groups-rings-fields

[hidden haven]

1 donut = 1 COFFIS CUP

[lethal cipher]

Question about exercise 2. So is that product supposed to be a sequence, like (12),(13),(14), up to (1n)?
Or the dots just some other transpositions?

[hidden haven]

there are commas in between

its a list

[lethal cipher]

So the former?

[hidden haven]

uh no so they are saying product of (transpositions of the form .....)

not products (of transpositions) of the form .....

or I don't understand your question

[lethal cipher]

Ohhh, so basically all elements in that product are of the form (1 k) for some k<=n

[hidden haven]

yes

[lethal cipher]

Hmmm

[hidden haven]

but they may be in any order and with repetitions

[lethal cipher]

Okay, so I could use some help seeing why.
For instance, what would (23) look like?

[hidden haven]

(12)(13)(12) I am guessing

ye

try to think of the first position as your hand and the remaining positions like the places on a bookshelf

this is like saying that you start with a book in hand and at each step you are exchanging the book in your hand with any book on the shelf

at the end you can get to any permutation on the shelf and any book in your hand

😌

[lethal cipher]

Okay, so I think I see how to do this. First break an m-cycle (i_1...i_m) into transpositions of the form (i_1 i_m)...(i_1 i_2). Then we can decompose those transpositions into (1 i_1)(1 i_k)(1 i_1) for each k

And that should do the trick

[hidden haven]

yep, nice

[lethal cipher]

Thanks for the help Moldilocks!

Also, I am pretty sure we can simplify this further by showing any transposition can be written in that form

[hidden haven]

isn't that what you asked earlier?

(ij) would be done in pretty much the same way as (23)

[lethal cipher]

Precisely. And that would be enough for the proof since all permutations are products of transpositions

[hidden haven]

ah I see what you are simplifying now

neat

[lethal cipher]

So I'm struggling a bit on #3, part 1.
I'm just struggling to figure out how to even begin to go about it

If sigma and the m-cycle are disjoint, it's easy. The issue is when they are not

[lethal dune]

suppose the m-cycle be π

then if π(i)=j then

$\sigma \pi \sigma^{-1}(\sigma(i)) = \sigma(j)$

[cloud walrus]

Ryuzaki

[lethal cipher]

But, does that deal with the issue. What if j is not any of the i's?

[lethal dune]

that means π(i)=i

doesn't mean π(i) is undefined

[wooden ember]

when $\pi(i)=i$ then conjugation gives $\sigma\pi\sigma^{-1}(\sigma(i))=\sigma(i)$

[lethal cipher]

Why does that mean pi(i)=I? I'm a bit confused

[wooden ember]

either pi moves elements in its cycle representation or it fixes them

lemme edit what i wrote lmao

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

It doesnt matter if pi o sigma are disjoint: we take an element i in {1,2,...,n} and consider pi's action on it. Then conjugation gives us an action of sigma o pi (i) on sigma(i) : we dont care if sigma and pi are disjoint or not

If you will, you can split ${1,...,n}$ into the image of $\sigma$'s restriction on ${i_1,...,i_m}$ and its complement. For an element $a_j$ in the image we can write $a_j = \sigma(i_j)$. Then applying the conjugation we send $a_j$ back to $i_j$, increment it to $i_{j+1}$ and then send this to $a_{j+1}$ through $\sigma$, so that $\sigma(i_j)$ got sent to $\sigma(i_{j+1})$. For an element not in the image, the inverse $\sigma^{-1}$ sends it to an element not in ${i_1,...,i_m}$ so that $\pi$ does nothing, and then applying $\sigma$ again changes nothing to the input.

[cloud walrus]

𝓛ittle ℕarwhal ✓

[lethal cipher]

So sigma(i) is fixed, got it.

[wooden ember]

(the increment is mod m but you get the gist of it)

[lethal cipher]

Yea, I see where this is going. Thanks for the help. I wasn't thinking to use the function sigma for the argument.

The other part is really easy, I won't need much help with that

[chilly ocean]

Maybe I'm stupid but

Prove or disprove: $(\mathbb{Z}_8,+,\times) \cong (\mathbb{Z}_8, \oplus, \otimes)$ as rings where $a \oplus b = a + b + 5, \ a \otimes b = 7ab + 3(a+b) + 2$

[cloud walrus]

識繒 (Shi Zeng)

[chilly ocean]

How do I solve this?

I'm having trouble with the \otimes operator

[mild laurel]

I haven't done the calculations, but one way would be to figure out how the additive groups are isomorphic, and then check if this extends to a ring isomorphism

[chilly ocean]

How do I do that?

I tried and got nowhere

[wooden ember]

okay so i think this isnt true but i may be spewing total bullshit

[mild laurel]

I mean, you know that 1 is a generator for the group on the left, so figure out what the generators are for the group on the right and map 1 to those generators

[wooden ember]

so you can check that the additive identity in the second group is 3 while the multiplicative identity is 2. Whatever $4\in (\mathbb{Z}_8,+,\times)$ is supposed to be mapped to (say $x$) it must satisfy $x\oplus x = 3$ which forces $x=7$. Then whatever 2 gets mapped to (say $y$) it must satisfy $y\oplus y=7$ which forces either $y=1$ or $y=5$. Both of these can be shown to not satisfy $y\otimes y\otimes y = 2$, which it should

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

unless ive made calculation mistakes i believe this works, or im having a complete misunderstanding

[chilly ocean]

Yeah this should work

I got the same values

[wooden ember]

just check the intermediate calculations cause i could have made a mistake somewhere

even if it works this is ugly

but i guess so is the question

[lethal cipher]

Okay, so I am close. My only issue is case 3. I am not sure what that is telling me.

[wooden ember]

i wouldnt proceed in this way personally

and i must admit im not sure how you worked though case 2

[lethal cipher]

If it's Z(S_n) then both equations (ij)B and B(ij) must give the same outputs

[wooden ember]

sure but B(i)=i doesnt guarantee B(j)=j

[lethal cipher]

The second case for ((ij)B) was just if B(i)=I.
But (B(ij))(i) is always B(j)

[wooden ember]

i guess where im getting at is im not sure i see what you're tying to achieve

are you trying to reach a contradiction

or force conditions on B?

[lethal cipher]

We need to end up showing case 1 and case 3 are impossible.

[wooden ember]

okay nvm im stupid

yeah sorry

[lethal cipher]

Case 1 is easy, since I doesn't equal j

[wooden ember]

||so for case 3 suppose B(j)=i, B(i)=j, and now consider the transposition (i j'). Then letting B conjugate on this you get (B(i) B(j')) = (j B(j')) =/= (i j') so B is not in the center||

this is why the n>= 3 is important

[lethal cipher]

What do you mean by conjugate?

[wooden ember]

B(i j)B^-1

B lies in the center iff B(i j)B^-1 = (i j) for all transpositions

[lethal cipher]

Oh, so we are doing that. Ah okay, this makes sense. BTW j'is not j right?

[wooden ember]

exactly

we pick it to be different from j and i

hence why n>=3 is necessary

[lethal cipher]

Ah, the only way they would be the same is if B(j')=I, which it doesn't

[wooden ember]

no even then youd get B(i j')B^-1 = (i j)

which is not the same as (i j')

[wooden ember]

or maybe i misunderstood what you meant here

[lethal cipher]

If B(j')=I, then (j,B(j'))=(i,j)

So we do need j' to be distinct (like you mentioned)

[wooden ember]

oh okay in that sense yes

[lethal cipher]

So wait, why does this mean it can't be in the center?

Oh duh!

[wooden ember]

because you found a transposition with which it doesnt commute if you assume case 2 is true, hence case 2 is false and so it doesnt commute with our first transposition

[lethal cipher]

(ij)B^-1 B= ij

I see now

[wooden ember]

yeah

[lethal cipher]

So the only possible case ends up being case 2, which is precisely the identity (since this needs to be true for every I and j)

[wooden ember]

a condensed argument would have started with conjugation straight away: then you just say B(i j)B^-1 = (B(i), B(j)) = (i j) iff B(i)=i, B(j)=j or B(i)=j, B(j)=i. If the former is true for all transpositions B is trivial. If not then the latter is true for some (i j) but then B(i j')B^-1 = (j B(j')) =/= (i j')

it's the same though

[lethal cipher]

Perfect. Thank you so much for the help

[wooden ember]

it's quite useful to think of centralizers and centers in terms of conjugation

since conjugation gives inner automorphisms you shift the perspective of your problem

[lethal cipher]

I see that now. That's a useful thing to keep in mind?

What is an automorphism?

[wooden ember]

an isomorphism from G to itself

so like a permutation but on a group: it also preserves group structure

dont worry about it if you havent heard of them yet though

dont want to confuse you or anything youll see it eventually

[lethal cipher]

Ah, that makes sense

[wooden ember]

bruh my r key is broken it only works with a probability of 1/2

it's annoying

[hidden haven]

buh

[lethal cipher]

So that's why you kept saying o instead of or :p

[wooden ember]

yeah haha

[wooden ember]

why do we change things up when D is congruent to 1 mod 4?

like sure with (1+sqrt(D))/2 we still get a subring but why did we do this in the first place?

the field norm can also be defined appropriately without having this so i dont understand where a complication might arise

[celest mantle]

you're asking how checking the congruence of D mod 4 is useful to know the ring of integers of a quadratic extension of Q ?

[wooden ember]

im asking why it's essential to define our extension differently when D is congruent to 1 mod 4

[celest mantle]

when you're trying to determine the ring of integers of a quadratic extension, you get to a part where you have to check when u² - dv² is in 4Z for u and v both integers, where u = 2a and v = 2b, where a and b are rationals such that x = a + bsqrt(d) is an element of the ring of integers
so you check that if d = 2 or 3 mod 4 then u² = 2v² or 3v² mod 4 and you get easily that u = v = 0 mod 4 so u = 2a and v = 2b gives you the thing you want, a and b are integers and we're happy
but if d = 1 mod 4 then its slightly different, you have to check for parity (when v is even there's no problem, but if he's odd then u² = d mod 4 and as d is squarefree you get that u² = 1 mod 4 and d = 1 mod 4)

[wooden ember]

is there a deeper reason for why this is the case?

it's a little unsatisfying to hear that it just turns out to be annoying when you works things out formally

surely there's a more fundamental reason to such a special case

[mild laurel]

The ring of integers of a extension of Q is supposed to mimic Z inside Q

[next obsidian]

When you’re congruent to 1 mod 4 you get extra elements that are integral over Z

[mild laurel]

You can define the ring of integers as the set of elements that satisfy a monic polynomial with integer coefficients, and then prove that the ring of integers of Q(sqrt(D)) take this form

[wooden ember]

not sure what you mean

[next obsidian]

The idea is that (1 + sqrt(D))/2 now satisfies the relation like x^2 + x + k

An integral element satisfies a monic polynomial over the base ring

Or something like that uh

I can’t figure out which polynomial it satisfies uhhh

[urban acorn]

i think it's quadratic

so just square it

do the elementary algebra

[next obsidian]

Okay

[urban acorn]

and try to express that as a linear combination of the previous powers

[next obsidian]

It’s x^2 - x - (D-1)/4

[urban acorn]

yeah, I remember something like that

[next obsidian]

The point is the ring of integers is supposed to the be the stuff in Q(sqrt(D)) which satisfies a monic polynomial over Z

This is almost like saying “take everything algebraic over Z”

[wooden ember]

so why do solutions to a quadratic monic polynomial in $\mathbb{Q}[\sqrt{D}]$ act like integers for $\mathbb{Q}$??

[cloud walrus]

𝓛ittle ℕarwhal ✓

[next obsidian]

It’s the integral closure of Z in that extension

This is a number theory thing

You’d have to ask like Zoph as to why they are useful but these are a central object of study in ANT

[mild laurel]

Since Z(sqrt(D)) is contained in Z(1 + sqrt(D)/2), you get into the theory of things called orders, and the ring of integers is the maximal order. Non-maximal orders you get bad properties that we don't want in algebraic number theory. Things like the set of (fractional) ideals isn't a group anymore and so for this reason we usually care about maximal orders

I mean, you want solution to monic polynomials in general, its just that Q(sqrt(D)) is only degree 2 so you only need to look at quadratics

[wooden ember]

i feel like DF made a mistake introducing this without context in the section "Basic definitions and examples" of the part on rings

[wooden ember]

wdym

you mean since we extend with a sqrt these can only be solutions to quadratics?

[urban acorn]

can you ELI5 (5 years undergraduate) what's an order?

[mild laurel]

basically yeah

[wooden ember]

but can Q(sqrt(D)) be explicitly constructed as solutions to certain quadratics?

like the subset of solutions with rational coefficients or smth

wait nvm every solution to a quadratic lies in some Q(sqrt(D)) right

[mild laurel]

Uh, there are a lot of equivalent definitions. Maybe the easiest is that its a subring R of the ring of integers O_K such that the quotient O_K/R is finite

The easiest examples are things that this question is pointing to

[urban acorn]

can you give me a definition not referring to the ring of integers?

cause you kind of used it to motivate the ring of integers

[mild laurel]

Z[1+sqrt(5)/2] is the maximal order aka ring of integers in a number field, whereas Z[sqrt(5)] is a non-maximal order

[mild laurel]

Yes this is true. I mean, you can define Q(sqrt(D)) as being isomorphic to Q[x]/(x^2 - D) so they're kinda solutions to certain quadratics

[mild laurel]

An order of a number field K is a subring of K that is a finitely generated Z-module (abelian group) such that you can pick generators for the order that are also a basis for K over Q

[wooden ember]

i think i just need to keep going or ill get confused lol

[mild laurel]

I mean I don't think you're really supposed to understand why these things are the way they are at this point. They're just nice examples of rings

[wooden ember]

yeah

itll probably come back in more detail later

[urban acorn]

btw since from context I'm picking up that you're starting to read about rings in dummit and foote

here's a nice motivation for not-necessarily-commutative rings with identity:

they're precisely the things that act on abelian groups

[wooden ember]

im listening

[urban acorn]

kind of like how groups act on "things", rings act on abelian things

meaning you can add two actions by adding them pointwise using the abelian structure

[wooden ember]

isnt a ring action like a module or smth

[urban acorn]

precisely

[wooden ember]

and a field action is a vector space right

[urban acorn]

yep

fields have a way to force things to be nice

[wooden ember]

so why does a ring acting on an abelian group bring about non-commutativity

[urban acorn]

thing if the ring of linear operators on a vector space

a vector space under addition is an abelian group

so something that acts on a vector space is in particular a ring

[wooden ember]

right

[urban acorn]

but obviously, since "a * b" is interpreted as "b then a", it may depend on the order

[wooden ember]

fair enough

[urban acorn]

that's not different from when considering arbitrary groups

by the way, there's one definition which - when considered over Set produces monoids, and when considered over Ab produces rings (unital, non-commutative)

[wooden ember]

what extra structure do we get by considering the set of linear operators on a vector space as a ring instead of a group

[urban acorn]

I mean, addition

we lose something too, that's invertibility

[wooden ember]

i never really understood the use for adding stuff in GL_n(F)

isnt the main point of transformations to compose them?

what do we get by adding them

[urban acorn]

GL_n(F) isn't an object I typically think about, so I don't have an example of it being considered as a ring being useful right now

but I'm sure someone here does

commutative rings are way nicer, and they're highly relevant to algebraic geometry and algebraic number theory

[mild laurel]

I mean, adding functions pointwise is pretty important too

[wooden ember]

that's fair

[urban acorn]

@wooden ember by the way, here's a neat short proof that every element of C satisfies a degree 2 polynomial over R:

{1, z, z^2} is a 3 element set over a 2d real vector space, and thus must be linearly dependent

[rustic crown]

wait, what is happening... how do define addition? whats the sum of T and -T?

[urban acorn]

you can't

we mean

add singular matrices to the mix

[rustic crown]

Mn(F)?

[urban acorn]

yes

GL_n(F) was not the correct notation

I noticed that after the fact

[wooden ember]

i have no clue what you are trying to show or what the proof is saying

[wooden ember]

oh yeah oop we do lose invertibility

[urban acorn]

formally the statement being shown is that given any complex number z, there are real a, b, c, not all zero, such that az^2 + bz + c = 0

[wooden ember]

ah no i see

the 2d vector space in this case is C?

[urban acorn]

yes

[wooden ember]

right so since its 2 dimensional we must be able to write z^2 = az + b

[urban acorn]

yep

[wooden ember]

so z satisfies that polynomial

[urban acorn]

yep

[wooden ember]

but arent a and b in C?

[urban acorn]

no

because we're considering C as a real vector space

[wooden ember]

oh nvm we identify C with R^2

[urban acorn]

so the coefficients in our linear combinations are from R

[wooden ember]

yeah

i got confused with a vector space over C

[urban acorn]

yeah

[wooden ember]

that's pretty neat yeah 👌

[mild laurel]

This is connected to the idea that everything in Q(sqrt(D)) satisfies a quadratic polynomial over Q

I mean, the proof is the exact same, you just need to figure out why Q(sqrt(D)) has dimension 2 as a Q vector space

[wooden ember]

i see

well i mean its easy to see every element can be uniquely written as a+bsqrt(D) so cant we just identify every element with (a,b) so that we identify Q(sqrt(D)) with Q^2?

[urban acorn]

yes

[mild laurel]

no

[wooden ember]

lmao

[mild laurel]

I mean, as groups, Q(sqrt(D)) is equal to Q^2, but not as rings

[urban acorn]

as Q-linear spaces

which is what we care about

because we're looking at the dimension over Q

[mild laurel]

ok sure

[urban acorn]

so this is valid

[wooden ember]

yeah i do see they arent equivalent as rings

i meant vector spaces

[urban acorn]

I guess what you had in mind though had to do with the construction as Q[x]/(x^2 - d)

[mild laurel]

how does Dummit and Foote define Q(sqrt(D))?

[wooden ember]

simply as {a+bsqrt(D)}

[urban acorn]

^

[wooden ember]

keep in mind this is just the first chapter on rings

[urban acorn]

yeah, and showing that your definition of Q[sqrt(d)] is equivalent to the one in the introductory examples in dummit and foote is really what "seeing why it's 2 dimensional" is about

[mild laurel]

oh okay sure

Usually, in field theory, we define Q(sqrt(D)) as the smallest field containing Q and sqrt(D)

since this makes it easier to define things like Q(sqrt(D), cube root of(N))

[wooden ember]

right

[urban acorn]

@wooden ember by the way, for any field F with Q as a subfield of F, and x in F such that x^2 - D = 0, you can uniquely embed Q[sqrt(d)] into F fixing Q and sending sqrt(d) to x

so it's "the smallest field containing Q and sqrt(D)" not just in the sense of subfields of C or something, but in a more universal sense

[wooden ember]

i suppose that makes sense

[urban acorn]

"universal" is kind of a pun here

[wooden ember]

is there actually a universal property at play here?

uniquely makes it sound like there is

wait dont answer

[urban acorn]

hint: a(n identity-preserving!) ring morphism F -> R with F a field is the same thing as an embedding of F in R

[wooden ember]

we have a canonical map from Q[x] to Q[x]/(x^2-D) and one from Q[x] to F by sending x in Q[x] to x in F and extending it to a morphism: this one has kernel x^2-D so it factors through to an injective map?

i must admit im a little confused at the quotient by x^2-D: it's an element in Q[x] so i dont understand how we can quotient by it

are we taking a quotient by the set of all its multiples?

[urban acorn]

yeah, this looks right

[urban acorn]

yes

the way you would say this is we're taking a quotient by (x^2 - D), the ideal generated by x^2 - D

[wooden ember]

right that makes more sense

ah okay

[urban acorn]

for a commutative ring, the ideal generated by one element is precisely the set of multiples of that element

[wooden ember]

right yeah i remember now: ideals are the analogue of normal subgroups in rings right?

[urban acorn]

yes

[wooden ember]

i see that makes more sense

[urban acorn]

ideals are precisely the objects you quotient a ring by

[wooden ember]

so the kernel of my map Q[x] -> F is (x^2-D) yeah

and now i see that it still factors through

[urban acorn]

they're the preimages of {0} under ring homomorphisms, and they behave "like 0"

[wooden ember]

we dont call them kernels?

"preimage of {0}" seems wordy

[urban acorn]

oh, yes

kernels

I just wanted to be clear that we're looking at the additive identity

[wooden ember]

right yeah

but we still have ideals in rng's right?

[urban acorn]

cause before I came across rings it wouldn't have been obvious to me a priori that we'd look at the additive rather than multiplicative identity

[urban acorn]

yes

but I don't have much of an intuition for rngs

so don't trust what I say about rngs

[wooden ember]

alright

i just know they cant be completely random

[urban acorn]

anyway, the analogy between a normal subgroup and an ideal also holds in that there's a description of ideals "from the inside", not in terms of maps

[wooden ember]

right

[urban acorn]

like how normal subgroups are defined to be subgroups fixed under conjugation, where it is not immediately obvious why that would be what kernels are

[wooden ember]

lemme try and see if i can figure out what it would be here

[urban acorn]

but the one place where the analogy is misleading is that ideals are not in general - and in fact almost never are - subrings

[wooden ember]

oh damn

[urban acorn]

you want subrings to contain your multiplicative identity, and there are no interesting ideals that do that

there's one ideal that does that, and that's the entire ring

[wooden ember]

i see

[proud bear]

multiplicative identity?

[urban acorn]

yes my bad

mistyped

[wooden ember]

wait so ideals are smaller in a sense than normal subgroups right? Cause if i take x that gets sent to the identity in a group homomorphism then <x> isnt necessarily its normal closure but for ideals <x> is precisely the kernel

[urban acorn]

what would <x> be?

[wooden ember]

in the first case the cyclic subgroup generated by x, in the second the ideal generated by x

those concepts are somewhat analogue no?

[urban acorn]

yes, but it's somewhat misleading

[wooden ember]

oh wait yeah no

[urban acorn]

because you might as well define "(x) in a group" to be the normal closure

the same way you define "<x> in a ring" to be the ideal generated by x

[wooden ember]

by "ideals is all multiples of x" you mean multiples by elements in R or the cyclic additive group generated by x?

[urban acorn]

the former

[wooden ember]

yeah okay i was mislead

[urban acorn]

yeah

my bad

[wooden ember]

so xR is not R in general 🤔

[urban acorn]

you can maybe even think of multiples by elements of R in this particular context as loosely analogous to cojugation in groups

[wooden ember]

which is very different to groups

[urban acorn]

yes, because rings are not invertible

for once, 0 can't ever be invertible unless it's equal to 1 and is the only element in your ring

and then if you require everything else to be invertible you get a field, which is an extremely special case of a ring

but, for example, in the ring Z, the ideal generated by 2 consists of all the even numbers

[wooden ember]

by invertible you mean is a unit?

[urban acorn]

yes

[wooden ember]

ah okay yeah i see

if x isnt a unit xR is clearly not R

but if x is a unit?

[urban acorn]

if x is a unit than (x) is all of R

[wooden ember]

ahhh

[urban acorn]

because you can write every element as a multiple of a unit

z is (zx^-1)x

[wooden ember]

and there's the difference between xR and Rx right? left and right ideals?

heard that before

[urban acorn]

I was talking in the context of commutative rings so far

[wooden ember]

fair

[urban acorn]

I'm not used to thinking about ideals in non-commutative rings

[wooden ember]

are some of the things we've said so far invalid in the non-commutative case?

[urban acorn]

yeah, I think so

like (x) isn't Rx in a non-commutative ring

[wooden ember]

i thought (x) was ambiguous for non-commutative rings

[urban acorn]

no, (x) is unambiguous

[wooden ember]

there's still a notion of an ideal that's not left or right?

[urban acorn]

so in non-commutative rings, some sets may be called left-ideals, and some sets may be called right-ideals

and then "ideals" so to speak, those sets that are the kernel of some ring homomorphism, are precisely those sets that are both left and right ideals

[wooden ember]

ah i see

but then why is (x) not Rx if x lies in an ideal?

[urban acorn]

for example, some elements of xR might not be in Rx but are definitely in (x)

okay, assume (x) = ker(phi), phi some ring homomorphism

phi(x) = 0 since x in (x)

so phi(xr) = phi(x)phi(r) = 0phi(r) = 0

[wooden ember]

oh i see

[urban acorn]

and
phi(rx) = phi(r)phi(x) = phi(r)0 = 0

so ideals always swallow under multiplication

they're not just closed under multiplication - meaning if both elements are in the ideal, the product is in it too

[wooden ember]

they absorb

[urban acorn]

they swallow under multiplication - meaning if either element is in the ideal, the product is in it too

[wooden ember]

that's the only intuition i ahd about ideals haha

[urban acorn]

and also they're additive groups

[wooden ember]

https://tenor.com/view/spongebob-squarepants-sponge-absorb-cry-gif-10458976

[urban acorn]

because obviously phi is also an additive group morphism so its kernel is also the kernel of a group homomorphism

hence a subgroup

[wooden ember]

so Rx lies in the ideal and xR lies in the ideal but (x)=/=Rx=/=xR

[urban acorn]

yes

[wooden ember]

but then how is (x) a left and right ideal

[urban acorn]

so, in the commutative case, the following are enough to classify ideals:
a set S is an ideal if:
- S is an additive group
- for every x in S, r in R, rx is in S (that is, S swallows under multiplication)

[wooden ember]

i see

[urban acorn]

and I think in the case of a non-commutative ring you just decide from which direction to multiply by r in the definition in order to define left or right ideals

and it's obvious that (x) - being an actual ideal and the kernel of a homomorphism - satisfies both properties

[urban acorn]

by this argument from earlier

[wooden ember]

yeah

[urban acorn]

ideals are sets that behave like zero, like how normal subgroups are sets that behave like zero

[wooden ember]

so if (x) is both a. left and right ideal can we say there exists y in (x) such that (x) = yR = Ry?

[wooden ember]

oooh

that actually makes a lot of sense

i never saw the act of conjugation like that

[urban acorn]

I believe we can't, but I don't know enough about non-commutative rings

[wooden ember]

normal subgroups "absorb" conjugation

that's a really nice thought

[urban acorn]

like, i'm not even 100% sure if I didn't miss a technical requirement here

[wooden ember]

that's fine

this is stuff ill see anyways but its nice to get another perspective and an intuitive head start

[urban acorn]

yes, like how the identity absorbs conjugation in groups, and 0 absorbs multiplication in rings

but let me connect this to something from earlier:

remember how R-modules are "ring actions" of R on abelian groups?

well, R is an abelian group under addition

and so R acts on itself by multiplication

so R is an R-module

[wooden ember]

right that makes sense

[urban acorn]

so then this is an action of R on itself that ideals absorb

just like how conjugation is an action of G on itself that normal subgroups absorb

[wooden ember]

so does multiplication by an element of R give Inner automorphisms of R?

[urban acorn]

no

multiplication by an element of R is not an automorphism of R

[wooden ember]

sad

[urban acorn]

the analogy doesn't extend that far

[urban acorn]

except multiplication by 1

[wooden ember]

i was thinking with all this in mind, is there some sort of concept of a set being quotiented by a normal subset under some action, where namely this is a subset fixed during an action of G on our set?

cause here normal subgroups and ideals arise from actions of groups and rings on themselves

this is probably stupid but im just wondering if there's some extension

[urban acorn]

there are "quotients of sets", but there are no "normal subsets" generating these quotients

quotients of sets are equivalence relations

[wooden ember]

yeah i do know that

i jsut thought there might be some way to make a set interact with subsets fixed under a G-action

i suppose the main problem is the set doesnt have some notion of an identity

[urban acorn]

any group homomorphism phi : G -> H gives a normal subgroup N of G such that phi is just an embedding of G/N in H

any ring homomorphism phi : R -> S gives an ideal I of R such that phi is just an embedding of R/I in S

any set map f : A -> B gives an equivalence relation ~ on A such that f is just an embedding of A/~ in B

[urban acorn]

yeah, something like that

[wooden ember]

the equivalence relation here is just a ~ b iff a and b are in the same fiber right?

[urban acorn]

yes

[wooden ember]

makes sense

[urban acorn]

i.e. if f(a) = f(b)

[wooden ember]

does this extend categorically then?

[urban acorn]

I don't know

but that's a good question

there are also quotients on topological spaces, given by an equivalence relation on the underlying set

there are module quotients

[wooden ember]

apparently you can take quotient categories but that's not quite the same thing haha

[urban acorn]

(and so in particular vector space quotients)

[wooden ember]

https://mathoverflow.net/questions/60241/why-dont-ideals-and-quotients-work-well-for-categories

dont understand what's going on but it seems there is such a generalisation

through "internal categories" and "coequalisers"

anyways thank you very much for this discussion @urban acorn it gives me a good intuitive head start

[urban acorn]

I enjoyed this discussion very much as well

[potent briar]

guys i think i have a good one:

If $(\alpha_0, \alpha_1, \alpha_2, ...)$ is an arbitrary sequence of complex numbers, and if $x$ is an element of P (polynomials), $x(t)=\sum_{i=0}^n\xi_it^i$ write $y(x) = \sum_{i=0}^{n}\xi_i\alpha_i$.

Prove that y is an element of P' (dual space of P), and that every element of P' can be obtanied in this manner by a suitable choice of $\alpha s$.

[wooden ember]

put your dollar signs around the math part

[potent briar]

yea

[urban acorn]

what is ${\xi}_i$ here?

[cloud walrus]

Ab

vik

[urban acorn]

are P polynomials in complex coefficients considered as a vector space over C?

oh okay xi_i are the coefficients of the polynomial x

[potent briar]

the \xis are the coefficients yeah

[urban acorn]

the notation confused me

[potent briar]

the book is from the 40s

[urban acorn]

well an element of P' is the same thing as a linear map P -> C which is the same thing as a choice of f(x^n) for every n

and alpha_n here is just where x_n goes

[urban acorn]

because the x^n form a basis

it's all pretty straightforward

[urban acorn]

and wow where did you get a math textbook from the 40s?

[potent briar]

i mean the print is new

it was written in the 40s

i think it's a fairly well known linear algebra book

[urban acorn]

author name?

[potent briar]

finite dimensional vector spaces by halmos

[urban acorn]

yeah, that sounds a bit familiar

[potent briar]

ok so i didn't understand

[urban acorn]

I think it was like an OG canonical treatment of linear algebra

[potent briar]

that's the content order

[urban acorn]

you need to show the following:
1. y(a + b) = y(a) + y(b)
2. y(l * a) = l * y(a)

this shows that y is a linear transformation P -> C

i.e. an element of P' by definition

[potent briar]

yeah i already proved that y(x) is in the dual space

idk what to do next

[urban acorn]

okay, so now you want to prove that every element of the dual space is in that form?

well, the powers of x are a basis for P

so any element of P' is completely determined by where it sends those

and whatever choice of where to take those, it immediately translates to your choice of the alpha_i's

[potent briar]

can i prove that there is a 1:1 correspondace

between the polynomials

and the "suitable choice of alphas"

[urban acorn]

I don't think that's gonna help you here

[urban acorn]

this book looks more like a reference than a textbook suitable for learning

and it doesn't seem to be in sync with the modern focus on linear maps as opposed to vector spaces themselves

[potent briar]

i actually find the prose quite witty and fun to read

[urban acorn]

well, that's nice

[potent briar]

it is from the 40s after all

[potent briar]

so its infinite dimnesional?

[urban acorn]

yes

[potent briar]

i dont understand this

if the dual space is the set of all linear functionals, there are presumably an infinite amount of these?

[urban acorn]

it's a basic fact of linear algebra: Given a basis A for V, a linear transformation from V to W is freely determined by where it sends A

[urban acorn]

yes, there are

[potent briar]

argh

i haven't seen transformations yet!

in previous questions everyone's like just take the determinant lol

but i can't ahaha

[urban acorn]

from a modern perspective it's unthinkable to learn about dual spaces before seeing transformations

[potent briar]

take it up with halmos hehe

[urban acorn]

no, I don't blame halmos, because he wrote the book in the 1940s

[potent briar]

wait ooh

so it's saying that ever element in P'

is of the form y(x)

thats nuts how do i prove that

[potent briar]

the preface does say "Frequently exercises are stated as soon as the statement makes sense, quite a bit before the machinery for a quick solution has been developed. A student who tries to solve such a "misplaced" exercise is likely to appreciate and to understand the subsequent development much better for his attempt."

[urban acorn]

there's a seemingly simple problem that for whatever reason I can't solve, it's algebraic geometry but I feel like it fits here much more than in #point-set-topology:

Consider k[x1, ..., xn] as functions k^n -> k. Prove that the ideal of functions that are zero at the point (a1, ..., an) in k^n is precisely the ideal (x1 - a1, ..., xn - an).

[small bison]

Isn’t this like nullstellenstaz or something

[urban acorn]

yes, but I can't use nullstellensatz for this, because I need it for the way there

[small bison]

Sadge

I remember something about using the evaluation map also

[urban acorn]

I think I figured it out.

For every polynomial p, I think I can show p has the same I-residue as a constant by induction on the degree of p.

and then I think that would prove k[x1, ..., xn]/(x1 - a1, ..., xn - an) is isomorphic to k so (x1 - a1, ..., xn - an) is maximal

which takes care of the hard-to-prove direction

[urban acorn]

clearly it's enough to show that a homogenous polynomial of positive degree n has the same I-residue as a polynomial of degree at most n-1

[urban acorn]

and then clearly it's enough to show that a monomial of positive degree n has the same I-residue as a polynomial of degree at most n-1

so the monomial is $p = c\prod_{i=1}^n {x_i^{m_i}}$

[cloud walrus]

Ab

[urban acorn]

and $j \in {1, ..., n}$ is such that $m_i \neq 0$ since this monomial is non-constant

[cloud walrus]

Ab

[urban acorn]

then consider $q = p - c(x_j - a_j)\prod_{i \neq j}{x_i^{m_i}}$

q is the difference between p and an element of I, and thus has the same I-residue

and it's of degree n-1

so, I have 2 questions:

1. Can someone verify the proof I came up with is correct?
2. Is there a more enlightening proof of this particular fact? This kind of just seems like bruteforce.

[cloud walrus]

Ab

[urban acorn]

OH, OKAY, I don't know how I missed this

I found an intuitive explanation for why this is true

adding xi - ai to the ideal allows us to replace xi with ai everywhere, so we can replace everything with constants, and show every polynomial is (equivalent to, up to the ideal) a constant

[cloud walrus]

dackid

[lethal cipher]

I got what I need. Ty

[lethal cipher]

So I need to figure out how many elements of order 4 and 2 are in S_6.
I am not too sure how to tackle this

[final pasture]

Being a vs automorphism is stricly stronger than being a group automorphism, so yes

[coarse stag]

is f(x) := det(x) a valid homomorphism from the general linear group of nxn matrices under matrix multiplication to the group of non zero real numbers under multiplication?

[gritty sparrow]

yes

[light sluice]

Nice

[lyric cape]

ive got no idea where to put this but i got an idea that id like to hammer out that has to do with the lucas numbers pascals triangle and f(x,y,n) = x^n + y^n
if anyone would like to discuss this new idea in mathematics voice, @ me
I'd love to iron it out with a second perspective

[azure plinth]

Hey so can someone help me with this?

Let $A$ be a ring, $M$ be an $A$ module, $I\subset A$ be an ideal, and let $N,N'$ be two submodules of $M$ such that $M=N+IN'$. If $N'$ is finitely generated, then show that there is $f\in A, f\equiv 1(\mod I)$ such that $f\cdot M\subseteq N$ and $M_f=N_f$

[cloud walrus]

AidenM27

[azure plinth]

And what role does N' being finitely generated play here?

[wooden ember]

Not sure how to prove the final converse here: id want to say that any subring is of the form Z[fw] for some f and then the conclusion follows easily but im not sure how to show that

Here w is sqrt(D) when D = 2,3 mod 4 and (1+sqrt(D))/2 when D= 1 mod 4

id imagine the idea is the same as showing the only subgroups of Z are nZ but i dont remember how that goes either

oh i think i might know what to do

[woven obsidian]

I'm having some trouble coming up with an equivalence

V_1 is the sign representation and V_2 is the standard representation of S_3

The most natural map is $$\lambda \otimes v_2 \mapsto \lambda v_2$$ but it is not intertwining

[cloud walrus]

AoiKunie

[gritty sparrow]

Hint:|| Nakayama' lemma (that should answer your question about why finitely generated is relevant||

[azure plinth]

Oh yes indeed, I used a related result to get that, or rather got that with the aid of some classmates. Thanks a bunch though!

[stoic rose]

How is it not intertwining?

If V1 is the trivial representation then your map seems perfectly fine to me

[woven obsidian]

V1 is the sign representation

If we call it $$\pi$$ it acts by $$\pi(\sigma)\lambda = \text{sgn}(\sigma)\lambda, \lambda \in \mathbb{C}$$

[cloud walrus]

AoiKunie

[woven obsidian]

So the map doesn't really work for 2-cycles

[wooden ember]

if I want to show $2+\sqrt{3}$ is a unit of infinite order in $\mathbb{Z}(\sqrt{3})$ can I just argue that the euclidian norm of $2+\sqrt{3}$ isn't 1?

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

or can I not use such a norm since we're working in $\mathbb{Z}(\sqrt{3})$ and not in $\mathbb{R}$

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

would i have to first give an explicit embedding, then construct the euclidian norm, and then show it's multiplicative?

or do you think its reasonable to skip those steps

[wooden ember]

or i could just generalize this since it works for all D>0 and then we automatically get that the gorup of units has infinite order in the ring of integers of $\mathbb{Q}(\sqrt{D})$

[cloud walrus]

𝓛ittle ℕarwhal ✓

[hidden haven]

Depends on definition of Z(√3)

Is it defined as a quotient or as a subfield of R or C

[wooden ember]

defined as {a+b\sqrt(3)}

[hidden haven]

Are you sure it's Z(√3) and not Z[√3]

And yeah that looks like subfield

Or subring in this case

Because you're doing this addition in C

Unless these are defined to be formal expressions with the obvious relations

[wooden ember]

oh yeah we write it $\mathbb{Z}[\sqrt{3}]$ but yesterday.i saw that people answered by notating with () and used [] notation for polynomial rings so I assumed that DF was using deprecated notation

[cloud walrus]

𝓛ittle ℕarwhal ✓

[hidden haven]

Round brackets mean subfield generated by whatever, square brackets are for subring

They also sometimes denote polynomial rings modulo some relations on the new variables

[wooden ember]

i see

[stoic rose]

And what is V2, is it the irreducible 2-dimensiobal representation?

[woven obsidian]

Yes

[stoic rose]

This comes down to writing a linear transformation $\phi:V_2\to V_2 $ such that $\phi(\sigma v) = \text{sgn}(\sigma) \sigma \phi(v)$

[cloud walrus]

radiateur-man

[stoic rose]

I'd say try to write down explicitely this condition for the matrix of phi and find out what phi can be

Hint: ||V2 can be thought as the plane x+y+z=0 in C^3, with the obvious action of S_3. Take as a basis (1,w,w^2) and (1,w^2,w) where w is a third root of unity. Write matrices for the generators of S_3 in this basis and figure out what the matrix for phi has to be.||

[lyric cape]

a = x+y, b = xy, and the whole thing = x^n + y^n

this way of writing it is meant to encapsulate writing x^n + y^n in terms of a and b

it might be useful because it has fermats last theorem vibes

you can write x^n + y^n in terms of a and b with a polynomial for any n, but the coeffecients in that series of polynomials were patterned enough that i decided to write it in a general form

that general form was as follows

where you remove all the terms that have a negative power of a
(it might be interesting to just consider the infinite polynomial without this rule)

And now an actual question

is there a better form to write this in? right now there is a product operator nested in a summation

also if anyone knows anything even tangentially related to what im doing here that would be great

[mild laurel]

How is this abstract algebra

[lyric cape]

well you see
i really dont know where to put this question

number theory?
maybe?

[lyric cape]

honestly i put this here because i thought abstract algebra might help but i wasnt sure how

would you know a good place to put this

its very general
its like what do i do with this

i meant simpler form

what makes you say that?

sure, but i meant like what that ive said makes you say combinatorics

this mostly started with me tryin to look at x^n+y^n in a different way, so that i could do algabra and number theory

but i dont have enough experience to know what to do at this point, and so i came here

alright

[hidden haven]

We don't really know the goal, so I don't know what we can say about finding the expression. As for finding a simpler way to write that expression, are you just looking for neater notation?

[upper pivot]

the neater notation is x^n+y^n

[lyric cape]

lmfao i was waiting for someone to say that

i got the direction i came for
I want to eventually apply the equality in other math
though now that i think about it abstract algebra was not the choice if i wanted that knowledge

and moldilocks yeah a neater notation would be cool if it exists

[hidden haven]

This seems like such a specific case that I am guessing that you'd have to define the notation yourself but maybe try asking some physics people, they know how to handle ugly notation

occasionally partake

nah still doing random stuff

[split flicker]

wuick

for ideals I,J

Is I subseteq IJ true?

counter example

[hidden haven]

Why quick, is it like a prof asked you a question in class

[split flicker]

no

[hidden haven]

(2) and (3) in Z

[upper pivot]

its the otherway around

[split flicker]

my intution is no

[upper pivot]

I >= IJ

[split flicker]

IJ in I cap J

which is in I

So thats my intution from set theory

Ye

I know it is false generally

and true when

uh idk when its true tbh

oh uh rarely my intution says

[hidden haven]

It's true only if J is (1) I think

[upper pivot]

not necessarilyt

like in a ring with an idempotent

[hidden haven]

Ah true

[upper pivot]

<e><e> = <e>

[hidden haven]

In domains I maybe?

[upper pivot]

hmm yeah in domains should be i think

cause if IJ=I then for each i we have ij=i for some j

and so j is identity

[hidden haven]

Neat

[split flicker]

if I maximal and J,K subset I. Is JK subset I when?

im thinking right

when I =JK only

wait rip

[hidden haven]

If they are just subsets then only I

JK ⊂ IR = I

[split flicker]

ok ic

[upper pivot]

in general u dont care for just subsets

[split flicker]

uh

rapid fire?

[upper pivot]

cause u can make them into ideals right

[split flicker]

ok boom

im tired af lol

want to get isekaied

[scarlet estuary]

isekai discussion goes in #point-set-topology thanks

[split flicker]

imma ask in like 69 minutes about pullbacks and pushforwards before I get isekaied

in topology and geometry

[chilly ocean]

ok, you're getting re-zero'd. have fun dying

[scarlet estuary]

isekai is japanese for "other world", its like a genre of mostly escapist/wish fulfilment fantasy japanese media

[upper pivot]

its cringe kek

stop polluting my pure algebra channel with ur filthy anime

tyvm

[scarlet estuary]

well yeah thats part of the point of isekai

watching cartoons ABOUT touching grass

watch re:zero though its good

99.9% of isekai ranges from "plain trash" to "trash thats also a crime in some countries" though

mushoku tensei...

[split flicker]

Given ideal I is maximal and a,b in R.
If ab in I then assuming for ideals J,K subset I, ab in JK implies I in JK implies I = JK implying ? Im trying to remember if maximal then prime and I am missing one step

[scarlet estuary]

why is there math discussion in my anime channel

[hidden haven]

wtf is that implication chain

[scarlet estuary]

also yeah thats unreadable lmao

[split flicker]

its also wrong

[scarlet estuary]

If ab in I then assuming for ideals J,K subset I, ab in JK implies I in JK implies I = JK implying ?

im gonna spring this on people

[split flicker]

Telescoping in, ab in JK

[scarlet estuary]

force them to make sense of it

[split flicker]

ab in I has that ab in JK

[scarlet estuary]

millionaire did you sign a deal with a demon that every second word from each of your sentences is deleted

[split flicker]

);

btw showing maximal implies prime w/o quotient

book has that exercise

I think I learned it a year ago

and I have to use absorption somewhere

[hidden haven]

So are you saying that you need help with showing that maximal ideals are prime?

[split flicker]

without quotienting at all

oh uh

I assume a or b in J which is a subset of I

and then without loss of generality

if a in J

then ab in I implies b in I

otherwise that would mean I subset J

and i think thats gas

juice gogo we winning

[hidden haven]

No clue what you said but you seem happy

[split flicker]

is my argument okay?

I,J are ideals

[hidden haven]

What are I and J

[split flicker]

I is maximal

J subset I

a,b in R

ab in I

[hidden haven]

How

[split flicker]

If a in J

and b is in J

wait a second

a in J

and ab in I

doesnt this mean that b is in I

by absorption or something

[hidden haven]

#groups-rings-fields-testing channel when

[hidden haven]

no lol

Norm

[hidden haven]

b could be anywhere in the ring

If a is in I

ab will be in I

That's what absorption means

[split flicker]

Ok

So we assume a in J and b in R-I

ab is in J by absorption

but we also have that ab in I in our hypothesis

[hidden haven]

Why are you even considering a J

Why not just inside I and outside I

[split flicker]

Trying to use maximality

[hidden haven]

That talks about ideals outside I not ideals inside it

[split flicker]

Yes I but if we take a or b to be outside I then trivially a and b are inside I because its saying that all ideals containing I are either equal to I or the entire ring.

Like If I choose J containing I

but I guess im not considering case where J is neither inside or outside

and just intersecting

[hidden haven]

If a and b are outside I then they can't also be inside I assuming existence of a maximal ideal in some ring and consistency of math

[next obsidian]

Word 🥗

[split flicker]

I meant if we assume they are outside with our given hypothesis of I being maximal and ab in I

[hidden haven]

Why without the hypothesis

You want to do maximal → prime

[split flicker]

I meant with our*

[hidden haven]

oh epic

[split flicker]

Ok so you making me see clearer now

because we can have an ideal that intersects I but neither contained or containing I

So I should start with assuming a in R-I

[hidden haven]

Yes, but I don't know how that is relevant

So you want to show that ideal is prime

You take elements outside it

And you prove their product is outside it

[split flicker]

I assume a and b are in R-I

[hidden haven]

Yes

[split flicker]

no im confused lol

their product cant be outside I

thats contradicting the hypothesis

[hidden haven]

How

[split flicker]

we have that ab in I

so ab in R-I cant be it

[hidden haven]

Why not

Why is ab in I

What

[split flicker]

An ideal I is prime if ab in I then a in I or b in I.

[hidden haven]

Yes

But how did you get ab in I

[split flicker]

Dont we need to assume that?

in the hypothesis

[hidden haven]

We assume
I is maximal
a and b are not in I

[split flicker]

oh

ok

[hidden haven]

And prove ab not in I

[split flicker]

So just contradicting?

[hidden haven]

Do you mean you want to try proof by contradiction?

[split flicker]

a and b not in I implies ab not in I is contrapositive of statement defining a prime ideal?

[hidden haven]

Yes

[split flicker]

Oh nvm this isnt contradiction

[hidden haven]

Contrapositive yeah

[split flicker]

would it be wrong to consider a or b as part of an ideal?

tbh probably

[hidden haven]

I can't figure out what that means

[split flicker]

like The next step would be assuming a belongs to an ideal J where I strict subset J

[hidden haven]

No

There's no such J other than the ring itself

[split flicker]

yea thats why I thought it was a bad idea

well uh

I was thinking something faulty again

[hidden haven]

I gotta sleep but I'll give a hint before leaving ||consider the ideals I+(a), I+(b), and their product||

[split flicker]

oh

(I+(a))(I+(b)=(I+(a))(I)+ (I+(a))(b)=I+(I(a))+(I(b))+(ab)

ig not helpful

I+(a) contains ab

similar with I+(b)

i want to show ab not in I

so I+(a) not in I?

I+(a) is R

no false

that would be if ab in I

ok

i folded and stack exchanged it

this makes me feel so bad..

[uneven folio]

[prisma thunder]

Quick question

Let $M$ be an $R$-module and let $\phi: M \to R$ be a homomorphism. If $R$ is a PID and $\phi$ is non-trivial, prove that there exists $x \in M$ such that
$$M \cong Rx \oplus \mathrm{ker}(\phi)$$

[cloud walrus]

eM

[prisma thunder]

Normally, I figured the proof would be similar to the structure theorem for finitely generated modules over PIDs but here, our M isn't necessarily free.

[mild laurel]

I think you can just the first isomorphism theorem?

[prisma thunder]

How would I write the quotient into a direct sum that's sensible? If M is cyclic, then I can see it but I'm not seeing off the top of my head.

[mild laurel]

I don't really know what you mean by cyclic. You look to show that M/ker(phi) is generated by one element

[prisma thunder]

I mixed up the placing in my head if M is cyclic. There's really nothing to show for that.

Generated by one element?

Hmmm.

[mild laurel]

Generated as an R-module by one element more specifically

[hidden haven]

0 → ker phi → M → im phi → 0 is exact and it splits because im phi is a free R module

[hidden haven]

I don't think this suffices, unless you show splitting in some way

[prisma thunder]

Is thinking of this in terms of exact sequences the way to go?

Been a while since I've done that but not bad.

[hidden haven]

It's convenient here certainly since this is immediate from the splitting lemma

But you could do everything without ses

So a surjective map f: M → N implies M = g(N) ⊕ ker f, if you can construct a map g: N → M such that fg = id

[mild laurel]

yeah ur right mb

[hidden haven]

Think of g as the natural inclusion into the direct sum, f as the projection

So it should be somewhat obvious that this is a necessary condition. Sufficiency is also a short proof

[next obsidian]

@prisma thunder if you know what a projective module is, free modules are projective. M -> R surjects onto the image which is an ideal, but any ideal is isomorphic to R as an R-module, because a principal ideal is isomorphic via R -> (x) given by multiplication by x

Now you have M -> xR surjective and this splits because xR is free, so projective

Then you get M ≈ xR (+) ker phi

This observation that any direct summand of a rank 1 free module is itself free gives a very clean proof that any finite projective module over a PID is free

You basically use this fact and induction on the rank of the free module you’re a direct sum of

If anyone wants to see it here it is lol. My friend wrote this up and learned it from his friend

I guess this is slightly stronger since it says submodules of free modules are free. It ends up being equivalent in the end but you don’t have to assume the module is finite projective, just a submodule of a finite free module

[next obsidian]

We are assuming (A,m) is a Noetherian complete local ring. Why can we take a system of parameters starting with p when the residue field has characteristic p? I think it has to do with the fact that p is in m, but I can’t figure out how that means you can take a system of parameters with p

Oh, A is also a char 0 integral domain in what I’m struggling with

[novel parrot]

this feels weird to me

is phi star inverse = phi

just with different domains

[hidden haven]

no, phi*^-1 is P(spec A) → P(spec B), P means power set

I mean I have just given you the domains

but like this isn't directly induced by phi itself

[terse crystal]

What book is it?

[gritty sparrow]

I think the main point is that p is not a zero divisor (obviously) or more specifically it isn't contained in a minimal prime ideal. so ht((p))=1 by krull's height theorem. Then you select an element of m that is not in any minimal prime ideal containing p by prime avoidance. now you get (p,y2) and this ideal has height 2 again by the height theorem. So you can keep doing this argument until you get an ideal (p,y2,...yn) of height n. This has to be a system of parameters as any minimal prime containing it has to have height n, so that means it must be m itself so this ideal is m primary.

This argument is in matsumura's commutative ring theory ch5 sec 14 thm 14.1

[novel parrot]

in part 4

is it asking to show homeomorphism between Y and V(Ker phi)

but a prime in B is a prime in A not containing Ker phi

[gritty sparrow]

A prime in B will map under phi* to a prime containing ker(phi)

[novel parrot]

B = A/ker

so B = A after we remove kernel elements

[gritty sparrow]

first of all im(phi) is A/ker

[novel parrot]

yea

[gritty sparrow]

but in any case prime ideals in A containing ker(phi) correspond bijectively to prime ideals in A/ker(phi)

this is the correspondance theorem

[novel parrot]

oh right ok

[gritty sparrow]

so you understand the question now, right?

[novel parrot]

yes but now im questioning my understanding of quotients

wait

so

any ideal in A/ker contains 0, and 0 = ker so contains ker huh?

and if it was in ker, its the same as 0

[gritty sparrow]

you really shouldn't write stuff like 0=ker, but I think you have the right idea. the point is that an ideal I in A/ker contains 0 so phi^-1(I) contains phi^-1(0)= ker(phi)

[novel parrot]

0=ker, its additive identity 🙂

but yeah i got the question now

[gritty sparrow]

Ah you mean like cosets

[novel parrot]

yep

[gritty sparrow]

ok then it's cool

[novel parrot]

[terse crystal]

Wait how did you solve it? I thought if such f exists then it belongs to ann(M/N) right ? But if I let A=Z, I=2Z, N be the direct sum of n copies of Z with index {1,…,n}, N’ be the direct sum of n copies of Z with index {n+1,…,2n} . Then M/N is isomorphic to Z^n as Z module, whose annihilator is zero I think.

[next obsidian]

OH DUH. I was thinking you could do something like this but concluded height n doesn’t tell you anything 🤦‍♂️

[next obsidian]

Commutative Ring Theory by Matsumura

[wooden ember]

it has been exposed that chmonkey's hands are in fact not chair arms

[next obsidian]

Those are just a little attachment to help me turn pages

[wooden ember]

i see

sounds useful

[terse crystal]

But I is an arbitrary ideal, not contained in Jacobson radical

[terse crystal]

The statement is IM=M then M=0 where I is contained in Jacobson radical and M is finitely generated right?

I don’t understand how Nakayama Lemma was used since nothing about Jacobson radical was mentioned

[next obsidian]

When was Nakayama ever used?

[terse crystal]

My bad I am too sleepy… now every Japanese name looks like Nakayama to me…😂😂😂

[next obsidian]

…

@gritty sparrow is there any chance you understand the proof of 26.10 in Matsumura

I think I almost understand it, I just don’t get how the rank of Omega_L/k is calculated to be 1 from the given computation of it

[chilly ocean]

If a prime ideal contains a finite product of ideals

Then it must contain one of the ideals

Im not too sure how to see this

I can see that for since the product of ideals A_I in our prime ideal P, that elements of the from a1a2…an are in P

And we can use the prime ness to get that some aj is in P

[hidden haven]

If none of them are contained in P, you can pick an element in A_i - P for all i

[chilly ocean]

Hmm

[hidden haven]

[chilly ocean]

Okay this makes sense

Oh and then

Call such an element b_i

The product of b_I should be in P

But my primeness

Yeah

[hidden haven]

yep

[chilly ocean]

Sorry do we really mean A_j-P

I don’t see why P must be in A_j

[hidden haven]

A-B doesn't require that B be a subset of A

You just remove elements of B from A

[chilly ocean]

Oh okay yeah

[hidden haven]

It's A ∩ (B complement)

[chilly ocean]

I was going to say should we really mean

Yeah

Thanks

[hidden haven]

[next obsidian]

You can strengthen this to the intersection of finitely many ideals

Look at Atiyah-MacDonald I think it’s in the first chapter

It has that plus prime avoidance, two super useful theorems

[chilly ocean]

Okay thanks

And also

Is inclusion always a partial order on ideals

[thorn delta]

Inclusion is a partial order on sets so yes?

[chilly ocean]

Ah yes

Partial order

We are allowed to have

“Incomparable” elements

[thorn delta]

Ye

[tiny pagoda]

IDK where to really ask this, but is my intuition right about distinguished triangles being a kind of generalisation of exact couples for spectral sequences

[chilly ocean]

If R is an integral domain, and every prime ideal is principal, then R is a principal ideal domain

Here is a proof

But I am struggling to see where R being an integral domain is used

[tiny pagoda]

Hmm

Is the last line showing that M is generated by a single element, contradicting the assumed non-principality?

[hidden haven]

Being an integral domain is part of the definition of a PID

The above just shows that all ideals are principal

[sturdy marsh]

No. Verdier's definition was probably motivated by the way things work for derived categories; i.e. to construct some sort of formalism that describes when you can get an "LES in cohomology" from a triple.

Derived categories are not abelian, so the usual SES to LES thing doesn't make sense

[tiny pagoda]

Yeah I got corrected in another discord, shift operator and all

[sturdy marsh]

but if you have an exact triangle, then all of the "usual" functors from the derived category give you long exact sequences

[spiral wolf]

Anyone know a good method for representing a group with permutations?

The group I specifically have to do is <x,y : x^7 = y^6 = e, xy = yx^3>

I started by representing X = (1234567)

But frankly... not sure how to go about y. I know it's got to be some sort of 6-cycle, but I don't know how to set it up to follow the group's rules

Please ping me with any advice!

[past temple]

how does one show that like

in an integral domain an element r isnt the product of arbitrarily many noninvertible elements

sry not rly sure how to phrase this

[gritty sparrow]

Do you want a technique for a specific ring and specific element or are you trying to show that this holds for any nonzero element in any integral domain?

Because it is not true in general

[lavish nexus]

probably some integral domain with a norm here

[past temple]

any noninvertible element r

[gritty sparrow]

It still isn't true in general

do you have some more conditions on the ring

[past temple]

noetherian

[gritty sparrow]

Then it is true

I think

[past temple]

not rly sure why it isnt true in just an integral domain

do u have any counterexamples in mind

[gritty sparrow]

make a huge change to this example sorry

instead consider the subring of Q[X] which is given by polynomials of constant coefficient an integer

now X=2x2x2x2x..x2(X/2^r) is what we need

I'll be deleting the faulty example if that is ok with you

[past temple]

ok i might need some time to digest this

im a little confused

it seems like ur definition is circular?

[gritty sparrow]

in what sense

[past temple]

the x's are like

the indeterminate x

not multiplicaiton

right

[gritty sparrow]

they are multiplication

[past temple]

ah okay

now it makes sense

shoot

[gritty sparrow]

I edited the ring to make things a little more clear sorry

[past temple]

this is very unfortunate i thought that result was gonna be true

but u said that its true for noetherian rings?

[gritty sparrow]

Actually I'm not 100% sure in that case either

for noetherian rings I can say this much: you can't have an infinite sequence of factorizations each of which is made by taking the previous factorization and splitting an element

but there is still the possibility of a collection of factorizations which are not related to each other in this way where an element has arbitrarily large factorizations

[steady crown]

group theory 🥵

[tropic spade]

So if A(x) (nonzero) is in Q[x] and (x-a) is in (R-Q)[x] is it possible for A(x)(x-a) to be in Q[x]?

It seems like it would be a contradiction.

Since a sum of a rational with an irrational is irrational likewise for a product given that the rational is nonzero.

So my thinking with A(x) is it has at least one nonzero coefficient hence by the thing about rationals/irrationals I mentioned above A(x)(x-a) should also have at least one irrational coefficient. :/

[gritty sparrow]

Let A(x)=a_0 +a1x+…+anx^n

[tropic spade]

I think I just saw it lol

[gritty sparrow]

Let i be the smallest number st a_i is nonzero

[gritty sparrow]

Nice

[cloud walrus]

DootDooter

[tropic spade]

Right?

Well, off by one actually.

[gritty sparrow]

What is a_n+1?

[tropic spade]

I meant a_{n-1}-aa_n

[gritty sparrow]

My point is that the coefficient of i will be a_ia which must be irrational

[tropic spade]

Seems like that would be the coeff of the leading term of (x-a)A(x)

[gritty sparrow]

That is not the leading coefficient

[tropic spade]

Ah dang

[gritty sparrow]

The leading coefficient is a_n

Do you understand what I’m doing in my proof?

[tropic spade]

Sorry if I'm being confusing I was meaning to say this is the leading coeff of (x-a)A(x)?

[gritty sparrow]

Oops I had a typo

[gritty sparrow]

a_n is still the leading coeff

You might want to pick an example of A(x) to understand what is happening

[tropic spade]

Yeah I think I'm being dumb and missing the coefficient I want by one lmao. I'm pretty sure I get what you're trying to do though

[gritty sparrow]

Also to be a stickler, x-a is not in (R\Q)[x] because the coefficient of x is 1 which is rational, so you may want to rephrase that as just a is irrational

[tropic spade]

Oh shoot, thank you for pointing that out.

[next obsidian]

I don’t see how to use Hensel’s lemma here

A_0 here is a complete regular local ring

You can take for granted that a_n in m_0 from the fact a_n in x_1A

m_0 is the maximal ideal of A_0

I can use Hensel’s lemma to show that a_n-1 is in m_0 but I don’t think I can go further

If I can show that x^n-1 + … + a_n-1 is irreducible I can do it inductively, but I don’t think this is true in general

[gritty sparrow]

Why change the polynomial?

As in if there exists any a_i for i<n that is nonzero mod m then the polynomial reduced mod m is reducible

Which is a contradiction by hensel’s lemma

[next obsidian]

I wasn’t making it work because I was struggling to show (x^n,x^m +…+a) is (1) in a polynomial ring over a field

Then finally I realized I can just look at the radical and then it’s easy as shit

😔

[gritty sparrow]

you could also say that x^i has divisors only powers of x and x^(n-i)+...+a_i can't be divided by any power of x that is greater than 1 right?

[next obsidian]

Yeah that works too

[bitter mauve]

what is hensels lemma?

[novel parrot]

suprenum of two elements is just the larger one yea?

[hidden haven]

yes, you can easily prove that

[novel parrot]

i will take it as definition

because algebra

what does distributive mean?

over the other

[rustic crown]

it's not necessary that a and b are comparable