindergarten dumb?

unknown.png
next obsidian
#

Like grab two vertical lines in opposite corners

robust pollen
#

I see, so like from the 1|4 edge to the 3|6 edge, and that is a transposition in S4 because S4 acts on the "long diagonals"

next obsidian
#

Idk

#

Maybe

#

Lol

winter thorn
#

so, an element in a finite group has finite order, right?

hidden haven
#

ye

next obsidian
#

KEEP RAISING IT IF IT FOESNR HIT e IT WILL GIVR A COPY OF Z

robust pollen
#

chdrunkey

chilly ocean
#

Let V be a finite dimensional vector space over a field K, and let W1, W2 and W3 be subspaces of V. By analogy with the Inclusion-Exclusion Principle for sets, and taking into account the dimension formula for a sum of 2 subspaces, I concluded reasonably that a dimension formula

dim(W1 + W2 + W3) = dim(W1) + dim(W2) + dim(W3) − dim(W1 ∩ W2) − dim(W2 ∩ W3) − dim(W3 ∩ W1) + dim(W1 ∩ W2 ∩ W3) (†) holds for the sum of 3 subspaces.

How can I prove by considering 3 distinct lines through the origin in $R^2$ that that (†) does not necessarily hold?

Many thanks

cloud walrusBOT
#

just_dom

Let V be a finite dimensional vector space over a field K, and let W1, W2 and W3 be subspaces of V. By analogy with the Inclusion-Exclusion Principle for sets, and taking into account the dimension formula for a sum of 2 subspaces, I concluded reasonably that a dimension formula

dim(W1 + W2 + W3) = dim(W1) + dim(W2) + dim(W3) − dim(W1 ∩ W2) − dim(W2 ∩ W3) − dim(W3 ∩ W1) + dim(W1 ∩ W2 ∩ W3) (†) holds for the sum of 3 subspaces.

How can I prove by considering 3 distinct lines through the origin in $R^2$ that that (†) does not necessarily hold?

Many thanks
906207760973201408.png
chilly ocean
#

maybe look at 3 different lines?

#

yeah but how to show that

#

show what? just calculate, its not hard with lines in R^2

#

look at x=0, y=0 and y=x

#

then all the dim intersections are 0

#

and I think 2 doesnt equal 3 in general

chilly ocean
#

because dim(one line + 2nd line + 3rd line)=2 when the lines are different

#

and dim(line)=1

#

and dim(line intersction other line)=0

#

I get that but how do I prove it based on the dimension formula above?

urban acorn
#

you don't, because the dimension formula above is not true

#

and what they are saying is the counterexample, K = R, V = R^2, and W1 through W3 are distinct lines in R^2 passing through the origin

chilly ocean
#

yeah it does not but I need to shoe that does not necessarily hold based on the 3 distinct lines in R^2

urban acorn
chilly ocean
#

Oh I am not sure what dim(W1 ∩ W2 ∩ W3) will be in this case of x=0 y=0 and x=y

urban acorn
#

it's the set of all points that lies on all of these lines, which points are these?

chilly ocean
#

not sure ;/

urban acorn
#

here's x = 0 in cyan, y = 0 in orange, and y = x in purple

Screenshot_2021-11-19_-_14-48-02.png
#

where do they all intersect?

chilly ocean
#

at 0

urban acorn
#

right, and what is the dimension of {0} as a vector space?

chilly ocean
#

the zero v.s. has dimension 0

urban acorn
#

so there you go, dim(W1 ∩ W2 ∩ W3) = dim( {0} ) = 0

chilly ocean
#

thank you so much so I will just show that
dim(W1 + W2 + W3) = dim(W1) + dim(W2) + dim(W3) − dim(W1 ∩ W2) − dim(W2 ∩ W3) − dim(W3 ∩ W1) + dim(W1 ∩ W2 ∩ W3) = 1 + 1 + 1 - 0 - 0 - 0 +0 =3

urban acorn
#

so you will show that dim(W1) + dim(W2) + dim(W3) − dim(W1 ∩ W2) − dim(W2 ∩ W3) − dim(W3 ∩ W1) + dim(W1 ∩ W2 ∩ W3) = 1 + 1 + 1 - 0 - 0 - 0 + 0 = 3, but dim(W1 + W2 + W3) = 2, and therefore the formula your originally asked about does not hold

#

W1 + W2 + W3 obviously can't be more than 2, since they are all subspaces of R^2 which is itself of dimension 2

chilly ocean
#

yeah now it makes absolutely sense many thanks again!

urban acorn
#

you're welcome

robust pollen
#

where do question related to algebraic manipulations of continued fractions belong?

hidden haven
#

probably analysis

robust pollen
#

ugh

rustic crown
#

wait you can algebraically manipulate continued fractions?

#

are there nice ways to find sum and product of two continued fractions?

#

i never thought about this

#

if this is true it's time to ditch decimal notation

robust pollen
#

Well, to be fair, I'm thinking of finite continued fractions of the form a - 1/(b - (c - 1/ ...)) where a,b,c blabla are natural, in particular they are rational numbers.

#

maybe should have clarified that pacman

chilly ocean
rustic crown
#

so real numbers are very bad already and decimal notations don't help much understanding them

#

like consider pi = 3.14159265358979323...

#

so this basically says that 314/100 is the best rational approximation with denominator 100

lethal dune
#

π=22/7hmmCat

rustic crown
#

but why would you use 100, that's so bad

#

the error is is like 1/1000

#

[3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14...]

#

on the other hand this is the continued fraction of pi

#

that bigg 292 says that you have an extremely good rational approximation for pi

#

[3; 7, 15, 1] = 355/113

#

and what's the error this time?

#

it's on the order of 10^-7

#

isn't that bizzare?

chilly ocean
#

fast convergence good for numerical stuff maybe

delicate bloom
cloud walrusBOT
lethal dune
chilly ocean
#

that's nice, but how can that be useful for someone who doesn't do applied math?

delicate bloom
#

pi comes up in pure math pretty often

rustic crown
#

continued fractions have crazy nice properties

#

for instance that's one of the only nice ways to solve the pell's equation

delicate bloom
#

you can use continued fractions to solve some diophantine equations

#

sniped lol

rustic crown
#

the continued fraction of sqrt(d) are periodic and in fact the period is palindromic

lethal dune
#

honestly I like this approach more than reverse contracting the euclidian algo

rustic crown
#

you can use continued fractions to prove 1 mod 4 primes are sums of 2 squares

chilly ocean
#

that sounds nice cuz i suck at number theory

rustic crown
#

i started liking abstract algebra because of number theory eeveeKawaii

lethal dune
#

I was once doing cryptography, that made me interested in number theory, fields and groups, not the rings / modules tho

#

they suck

#

but PID/ED/UFDs are nice

rustic crown
#

i've heard that if the decryption key somehow happens to be very small in the RSA algorithm, then continued fractions give a very effective way to break it

lethal dune
rustic crown
#

by very small i think it was like d^4 < n

#

but n is like product of those two huge primes

#

so even the 4th root is crazy big

chilly ocean
#

Let V be a finite dimensional vector space over a field K, and let W1, W2 and W3 be subspaces of V. By analogy with the Inclusion-Exclusion Principle for sets, and taking into account the dimension formula for a sum of 2 subspaces
dim(W1 + W2 + W3) = dim(W1) + dim(W2) + dim(W3) − dim(W1 ∩ W2) − dim(W2 ∩ W3) − dim(W3 ∩ W1) + dim(W1 ∩ W2 ∩ W3) (†) holds for the sum of 3 subspaces.

This formula does not hold because I proved that it does not, but I need to state general assumptions on the subspaces of W1,W2,W3 which guarantee that the formula does hold and prove it under these assumptions.

But how do I do that?

Many thanks

lethal dune
#

I was trying to solve a CTF challenge which was about crypto. the ring that was used was Z[i] (mod n) component wise. the right wasn't even a ID. that was pain, couldn't do it

chilly ocean
cloud walrusBOT
chilly ocean
#

but how can I start the proof?

rustic crown
#

by second isomorphism theorem, the formula is true for two subspaces

cloud walrusBOT
chilly ocean
#

but based on which assumptions I have to prove it?

chilly ocean
rustic crown
#

that's called the second isomorphism theorem, which is like a standard theorem in abstract algebra that applies to all sorts of algebraic structures

#

in our case, taking dim of both sides gives us
dim(W1 + W2) = dim(W1) + dim(W2) - dim(W1 n W2)

#

i was saying to use this formula twice

rustic crown
chilly ocean
#

oke I try that thanks for ur time

barren sierra
#

"So, to show that f is an isomorphism it is enough to show that f is injective"

#

why?

#

why do we not need to show surjectivity

hot lake
#

are the groups finite and do they have the same size ?

barren sierra
#

well for my questions

#

I need to show a group of size 9 with no elements of order 9 is isomorphic to Z_3 x Z_3

#

trying to construct the isomorphism

#

so I have <h>, <k> such that h, k have order 3

#

h =/= k

#

so I want f(h^i k^j) = ([i], [j])

#

just want to show that it is bijective

#

my other idea was use this theorem
If H, K are subgroups such that
H, K normal
H ∩ K = {e}
HK = G
Then G ≃ H × K

#

so I need to show HK = G

#

H = <h> and K = <k>

#

just stuck on that

hot lake
#

what if h = k²

barren sierra
#

so I already showed H ∩ K = {e}

#

I just need to show hk, h^2k, hk^2, h^2k^2 are not in <h> and <k>

#

which I'm not sure how to do in a clean way

hot lake
#

how did you show that H inter K is {e}

barren sierra
#

so we have h =/= k, e

#

so that part is trivial

hot lake
#

???

barren sierra
#

we define h =/= k

hot lake
#

what if h = k² ?

barren sierra
#

well then we have hk = k^3 = e

#

wait hm

#

fuck

#

this problem shouldn't be this hard idk why it's been tripping me up

#

here's the full problem

unknown.png
#

4b specifically

hot lake
#

oh you assumed that G was abelian

#

that makes it kinda trivial

barren sierra
#

yea so we get normal for free

#

so I guess I need the other two conditions

#

H ∩ K = {e}
HK = G

hot lake
#

you need to show that HK has exactly 9 elements then

barren sierra
#

unless there is an easier way

#

yea

rustic crown
#

|HK| = |H||K|/|H n K| so that's done if you show one

hot lake
#

once you know that you might as well define an isomorphism directly

weak oriole
#

HK either has 3 elements or 9

#

If it has three elements then H=K

#

But since you have 9-1 total elements of order 3

#

U can make sure H=K doesn't happen

barren sierra
#

yea

barren sierra
#

yea that is easier

rustic crown
#

yep

barren sierra
#

I was trying to do subset inclusion stuff

#

so yea I just need H ∩ K = {e}

rustic crown
#

so how did you get H?

#

you picked any non-identity element h and took H = <h>

#

how should we choose K?

#

only thing we want is H n K = {e}

barren sierra
#

yea so pick k from G \ <h>

rustic crown
#

yep!

barren sierra
#

so k =/= h, h^2

rustic crown
#

and e

barren sierra
#

yea

#

was about to add that

#

so then need k^2 =/= h or k^2

rustic crown
#

so H n K is a strict subgroup of K, indeed k is in K, but is not in H n K

rustic crown
rustic crown
#

a few days ago my prof mentioned that only groups that occur as galois groups are pro-finite groups. does anyone know nice place were i can read about this? i don't know any infinite galois theory

waxen hedge
#

That's because Galois groups of infinite galois extensions are defined as profinite groups ! For a reference, you can try "Galois groups and fundamentals groups" by Tamás Szamuely

rustic crown
#

wait aren't they defined as automorphism groups of normal and separable extensions?

#

i can kinda see why it will be projective limit of finite groups

#

but he gave a more topological criterion

waxen hedge
#

basically, if you have a Galois extension L/k, you define his Galois group as the projective limit over the Gal(K/k) where K/k is a finite Galois extension contained in L

waxen hedge
#

But the projective construction makes the topology natural on it

rustic crown
#

oh

waxen hedge
#

(Maybe I misunderstood your question. I don't know if every profinite group is a Galois group)

waxen hedge
rustic crown
#

i think that's what he mentioned, just like every finite group can be realized as a galois group of some finite galois extension same holds for infinite but need to look at profinite

#

seemed very interesting as he mentioned that using those topological facts, compact, hausdorff and totally disconected one can show it's either finite or uncountable, so Z is never a galois group

waxen hedge
#

Oh that's nice, I never thought of that

cosmic raft
#

how can i use a hint like "p(x) = x^4p(1/x)" to help factor a polynomial over a field?

delicate bloom
#

maybe that lets you put it into a form where you can recognize a factorization easier, or show that it's not factorable by eisenstein

#

that transformation will, assuming it's degree 4, reverse the coefficients on the polynomial

cosmic raft
#

hmmm

upper cape
#

How would I go about showing that the ideal (2,1+sqrt(-11)) is principal in Z[(1+sqrt(-11))/2]?

next obsidian
#

Compute the gcd of the two elements

#

That’ll generate the ideal

#

By property of being a gcd 2, 1 + sqrt(-11) will be in the ideal generated by the gcd cuz well, they’re multiples of the gcd

#

And you should be able to find a way to express the gcd as a sum of multiples of 2 and 1 + sqrt(-11), this is like Bezout’s lemma for Z

#

[note that all of this hinges on the ring you have being a UFD, this guarantees the existence of a gcd and everything else I said. I still think even if the ring isn’t a UFD if that ideal is principal then it’s generated by the gcd but I’m not 100% sure]

upper cape
#

That follows from working over a Dedekind domain right?

next obsidian
#

Uhhhh

#

Probably?

hot lake
#

I think you should just look at elements of it of small norm and see if they generate the whole ideal

#

like

#

uh

#

2

orchid rover
#

Hello

#

Rep theory discord

gleaming bronze
viscid pewter
#

yes

gleaming bronze
#

any subgroup of containing must contain all the powers of by closure ,why this is true ?

#

i cant get

#

okay i think i get it ,but i had to think a lot i dont know why

chilly ocean
#

i think your math symbols are missing

#

if g is in a subgroup, then so is g^2 = gg, since the subgroup is closed under the group operation which i've written as multiplication. similarly for g^3 = ggg, etc

gleaming bronze
chilly ocean
#

no its just the word im using for it

#

i mean i wrote it like multiplication

#

im explaining my notation for the group operation, which could be anything

gleaming bronze
#

i don't understand why every subgroup containg a must containg all the power of a

chilly ocean
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i just explained it

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what part is unclear?

gleaming bronze
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the part that the subgroup is closed under the group operation

chilly ocean
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the group operation applied to any two things in the subgroup produces another element of that subgroup

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if a is in the subgroup

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then aa = a^2 is in the subgroup

gleaming bronze
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okay i get it

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by the definition of binary operation

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i think

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thats the case

chilly ocean
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by "closure"

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"subgroups are closed under the binary group operation"

gleaming bronze
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learning group theory, i am getting the same felling when i was learning basic category theory a long time ago

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i don't know ,its like thinking different all the math

chilly ocean
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category theory before group theory catThink

fossil shuttle
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a group is just a one-object category where all the morphisms are isos

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a group homomorphism is just a functor between groups

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what's the problem??

sturdy marsh
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prove the sylow theorems

gleaming bronze
gleaming bronze
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but i remember these words

fossil shuttle
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Category theorists when they're trying to construct a map into a sheafification

gleaming bronze
gleaming bronze
gleaming bronze
chilly ocean
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anything in the subgroup

dire summit
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i have a question that i believe can be answered by the g/z(g) theorem, but i'm not sure how to get to that step:

Let $p$ be a prime and let $G$ be a group of order $p^n$ for $n\geq1$. Suppose that $|Z(G)|\geq p^{n-1}$. Prove that $G$ is abelian.

cloud walrusBOT
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Snodlop

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dire summit
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some thoughts on the problem before i ping for help, just to help me see some things in writing

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if |Z(G)| is strictly greater than p^(n-1), then we would have to have Z(G)=G (by lagrange's theorem the order of each subgroup must divide the order of a group, and since |Z(G)| is larger than all of |G|'s other factors it must be p^n)

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so the case we really care about is |Z(G)|=p^(n-1)

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in this case, we'd have |G/Z(G)|=p, which would mean G/Z(G) is cyclic, which means G is abelian

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is this correct?

final pasture
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The part that deserves a bit more details is "G/Z(G) is cyclic so G is abelian" but this is true (and not particularly hard to prove)

dire summit
final pasture
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Any group of prime order is cyclic. To see that, pick any non-identity element x, what are the possible values o(x) can take? @dire summit

dire summit
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oh that's right duh

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thank you

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i've been stuck on this problem set for a while now lol, that was my last problem

tropic spade
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I'm working this exercise out of my book and I'm stuck.

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I worked up to 4 and I think I've gotten it figured out

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Except I'm not exactly sure why c can't be zero?

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It seems like 4 is just untrue if c is 0?

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(By "is a root field" I think the author just means "is a splitting field")

rustic crown
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You can vary b until you get a non-zero c

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That requires knowing linear independence of characters

tropic spade
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Ah dang I do not know that.

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I was thinking maybe since pi generates Gal(F_i:F_{i+1}) that "pulling out" all those powers of pi could let me use a group identity?

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There's that weird one where if ur group has an x that is its own inverse then the prod of all elts in the group is e or something?

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Idk if that helps either though

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Sorry it's if there is no such x

rustic crown
tropic spade
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I mainly did that by rearranging the sum.

next obsidian
tropic spade
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Maybe I used it implicitly.

rustic crown
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if you picked b = 0, then you'll definitely get c = 0

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so problem is with b, not with pi

tropic spade
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Yah true

next obsidian
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Eevee boothy

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Also I swear there’s an easier way to do this

rustic crown
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try to show that if G is a group and k is a field, then Hom_{Grp}(G, k*) is linearly independent over k.

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the proof is an exact copy of "eigenvectors with distinct eigenvalues are linearly independent"

tropic spade
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The notation Hom_{Grp}(G,k*) here refers to what?

rustic crown
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set of group homomorphisms G --> k*

tropic spade
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Ah I see.

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I'll give that a shot.

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Thanks!

sour plume
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If $A$ is a unital $\mathbb{K}$-algebra for some field $\mathbb{K}$, and $M$ is a free $A$-module, is then $M \otimes_{\mathbb{K}} M$ a free $A$-module?

cloud walrusBOT
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Lartomato

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sour plume
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Hmmm wait maybe i'm not thinking about this right

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I have a specific problem that I was trying to phrase in some generality but I think in this generality it's not even clear how A can act on M \otimes_{\mathbb{K}} M

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Rephrasing: Let $\mathfrak{g}$ be a Lie algebra over a field $\mathbb{K}$, and $U(\mathfrak{g})$ its universal enveloping algebra. Further, let $M$ be a free $U(\mathfrak{g})$-module. Is then $M \otimes_{\mathbb{K}} M$ a free module over $U(\mathfrak{g})$, with the action being induced by (i) $1 \cdot (m \otimes n) = m \otimes n$, (ii) $g \cdot (m \otimes n) := (g \cdot m) \otimes n + m \otimes (g \cdot n)$ for $g \in \mathfrak{g} \subset U(\mathfrak{g})$ and (iii) the Leibniz rule for higher tensors in $U(\mathfrak{g})$?

cloud walrusBOT
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Lartomato

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robust pollen
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$M \otimes_k M$ is isomorphic to $\dim_k M$ copies of $M$

cloud walrusBOT
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expectTheUnexpected

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robust pollen
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oh, you're doing infinite stuff

sour plume
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Yeah, and I think the fact that my action mixes the two copies also means I can't just view M otimes M as a big direct sum

lethal dune
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Is there a intuitive description of invariant factor decomposition?

simple valley
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every left R-module is an (R, Z)-bimodule right?

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by the virtue of being an abelian group

chilly ocean
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na just means a+...+a n times and analogous for negative n

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wikipedia confirms btw

robust pollen
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yes, that's just true. Abelian groups are the same as Z-modules. A ring is just an algebra in the category of abelian groups. Every left R-module is precisely a left R-module of the algebra R in Z-mod, so it in particular also comes with a right Z-module structure