groups-rings-fields

[oblique river]

so it does turn out after the fact that like

[maiden ocean]

so i can send it to the size of the free part?

[oblique river]

"length of the Q(A) module Q(A) \otimes M"

[maiden ocean]

using the structure theorem

[oblique river]

is an additive function on finitely-generated A modules

which does give you the isomorphism K_0(A) to Z

but im not sure if thats the best way to phrase the argument

[oblique river]

yeah that's right

[maiden ocean]

ok this makes much more sense

[oblique river]

by our observation above about the A/p for nonzero p

you know that any f.g. module M is equivalent in K0 to A^n

for at least one n

and then I suppose you just want to make sure that you don't have any collisions

like you don't want A^2 = A^3 in K0

one way to see that is to tensor with the fraction field and then appeal to linear algebra I suppose

[maiden ocean]

yeah which is going to not be true just by taking A^n -> A^m -> A^m-n

[oblique river]

yeah

[maiden ocean]

or it suffices to show that its additive on the class of finitely generated A-modules which i think i can figure out

thank you buncho

[oblique river]

np glad we could get it sorted out!

[empty valve]

I am currently looking for some examples of statements / definitions in non-commutative algebra (/ring theory / group theory) which become trivial if one would only consider the commutative case.

To name a few that came to my mind so far:

• normal sub-groups become trivial in the case of abelian groups

• The commutator subgroup is trivial and dividing it out does not does not change anything

• Morita equivalence of commutative Rings is trivial

anything else that comes to mind?

[chilly radish]

Classifying finitely generated groups? It's not trivial in the abelian case but it's definitely magnitudes easier and actually solved

[crimson falcon]

Easy peasy "checking if i understood" question

The unique morphism is particular to s and each n-tuple here right? so a different n-tuple will have a different morphism?

[vestal snow]

Yes

Well

The unique morphism is particular to \phi_S and each n-tuple

[crimson falcon]

right, but the morphism to S stays the same as we go to different n-tuples, right?

Thanks! 😄

and the R-Algebra contains exactly n elements?

[crimson falcon]

wait no, if it has only n elements we cannot make it commute in the diagram, so is n elements with that property + all the others, right

[gritty sparrow]

If you’ve seen polynomial rings before, then you might recognize T as just the same as R[t1,..,tn]. So yes, there are many more elements, namely all the R polynomials in the variables t1,..tn

[crimson falcon]

I saw them before, about yesterday :p Thanks for confirming!

[gritty sparrow]

Np

[ivory dust]

How to show Q(sqrt(2),sqrt(3),sqrt(5)):Q(sqrt(2),sqrt(3)) = 2

the algebra is ... ugly

if i assume sqrt5 is in Q(sqrt(2),sqrt(3))

to arrive at contradiction

is there easier way w

Galois theory

[prisma ibex]

use the fact that Q(√2,√3,√5) is the compositum of Q(√2,√3) and Q(√5)

[oblique river]

By galois theory the only quadratic subfields of Q(sqrt2, sqrt3) are Q(sqrt2), Q(sqrt3), and Q(sqrt6). None of those contain sqrt5 so therefore sqrt5 is not in Q(sqrt2, sqrt3).

That sublemma can be proven directly or by proving the more general lemma that Q(sqrt(a)) = Q(sqrt(b)) if and only if ab is a square in Q

[ivory dust]

tyy guys

Gal(Q(sqrt(2),sqrt(3),sqrt(5))/Q) = {phi1, ... phi8}

whats the group structure

is it Z2 x Z2 x Z2?

[prisma ibex]

Yes; if you adjoin algebraically independent quadratic elements like this, this is the group you get

[eager willow]

here's an abstract question is there a name for this kind of reordering of the abelian group of a ring?
The ring Z/8Z is also a group under the operator ^ defined by i^j = 4ij + i + j, and this group is cyclic order 8. It's even generated by 1, but the cycle is 1 6 7 4 5 2 3 0, by definition it's not an automorphism but it's still a special kind of isomorphism because the operator form is seen as an element of the ring Z/8Z [i, j] so is this some kind of named permutation of the group Z/8Z?

it appears also quite naturally as a Galois group

[empty valve]

Ah yes, also a good one, thanks

[wooden ember]

bruh i was so hyped to see the proof in reading the theorem: this is downright disappointing

[chilly ocean]

i think this is common, at least i think i recall lang's algebra blueballed you like this too. or maybe not, tbh i cannot recall

[wooden ember]

i am the big sad

[rustic crown]

~~prove it yourself ~~

[void cosmos]

this is fundamental theorem of finitely generated abelian groups

right?

tell me im right cuz im smart

this is where the author thinks he is slick by proving the general case for finitely generated modules

where abelian groups are just Z-modules omg so math mao

[chilly ocean]

Wrong.

[wooden ember]

yeah it is

[wooden ember]

probably

[thorny flame]

Is there a way to calculate the order of a number n_1
n_1 = n_2 (mod n)?

[lofty vortex]

Hey, do I need group theory for ring theory and fields? Long story short, I'm switching faculties and I'm considering if I should retake group theory class since I didn't learn stuff like butterfly theorem, sequence of groups, free groups, finitely generated Abelian groups

[nova plank]

Not really

Until you get to advanced stuff, they're mostly separate

A lot of algebra books even do rings before groups

[lofty vortex]

Thats great news

[void cosmos]

just basic stuff imo but no you dont need any of those theorems u listed for normal ring theory

[wooden ember]

is this argument correct? Can I say $A/\operatorname{im}\phi \cong \operatorname{ker}\phi$ where $\phi:A\to A$ is a homomorphism by arguing that $A/\operatorname{im}\phi \cong (A/1)/(A/\operatorname{ker}\phi)$ by the first isomorphism theorem, and $(A/1)/(A/\operatorname{ker}\phi) \cong \operatorname{ker}\phi$ by the 3rd isomorphism theorem?

[void cosmos]

ofc

why not man

[cloud walrus]

Little Narwhal ✓

[wooden ember]

im just feeling fishy about manipulating isomorphisms like equality ive made mistakes because of it before so just wanted to make sure

[void cosmos]

ur not manipulating anything

u used third iso theorem on A

[wooden ember]

i felt mainly uncomfortable about saying $A/\operatorname{im}\phi \cong (A/1)/(A/\operatorname{ker}\phi)$

[cloud walrus]

Little Narwhal ✓

[void cosmos]

yea u probably meant A/ker

not /im

[wooden ember]

the third isomorphism theorem part felt fine but this feels weird

no i meant /im

since A/ker is isomorphic to im

by first iso theorem

[void cosmos]

yea

yea isomoprhisms are like equivalence relations

u can do transitive shit with them

ur good

[wooden ember]

but "substituting" A/ker for im in the quotient feels wrong

[void cosmos]

A is iso to B

B is iso to C

A is iso to C

i think thats what u did

dont u

[wooden ember]

i dont think so

since im not changing the group as a whole but only what we're quotienting by

[void cosmos]

oh wait so

okay can u rewrite what u did

clearly

[wooden ember]

like, it's not neccessarily true that A/B is iso to A/C when B is iso to C

[void cosmos]

yea

[wooden ember]

was this unclear?

[void cosmos]

yea ur good

[wooden ember]

aight

[void cosmos]

but yea

if A/B = C/D and B=D ( = here means iso )

A is not necc = C

[wooden ember]

for context it was for this q where i wanted to show the last part by showing ker was isomorphic to the elementary abelian group of order p^n and then showing A/im is iso to ker

[void cosmos]

yea

ur correct

[wooden ember]

Coolio. How could i have known for sure i was correct on my own?

is it always true that if B is iso to C and both are normal subgroups of A then A/B is iso to A/C?

[void cosmos]

we know for sure its not the other way around

[wooden ember]

yeah

[void cosmos]

but yea

thats not true

2Z is iso to 3Z

yet Z/2Z is not iso to Z/3Z

[wooden ember]

right

so im still not feeling confident about my scenario then

[void cosmos]

let me reread it maybe i did a mistake

[wooden ember]

the thing is the conclusion is definitely correct just from the exercise reading, it's the method i dont like

[void cosmos]

the conclusion is correct

haha yea its stated in the exercise

but i dont know why

you tried to do it in general

instead of trying to compute the actual ker and im

u have a defined map

[wooden ember]

out of curiosity

also because i dont like computing quotients lmao

though this one's easy

[void cosmos]

no one does

thats why iso theorems are helpful

and for the problem

[wooden ember]

ig ill just compute A/im and then show iso to E_p^n

[void cosmos]

can u construct an onto homo f with ker(f) = im(phi) ?

f being from A to E_p^n

[wooden ember]

sure

but the thought process is the same as computing A/im in the end so eh

it's fine ill just do it that way, thanks for the help

[void cosmos]

np

[viscid wyvern]

I have a question about tensors

I was learning about them and saw this chart

All of them have the transformations on the left hand side of the tensor, except for the (1,1) tensor for linear maps, which has the forwards transform on the right and backwards transform on the left

then this rule is given

but the (0,1) and (0,2) tensors both have the forward transforms on the left instead of the right like this rule says

so, given the type of a tensor, how do you know whether to apply the transforms to the left or right side?

for example, if I had a (1,2) tensor and wanted to change coordinates, I’d need one backwards and two forwards transforms, but how would I know where to put them?

[cursive temple]

does anyone have any hints/ideas to the part where you have to show that if \sum r_{ij}e_{ij} = 0 then r_{ij} = 0

im pretty stumped

[hot lake]

what does the C mean ?

[cursive temple]

center of the set

[hot lake]

you can probably multiply sum rij eij by somethings to make it into rij

[safe bone]

can someone help me in solving this sum i dont hv any idea

[cursive temple]

you can multiply by eg e_11 to just sum over r_{i1}e_{i1} but im not sure how to kill the rest of the terms

i feel like i need some sort of a commutative relation between the e_{ij}:s

[hot lake]

have you tried multiplying on the right and on the left

[cursive temple]

not really

i dont see any nice cancellations with multiplying from the left, except for maybe getting it into the same form as the a_ij terms

[hot lake]

you know that rij commutes with every ekl

[cursive temple]

oh right

damn

yeah so the problem is pretty much trivial lol

thanks for your help

[urban acorn]

R[x,y]/(y^2-x) is isomorphic to R[x], right?

[uncut girder]

Yeah the isomorphism is y -> x.

[oblique river]

yeah, maybe it's easier to visualize it if you say that R[x,y]/(y^2 - x) is isomorphic to R[y]

(which is isomorphic to R[x] -- whatever you call the variable doesn't matter)

in R[x,y] you have two variables, x and y

but then quotienting by y^2 - x is just the same thing as setting y^2 - x = 0, that is, setting x = y^2

so we have two variables, x and y

but secretly x = y^2, so x is redundant, we dont need x anymore

all we need is the variable y

so that's why R[x,y]/(y^2 - x) is isomorphic to R[y]

which is then isomorphic to R[x]

@urban acorn

[urban acorn]

yeah, that was my intuition, I just asked here to make sure

and now I think one easier way to see that is that R'[x]/(x-r) with r in R is clearly isomorphic to R', and when you set R' = R[y], r = y^2, you get that

[maiden ocean]

If A is noetherian and we have r(q1) = r(q2) = m maximal, where the qi are in a minimal primary decomposition of an ideal a, does q1 = q2?

i know that if we have a = bigcap q_i and r(qi) = r(qj) then m^n subseteq q_i subseteq m for n, likewise with m^k and q_j

but i dont see how we can use minimality to show that q_i = q_j here

or maybe in general theres something we can say about primary ideals with maximal radical

i guess put another way if A is noetherian and q1, q2 are m-primary does q1 = q2?

or can we at least say that q1 subseteq q2 or q2 subseteq q1?

[gritty sparrow]

In a minimal primary decomposition aren’t the radicals of each qi distinct by definition?

In general equality is definitely impossible, (pZ and p^nZ in Z) but let me think about the subset thing, my feeling is that that won’t hold either

[maiden ocean]

i think this is supposed to be true but AM does not define it this way

Hell

Pain.

[gritty sparrow]

I think this should work for the radical thing, but I haven’t checked properly, in C[x,y]/(x^2,y^2). The radical of (y) and of (x) should be (x,y) I think

[maiden ocean]

Or maybe they do its just confusing because they also feel the need to justify this

Like

[gritty sparrow]

In any case, the intersection of p primary ideals is p primary so we can always get a primary decomposition where each term corresponds to different primes

[maiden ocean]

yeah its more or less fine the details of the argument just dont work without this

(AMs argument for this proof)

Idk its weird they "prove" that the radicals are distinct but the "proof" is literally just p_i supseteq q_i supseteq a neq 0

[gritty sparrow]

What exactly is AM trying to prove here?

[maiden ocean]

Noetherian and dim 1 implies unique product of primary ideals

[gritty sparrow]

Oh i think that he did that to show that the pi’s are maximal, not that they are unique

Like, by showing that pi is not 0

[maiden ocean]

[clever dust]

What kind of answer do you think the problem is looking for?

[maiden ocean]

so theres this theorem that if A is a local domain, its a DVR iff every non-zero fractional ideal is invertible

this is assuming A isnt a field right?

[oblique river]

I would say that "local" already implies "not a field"

so you dont need to assume anything more about A

[maiden ocean]

[next obsidian]

Ah, dealing with DVR but field?

This stuff is so cringe lol, you lose no matter what you do

IIRC Matsumura is technically wrong in something because some theorem doesn’t specifically specify non-field and he doesn’t specify a valuation ring is not a field and blah blah blah. It’s all 💩

[novel parrot]

i dont particularly understand how the operations can make A into C

since A is m x n and C is m x 1

[chilly ocean]

my guess is that they're using C to refer to a different matrix (one that would be m x n), as opposed to the one that they already called as C. but this is a really shitty writing if that's what the author meant

[novel parrot]

wdym?

[chilly ocean]

like they probably meant to write "If a matrix A can be transformed into a matrix D by ..."

or some other letter that is not C

[novel parrot]

so how is it actually done?

what are we doing to matrix A?

[chilly ocean]

the second set of (1),(2),(3)

interchange rows, etc.

[novel parrot]

but transforming it into C how?

like

what does transform mean here

[chilly ocean]

just to change it. like if i started with a matrix [[1, 1], [2, 2]] and i subtract the first row from the second row, then i get [[1, 1], [1,1]], i just transformed the first matrix (A), to the second matrix (C)

[novel parrot]

but the matrix have different size

[chilly ocean]

i'm saying the author meant to use a different letter than C, probably

so really whenever C is a m x n matrix, say D instead of C

[novel parrot]

but C is m x 1 always

or not?

[chilly ocean]

initially C refers to an m x 1 matrix, later the author forgets to switch variables and uses C to mean an m x n matrix

i think it's an error in the textbook, is what i'm saying

[novel parrot]

so in question 14, C is not the column matrix of what it was before, but rather something also m x n?

[chilly ocean]

yeah, something else that is m x n

[novel parrot]

okay

understood

[chilly ocean]

there are many shit tier textbooks where there are many errors and typos and what not

[novel parrot]

this is dummit and foote so an error was not expected from me :c

[chilly ocean]

ah, i should have recognized the font

dummit and foote does have errors though

[novel parrot]

yeah

[chilly ocean]

although i didn't read the vector spaces chapters

[novel parrot]

i checked the errata and its not mentioned there

guess it just slipped by

[clever mountain]

I have proved 2 is an irreducible element in Z[sqrt{-5}].
I did this via the norm N(a+b sqrt{-5}) = (a+b sqrt{-5})(a-b sqrt{-5}) = a^2 + 5b^2.
But I don't know if it would still be true for any other norm.
I don't know how to ascertain that 2 is irreducible in Z[sqrt{-5}], independent of choice of norm.
Does it depend on norm? It shouldn't, one would think. Z[sqrt{-5}] can never be a Euclidean domain, no matter
how we choose the norm.

[void cosmos]

okaay

so u define a norm

and then do factorization over this norm

@clever mountain

[prime minnow]

The irreductibility does not rely on the norm but a well chosen norm can help you prove the irreductibility

[clever mountain]

@void cosmos yes I did that
@prime minnow yes I managed to do that with the convenient norm a^2 + 5b^2. But I don't see a simple argument for it that is norm-independent

[prime minnow]

Its always the same ring.

2 = ab implies a or b invertible has nothing to see with the norm

[still stream]

So irreducibility is independent of norm?

[prime minnow]

It's the law which is important here, the norm is only an application on an already created ring

[clever mountain]

@prime minnow I wonder how one can show that 2 = ab forces a or b is a unit in Z[sqrt{-5}] without
the aid of an imposed norm

[prime minnow]

2 = (u + iv.V5)(x + iy.V5) gives you

ux - 5vy = 2
uy + vx = 0

If you multiply 1st equality by x and seconf by 5y you obtain

u(x^2+5y^2)=2x

x=/=0 and u=/=0. You obtain y =0 otherwise you have not the equality

Same with v

And 2 = u.x so either u or x is invertible

[clever mountain]

Hm, I get
u(x^{2} + 5y^{2}) = 2x

[prime minnow]

Argh sorry

[clever mountain]

how did you conclude from 2 = u.x that either u or x is a unit?

[prime minnow]

u and x are integers

So one of them is equal to 1 or -1

Since 2 is prime

[maiden ocean]

can anyone give a hint on showing that a noetherian valuation ring has dimension 1?

Ofc its local because its a valuation ring but im not really sure where to go from here

[sturdy marsh]

@maiden ocean what is your definition of valuation ring

[maiden ocean]

Uh ring given by the set of x st v(x) geq 0

[sturdy marsh]

okay

[maiden ocean]

Where v is a discrete valuation on a field K

[sturdy marsh]

try proving that all ideals are principal

[maiden ocean]

I have an idea of how to but im not sure where Noetherian comes in

Like if we can prove m is principal we r done right

[sturdy marsh]

noetherian implies that all ideals are finitely generated

I think you need this

[maiden ocean]

I see

Im not at home but i think i can try

Lemme think abt it

Oh u can evaluate the valuation on the generators maybe

[sturdy marsh]

yup, just pick the smallest one

there cannot be any elements of lower valuation in the ideal

[maiden ocean]

Yea i see

Pog

Thanks brofibration

[chilly ocean]

let a+bsqrt(r) be an element of F(sqrt(r)). then a,b in F, so a,b in E. since sqrt(r) in E, a+bsqrt(r) is in E as well because E is closed under addition and multiplication.

[maiden ocean]

an exercise in AM says to show that if M is finitely generated over a dedekind domain A and every element of M has torsion then M can be represented as a unique finite direct sum of A/p_i^e_i for p_i prime

[fickle tapir]

yeah that makes sense thanks

[maiden ocean]

the hint says that you can pass from M to M_p and then apply the structure thm for modules over a PID which make sense

but im not sure how to reconstruct a decomposition of M from a decomposition of all of the M_p

[fickle tapir]

is the intuitive explanation for E containing F(root(r)) shows that F(root(r)) is the smallest number field because E can also be F(root(r))?

[gritty sparrow]

I would imagine that you would take the direct sums of the form R/p^i where R_p/p^i appears in the expansion of M_p as an R_p module

[maiden ocean]

Would that necessarily be finite if you have infinitely many prime ideals?

[gritty sparrow]

I think by the finitely generated condition, only finitely many of these things will be nonzero, but I haven’t worked it out properly

[maiden ocean]

sadge

comm alg details can be a pit of a pain

[unreal portal]

It's the same sort of logic used in the definition of gcd, where d is the gcd of a and b if (among other requirements) for any c that divides a and b, c divides d. This implies d is the greatest such divisor in the same way that logic shows that's the smallest number field

[gritty sparrow]

True these sorts of verifications can get quite annoying

[maiden ocean]

@gritty sparrow late cause i took a break but i bet you can just use minimal primes

and noetherian implies there are only finitely many

[gritty sparrow]

||I was thinking of looking at the prime factorisation of the ann(M). Localising M at primes outside of the primes in the annihilator should always be 0 because we can find a t in the annihilator that annhilates all elements, and won’t be in that particular prime. This should be related to the minimal primes idea as well because the primes that should appear in the factorisation of the ideal would be primes containing the annihilator, which would be minimal in R/I|| this would be a bit of a spoiler to the problem, but I think your idea could work, so I wouldn’t suggest opening it. (Hopefully the spoiler stuff works out, but again I didn’t properly check)

[maiden ocean]

@gritty sparrow i had basically an equivalent idea where i tried using the associated primes of M

then if the associated primes are p1, ..., pn you can let N be the direct sum of A/p_i^e_i and localizing at any maximal m you see that M_m = 0 iff N_m = 0 iff m is not among the pi

but then when m is among the pi M_m = N_m clearly

so M = N

[gritty sparrow]

There is a slight issue with the last point, two modules with isomorphic localisations need not be isomorphic. The correct statement is that if you have a map all of whose localisations are isomorphisms then the map is an isomorphism. The issue is that given local isomorphisms, they may not be “compatible” and you won’t be able to piece together a global isomorphism from them. I’ve been trying to find an example where this is a problem, but it is taking me a while. In our case, there should be natural inclusions from each of the p-primary components of M (the elements of M annihilated by some power of P) and M itself, so I don’t think it will be much of an issue.

[maiden ocean]

well yes but in this case we have natural inclusions

so

its not a problem

[gritty sparrow]

True

[thorny flame]

How do I prove that o(gag^-1) = o(a)? for every a and g?

[gritty sparrow]

(gag^-1)^2=ga(g^-1)ga(g^-1)=g(a^2)(g^-1). Try to say something similar for higher powers too

[thorny flame]

oh right than
g(a^o(a))g^-1=g id g^-1 = g *g^-1 = id
that means o(g a g^-1) = o(a) due to g and g^-1 would cancel each other. right?

[dusty river]

Yeah but that only proves that the order of the conjugate is smaller

Or equal

[thorny flame]

order of the conjugate?

[unreal portal]

o(gag^-1), that's often called the conjugate

[thorny flame]

I know

I just don't understand how it proves it's smaller

It needs to prove it's equal

[dusty river]

Order is the smallest positive power that makes your element identity

You proved that o(a) is such a power

You need to also prove that it is the smallest such

[thorny flame]

what about?
g(a^o(a))g^-1=g id g^-1 = g *g^-1 = id = a ^ o(a) that fixes it?

[gritty sparrow]

Maybe a cleaner way to rephrase your argument in line with what moldi is saying would be to show that for all n
a^n=1 <=> (gag^-1)^n =1 and conclude from there. Your argument is essentially the forwards direction while moldi wants you to prove the backwards direction

[thorny flame]

hmm. what would be the backward direction?

[unreal portal]

show that if (gag^-1)^n = 1, then a^n = 1

[thorny flame]

(g a g^-1)^n = g a g^-1 g a g^-1... n times
We cancel all g ^-1 g
= g a^n g ^-1

Now what?

[unreal portal]

and we're assuming that this equals 1

want to show a^n=1

[thorny flame]

we don't even know if wer'e talking about a group of numbers TBH so we can't tell

unless 1 means something else

[dusty river]

1 means the identity of the group

[thorny flame]

oh

g a^n g ^-1 = 1
g a^n g^-1 g = g
g a^n = g
g^-1 g a ^n = g^-1 g
a^n = 1

just put another element from each side to cancel the g and g^-1

[unreal portal]

yeah looks good

do you understand why doing this 2nd part was necessary?

[thorny flame]

to isolate a?

[unreal portal]

2nd part as in showing that if (gag^-1) = id, then a^n = id

[thorny flame]

no. I don't?

[unreal portal]

so imagine, hypothetically, that o(a) = 4 and o(gag^-1) = 2

[thorny flame]

ok

[unreal portal]

you calculated g(a^o(a))g^-1 = g id g^-1 = g g^-1 = id

nothing in that first direction showed that this hypothetical case is impossible

but for the second direction, if we set n = 2 (which is o(gag^-1)), you'd get

g a^2 g^-1 = 1, and then g a^2 = g

but a^2 is some arbitrary element of the group, you can't simplify any further

it's this second direction that shows the hypothetical situation is impossible

[thorny flame]

right. I see

[trim grove]

is every mapping with fof=f is projection?

[dusty river]

What's the definition of projection, and what is f?

Are you in a vector space

[rustic crown]

i think some people just use that as the definition of projection 😛

[trim grove]

f is a map form some set to some set,

[rustic crown]

no vector space/module? weird.

[dusty river]

Yeah then you'd need to say how you are defining a projection

[trim grove]

for vector space

[dusty river]

Ye then f is a projection onto its image iff f² = f

Usually that's how it would be defined as det said

Are you defining it using direct sum decomposition

[trim grove]

things are getting weird as i'm going deep into concepts

[rustic crown]

basically the idea would be v = (v - f(v)) + f(v)
notice that f(v - f(v)) = f(v) - f(f(v)) = 0

so you can say f is a projection if V = im(f) + ker(f)

i'm saying it wrong, and i know it

[dusty river]

Yeah every exact sequence should split for vector spaces

Wait no what

I think you're right lol idk

As long as you take direct sum

Instead of just sum

[rustic crown]

direct sum would be wrong? as every exact seq splits

[trim grove]

is't this is same for every function defined on vector sapce? or just for porjection?

[dusty river]

im(f) can intersect ker(f)

[rustic crown]

not really

[dusty river]

This is slightly different from splitting condition

I think

[rustic crown]

lol that's what we're thinking

[dusty river]

0 1
0 0
Has (1,0) in kernel and in image

[rustic crown]

normally f o f = f is the definition of a projection

[dusty river]

If you are thinking of the rank nullity theorem then that also has dimensions in it

[rustic crown]

yea so this isn't a projection. we want V = ker(f) + im(f), if they intersect dim of right would decrease

[trim grove]

hmm rank f+ nullity f = dimension of sapce

[dusty river]

Wait then why is direct sum wrong

[rustic crown]

because it's true for every f

[dusty river]

No that's just what I gave a counterexample to

[rustic crown]

V/ker f is iso to im f

[dusty river]

Ker f and im f can intersect

Oh isomorphism

I was thinking of equality

Internal direct sum

[rustic crown]

oh

yea, i think saying V = im f + ker f implies they don't intersect

[dusty river]

Only if you use the direct sum tho

[rustic crown]

why

[dusty river]

• allows intersection

⊕ doesn't

[rustic crown]

but the first condition is V = im f + ker f

[maiden ocean]

I would assume + here would be the sum of subspaces/submodules

[dusty river]

Ye

[rustic crown]

if they intersect, then dimension of right is smaller than dim V

[dusty river]

Only in finite dimensions

I guess

If you have countably infinite basis e_0, e_1, ..., you can map e_0 to 0 and e_i+1 to e_i

Surjective map with non zero kernel

[rustic crown]

yea right

so whats another way to define projection?

[dusty river]

f is a projection iff A = B ⊕ C, internal direct sum, for some B and C, and f(b,c) = b

I think exactly what you said

But with internal direct sum

V = im f ⊕ ker f

[rustic crown]

yea makes sense, i was interpreting the right as an abstract sum, so it didn't make sense to say V and that thing containing pairs are equal

[dusty river]

Ye

[trim grove]

any conditon of B and C ( like sets or subspaces)

[dusty river]

Subspaces

[trim grove]

is this is the only way to define projection?

[rustic crown]

nope

if f satisfies f o f = f, then take B = im f and C = ker f

any v in V can be written as f(v) + (v-f(v)) the first thing is in B and the second in C

so V = B + C. they can't intersect non-trivially. as if w in B and C then w = f(v) => 0 = f(w) = f(f(v)) = f(v) = w

[trim grove]

i don't know to which group this questions belongs( i think linear algebra) if $A \in M_n(\mathbb{C})$ is such that $<Ax,x>=0$ for all $x \in \mathbb{C}^{n}$ , then $A=0$ true or false, ? any hint?

[cloud walrus]

honey99

[dusty river]

<Ax, x> = 0 means x and Ax are perpendicular

Try to think of this in different dimensions

Wait you can't really do that since C

[rustic crown]

just eigen bash it

if x is an eigen vector, with eigen value k then <Ax, x> = <kx, x> = k <x, x> = 0 => k = 0

all eigenvalues are 0 😄

[upper pivot]

theres non-zero matrices like that unfort

[maiden ocean]

non invertible ones

but if its not in GL_n whats the point tbh

[hidden haven]

But not every A would have an eigenbasis would it

[rustic crown]

yea i assumed C would be nice enough

[upper pivot]

maybe we should upper triangularize

we know the diagonals have to be like, 0

[trim grove]

so it means all its eigenvalues are zero?

[upper pivot]

I think this should be false? || we are mapping M_n (C) linearly into the dual space of C^n, and by dimentionality reason nontrivial kernal?||

[stone fulcrum]

Okay nvm I wasn't too late. Can't we just use properties of the inner product?
<Ax, x> = A<x, x>
Since <x,x> ≠ 0 for x ≠ 0, we must have A = 0

[trim grove]

sorry to say but answer is true

[stone fulcrum]

wtf my name is blue

[upper pivot]

oh hmm

can someone find the flaw in my arguement

[hidden haven]

Kaynex A is a matrix

[stone fulcrum]

Dang it

Yes I assumed a scalar for no good reason

[upper pivot]

we are mapping M_n (C) linearly into the dual space of C^n, and by dimentionality reason nontrivial kernal

I dont see why this isnt correct.

wait nvm just realized

[hidden haven]

is the map A maps to <Ax,x>

[upper pivot]

the inner product isnt linear in x

lmao

yeah

[hidden haven]

F

[stone fulcrum]

F(x)

[hidden haven]

Also this doesn't capture the behaviour for all x right?

It would just say that there is some x in the kernel of that map

[upper pivot]

I was thinking of sending A to the function <A-,->

which if it were linear would be in the dual of C^n

[hidden haven]

Ah right

[rustic crown]

i think i got it

0 = <A(x + y), x+y> = <Ax, x> + <Ay, y> + <Ax, y> + <Ay, x> = <Ax, y> + <Ay, x>
replacing x with i* x, we get
i<Ax, y> + (-i)<Ay, x> = 0
both these gives <Ax, y> = 0

[upper pivot]

oh neat!

[chilly radish]

I have a counter-example

[rustic crown]

now plugging x = ei and y = ej tells that i,j entry is 0

[chilly radish]

I think

wait nvm

I did the inner product over R by mistake

go on det

[hidden haven]

oof

nice solution

[upper pivot]

yeah v clever

[trim grove]

that was the first part and i solved that

[rustic crown]

[upper pivot]

im guessing on R any anti-symmetric matrix would work

[chilly radish]

Wait why would this part be true

[chilly radish]

yup

[rustic crown]

i kinda assumed that its standard inner product ish

but shouldn't be too hard to fix

[hidden haven]

I mean you got <Ax,y> = 0 for all x and y

[rustic crown]

yea

so <Ax, Ax> = 0

lol nice

[hidden haven]

Yeah

[stone fulcrum]

Ah neat

[chilly radish]

hold on, why does this give <Ax,y>=0. Wouldn't it just give <Ax,y>=<Ay,x>? I feel like i'm missing smth

[trim grove]

little bit confusion , in "both these gives"

[rustic crown]

<Ax, y> = <Ay, x> from the second equation and <Ax, y> + <Ay, x> = 0 from the first

[chilly radish]

ah right

[rustic crown]

hopefully C doesn't have char = 2

[chilly radish]

nice

lmao

[upper pivot]

char 2 C moment

[maiden ocean]

Based

[gritty sparrow]

you’re original idea could work as well. if you think about JCF of a matrix with all 0 eigenvalues, the vector of all 1’s should be something with a nontrivial inner product if the matrix is nonzero

[maiden ocean]

Just base change 4head

[upper pivot]

but jordan normal form is

[stone fulcrum]

For char of C approaching 2

[chilly radish]

imagine using jordan form ever

[rustic crown]

had to use it in my exam

[chilly radish]

same

[maiden ocean]

just dont do the exam

[upper pivot]

I literally made a blog post cause I hate jordan normal form so much.

[chilly radish]

I'm glad to never touch linalg again

[gritty sparrow]

I like the jcf, what’s with all the hate

[chilly radish]

john what's your opinion of rcf?

[upper pivot]

rcf?

[rustic crown]

much better

[chilly radish]

rational canonical for

form

[rustic crown]

anything rational is nice

[hidden haven]

I like JCF after having seen how Saketh uses it in his solutions

[maiden ocean]

gaslit into liking cringe

[upper pivot]

this looks like a LA meme i dont know rcf kek

[gritty sparrow]

Jcf for when your field doesn’t split the polynomial, basically. The problem is that it is not as powerful

[hidden haven]

Real closed field 😌

[upper pivot]

ic

[chilly radish]

but it's applicable over every field

it's much less nice tho

[upper pivot]

i mean just consider it in algebraic closure

[chilly radish]

also proving it is pretty hard iirc

[gritty sparrow]

No, you can prove it on the way to jcf

[chilly radish]

only over C

[rustic crown]

follows immediately from structure theorem of mod/pid

[hidden haven]

Easier than JCF

[chilly radish]

I mean for non algebraically closed fields

[upper pivot]

Structure theorem for pid modules (and in general dedikend domain modules) is something i black box

the proof had too many matrices.

[gritty sparrow]

You can do it without matrices as well

You might wanna look into that

[rustic crown]

yea, i kinda like the non-matrix proof more

[upper pivot]

I probably should lol

artin moment

[maiden ocean]

if you assume it for PID you dont need to use matrices for dedekind

[chilly radish]

yea the general proof uses the structure theorem over PID

[maiden ocean]

probably a non matrix proof for PID as well

[upper pivot]

right moth

[novel parrot]

why does a1y1 = y?

[maiden ocean]

i did the general dedekind proof earlier today

saketh helped but i did not read his solution

ok i did but only after i did it myself first because i came too late to read it

sad!

[upper pivot]

probably time to move to #discussion lol

let the guy have his q answered

[maiden ocean]

Yes

ironically its related to the structure theorem

and does not use matrices

[novel parrot]

yes

trying to prove the ablian group theorem

finally

[gritty sparrow]

a1y1 = sum(a1bixi). Now a1bi=pi_i(y). So a1y1=sum(pi_i(y)xi) now by definition of the projection map (it takes y to the coordinate of xi is the expansion of y) this element is the same linear combination of xi’s as y is. Hence by linear independance they are equal

[novel parrot]

by coordinate it means the i'th module element?

or ring element?

[gritty sparrow]

So if y=sum(l_ixi) for some elements l_i in R, then we know that these elements are uniquely determined. What I mean by the coordinate of xi in the expansion of y is the element of the Ring l_i

I am so sorry, I had a typo

I meant element of the ring, not of the module

[novel parrot]

its okay

understood

[gritty sparrow]

Nice

[novel parrot]

thank you

[trim grove]

I have solved this till x3=0, x1=x2 , whre x1,x2,x3 is vector x, but not getting how to find funtional , any hint?

[rustic crown]

so you have shown W is precisely all the scalings of the vector (1, 1, 0). you want these to die.

do you see how you can construct a phi?

map to 0 😓

[dusty river]

😵

[rustic crown]

henlo moldi

[dusty river]

henlo det

[rustic crown]

y r u dz?

[dusty river]

That's the dead emote

[rustic crown]

ah, the mouth looks like a 0

[dusty river]

xox

[trim grove]

i know this , but how mapping zero can help?

[rustic crown]

say the functional had the matrix [a b c]

what do we want it to satisfy?

[trim grove]

null space?

[rustic crown]

hm? the two conditions are phi(x0) = 1 and phi(x) = 0 for every x in W

[trim grove]

opps

i can't be null space

but x_0 is not in W i think

[rustic crown]

yea its not, W is multiples of (1, 1, 0)

[trim grove]

we can see W is kernal of Phi?

[rustic crown]

W should be contained in the kernel of phi

but not same as the kernel

R^3 --> R^1 loses 2 dimensions, so kernel should be dim 2

[trim grove]

hmm but here W is one dimenaional

[rustic crown]

yep, that just means that there are lots of such phi

you just want one

[trim grove]

how to procede ?

[rustic crown]

just think about what we want

[a b c] is the functional, then we want a + 2b + 3c = 1 and a + b = 0

[trim grove]

what about a=0,b=0,c=1/3

[rustic crown]

yep, that works!

[trim grove]

so what will be functional now? is it [0,0,1/3]

[rustic crown]

in matrix notation, yep

[trim grove]

i know only one defination of functional i.e function of functions

[rustic crown]

isn't a functional on R^3, just a map from R^3 --> R?

[trim grove]

okk so z/3 is functional

[rustic crown]

yea, phi((x, y, z)) = z/3 will do

[trim grove]

hm , because it satisfies all the given conditions ,

[chilly ocean]

det when u learned jordan form did you learn it as upper or lower triangular matrix

[rustic crown]

upper

[chilly ocean]

good.

[gritty sparrow]

I learned it while doing cartwheels, so both

[rustic crown]

i first proved it using induction, don't like induction since then

[chilly ocean]

i love induction

[hidden haven]

proofs and types course next sem

[rustic crown]

i hated induction the most during the ToC course

proofs and types would be next

[chilly ocean]

whats ToC

[hidden haven]

[chilly ocean]

you were reading tables of content?

[rustic crown]

computation, automata, grammar

[gritty sparrow]

Theory of computation

[rustic crown]

btw why did you beocme Ledog from Godel?

[hidden haven]

wasn't godel someone else

[rustic crown]

they looked isomorphic, so assumed they're same

[chilly ocean]

nope

[hidden haven]

[rustic crown]

Godel had right side up cat

[chilly ocean]

whats up cat

[rustic crown]

right-side-up

[chilly ocean]

$\Ledog$

[cloud walrus]

Ledog

[terse crystal]

Consider the free group generated by two words x and y :F=F(x,y). are the two quotient groups G1 and G2 isomorphic to each other? where G1 is the quotient of F over <xyxy^-2> and G2 is the quotient of F over <xy^2xy^-3>

[deft oasis]

It could also be R^3 -> Q, for example :^)

[dusty river]

I don't think those subgroups are normal, so you might have to take normal closures first

[terse crystal]

My bad, I used <xyxy^-2> to represent the minimal normal subgroup of F containing xyxy^-2

[urban acorn]

what's up cat?

[chilly ocean]

Don't worry about it.

[urban acorn]

(what's up, cat?)

[chilly ocean]

,w up

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Ledog#7947

[wooden ember]

Want to show every group has a unique maximal perfect subgroup that is normal. I’ve shown that if we have at least one maximal perfect subgroup it must be normal and unique but I’m having trouble proving existence

[chilly ocean]

Try induction.

For small group sizes it is easy to verify.

You can even show a stronger result: Let G be a finite group. Suppose G contains a unique maximal (proper) subgroup. Show that G is cyclic.

[wooden ember]

Induction will only work for finite groups though won’t it?

[chilly ocean]

Oof, my bad, assumed finite.

[urban acorn]

some induction-based results can be extended to the infinite case using transfinite induction (induction on ordinals)

and other results that use induction on finite objects to prove the maximality of something can be extended to the infinite case using Zorn's lemma

[wooden ember]

Yeah I was thinking of using zorn’s lemma but I’ve only used it in a proof once before and this textbook told you to use it so I assume there’s a way to show this without it

[urban acorn]

by "maximal perfect subgroup" do you mean a maximal subgroup which is also perfect, or a perfect subgroup which is maximal among perfect subgroups?

[wooden ember]

Maximal among perfect subgroups

Otherwise it would be false

[urban acorn]

okay, so given a group G, let S be the set of all proper perfect subgroups, and partially order it by containment. Let C be a chain in S (so, a subset which is totally ordered by containment), can you find another element in S which is an upper bound of C?

[wooden ember]

Hmmm

[urban acorn]

fyi i don't know that zorn's lemma is what you need here, i haven't worked with commutators for a while and i needed to google what a perfect group is

[wooden ember]

I’m certain there’s an alternative to zorn’s lemma by the structure of the book

I would assume if we had to assume the axiom of choice the exercise would say so

[urban acorn]

okay, I think I know a path that will eventually lead you to the answer

if A, B are perfect subgroups, can you find a perfect subgroup containing both A and B?

[wooden ember]

I can

The subgroup generated by A and B is perfect

[urban acorn]

okay, what about any arbitrary collection of subgroups, can you find a perfect subgroup containing all of them?

[wooden ember]

Yep, same thing just generate a group out of them

[urban acorn]

okay, so let the collection be all perfect subgroups

[wooden ember]

Ah so it gives me an upper bound for the chain

[urban acorn]

it doesn't even require Zorn's lemma

just let P be the collection of all perfect subgroups of G, then let H be the subgroup generated by the union of P

[wooden ember]

It doesn’t tell me it’s maximal though?

[urban acorn]

by construction, it contains all other perfect subgroups

it was generated by the union of all perfect subgroups

[wooden ember]

Oh right including itself I’m stupid

[urban acorn]

yeah

[wooden ember]

My mind was russel paradoxing

[urban acorn]

lmao

[wooden ember]

Thnx

[chilly radish]

Russel talks about membership, not inclusion

[urban acorn]

wdym by membership? set containment?

[chilly radish]

Set membership

$\in$

[cloud walrus]

k The Spring Constant

[urban acorn]

that is inclusion

[chilly radish]

No

Inclusion is $\subseteq$

[cloud walrus]

k The Spring Constant

[urban acorn]

that is not the standard meaning of inclusion

[chilly radish]

I've never seen set membership referred to as inclusion

[urban acorn]

well, it's semantics anyway

[chilly radish]

Well yea

My point still stands

[unique juniper]

Can I ask a question

[urban acorn]

Russel's paradox does indeed talk about $\in$ not $\subset$

[cloud walrus]

Intel

[urban acorn]

yeah, sorry for interrupting

[unique juniper]

np

If K/F is a cyclic extension, $\sigma$ the generator and $N_{K/F}(\alpha) = 1$ show $\alpha = \frac{b}{\sigma(b)}$

[cloud walrus]

Yes ツ

[unique juniper]

how shall I start on this?

[crimson falcon]

Is there any intuition behind all maximal ideals being prime ideals? I follow the proof, but i don't quite catch the intuition behind it :/

(which also means that if i were to try and prove it myself i would just smash definitions till something comes up rather than see how they are linked xD)

[unreal portal]

If $M$ is a maximal ideal and $ab\in M$ but $a\notin M$, then the ideal containing both $M$ and $a$ is the entire ring $R$. Therefore, 1 is in this ideal and so we can write $1=ax+m$ for $x\in R, m\in M$. Then we have $b=abx+bm$ which must be in $m$ because $ab, m\in M$. That's the general idea. You can also see it come from the fact that the quotient ring $R/M$ is a field. If $ab\in M$, then if we consider the cosets $\bar{a}, \bar{b}\in R/M$, we'd get that $\bar{a}\bar{b}=\bar{0}$. But in a field, this is only possible if $\bar{a}=\bar{0}$ or $\bar{b}=\bar{0}$, and by definition of quotient ring, either $a$ or $b$ is in $M$.

[cloud walrus]

cgodfrey

[crimson falcon]

hummm i'm gonna give it some thought. thanks! It still fills a bit like "just happens to work like that" but maybe i'm just too tired to see the big picture

[unreal portal]

It can also help to think specifically of these ideals in $\mathbb{Z}$. A maximal ideal $M$ in $\mathbb{Z}$ has the property that if you add any other number as a generator of the ideal, you end up with all of $\mathbb{Z}$. This is the same as saying that $1\in M'$, this modified ideal with the additional element. But since $\mathbb{Z}$ is also a PID, the ideal generated by $(n_1, \dots, n_k)$ is the ideal generated by $\text{gcd}(n_1, \dots, n_k)$. So if the maximal ideal $M$ has the generator $m$, we want the property that if we consider the ideal generated by $(m, n)$ (where $n$ is not a multiple of $m$), we get the entire ring. But this exactly happens when $m$ is prime, and thus $M$ is a prime ideal

[cloud walrus]

cgodfrey

[crimson falcon]

PID?

[unreal portal]

Principal ideal domain

every ideal can be written as generated by a single element

[crimson falcon]

Ah gotcha

[unreal portal]

This idea doesn't work in general, but I think it helps to see the general idea

If $M$ is generated by $m$ and $n\notin M$, $\text{gcd}(m, n)=1$ for all such $n$ is only possible if $m$ is prime since the gcd of a prime number and any nonmultiple is 1

[cloud walrus]

cgodfrey

[crimson falcon]

ohhh that last thing is starting to make things fall into place

I'll do some sketches and examples on Z, thanks a lot!

[unreal portal]

Just be careful not to get too focused on Z. Because it turns out in Z, all prime ideals are maximal but that's not true in general

It can be easy to do things that only work one way that don't hold in general

[crimson falcon]

Gotcha gotcha, will start from there and then move to general case 🙂

[unreal portal]

👍

[urban acorn]

The definition of a prime ideal is very naturally interpreted as requiring that the quotient be an integral domain, and the definition of a maximal ideal is very naturally interpreted as requiring that the quotient be a field.

• I is a prime ideal if for all a,b in R such that ab is in I, either a or b are in I
  • Interpreted in R/I, I is a prime ideal if for all a,b in R/I such that ab = 0, a = 0 or b = 0
• M is a maximal ideal if there is no ideal I of R strictly between M and R
  • Interpreted in R/M, M is a maximal ideal if there is no ideal I of R/M strictly between 0 and M

Exercise 1: Prove that for a ring R with ideal I, the ideals of R/I precisely correspond to ideals of R that contain I.

Exercise 2: Prove rigorously, and provide an intuition why, a ring R is a field precisely when it has no non-trivial proper ideals. (technical note: you also need 1 =/= 0. I consider rings to necessarily be commutative and with identity)

For further perspective relating to how an integral domain is a special case of a field, think of it this way:

• A ring R is an integral domain precisely when nonzero elements are not zero divisors.
• A ring R is a field precisely when nonzero elements are units
• Being a unit is stronger than not being a zero divisor

Exercise 3: For an element a of a ring, we get a map x -> ax. Prove that this map is injective precisely when a is not a zero divisor, and bijective precisely when a is a unit.

@crimson falcon this is more than enough intuition for why maximal ideals are prime

[cloud walrus]

Hausdorff

Hausdorff

[delicate orchid]

are the functions in these sets from C -> C?

[urban acorn]

I think they're continuous maps [0,1] to C

[delicate orchid]

ah then my idea doesn't quite work, ||I was thinking of showing that each coset in X/M is represented by {f \in X : f(0) = a} for some a \in C => X/M iso to C but I don't know if that works if the domain isn't C||

oh yes of course mb

[urban acorn]

@median pawn f,g are elements of C[0,1], and you know f(0), g(0). How do you determine whether phi(f) = phi(g) where phi is the natural projection X -> X/M

[urban acorn]

put that in a spoiler, they asked for a hint, and what I said provides them less of a solution and lets them do more solving themselves, so if that won't be enough for them they can look at your spoiler

[delicate orchid]

sorry lol

[urban acorn]

no problem lol

[delicate orchid]

I didn't realise it was actually a solution or I would've

[lunar spruce]

If I have a set S of tuples {(g1,...,gp) element of G^P | g1*... *gp = 1}. And defined a relation ~ with (g1,...,gp)~(gi+1,...gp,g1,...,gi) for 1<=i<=p. Why exactly is g1 = gi+1 = g2i+1 ... ?

[rustic crown]

the question isn't phrased nicely. i can take any set and definite any relation on it. that won't tell you much about the elements.

[south patrol]

i assume this is related to some proof of cauchy's theorem lol

[lunar spruce]

Yes

[rustic crown]

yep, that is clear. but for that you usually define an action of Z/pZ on that set S

if you're trying to show that the elements with orbit of size one are precisely the elements of the form a^p = 1

notice that they should be fixed by rotation by 1 unit

so g2 = g1; g3 = g2; ...

this shows all gi = a and since product of gi = 1, we get a^p = 1

[urban acorn]

that's not exactly true, rather, there's an action of the cyclic group of order p on the set S

and this relation precisely corresponds to orbits of the action

and by the orbit stabilizer theorem and the subgroup structure of the cyclic group of order p, the sizes of orbits are either p or 1

so then, since the size of S is divisible by p, the amount of orbits of size 1 must also be divisible by p

and these are precisely solutions to a^p = 1

since 1 is one solution, there must be at least p-1 more

[cursive temple]

i have no clue on how i should completely characterize the ideals of T

i have proved that the ideals of a matrix ring are completely determined by the ideals of the base ring

but that doesnt seem very applicable

[urban acorn]

ideals of rings restrict to ideals of subrings, so that already gives you some ideals to look at

then try to see whether there could be more ideals than those ones

like, those matrices in T where a=0 form an ideal, I believe (verify this)

and then intersections between these and the ideals given by ideals of Z are also ideals

my intuition is that these probably are everything, so try to see if you can show that

hopefully that gives you a direction to work towards

[sly crescent]

How come the symbols for Coxeter groups skip J through O?

[quaint tree]

Hey, just to confirm that I'm not being stupid with how I'm answering a problem, if we have two abelian groups G and H, the direct sum G⨁H is the same as the direct product G×H, right?

[urban acorn]

@quaint tree yes

[quaint tree]

Cool, I just wanted to be sure that it was just a name-change because additive notation instead of something else that looks kind of similar.

[urban acorn]

this looks like weird notation, but it's to do with the fact that G x H is constructed such that a homomorphism (from some group) into G x H is the same thing as a pair of homomorphisms (from that group) one into G, and one into H.

A direct sum, however, is defined such that a homomorphism from G⨁H (into some group) is the same thing as a pair of homomorphisms (into that group) one into G, one into H

[chilly radish]

Note that the two differ in the infinite case

[urban acorn]

and in the case of abelian groups, these are the same

[chilly radish]

Also yea, what Intel said, in Ab the product and coproduct coincide

[urban acorn]

they do? what am I missing?

[quaint tree]

Oh so I was right these are two different things instead of just a notation change due to using additive notation.

[chilly radish]

In the infinite case you define a direct sum to be nonzero in only finitely many coordinates

[urban acorn]

so in general they are different things, but for abelian groups they are the same thing

[urban acorn]

oh, right, my bad

@quaint tree yeah, so for infinite abelian groups I was wrong

[quaint tree]

Okay cool so I definitely need to read up on direct sums at sum point then, but for this problem I can probably just use what I know about direct products. Thanks.

(my abelian groups are finite)

[urban acorn]

then we're good

if you wanna read about the free product on groups, then that's the "direct sum" of groups in general

[chilly radish]

Not to be pedantic but It's actually for infinite Sums, the groups themselves can be finite

But you're just doing 2 groups

So it doesn't matter

[urban acorn]

oh, right

that's what had me confused too

[upper pivot]

in a direct sum of a family of groups (G_i lets say), elements are (g_i) where all but finitely many g_i are 0

in direct product we dont need that last condition

[rain oxide]

Not sure what topic in math this is from... But this is in a cryptography tutorial that has a bit of abstract algebra adjacent math. Does anyone know what function distance is

sorry fractional distance*

This is in the context of mapping a univariate function to a bivariate function to show that it disappears on a subspace. No clue what any of it meant, especially not the fractional distance from a polynomial in its first variable and a second polynomial in its second

[urban acorn]

I don't know what the fractional distance between Q and some polynomial f is, but I can explain the rest in terms of that

For some choice of d1, d2, this maps Q to the minimal "fractional distance" that Q has two some polynomial in two variables with coefficients in F, with d1, d2 being bounds on its degree with respect to both variables respectively

[terse crystal]

Hilbert’s theorem 90

[frank fiber]

Let $G_{n+1}=[G_n,G]=\langle \lbrace h^{-1}g^{-1}hg |h\in G_n, g \in G \rbrace\rangle$, then why $G_{n+1} \subset G_n$?

[cloud walrus]

Or x1

[frank fiber]

wher $G_n$, $G$ and $G_{n+1}$ are groups

[cloud walrus]

Or x1

[urban acorn]

It is not in general true for a subgroup H of G that [H, G] is contained in H, however, what you're trying to do is to prove a certain special case

[urban acorn]

critically what you forgot here is that you initially define G_0 = G

then you can prove this by induction

When H1 contains H2, [H1, G] contains [H2, G] (verify this), and you can use this for the induction step

[frank fiber]

thanks

[urban acorn]

Is there any nice structure on the Hom set between two (commutative, with 1 and unity preserving morphisms) rings?

I remember having a brainfart in the past where I thought for some reason that it was an additive group, but didn't realize that doesn't preserve multiplicativity

But is there any structure on it that might be interesting, besides just the structure of a set?

[frank fiber]

https://math.stackexchange.com/questions/982986/is-it-possible-to-turn-the-set-mathbbhomr-s-of-ring-homomorphisms-from-r

it says that homRing(𝔽2,ℤ)=∅ .

[frank fiber]

hom set you mean set of homomorfism or the set of functions?

[urban acorn]

of unital ring homomorphisms

[urban acorn]

well, maybe there's a structure that allows for the set to be empty

i.e. topological spaces can be empty

I'm looking for any interesting structure on them

[terse crystal]

I think any ideal I of T has the form I=I_r,s,t={A from T: A_11 is from rZ, A_22 is from sZ A_21 is from tZ} where t divides both r and s

[oblique river]

I don't think you're going to find any, unfortunately.

if you broaden your horizons a little bit and say things like

"look at all the ring homomorphisms from R to some field"

that's sort of the same as looking at all prime ideals of R

and that is a set which you can put a topology on

[urban acorn]

all maximal ideals

[oblique river]

no

[urban acorn]

or you mean an integral domain

[oblique river]

I meant what I said haha

I didnt' ask for the homomorphism to be surjective

[urban acorn]

ohhh

right

[terse crystal]

Clearly there exists a natural isomorphism mapping f to f(0)

[urban acorn]

well, still not quite exactly, I think

[oblique river]

if you want you can restrict to just maximal ideals and put a topology on that

[urban acorn]

because notice that we're fixing some field

[oblique river]

no I'm not

you were in your original question

I'm saying look at homs from R into any field

[urban acorn]

oh, okay

[oblique river]

in any case if you do want to restrict you can look at something like

finitely generated C-algebras

C = complex numbers

so like, quotients of C[x, y, ...]

[urban acorn]

alright

[oblique river]

all maps from those kinds of things to a field factor through C

[urban acorn]

makes sense

[oblique river]

errrr that's not quite true but like

basically all of them

all quotients of such rings by maximal ideals are isomorphic to C

is what i want to say

[urban acorn]

okay

I don't see your point though

[oblique river]

but yeah sorry this "special case" is just like "if you wanted to fix the codomain"

my point is that you can topologize "the set of prime/maximal ideals of a ring R" in an interesting and useful way

and that set is very related to "the set of all homomorphisms from R to some field"

so I'm just saying that while you can't put any interesting structure on Hom(R,S), you can with "homomorphisms from R to fields"

[urban acorn]

I actually want a structure which isn't a topology.

[oblique river]

fine, it's a scheme ;P

[urban acorn]

hmm, that might actually work

[oblique river]

https://en.wikipedia.org/wiki/Scheme_(mathematics)

[urban acorn]

idk any algebraic geometry yet

[oblique river]

what is your goal with this question? and why does "it's a topology" not work for you?

[urban acorn]

I basically just know "schemes are geometric objects that are 'locally like' affine schemes, which are the spectrums of rings"

[oblique river]

that's about all you need to know :P

[urban acorn]

I don't know the nature of how they are "locally like" affine schemes though

[oblique river]

that question was aimed at your initial inquiry

you asked if there was some kind of structure on Hom(R,S)

[urban acorn]

okay, anyway

[oblique river]

but topology was not an acceptable answer to you haha

so I was trying to ask what kind fo answer you were looking for

[urban acorn]

so, you know how given a topological space

and considering continuous maps into R

you can learn about the space by looking at the ideals of this ring

[oblique river]

(in the same way that a manifold is locally like R^n, if you know what manifolds are. if not, imagine the sphere. If you cut out a little patch of it and stand on it, you would just think you were standing inside R^2)

[oblique river]

yep

[urban acorn]

I was trying to find the right restrictions on a space (maybe compact and Hausdorff?) and "get back" some of the structure of rings on those spaces

[oblique river]

what do you mean get back?

[urban acorn]

so like, I originally had a brainfart where I thought ring homomorphisms were additive

so I said, we have a contravariant functor F : Top -> Ring

[oblique river]

I think compact hausdorff is what you need in order to say there's a bijection between X and maximal ideals of C(X, R)

[urban acorn]

then I was trying to figure out a way to sort of make the maximal spectrum functorial

which it typically isn't

but I was trying to see whether I can look at some nice case

or relax what it means a little (like add the entire ring as a potential point)

[oblique river]

just the spectrum is functorial though

[urban acorn]

yeah

[oblique river]

and you can always recover maxspec as the subspace of closed points

[urban acorn]

hmm

that might be the right way to look at it

[oblique river]

meaning if p is a prime ideal of R, then {p} is a closed subset of Spec R iff p is a maximal ideal of R

also like, what you are trying to do is basically what algebraic geometry is about, in some very coarse sense

[urban acorn]

except I'm looking at continuous functions instead of polynomials

and obviously I'm aware that many people smarter than me looked at this before

[oblique river]

like, there is an equivalence of categories between rings and affine schemes

[urban acorn]

I'm just trying to get a handle on it myself

[oblique river]

im suggesting that you shouldnt look at continuous functions and instead should look at polynomials :P

but also not even that, like, if your ring is C(X, R) for some topological space X

which is compact hausdorff

then you can recover X from Spec C(X,R)

well... not as a topological space....

just as a set

[urban acorn]

you can get a topology on it that is coarser than the original one

[oblique river]

while the main examples you'll see of schemes are like, spectrum of a quotient of a polynomial ring

[urban acorn]

maybe with a different ring than R, we can capture more closed sets as zero sets

[oblique river]

the whole point of schemes is that we dont have to restrict ourselves to polynomials

we can do it on any ring

[urban acorn]

hmm

I really wanna work towards learning algebraic geometry

but it strikes me as a bit difficult to get into

[oblique river]

I think it has a bad reuptation

but like, I think the geometry of varieties over C (and how that relates to the algebraic structure of their coordinate rings) isn't too difficult, and that's a good springboard to get into more abstract stuff

like, I wouldnt recommend just picking up a book like hartshorne unless you really know what youre doing

[urban acorn]

what's a good book for getting into it?

[oblique river]

hmmm

maybe something like this? http://www.math.lsa.umich.edu/~wfulton/CurveBook.pdf

ive never used it but ive heard it's good and a lot of the content looks pretty classical

i.e. not crazy abstract

the scope is also pretty narrow, instead of talking about varieties in general it focuses on curves

but once you have some background in the terminology (you dont even need the whole book for this) you could kinda jump into learning about schemes from another source

[urban acorn]

is a curve a variety in 2d?

[oblique river]

a curve is any 1-dimensional variety

not all of which can be embedded in 2 dimensions

[urban acorn]

what notion of dimension?

[oblique river]

the usual one

[terse crystal]

Speaking of algebraic geometry , I am reading algebraic geometry and arithmetic curves written by Q. Liu

[urban acorn]

topological?

[oblique river]

I mean, there is a precise definition of dimension, I'm just trying to communicate that for things like R^n it agrees with what you think

a 1-dimensional variety over the field R is going to look like what you think of as a "curve"

a 1-dimensional variety over the field C is going to look like what you think of as a "surface"

[urban acorn]

so, for example, x^2 + y^2 = 1 is a "curve" over C even though topologically it's 2 dimensional

right?

[terse crystal]

dimX=sup dim_x,X where dim_x,X=inf dimU where U is an open neighborhood of x.when U is affine then dim U is the Krull dimension of O_X(U)

[oblique river]

yes, $\lbrace (x,y) \in \bC^{2} \mid x^2 + y^2 = 1\rbrace$ is a complex curve

[cloud walrus]

Buncho Bananas

[oblique river]

it's the maxspec of the ring $\bC[x,y]/(x^2 + y^2 - 1)$

[cloud walrus]

Buncho Bananas

[terse crystal]

For a special case when X is an integral algebraic variety over a field k then dim X = trdeg_k(k(X))

[oblique river]

i'm impressed that you know such big words but I don't think that you're being very helpful right now.

[urban acorn]

okay, so I can see how that ring in some sense represents "polynomials defined over V(x^2 + y^2 - 1)", in the sense that it's polynomials over C^2 up to being equal on that set

[oblique river]

that's right!

[urban acorn]

what's a general statement we can make that explains how such rings have a maximal spectrum that gives us back the variety?

[oblique river]

https://en.wikipedia.org/wiki/Hilbert's_Nullstellensatz

basically, you can go in two directions: first, given an ideal I of C[x, y, ...] you can look at the variety it defines, which is basically maxspec(C[x, y, ...] / I)

second, if you are given some variety in C^n, you can look at the ideal of polynomials which vanish on that variety

the nullstellensatz basically says that those two processes are inverses of each other

(the "basically" in that sentence is covering up a slight technical detail, essentially that if you were silly and took the ideal I = (x^2) in C[x] to start, the variety you get would be just the point {0} in C, and then if you tried to go backward to get an ideal you would get the ideal J = (x), not I = (x^2))

(the ideal J is the radical of the ideal I, though, and that's the technical detail you need to make the statement work)

[urban acorn]

okay, so one of the compositions is the identity, and the other is the radical?

[oblique river]

that's right

assuming you actually started with a variety in the first place

[urban acorn]

which, if not, you would take a "closure" into a variety?

[oblique river]

yeah that's a good way to think about it

smallest variety which contains that set

so if you did something like, tried to input Z as a subset of C

the smallest variety containing Z is in fact all of C

i.e. there is no nonzero polynomial which vanishes on all of C

[urban acorn]

it's literally a topological closure in the Zariski topology, right?

[oblique river]

that's right

[urban acorn]

and this does give a bijection between radical ideals and varieties, right?

[oblique river]

that's right

[urban acorn]

okay, and this holds over affine spaces over fields

[oblique river]

that is on page 11 of that curves book i linked earlier

:)

[oblique river]

i think you want the field to be algebraically closed

but otherwise, yes

[urban acorn]

so if I take V(x^2 + y^2 - 1)

is it in some sense "isomorphic" to a - possibly lower dimensional - affine space over a - possibly different - field?

[oblique river]

V(blah) isn't defined unless you specify which field

[urban acorn]

C

[oblique river]

as a whole it wont necessarily be -- just imagine replacing C with R

the unit circle isn't isomorphic to R

over a field like R or C, varieties will be manifolds with singularities

if you use the standard subsapce topology of the standard topology on R^n or C^n

[urban acorn]

okay, so, I have a question about schemes

[oblique river]

hit me

(meaning: let me hear it!)

[urban acorn]

so affine schemes are spectrums of rings, and they're considered in the Zariski topology, and with a structure called a "sheaf" (which I don't understand yet), that describes how a ring is associated to each open set

so, first thing, are these rings on the complement of V again R/(subset that induces V)?

[oblique river]

no, they'll be more complicated

[urban acorn]

okay, so they'll properly contain them?

[oblique river]

no it will be unrelated to that quotient of R

the ring associated to some open set U will be the "algebraic functions on U"

as an example, let's take R = C[x]

whose spectrum is just C

and let's take the variety V = {0} which comes from the ideal I = (x)

and U is the complement of that

[urban acorn]

with an extra point for the 0 ideal, right?

[oblique river]

ah, yes, that's right, although topologically it's sort of a weird point

it's not a closed point and so you shouldnt really visualize it as living in one particular point of space

think of it like a bowl of pudding with raisins

the raisins are the closed points corresponding to the maximal ideals (x-a) for a in C

and the ideal (0) is the pudding which holds all the raisins together

[urban acorn]

okay

[oblique river]

but yes that is there, but to be clear, I'm talking about the complex number 0 which corersponds to the ideal (x)

I'm not talking about the "generic point" which corresponds to the ideal (0)

[urban acorn]

yeah

understood

so U corresponds to the nonzero complex numbers