groups-rings-fields

[barren sierra]

if a + I = I then no such c exists

[lethal cipher]

Yea, that makes perfect sense.

Okay, so we'll say J is an ideal containing I.

So, let's take an element in J, call it a for old time's sake, and see what we can get from a+I.

Well, since R/I is a field, a+I has a mult. inverse, call it c+I.
Then, using the ideas from before, we have that ac-1 is in I.

However, from here this tells us that there is an i such that ac-1=i -> ac-i=1. But ac is in J and -i is in J, so this means 1 is in J

So J must be R

[barren sierra]

all correct, minor issue

you can't just take any a

a must be an element in J \ I

[lethal cipher]

a in J not I

Yea, thank you for correcting me there

[barren sierra]

otherwise yes that's correct, so now you've shown that I is maximal iff R/I is a field

[lethal cipher]

This was very helpful, thank you very much Spamakin

[barren sierra]

ok so back to this

can you show that k[x_1, ..., x_n] / (that ideal) is isomorphic to a field?

basically what was said here

[lethal cipher]

Ah, that is the final goal of the problem in the book is to show it's isomorphic to k

Which would imply that that ideal is maximal through what we discussed

Okay hmm. So let's start with n=1, maybe that will help. Obviously, the point a_1 maps to I

[barren sierra]

can you be more explicit?

ok so n = 1

what is your map k[x] -> k?

[lethal cipher]

I'm trying to figure that out tbh.

[barren sierra]

well what do you want your kernal to be?

you want to show (x - a) is maximal

[lethal cipher]

What about x maps to I(x)+I?

[barren sierra]

what is I? we're mapping to k

k has no proper ideals except (0)

[lethal cipher]

It's just variable confusion. I meant x in k.
Let K in k, then my idea is to map K to I(K)+I (I(K) is the set of poly's that vanish at K)

[barren sierra]

what is I

[lethal cipher]

Oh yeah, I is our ideal

[barren sierra]

how can you map a single element to an ideal

you need to map elements of k[x] to elements of k

elements of k aren't ideals...

[barren sierra]

answer this first

you want a surjective map k[x] -> k with what kernel?

(think isomorphism theorems and what we want to show)

[lethal cipher]

Well, if we're going iso, we definitely want I to map to 0 (I forgot k was a ring for a sec)

[barren sierra]

you want elements in (x - a) to map to zero yes

ok ok what do you need to do to define a k linear map k[x] -> k?

[lethal cipher]

Sorry, I'm back. Umm, one thing we could do is sum the coefficients. I dunno if that plays well with what we're doing though.

[barren sierra]

answer this

or ok let me be more direct

what must constants in k be mapped to?

[lethal cipher]

Oh, well, the natural thing would be to have them map to themselves

[barren sierra]

yea they have to

since you need 1 -> 1

and the map, call it f: k[x] -> k, has to satisfy f(k) = k * f(1)

ok so now all that's left is to decide where to map x, correct?

what do you want your kernel to be?

(think isomorphism theorems and what we want to show, that k[x] / (x - a) is a field)

[lethal cipher]

Wait, one sec. Why does f(k)=k*f(1)?

[barren sierra]

well it has to be a k-linear map

linear over the field k

or I guess it doesn't have to be

[lethal cipher]

Ah, okay I see. You pulled out the scalar

[barren sierra]

yes that's all I did

but ok lets say elements of k map to themselves

note this means that our map f: k[x] -> k is already surjective, a useful thing because again I want to use isomorphism theorems

[barren sierra]

so answer this

what do we want our kernel to be

and use that to decide what to map x to

since if we decide where to map x, and we decide where to map scalars, then we've completely determined our homomorphism f

[lethal cipher]

It feels like it'd make sense to map x to a

[barren sierra]

why

[lethal cipher]

I'm still trying to figure out a good reason for it...

[barren sierra]

what isomorphism theorem do you want to use

[lethal cipher]

Probably first iso. We want k[x]->k to be a homomorphism. We already determined our map is surjective, so the map from the quotient field to k will be an isomorphism.

[barren sierra]

quotient of what by what will be isomorphic to k?

[lethal cipher]

k[x]/(x-a) will be iso to k

[barren sierra]

yea since the kernel will be (x - a)

ok so overall what have we done

we defined a surjective homomorphism f: k[x] -> k by mapping scalars to themselves and x to a. This means ker(f) = (x - a). So like you said k[x] / (x - a) is now isomorphic to k

so what can we say about the ideal (x - a)?

[lethal cipher]

Wait wait wait, have we shown it's a ring homomorphism yet? Or are we just saying f is one (whatever f may be)

[barren sierra]

I mean it's not hard to show

I'll leave that to you

[lethal cipher]

So are we mapping x to a? You didn't really elaborate on that

[barren sierra]

yes

because to define a ring homomorphism k[x] -> k you just need to specify where scalars go and where x goes

since the homomorphism property then allows you to just map arbitrary polynomials

[long geyser]

do you not have access to a theorem like this? super useful

[lethal cipher]

It's gonna be maximal

[barren sierra]

for example f(x^2 + 1) = f(x^2) + f(1) = f(x)^2 + f(1) = a^2 + 1

[barren sierra]

yup!

ok so now you've shown that (x - a) is maximal

do the n variable case

it's extremely similar

[lethal cipher]

So follow up, did the choice of what x mapped to actually matter

[barren sierra]

I'll leave it to you

[barren sierra]

yes because we need kernal of f to be (x - a)

[lethal cipher]

Yeah, the n case would be to map each x_i to a_i and leave the rest the same

[lethal cipher]

Okay, that's what I was thinking too

Thank you again Spamakin, this has all been very helpful

[noble lynx]

which proof of quadratic reciprocity is the most intuitive?

i am having trouble understanding the motivation of the gauss sums one

[delicate bloom]

I think the intuition is ||Gauss was just a literal wizard and did magic sorry||

[rotund aurora]

Growing up is realizing that proofs need not be "intuitive"

[delicate bloom]

eh I wouldn't say that

but it might not be worth your time to really drag yourself into a place of feeling intuition

[cobalt heath]

Honestly, reciprocity is a magic, really

Otherwise we wouldn't have as much problem on it, I think?

[delicate bloom]

Well, I only condone magic in so far as it is motivating to learn more about how it works, not as a way of giving up as something not being understandable

[dim widget]

The point of the gauss sums one is that if p is an odd prime then the quadratic gauss sum g_p generates the unique quadratic subextension of Q(\zeta_p)

By standard facts about gauss sums this turns out to be $Q(\sqrt{p^})$ where $p^ = (-1)^{(p-1)/2}p$

[cloud walrus]

Math_Discord_Final_Girl

[dim widget]

then if q is another odd prime the galois group $Gal(F_q)$ acts on $Q(\zeta_p)$ by $\zeta_p \to \zeta_p^q$. This fixes the Gauss sum if and only if $q$ is in the unique index 2 normal subgroup of $Gal(Q(\zeta_p)/Q)$ if and only if q is a square mod p. On the other hand by Galois theory $Frob_q(\sqrt{p^}) = \sqrt{p^}$ mod q if and only if $p^*$ is a square mod q.

[cloud walrus]

Math_Discord_Final_Girl

[dim widget]

this is how I think about what’s going on with that proof of QR anyway

[cedar vault]

how did they go from $f(x)f(k)f(x^{-1})=f(k)$?
Shouldnt the simplification look like:
$f(x)f(k)f(x^{-1})=f(x)f(x^{-1})=e$

[cloud walrus]

bathroom mug

[spice whale]

yeah

[cedar vault]

Oh, and how do i prove the part that says any permutation sigma is the product of $i$ transpotions, with $i$ being unique upto parity? Im thinking i let n be the least number such that $x_1 x_2...x_n=\sigma$. Then any other sequence of transpositions $y_1 y_2...y_k$ that you can add on to $x_1 x_2...x_n$ such that $x_1 x_2...x_n y_1 ... y_k=\sigma$ will have to first take a number out of its place and put it back. Since every number will be moved an even number of times by $y_1 y_2...y_k$, the parity of n will be preserved, since n+2j = n mod 2.

[cloud walrus]

bathroom mug

[cedar vault]

Any flaws with the argument?

One final question. How did they figure out that <(123)> ≤ S3 is the kernel of the sign homomorphism f(x)= (-1)^{i} where i is the number of "factors" of x?

Actually i can see why it makes sense. All elements in <(123)> require an even number of transpotions to achieve, and all the other elements of S3 require an odd number of transpotions to achieve, so it makes sense to map it to (-1)^i. But why is it that all permutations generated by <(123)> require an even number of transpositions to achieve, other than "it is true by computation"?

[dull ginkgo]

Product of two nonequal, nondisjoint transpositions is a 3-cycle

like (i j)(i k)

But also has even parity

[cedar vault]

makes sense

thaks!!

[dull ginkgo]

You’re welcome

[cedar vault]

why must the red box be true? suppose we have
exp:(C,+) ---> (C,*) exp(z)=e^z
and consider the subgroup H= {x:x in R}. H is not contained within the kernel (in this case the set{0}) but it is mapped to the subgroup H'={e^x: x in R}

[sage lodge]

are you sure the kernel is {0}?

[cedar vault]

{2*pi*i*k, k in Z} is the kernel

so lets change the domain and range to (R,+) and (R,*), with exp(x) ---> e^x

point still holds, unless im missing something

i mean for nontrivial G, G itself is a subgroup of G, and it isnt contained within the kernel. yet G can be mapped to the subgroup f(G) of G'

[sage lodge]

maybe it means not containing the kernel?

[cedar vault]

~~thats impossible (subgroup must contain e which is in kernel) ~~ nvm, you said not containing the kernel

[fading field]

the aforementioned group also doesn’t contain the kernel

idk why they said this tbh, what book are you using?

[cedar vault]

these are lecture notes provided by mit. i think i should change sources because this is the second mistake im seeing on my first day of learning...

[fading field]

like subgroups of G are literally always mapped to subgroups of G’ by homomorphisms

[sage lodge]

wdym

in the example with C as the domains?

then H isn't mapped to by H', f^{-1}(H') will be larger than H (e.g. since it contains 2*pi*i)

[fading field]

oh mapped to by any subgroup of G’

[rocky cloak]

I think they meant to say 'any subgroup not containing the kernel'. Now it sounds like they're saying it has to be contained in the kernel...

[sage lodge]

not containing the kernel seems right

because they show that ker(f) is contained in every such preimage f^{-1}(H')

[cedar vault]

yeah thats what i thought too

i think this might actually be helping me in learning because im having to verify everything written in here 😆

[tardy hedge]

Yes, that is pretty intuitive to me!

But that way doesnt work with just any old cosets

Non normal coset

[south patrol]

It does work in general

indeed this works for any equivalence relation

you only need a set theoretic map G -> G/H (or G/~ for a general relation)

@tardy hedge

[sage lodge]

Is there a way to see that the set of elements whose a-exponents and b-exponents both add to zero is the commutator subgroup of F2 (the free group generated by two elements a and b)?

or for example, how to write the example element a^7 b^(-5) a^(-10) b^5 a^3 as a product of commutators of F2?

[tardy hedge]

oh yeaaah doing that in general doesn't necessarily have to do with kernels

[dull ginkgo]

@sage lodge the commutator subgroup is the kernel into the group’s abelianization, i.e the “most general” abelian group it maps into.

I.e if G maps into Abelian group H, then there is a map from G’s abelianization into H

Now. If the exponents of an element all sum to 0, then consider any Abelian group it maps into. Since the map allows the image to commute, all the powers of that element in the string will cancel out. So if this happens to every element in the string, then the string is sent to 0, I.e is in the kernel of that map into the Abelian group

And this happens for every abelian group the original group maps into, thus strings where the sum of powers of each element is 0 is in the kernel of every map into an abelian group

So therefore those special strings are in the commutator subgroup

But what I leave to you is showing those are the only elements in the commutator subgroup :3

(Assume the sum of powers for one element in the string ISN’T 0 but is in the commutator subgroup, what happens to it under a map into an abelian group?

[tardy hedge]

but i dont see how just a set theoretic map tells us why non normal subgroups partition G

[sage lodge]

I'm trying to understand why this part follows

the preceding sentence makes sense to me

[dull ginkgo]

Well they are sent to the identity for every map into an abelian group

And quotienting out by the commutator subgroup sends it to an abelian group

so poof, they go to the identity and are in the commutator subgroup

[sage lodge]

ohh got it

[dull ginkgo]

You can also define the commutator subgroup to be the intersection of normal subgroups such that their quotient is abelian

[rocky cloak]

A fairly explicit way to do this is that whenever you have a product multiplying by the commutator swaps them.

For example in your case if we multiply
a^7 b^-5 a^-10 b^5 a^3
by
b^-5 a^7 b^5 a^-7
we get
b^-5 a^-3 b^5 a^3

If you continue multiplying by commutators until you reach the identity, then what you started with is the inverse of what you multiplied by

[dull ginkgo]

oooooh that’s smart

[sage lodge]

wow

that's what I was trying to come up with

that makes so much sense

up to multiplying by commutators, we can just reorder the elements, so of course they will cancel out to the identity

[dull ginkgo]

Excuse my extremely shategoric way of doing things lol

also I realized I basically used the same thing jagr did to prove the reverse implication lol

So I could’ve started with that

[rocky cloak]

If your want to prove it, I think your argument is nice. But if you actually want to know how this is a product of commutators, then you need a more concrete approach

But if you want to turn this into a proof you need some ugly induction

[dull ginkgo]

My first exposure to the commutator subgroup was when I was tasked with funding the intersection of all normal subgroups such that G/H is Abelian lol

[tribal moss]

That must take some fancy grantwriting.

[mighty kiln]

Universal property of abeluanization

[dull ginkgo]

what

[tribal moss]

Funding.

[mighty kiln]

Wait my typo was unintentional

[dull ginkgo]

Funding for what?

[tribal moss]

The intersection of all normal subgroups such that G/H is abelian.

[dull ginkgo]

It’s easy

[rocky cloak]

You made a typo in your original comment

[mighty kiln]

I wish funding was easy

[cedar vault]

the fact that f is surjective is used in ensuring that f^-1(H') has a preimage. what is this revising the statement that they are talking about?

forgive the typos i am heavily intoxicated

[tribal moss]

I think the revision they have in mind is to replace H' subseteq G' by H' subseteq img(f).

[cedar vault]

ah okya thats actually written in the line above i thoight they had some other condiiton

also i think in the line above they shouldve said subgrups of G' not subgroups of H'

[tribal moss]

Yeah.

(That does makes it a bit strange that the author would tease the revision in a footnote and then without comment make that very revision in the prose following the formula ...)

[dull ginkgo]

OOOOH I see

[cedar vault]

Maybe because the authors were students

[tribal moss]

That would make sense.

[upper vigil]

abuelanization

[dull ginkgo]

Tldr:

By definition the commutator subgroup, K, is the smallest normal subgroup containing all the commutators.

But let H be a normal subgroup, G/H be abelian, and h be the projection. Then for each a, b, h(ab) = h(ba), therefore h(ab)h(ba)^-1 = h(ab a^-1 b^-1) = e, ergo [a,b] is in H so H must contain all the commutators, and therefore must contain the commutator subgroup.

Thus the intersection of all normal subgroups such that ditto contains the commutator subgroup. However, the projection k for G into G/K is abelian for the same reason and thus must contain the intersection. Thus they are equal

[velvet steeple]

How much of linear algebra is covered in abstract algebra again?

[terse crystal]

Everything

Like basic algebra by Jacobson, chapter 3,6,7 cover basically all basic stuffs in linear algebra as special cases

(Volume 1)

[velvet steeple]

Is this also when taking a course on AA?

[terse crystal]

Idk

Linear algebra is just special case

[velvet steeple]

One might as well start at abstract algebra then

[terse crystal]

Yeah

More general

[velvet steeple]

How is it a special case? In what sense?

[terse crystal]

Like chapter 3, structure of finitely generated modules over PID, one special case is when the pid is F[x], so standard form in terms of similarity of matrices is just a special case of the result

Chapter 6, quadratic forms, similar, linear algebra just contains few special cases

[velvet steeple]

You mean like Frobenius and Jordan normal forms?

[terse crystal]

Yeah things like that

[velvet steeple]

Alright, thanks

[velvet steeple]

So abstract algebra is very similar to linear algebra?

[terse crystal]

Abstract algebra is algebra

Linear algebra contains special cases of it

[limpid ferry]

How should I understand the last part of this question? :"Find all subgroups of D_4 (dihedral group) and determine which one is normal. What are all the factor groups of D_4 up to isomorphism?"
I did not pay attention in class at some points so maybe this is where I missed.

[prisma ibex]

factor group is a funny way to say quotient groups

so you're just working out what the normal subgroups and the corresponding quotients are

[ornate tiger]

can someone remind me why S_2 = {id, (12) }

um

oh ok what about S_3

and why A_3 = {id,(123),(132)}

but (123) and (132) are not pairs

ok 1 goes to 2 but where does 2 go

according to (123) it goes to 3

but in the right side

we start from the number on the right ?

like 2 from the right

😭

oh i remembered something

like if we had (1234567) = (17)(16)(15)...(12)

thx

[limpid ferry]

What is "up to isomorphism" in this question? I dont know what that means

[coral spindle]

Up to {equivalence relation} means that you will consider things related under that relation to be the same

[prisma ibex]

you're just having to list which quotients appear, not necessarily how many times they appear distinctly

[coral spindle]

So e.g. despite there being infinitely many distinct groups of order 3, there is only one up to isomorphism

[prisma ibex]

like in Z/2Z x Z/2Z the quotient Z/2Z appears in several different ways but they are all isomorphic as quotients

[kind niche]

reading an article and this theorem has no proof, the book it uses as reference cant be found as a pdf
does anyone know a proof / a free place i can find it

[dim widget]

Yes: the point is that if S is a finite cardinality n multiplicative subgroup of a field F, then (X^n - 1) has exactly n roots over F. But we know that amongst the roots of X^n - 1 is a primitive nth root of unity r (by the theory of cyclotomic polynomials) where r has exactly order n, thus S is cyclic and generated by r

In the special case that F is finite and S is F - {0} this gives your result, and alpha is the same as the r that I mentioned

To elaborate on the “theory of cyclotomic polynomials” part: because S is a finite abelian group there is an element e such that the order m of e is divisible by the order of all the other elements in the group. But then all elements of S satisfy X^m - 1, which is impossible by counting unless n = m

[tribal moss]

Alternatively: By the structure theorem, the multiplicative group F^× is a product of cyclic groups, and if F^× is not cyclic, there will be two of those factors whose order is a multiple of the same prime p. But then there will be at least p² solutions of x^p=1, which is not allowed in a field.

[barren sierra]

How is $(\mathbb{Q}_{> 0}, *)$ is a free $\mathbb{Z}$ module since say $2 \cdot \frac{1}{2} = 1$

[cloud walrus]

Spamakin🎷

[barren sierra]

Right because I'd want to show the only torsion element is 1 since that's the identity of the underlying group

but haven't I just "shown" that 1/2 is torsion?

obv there's a hole in my logic but where

[rocky cloak]

You seem to be getting confused by the difference between multiplicative and additive notation

[barren sierra]

How so? the underlying group is positive rationals under multiplication

the identity is 1, not 0

infact 0 isn't even in Q_{> 0}

[rocky cloak]

2 * 1/2 does not mean apply the operation to 1/2 two times. That would be (1/2)^2

[barren sierra]

Ahhhh

ok ok that makes sense ty

[limpid ferry]

So for the question posted. I found 4 normal subgroups of D_4, let just denote as A,B,C,D. Am I good to just answer the last question as:" The factor groups are D_4/A, D_4/B, D_4/C, D_4/D " ?

[prisma ibex]

yes but you should work out what those quotients actually are more explicitly

[prisma ibex]

also assuming D_4 here is the dihedral group with 8 elements, you should have 6 normal subgroups

I guess you have found the 4 normal subgroups that aren't just 0 or D_4 itself

anyways one of these should have quotient the Klein four group Z/2Z x Z/2Z, the other three should have quotient Z/2Z

[long geyser]

is there an error in their example?

I agree that J is equal to matrices that are 0 in the second column

but JM =\= J like he states

JM should be, matrices that have linearly dependent columns is that correct?

actually, idk if that is even a submodule so maybe I'm screwing up somewhere

anyway, JM should be JM as defined, set of am with a in J and m in M, but that can't possibly be equal to J

[south patrol]

JM=J just because it's an ideal

Them writing M for R is annoying lol

[long geyser]

but it's a left ideal no?

[south patrol]

But note JR is contained in J as it's an ideal, but also 1 is in R

Ooh

Oopies

[rocky cloak]

JM = M here.

There's also the mistake that JM should be defined as the submodule generated by am, not just the set am

[south patrol]

Mb sorry

[long geyser]

I think in this example the set am works because it is a specific construct of a left ideal

I checked

[south patrol]

Still doesn't work I'm pretty sure

[long geyser]

this is the following

[rocky cloak]

Yeah, it works because J is principal, but not in general

[south patrol]

in general anyway

[long geyser]

hmm

wait

[rocky cloak]

I see, I thought they were defining it. But they were just saying they were equal I guess

[long geyser]

when I checked I used that r is in the center, but it is not

let me reset

okay maybe you are right, I'm not sure if the set of am is a submodule

and yeah, it really does equal M

well, that example has like 4 mistakes

not good!

[true ingot]

Could someone find a mistake regarding my reasoning about projective linear groups?

So we know $PGL(n,F) = GL(n,F)/Z$ (where Z are nonzero multiples of the identity).
But we also know $PSL(n,F) = SL(n,F)/(Z \cap SL(n,F)$. However $SL(n,F)$ is normal in $GL(n,F)$, so by 2nd isomorphism theorem we should have also $PSL(n,F) \cong <SL(n,F),Z> / Z$. But isn't $<SL(n,F),Z> = GL(n,F)$, from which it would follow that always $PSL(n,F) = PGL(n,F)$? Where did I make a mistake?

[cloud walrus]

Faputa

[rocky cloak]

<SL(n, F), Z> doesn't necessarily equal GL(n, F)

Like if F is the real numbers and n=2, then the first only consists of matrices with positive determinant

In general <SL(n, F), Z> will consist of those matrices with determinant an nth power in F

[true ingot]

Ahhh! Okay thank you that makes sense

[long geyser]

I proved that finitely generated implies the chain thing, how would I prove the chain thing implies finitely generated?

[rocky cloak]

Say M = (m1, m2, m3, ...) and consider
Ai = (m1, m2, ..., mi)

[long geyser]

(m1, m2, m3, ...) goes on how long?

[rocky cloak]

Long enough to generate M. I guess this only works if M is at most countably generated though...

[long geyser]

yes

I considered stuff like well-ordering, but then you are no longer a limit of finitely generated so it wouldn't work

[rocky cloak]

Wait, no it works fine

Because any decreasing sequence of ordinals terminate

[long geyser]

but if you are past countable land, A_kappa for a cardinal kappa no longer is (necessarily) finitely generated either, no? how can it work?

[rocky cloak]

So M is the union of A_i, since the union of proper submodules is proper, A_kappa = M. By reordering finitely many elements we may take kappa to be a limit ordinal. But then A_kappa is the union of A_i : i<kappa

Rinse & repeat

[long geyser]

hmm, I get the idea

though this book doesn't even cover transfinite induction until 50 pages later, and even then I doubt the book incorporates a fact like "decreasing sequence of ordinals terminate"

tbh I've seen enough mistakes from the book that maybe this is the solution

just can't think of anything that doesn't use AoC

[tribal moss]

It's weirdly worded that the problem defined a "chain" as an arbitrary totally ordered family that explicitly doesn't have to agree with an ordering of the index set -- but then suddenly jumps to speaking about "every ascending chain" as if that is to be understood as a special kind of chains.

[long geyser]

well, it is ascending in terms of inclusion, I don't think the index set has any structure here

[tribal moss]

But would it mean anything different just to speak about "every chain of proper submodules"?

[long geyser]

nope

or

it wants to talk about a situation like A_i_1 cc A_i_2 cc ...

I guess, if it was descending, then it is automatic

[tribal moss]

But saying "i_1, i_2 ..." there would seem to refer to an assumed structure on I, right?

[long geyser]

no, I'm just looking at a subfamily of the (A_i) where we have a minimal element, call it A_i_1

and the rest of the subfamily also has a minimal element

call it A_i_2

is this not a legitimate thing?

well, I don't think it matters and (I believe) you are right, you could talk about any chain here

oh, I don't think minimal works here, call it minimum

or least

[lethal cipher]

Okay, so I have questions about the last paragraph of this proof. By bringing the degree lower, i do understand that difference is in I'.
However, why do we choose to subtract the Q_i F_i's in the first and the Q_j F_mj's in the 2nd.
Doesn't the F_i one accomplish the goal in both cases?

[long geyser]

you want to cancel out the leading term in G, if the leading term has degree less than N, it may have degree that is smaller than the degree of some F_i (this is how N was chosen)

and that F_i cannot contribute to the linear combination, because multiplying by a Q_i can raise degree but not lower

and since you may need to use the leading coefficient of F_i to match the leading coefficient of G's leading term, you are in trouble

[lethal cipher]

Ah, i see. And N being larger than the degree of all of them makes the first case when deg(G)>N possible. That makes sense now. Thank you

[wraith swan]

Got some time again 🙂 I am working on an exercise to show that the free group on a finite set of n elements is isomorphic to the coproduct Z^n. And the exercise mentions to just look at the universal properties. I have also checked a solution that just says oh this and this commutes and done. But I am missing something in that reasoning. Can someone check this for me?

I have already showed that Z is the free group on singleton.

[slim kayak]

Yeah, the argument is often a bit poorly explained

[wraith swan]

How do i go from gamma_k(1) = psi phi_k(1) to gamma_k = psi phi_k?

[slim kayak]

The coproduct X is defined by its universal property. So once you showed the free group over a finite set Y has the same universal property you can then do the following:
You get a unique map X to Y commuting with the inclusion maps. And switching places you get an unique map Y to X commuting with the inclusions

Call these maps f and g, then g o f : X -> X is a map that commuted with the inclusions into X. But the universal property says such a map is unique

Notice then, that the identity also commutes with the inclusion. So g o f must be the identity. Same in reverse and you get that f and g are actually isomorphisms

[wraith swan]

Ok let me check if my proof follows it

I just saw that I use Z^n for the coproduct but that also works in Ab. Is \amalg^n Z notation used for such a coproduct? or \sum^n Z?

[slim kayak]

$\amalg$

[cloud walrus]

Kerr

[slim kayak]

Yeah that

[wraith swan]

ok. Ah there is just a bigger \coprod

[slim kayak]

To throw in my own question: If I know that if B is a finitely generated A-algebra and $B \otimes_A S^{-1}A$ is finitely generated as a module, can I say the same about B?

[cloud walrus]

Kerr

[slim kayak]

In this case S is just one set, say the right hand side is the fraction field if A is integral

[wraith swan]

Oh I get what you mean now. Problem is I haven't show yet that the free group has the same universal property as the coproduct

[slim kayak]

Well, do that then

That's usually the first thing you show after defining the free group over some generating set

[wraith swan]

That's where I am stuck on the second to last line

The solution I read just doesn't mention that part and says well they commute.

[slim kayak]

(Nvm, take A=Z and B=Q)

[tulip robin]

Let $R$ be a ring such that we either have $a^2 = 1$ for all $a \in R$ or there is some $n \geq 1$ such that $a^n = 0$ for all $a \in R$.

[cloud walrus]

reach

[tulip robin]

Isn't R = {0} always? Since in the first case we have 0 = 0*0 = 1.

[rocky cloak]

Yes, and in the other case you have 1 = 1^n = 0

[tulip robin]

I'm confused because i have three exercises to prove something about the group of units of R

[slim kayak]

Maybe? You'd have to draw out a different diagram afaik.

I want to note that the argument I laid out above works for any universal property, so I suggest maybe going that route. The argument shown does seem to make use of the universal property of a free group so you probably defined such

[rocky cloak]

Are you sure the question isn't saying that a^2 = 1 for each unit in R then...?

Or perhaps that for all a either a^2 = 1 or a^n = 0

[wraith swan]

yeah this is my argument. The solution I saw went the same way but just says states that the diagram commutes and done...

Another solution just defines all morphisms and shows there is a unique morphism. It doesn't really use the universal property

[tulip robin]

ahhh yes that's what is meant for sure

@rocky cloak so for each a either a^2 = 1 or a^n = 0

[rocky cloak]

Yes, so here you have examples like Z/2, Z/3, Z/2[x]/(x^2), Z/4

[tulip robin]

@rocky cloak I see, tyvm

[slim kayak]

Alright. I suggest you check that the free group fulfills the universal property of a coproduct given a Z for each x_i mapping 1 to x_i in F(X)

[wraith swan]

ok

[slim kayak]

Once you have verified that you can convince yourself that any two objects sharing the same universal property must be isomorphic. Emphasis on the specific universal property not being important here

[wraith swan]

yeah that part I have already proven earlier in general. Or that is by definition

[slim kayak]

Kind of, it might not be immediately obvious

Okay looking back at the image that's what you more or less what you tried

You might want to draw out the diagram sequentially. Given for x in X a map Z to F(X) consider a collection of maps into some G.

[wraith swan]

Aren't those the phi_k?

[slim kayak]

Then construct a map from X to G with these, then obtain the unique morphism from F(X) to G.
Then check it is compatible with the diagram of the Z. The uniqueness of the map follows the map from X into F(X) specified is unique, and then the induced map is unique

[slim kayak]

Yeah

It's a bit weird to write them with a dotted morphism, since you are basically starting with them

[wraith swan]

No. it's the unique morphisms from F({*}) ~= Z to F(X) by the universal property of F({*})

[slim kayak]

Yes, but you want to show that F(X) has the universal property of the coproduct

So you start with maps Z into F(X) and then extend that diagram into the free group diagram

I mean, sure

[wraith swan]

Oh like that. Ok let me try that

[slim kayak]

That's where you got those maps into F(X) in the first place, but you still try to use the X, {x_k} and so on as "scaffolding"

[wraith swan]

This is the solution I found online. For F({x, y}) isomorphic to Z * Z

It also shows g is forced by the diagram but then it really quickly shows psi is unique sucht that the above morphisms commute

I don't get how he concludes "Then we have a unique homomorphosm ..."

[slim kayak]

I figured this is faster

I love my convoluted glued together diagrams, but I am not sure it adds much in this case

@wraith swan does this work for you?

[wraith swan]

I'm trying to understand what you did different than what I did. Thank you for working this out for me.

Oh I see it now. Thank you Kerr.

[slim kayak]

Mostly just explicitly checking in the end. beta(x_i) = alpha_i(1) is just stating that the diagram below commutes

But the universal property already said that alpha is the unique map making it commute. So alpha=beta

[wraith swan]

oh ok

That looks the same as my

And thus gamma_k(1) = psi(x_i)

[slim kayak]

Maybe? I can't honestly tell what that step was for

[opaque wind]

Hi, can somebody explain how this is a decision procedure for the word problem for a finitely generated free group F_n? It seems to me like it's only a decision procedure for the question "is the reduction of w trivial?"

A link to a better proof would also be fine

[rocky cloak]

Two words v and w can be reduced to each other iff vw^-1 can be reduced to 1.

[opaque wind]

right but the above algorithm searches for an expression x_i*x_i^-1 in w though, is it missing the first step where it multiplies both sides by rhs^-1?

[rocky cloak]

The point is that determining if a word is 1 is completely equivalent to determining if two words are equal. So which one you call "the word problem" is not particularly relevant. This is a solution to the former

[opaque wind]

well let's say you want to determine if xyz = xyz, the algorithm goes directly to step 3 because no subexpression of w contains a generator followed by its inverse, and it halts with the answer "no"

so I don't see how it is a solution

unless you first multiply on the right by the inverse of the rhs

[rocky cloak]

So there's two different problems you might call "the word problem"

'Given v and w are they equal?'

'Given w does it equal 1?'

A solution to one immediately gives you a solution to the other, so solving either is sufficient. This is a solution to the second one.

If they explicitly said they were only considering the first problem, then this is a minor mistake yes

[opaque wind]

let's see

this is how they define a solvable word problem

[rocky cloak]

Right, so here they're only giving you the solution for the case g=1. But you can easily extrapolate the solutions for the other values of g.

[opaque wind]

ah, I guess, I didn't look at it that way

Anyway, I get it now, thanks

[true ingot]

Posted in #combinatorial-structures but I am not sure if it's perhaps better suited here since it's about basic group actions:

Consider the affine plane AG(2,3) (affine plane F^2, where F is the field GF(3)).
The group AGL(2,3) of affine transformations acts on AG(2,3).

A textbook I am reading claims that any quadrangle (set of 4 points, no 3 collinear) in AG(2,3) can be mapped to any other quadrangle under AGL(2,3).

How do we see this?

[tribal moss]

This is specific to the underlying field being F_3 in particular.

It is clear that we can (for any field) map any such set such that 3 of the points are (0,0), (0,1), (1,0).

So we only need to know that all quadrangles that contain those 3 points can be mapped to each other.

But there are not may possibilities there! Of the 9 points in the plane, 3 are already taken and 3 others are excluded by not allowing 3 collinear points. So there are only three places to put the third point, and two of those obviously give symmetric situations.

So all you really need is to show that {(0,0),(0,1),(1,0),(1,1)} and {(0,0),(0,1),(1,0),(1,2)} are affinely related.

[cobalt heath]

Sorry to interrupt, but so Is this one of exceptional behavior of small fields?

[tribal moss]

You could say that, yes.

[cobalt heath]

Maybe too trite/uninteresting of a phenomenon

[true ingot]

Sorry just to clarify. Here you are saying that we can map any set of the form {(0,0),(0,1),(1,0),x} (where x is an arbitrary point not equal to the other three) to any other set of this form?

[tribal moss]

It's not like it's a crisply delineated concept anyway.

Yes, except "arbitrary point not equal to the other three" should also have "and not collinear with two of them".

[true ingot]

But assuming the former is clear doesn't that immediatelly follow?

[tribal moss]

That's another way of saying what you just did.

Ah, no I see the difference.

[tribal moss]

Here I was saying that an arbitrary quadrangle can be mapped to one of the special {(0,0),(0,1),(1,0),x} quadrangles. Next I'm arguing that the special ones can be mapped to each other.

[true ingot]

Ah okay. Then I agree with the first fact that we can map any quadrangle to one of the special ones (I guess the idea is, that we can map any two linearly independent vectors to another pair of linearly independent vectors, meaning we can make the two sides of a quadrangle parallel with x,y axis and then just add a translation so the corner is actually at the origin?)

[limpid ferry]

Let G = Z_12, H=<4>. The factor group G/H is isomorphic to the groups Z_4 and rotationccw. Specifically and generally, how do we find groups to which a factor group is isomorphic? Any handy tool or theorem used?

[tribal moss]

Yes.

[tribal moss]

In the case there's the general fact that quotients of a cyclic group is always cyclic.

[limpid ferry]

Yes this one is understandable to me

[tribal moss]

But there's no general procedure -- in fact, for many important groups the nicest description of them we have is as such-and-such quotient.

So being able to find something even nicer that the quotient is isomorphic to is the exception rather than the rule.

[limpid ferry]

For example: D_4{(1), r^2} is isomorphic to the Klein 4-group. How do we work it out typically? Or just such and such quotient?

[limber sequoia]

Assume that E is an algebraic extension of F, so that $E\neq F$. Then E's degree over F is larger than 1.

[cloud walrus]

tomer_k

[limber sequoia]

is that how you'd say that in English?

(E,F are fields)

[rocky cloak]

Yup

[limber sequoia]

and you'd say "items"?

for the various $a\in F$?

[cloud walrus]

tomer_k
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[rocky cloak]

Elements

[limber sequoia]

thank you, obliged

[coral spindle]

This gets interpreted as \inF not \in F btw, that's why there is an error (if you were wondering)

[limber sequoia]

thank you i was

[true ingot]

I have a multipart question I would like to solve. Let $G = PGL(2,5), H = PSL(2,5)$.
First part asks you to show that the equivalence class of the matrix [1,0 ; 0 -1] from G is also in H. I managed to do that.

[cloud walrus]

Faputa

[true ingot]

Second part asks to find a Sylow subgroup P of H that contains this matrix.
Now since $|PSL(2,5)| = 60 = 2^2 3 5$ it suffices to find a group of order 3 that contains it.
I think the group in question is simply the one consisting of the identity, the matrix [1, 0 ; 0,-1] and [-1, 0 ; 0, 1].

[cloud walrus]

Faputa

[true ingot]

I get stuck on the third part which asks to show, that the normalizer $N_H(P) \neq P$. I have no idea how to approach it

[cloud walrus]

Faputa

[rocky cloak]

[1, 0; 0, -1] has order 2, so you want to find a subgroup of order 4. Also the two matrices you've listed are the same element in G

[true ingot]

... of course how could I miss that... Thank you

Then perhaps... [1 0; 0 1], [1 0; 0 -1], [0 1; 1 0], [0 -1; 1 0] as a subgroup of order 4 (Sylow 2-subgroup then)

But the third part is still unclear

Perhaps by trial and error to find an element x not in P such that x^(-1) P x is in P..

Ah [2 0; 0 3] seems to do the trick

[tribal moss]

That's not a subgroup: [01;10] [1 0; 0 -1] [01;10] = [-1 0; 0 1].

Wait, we're in PGL? Then ignore me.

[true ingot]

The following question asks how many Sylow 2-subgroups PSL(2,5) has.
From Sylow theorems I get that the possible number is 1,3,5 or 15. Now it can't be one because PSL(2,5) is simple, but Im not sure how to rule out the remaining possibilities

[tribal moss]

Can you learn something by counting elements of order 2?

[rocky cloak]

The action on the sylow subgroups defines a homomorphism to Sn, so since 60 doesn't divide 3! It can't be 3 either

[rustic crown]

try looking at other primes maybe? there are 1, 4 or 10 sylow-3s, and 1 or 6 sylow-5s. since 60 > 4! it can't be <= 4. so 10 sylow-3 and 6 sylow-5.
so this means there are 20 elements of order 3 and 24 elements of order 5. That leaves us with 16 elements to play with.

worst case scenario, realize that A_5 is the unique simple of order 60. and look at all order 2 permutations

[rocky cloak]

It's still a little tricky, because you have account for how the sylow subgroups might intersect each other

[rustic crown]

hmm

[rocky cloak]

Thought I guess if it's been 5 and 15, that difference is quite noticable anyway

[next obsidian]

I mean…

It can’t be 5

Wait

5! Is not 60

[rocky cloak]

5! is 120

[next obsidian]

My final message. Goodb ye

Oh wait

[rocky cloak]

But I agree it can't be 5

[next obsidian]

You just have to rule out it being the alternating group then

Which you can probably do by counting orders of elements or some shit

[south patrol]

Real

[rocky cloak]

No wait, I don't

[rustic crown]

it is the alternating group tho

[next obsidian]

It is?

[south patrol]

Chad

[rustic crown]

yea unique simple of order 60

[next obsidian]

Oh

Yeah true

I should go to sleep

[south patrol]

Is PSL(2,5)=PSL_2(F_5) lol

[rocky cloak]

That is what it is

[south patrol]

This felt very french

What is it that it is

[rocky cloak]

C'est la vive

[tribal moss]

If I count correctly, there are 10 elements of order 2. That's just enough to make 15 copies of V_4 if they all commute.

[rustic crown]

there should be an odd number of elements of order 2

[rocky cloak]

I count 15

[tribal moss]

Hmm, wonder what I did wrong there.

[true ingot]

It is 5 I think

i solved it with a different approach. By using PSL(2,5) = A5, where we know how groups of order 4 look like

[rocky cloak]

Idk, but my count is
(ab)(cd)(e) = 5!
divided by swapping ab, swapping cd or swapping the two transpositions gives 5!/8 = 15

[true ingot]

Hm did I mess up?

[rustic crown]

idts

[true ingot]

But... The next (and last) subexercise in this is "Prove PSL(2,5) is isomorphic to A5" so this isnt how its meant to be solved

[tribal moss]

Oh, I was actually trying to count matrices modulo 5 :-)

[true ingot]

I thought V4 in A5 are of form: {1,(ab)(cd), (ac)(bd), (ad)(bc)} and there are 5 choose 4 choices for that

But... it seems there has to be another way of seeing so. Perhaps using the previous exercise (that the normalizer of P is not equal to P for that one specific Sylow 2-group?)

[rocky cloak]

Yeah, it's quite natural to go the other way. Ones you know it has 5 sylow subgroups, then the action gives you an isomorphism with A5

[rocky cloak]

Oh right, ||the index of the normalizer is the number of subgroups||, so it can't be 15

Then only 5 remains

[true ingot]

Ahh I didn't know that

[rustic crown]

why can't the normalizer be the sylow-2 itself?

[true ingot]

The previous exercise was to prove that this doesn't happen for one of the Sylow 2 subgroups

[rocky cloak]

It comes from the normalizer being the stabilizer when your group acts on the sylow subgroups

[true ingot]

Okay I will digest this all. Thank you all :)

[tribal moss]

OK, counting more carefully does give me 15 elements of order 2 in PSL(2,5) too.

[next obsidian]

Wait

This is fucking stupid

The only other option was 15 right?

[rocky cloak]

Yup

[next obsidian]

Oh nvm these are Sylow 2

[round jay]

Can someone give me a hint for 1.11? There doesn't seem to be enough information to prove that each conjugacy class has one element

[charred iris]

It's a fact that $|C(g)| = {|G| \over |C_G(g)|}$ where $C(g)$ is the conjugacy class, and $C_G(g)$ the centraliser of $g$.

[cloud walrus]

Edward II

[round jay]

Yes i know, but I don't have info about the size of G

Also having trouble computing the size of the centralizers

I know there are at least r elements in the centralizers of each g_i

Not much more than that though

[charred iris]

you also know the exact size of one of the conjugacy classes

[round jay]

Oh yeah it's 1

[charred iris]

and that each element of G is in some conjugacy class

[charred iris]

and this gives an upper bound for the size of the other conjugacy classes

[round jay]

In terms of the size of G

[tribal moss]

In terms of |G| and r.

[round jay]

Yeah

Don't see where to go from these observations

[tribal moss]

It sounds crazy, but try to derive an upper bound for the size of G in terms of |G| and r ...

[round jay]

I got that's it less than r squared times its size

[tribal moss]

Uh, no. What was your upper bound for each conjugacy class?

[round jay]

r times the size of G

Wait i messed up

Algebra mistake

[charred iris]

... technically true though can't see the issue

[round jay]

It's |G|*1/r

[tribal moss]

Eyup. So the upper bound on |G| we get out of this is?

[round jay]

I got that the order of G is r

[tribal moss]

We'll need to conclude that eventually if G is to be abelian, but how did you get it just now?

[round jay]

Just some basic algebra gave me that order of g is at most r

It's also at least r

Because r different classes are assumed

So it's r

Since the classes partition G each must have size 1

The class formula gives us that the order of the center of the group is r

[charred iris]

you don't even need the class formula because here you essentially found that g1, ..., gr are all the elements of G, and it's given they all commute

[round jay]

Oh yeah true

Thanks guys

[tribal moss]

"Just some basic algebra" does not fill me with confidence, but if you're happy, I suppose that's as far as we'll get.

[true ingot]

One question about this: The homomorphism in action is G acting on the set of Sylow 2-subgroups by conjugation right? But do we know if this action is faithful?

[tribal moss]

If the action wasn't faithful, it would have a nontrivial kernel -- but you know already that G is simple.

[rocky cloak]

Well, you said you knew it was simple

[true ingot]

Ahhh right... Yes yes

[old hollow]

Determine $\text{Aut } \mathbb{Q}[\sqrt2]$.

[cloud walrus]

a.b.s._.0.

[old hollow]

i'm trying to do this problem

first we did a lemma showing that Aut Q is trivial

so now i'm thinking, if i have an automorphism of Q[sqrt2], it surely fixes Q, right?

[rocky cloak]

Yes, you should be able to argue that in the same way as in the lemma

[old hollow]

ok

[old hollow]

but now I don't understand what to do with sqrt2

seems independent in some sense

I feel like I should be looking for linear maps of Q[sqrt2] as a vector space which fix the Q axis

does that make any sense

[rocky cloak]

Yes, that's absolutely correct

[old hollow]

once I decide where sqrt2 goes, then we're done

i can't send it to any rational number

[rocky cloak]

Indeed

[old hollow]

but i can send it to anything else?

so I can send sqrt2 to the number 3 + 5sqrt2?

[rocky cloak]

One important observation is that (sqrt2)^2 is rational

And an automorphism would have f((sqrt2)^2) = f(sqrt2)^2

[old hollow]

T(sqrt2 * sqrt2) = T(sqrt2)^2 = T(2) = 2

oh shit

so then T(sqrt2) = sqrt(2)

so it fixes it?

wait

[rocky cloak]

Well, maybe.

[prisma ibex]

no

[old hollow]

did i fuck it up lol

[rocky cloak]

All you know is that T(sqrt2)^2 = 2

[old hollow]

ok

[prisma ibex]

you could also have T(\sqrt{2})=-\sqrt{2}

[old hollow]

OHhhh

forgot about that case...

[prisma ibex]

those are the only two possibilities, your automorphism group is Z/2Z

[old hollow]

but there are only 2 solutions to x^2 = 2

so

yea we're done

okay cool

[prisma ibex]

yeah it's very similar to the automorphisms of C fixing R, the identity or complex conjugation

[old hollow]

i'm curious how this would work for Q(sqrt2, sqrt3) though

[prisma ibex]

you're either fixing \sqrt{2} or conjugating it

[old hollow]

yeah that's what I was thining

when u said that

[old hollow]

hmm

[rocky cloak]

It works quite similarly

[old hollow]

it seems like, if the only rule is that Q is fixed

you could swap sqrt2 and sqrt3 as basis vectors

and then also conjugation

[prisma ibex]

it's similar, such an automorphism would be determined by where it sends \sqrt{2} (either to \sqrt{2} or -\sqrt{2}) and where it sends \sqrt{3} (either to \sqrt{3} or -\sqrt{3})

[old hollow]

oh

shit

[prisma ibex]

so the automorphism group is Z/2Z x Z/2Z

[old hollow]

hmm, what about Q(cbrt2) then

oh wait

isn't that 3 automorphisms

cuz you can send cbrt2 to the other roots of unity

am i tripping

wait no hold on

[prisma ibex]

it's more complicated than that

the automorphism group in this case is S_3

[old hollow]

hm

[prisma ibex]

err for the Galois closure that's right

[old hollow]

oh shit

wait it's 3 dimensional

uifdguidjfoig

[prisma ibex]

if you adjoin the roots of x^3-2 to Q you get a 6-dimensional Q-vector space

[old hollow]

no no

just adjoining the real root of x^3 - 2

[prisma ibex]

right

then yeah you just get something that is 3-dimensional

[old hollow]

so wouldn't it look like Q(cbrt2) = {a + bcbrt2 + c(cbrt2)^2}

[prisma ibex]

yup

[old hollow]

yeah

so then, intuitively, a linear automorphism fixing Q would then be determined by where it sends cbrt2 and (cbrt2)^2

right?

[prisma ibex]

I mean in this case you don't really want to talk about the automorphism group

you really only want to talk about automorphism groups fixing Q when the extension is Galois

[rocky cloak]

Notice T((cbrt2)^2) = T(cbrt2)^2, so really it's determined by where you map cbrt2

Just like before cbrt2 would have to map to another cube root of 2, so the question is how many of those are in Q(cbrt2)

[old hollow]

ohh

man I gotta keep in mind it respects multiplication

[rocky cloak]

I feel like you will miss out on the importance of Galois extensions if you never try to calculate the automorphism group of other extensions

[old hollow]

that sounds like it makes sense

[old hollow]

well, none right? because the other cube roots of 2 are imaginary

[rocky cloak]

Bingo

[old hollow]

man

this is uncharted territory

it's kinda unintuitive to me

most group theory felt obvious or like I could intuit it in some way

but this is almost computational

[prisma ibex]

another fun example is to compute the Galois group of Q adjoin n-th roots of unity (so adjoin all roots of x^n-1)

also good to work this out for finite fields

[old hollow]

hmm

nth roots of unity are cyclic, right? so wherever the generator is sent determines the whole automorphism

the generator can be sent to any other generator

so I assume it's (Z/nZ)*...

[chilly ocean]

Im confused on how I could prove this

[primal beacon]

@chilly ocean Can you find a group automorphism that is trivial in an abelian group ?

[chilly ocean]

Automorphism is an isomorphic mapping of a group to itself right?

[old hollow]

what the hell does this mean

sorry

[chilly ocean]

Sorry, maybe this was important. I was learning about equivalence classes when this question came up, so I guess the author is talking about equivalence classes here

In the book, an equivalence relation is defined by $g' \sim g$ such that $g'=f^{-1}gf$. So if the group is abelian, we can write $g'=f^{-1}fg=g$ so that the equivalence relations that exist with g' are with itself and it exists in it's own class. Would this be a valid proof?

[cloud walrus]

3.14 rate

[old hollow]

Well, it’s an explanation, but not a rigorous proof

You want to say, let [g] be the equivalence class associated to g. Then blah blah blah, so [g] = {g}

How do you do this rigorously? You can say, suppose g’ in [g]. Then …, so …, hence g’ = g.

Therefore [g] is really only one element

[chilly ocean]

Will the logic here work though?

[old hollow]

It’s not worded very well

“the equivalence relations that exist with g’” doesn’t make sense

equivalence relations don’t exist with a particular element. I’m being pedantic, but if this is your proof, it’s not comprehensible enough

[chilly ocean]

I'll write it down like this. thanks

[chilly ocean]

No, thanks for pointing it out. I'd have my marks cut if I was writing this down for an exam

[primal beacon]

yeah, what I said wasnt very precise on purpose but I can detail more
. when you have an automorphism you can form a quotient group related to the equivalence relation from the morphism

[hasty lynx]

I like math texts that only e.g. define matrices after establishing that a data structure of that form would be useful to nail down linear maps. Can groups/rings/fields be handled similarly? That is, first we identify that precise concept falling out "in the wild", and then we go "Well, if we're going to have to study this pattern, we might as well name it."?

[kind temple]

like, exponentiation by an integer, anything with cycles or permutations, symmetries of the plane, symmetries of polygons, RSA encryption, modular arithmetic, finding rational solutions on a curve (e.g., for classifying pythagorean triples) demonstrating the power of group theory, etc

[hasty lynx]

I don't mean merely a list of examples before the definition - what might count is a problem statement to which the solution class is precisely the class of groups

[rocky cloak]

One of the early historical motivations for both groups and fields were about permutation groups acting on the roots of polynomials. What we today call the Galois group of a polynomial.

So assuming you care about solutions to polynomial equations this motivates (number-)fields and (finite) groups

[hasty lynx]

Can you get every (finite) group that way?

[fading field]

this doesn’t seem like the right perspective no matter the answer to this question

[kind temple]

you can get every finite group as the symmetry group of some object i think

[hasty lynx]

"Permutation groups acting on the roots of polynomials" seems too... late in the tech tree? to be the... "true address" of groups in Plato's idea-space

[rocky cloak]

Yes, every finite group is the Galois group of some Galois extension (of number fields). But it's an open problem whether the base field can always be Q

[hasty lynx]

that, on the other hand, sounds workable. does this theorem have a name?

[fading field]

cayleys theorem

[kind temple]

something something cayley graphs
here is a link you might find useful
https://math.stackexchange.com/questions/3254492/can-all-groups-be-thought-of-as-the-symmetries-of-a-geometrical-object

[hasty lynx]

"every group is isomorphic to a subgroup of a symmetric group" ...that sounds much less impressive

[rocky cloak]

You can word things differently to make them more or less impressive if you like

[fading field]

i think you’re missing the point, geometric objects are defined by their symmetries, and taking a subgroup of the permutation group perfectly captures such a notion of a “group of symmetries”

[hasty lynx]

I'm trying to get to the concept of group here without starting from it

[fading field]

okay, “what are geometric objects” then arrive at subgroups of permutation groups and then groups in general.

https://tenor.com/view/aristotle-ballin-gif-22706788

groups literally are geometry, i promise you’ll come to see that you don’t need this kind of question to realize that groups truly are platonic

[rocky cloak]

If you take some examples of symmetry groups of, for example, polygons and polyhedra and write down common properties, the axioms of a group fall out quite naturally.

But after that you'd want to motivate groups as an abstraction of this idea.

3b1b has a video on this very topic
https://youtu.be/mH0oCDa74tE?si=fEgJ3uFJ0u0n6cCt

[hasty lynx]

If that is true, it ought to be formalizable.

[fading field]

okay then go read about it, it’s already been formalized enough probably in the way that you’re asking. but you can’t really formalize the platonic realm through these questions.

[hasty lynx]

How do you know? It sure would be convenient to formalize the platonic realm and finish what Leibniz started.

[fading field]

have fun with that

in the meantime i’m convinced that you’ll end up thinking that groups are platonic once you know more about groups. also, not to be confrontational or anything, but i’m not sure why you brought up matrices when matrices are not platonic.

i also said that in a confrontational way, but i really want you to know that i don’t mean any snark by it, i’m just a bit tired.

[hasty lynx]

Why do you think matrices are not platonic? The data you need to nail down a linear map between vector-spaces-with-basis sure forms a matrix.

[fading field]

yeah but we only came up with matrices to describe the actual thing itself, which is the linear transformation, and when we didn’t understand those, we came up with matrices as a bookkeeping tool to specify that there is a unique linear map sending this (arbitrarily chosen) vector to this one, this (arbitrarily chosen) vector to this one, and so on, and ALSO in which each image was the specified linear combination of the arbitrarily chosen (by us) basis for the target vector space. the property that this data can be arranged into a square grid is the result of our clever bookkeeping and our careful study of the niceness of linear maps, and not the result of the fact that matrices somehow are the actual transformations themselves.

i might have misunderstood you if you really did mean linear map though

which is possible

[hasty lynx]

What else might I have meant?

[fading field]

matrix

[hasty lynx]

1. Sure, matrices are merely a bookkeeping-reification of linear maps; I yet find it possible that groups will merely turn out to be the bookkeeping-reification of actions

2. By platonic I mean ~"appears reasonably high up in the periodic table of math" - and "the canonical bookkeeping-reification" sounds like a fair concept to me

[fading field]

it’s not canonical

[fading field]

i don’t know whether i agree with you or not. i currently do believe these truly are the same

by that i mean groups and actions (to a certain extent)

[hasty lynx]

maybe we mean slightly different things by canonical? consider the free-forgetful adjunction between Set and Vect. its hom-iso form directly says that the maps S->V correspond to the linear maps R^S->V

[fading field]

by canonical i meant that for a map V -> W the choice of matrix to represent transformations between them depends on a choice of isomorphism V -> R^n and W -> R^m

which you might tell me is bad but i think is important

[hasty lynx]

by canonical I meant that if you try to get a bookkeeping-reification, you straightforwardly get the notion of matrix. maybe you mean that the choice of basis is not canonical? (but then surely you should also reject that every vector space has a basis at all)

[rocky cloak]

You should reject that every vector space has a canonical basis yes

[hasty lynx]

Alright, we seem to understand each other's position here. I claim that, for figuring out what the periodic table of math looks like, my notion of canonical is more appropriate than yours; agree?

[fading field]

no

but i think we can leave it here for now, i enjoy talking to platonists

but i’m very tired

so maybe later

[hasty lynx]

yes, there's more than one basis; but if one grants that every vector space has a basis (which requires the axiom of choice), i claim one might then as well go ahead and choose a basis for every vector space

(hmm, i guess one might have a weaker axiom of choice which only applies to small sets, which is enough to choose a basis for one vector space, but not for each of them at the same time)

[delicate bloom]

I empathize with you here. I know I gave up pretty early on that path myself, best I could really tell myself is that in some sense it's a study of symmetry itself, whatever that means.

[rocky cloak]

Sure you can.

But I think I've lost the thread of what you are actually after. Like you want to build a periodic table of math based on a philosophical hierarchy?

Or are you after a more pedagogical route to learn algebra?

[hasty lynx]

Sure, though the philosophy is instrumental. In other words, a language in which the short phrases correspond to the useful concepts. It's not about pedagogy, in fact "is this useful to recruit human neurons" is a useful question to find weeds that need removing.

The point is to thus become more able to fill in gaps and transfer results.

[rocky cloak]

Allright, then I don't think I have much to contribute

[mighty kiln]

What's a bookkeeping-reification

[hasty lynx]

My archetypal example is that the matrices reify the linear maps.

[mighty kiln]

Represent?

[tawny magnet]

Yh what does reify mean

[hasty lynx]

There may well be existing terms for what I mean! ^^

[kind temple]

make an abstract thing more concrete

[hasty lynx]

"Hilbert spaces can be _'d as the L2 functions on some measure space."

[tawny magnet]

Doesn’t it have to be separable

[mighty kiln]

Are groups _'d as permutation groups

[delicate bloom]

hmm do matrices really reify linear maps? Similarly, I feel like laying coordinate lines down on a 2D plane doesn't reify it either.

maybe I'm using a different idea for 'reify' here

the plane was already real, we're just putting lines on it for convenience

[rocky cloak]

No, it should work either way. But the measure space might be big

[hasty lynx]

Oh well, if my word has failed to hone in on a neuron in each of y'all's heads, we can skip that; to recap the state of how answered my original question is: groups are to "automorphisms of an object" as rings/fields are to what?

[rocky cloak]

Or endomorphism rings of objects in additive categories

[tawny magnet]

Rings are rings in the language of rings 🤣

[mighty kiln]

Right that

[hasty lynx]

"groups are groupoids with one object" is the kind of answer that i think i might be able to formally rule out by going "that definition would let you define large groups; what definition would only let you define small groups?"

[delicate bloom]

yeah this question should be answered in chapter 1 of any intro algebra book, but I don't think it ever is

[mighty kiln]

Locally small groupoid with one object

[hasty lynx]

as i would have thought goes without saying: your definition isn't allowed to say "small", either.

[fading field]

honestly idk if rings really are platonic, they are structures that you enrich some hom sets with, among other things, and maybe it just has a nice name because it’s a particularly nice example of the platonic thing

[tawny magnet]

Add in the pseudosentence ‘it is small’

[tawny magnet]

Nevermind lol 🤣

[mighty kiln]

Is that even possible

Surely you can't just get smallness without using specifying it somewhere

[hasty lynx]

to have a definition that only works for small without saying small? the example i have in mind is "a category is a monad in the spans on Set"

[mighty kiln]

Oh then groupoid with one object such that there exists a faithful functor to Set then

[tawny magnet]

Take first order logic then say The model of the language of groups with order n or less where you pick n surely is fine

[fading field]

i’m gonna regret saying this but my mind is a cloud rn

[mighty kiln]

I think this is correct, rings are preadditive cats with one object

[rocky cloak]

Rings are just fancy quivers

[fading field]

yeah but i feel like rings might already be geometric in a sense, but i don’t know enough about the perspective of people who would say such a thing to conclude whether they are or are not

[hasty lynx]

if that works, my formal rule was insufficient. perhaps it helps to say - categories are only supposed to be defined in the metalanguage as of yet, they only end up part of the language at some later point in the tech-tree

[tribal moss]

All of mathematics is just applied quiverology.

[mighty kiln]

CRing^Op → Sch is fully faithful I think

[fading field]

i don’t understand what this functor is, nor the implication of it being fully faithful on my religious beliefs

[rocky cloak]

A scheme is a geometric object and a commutative ring is just an affine scheme, I guess is the take away you're supposed to have

[fading field]

i’m not currently convinced that schemes are geometric, because i don’t know what a scheme is

not that i’m unwilling to learn but it’s probably just more efficient to read about it later than to try to do it now

[mighty kiln]

Schemes are manifolds but you glue crings instead of R^n

[fading field]

unless you want to talk and then i’m happy to listen

[mighty kiln]

Where crings = affine schemes

[rocky cloak]

Yeah, the reasoning is somewhat circular as a scheme is just a bunch of affine schemes glued together

But the main idea is if you take some affine variety in C^n (some common solution set to some polynomials), and look at the polynomial functions on that space. The space can be reconstructed just from the resulting ring.

Then you just apply that same construction (roughly) to arbitrary commutative rings, and what you get is an affine scheme

[hasty lynx]

doesn't "polynomial" come after "ring" in the tech tree

[mighty kiln]

I'm still not sure what your tech tree is

[rocky cloak]

The "tech tree" isn't really a well defined thing is it? But polynomials are certainly a motivation for defining rings

[mighty kiln]

Polynomials using universal property is morally correct

[hasty lynx]

the tech tree is just the graph in which circles correspond to circular reasoning ^^

[mighty kiln]

tech tree is neither tech nor tree

[rocky cloak]

Like should abstract stuff come first or last in the tech tree?

Like what comes first out of fields, rings and the complex numbers

[cobalt heath]

What was the joke thread

[fading field]

but now i think i have run into the same problem, i’m stuck trying to understand why doing this for arbitrary commutative rings give us geometry in the same way that polynomial rings or C^infty(C^n)(?) (i was told not to do this, but i was raised wrong as a kid and have to wonder about this, i can’t help myself) do

[rocky cloak]

It just a quacks like a duck situation. We are the ones defining what geometry is after all

[mighty kiln]

It turns out the construction of A^n from C[x1, …, xn] is purely algebraic, and we can do the same thing for any cring

[cobalt heath]

Spec Z is geometry of usual places

[mighty kiln]

Also good category of weird objects > weird category of good objects

[rocky cloak]

You might also say "what's geometric about linear algebra?", "Sure, in 2D and 3D it looks geometric, but linear algebra is also about n-dimensional spaces for n>3"

[mighty kiln]

And fields that aren't R

[fading field]

stack

[fading field]

i guess if i spent more time with the objects i would be able to see geometry in schemes better

[worldly sedge]

I remember someone once said 'if you want to imagine 14th dimention then imagine 3 dimentions and scream loudly '14 DIMENTIONS''

[fading field]

yeah that’s how i do it

[rocky cloak]

Maybe, I'm not claiming I do, but I'm not a geometer either

[fading field]

lol

[worldly sedge]

Yes. Geometry of quotient set with an operation where each point on x axis represents a set same goes with y axis or x axis with each point being a set and y axis where each point being an element from original set.

[fading field]

only one way to find out 👀

[worldly sedge]

I really tried to apply geometric approach when studying field theory but my head started getting hot. Like car engine.

I did got some intuitions but they were too much to process in short time.

[cobalt heath]

I mean, traditional varieties are pretty geometric, aren’t they

You just restrict your interest to polynomials

From what I can gather, ring spectrum is just extension of this, and a scheme is just a gluing of specs

[slim kayak]

we abduct chmonkey and force him to finally make that reading group?

[dull ginkgo]

locally winged space

[slim kayak]

algebraic geometers can draw some pictures for arbitrary ring spectra, is that enough to make it "geometric"?

[south patrol]

Lol minor thing, ring spectra has a specific technical meaning in homotopy theory and I'm pretty sure AGers say spectra of rings or whatever to avoid saying that

[mighty kiln]

Prime spectra

[slim kayak]

Poor Lurie having to deal with spectra of spectra

[bitter locust]

speculative question

wouldnt calling "prime numbers" "maximal numbers" instead make more sense? im saying that because of the (0) ideal that is prime but 0 is not a prime number

ik that this is a convention, but if renaming math was possible would this be sensible? or are there situations which clash with a proposed "maximal numbers" term

[dull ginkgo]

In a way yes and prime powers would be “prime” under this definition

[mighty kiln]

Why are prime powers prime

It's just zero and negative primes I think

[coral spindle]

(p^2) is not a maximal ideal of Z for p prime. It is contained in (p) after all.

[mighty kiln]

I think one could also say positive prime

But usually "prime number" vs "prime in Z" gets the point across

[bitter locust]

wait negative primes? wouldnt they be contained in "positive prime ideals" or other general prime ideals

[coral spindle]

I think I would find it confusing when I wanted to talk about maximal wrt some property. “The maximal number with property xyz is maximal (prime)” 😖

[mighty kiln]

(-3) = (3)

[bitter locust]

yeah

[mighty kiln]

So -3 is prime

[bitter locust]

lol fair point

[coral spindle]

Also, there are prime ideals in rings that are not maximal

It’s arguably a coincidence in Z

So I guess it wouldn’t be wrong per se but perhaps not telling the whole story

[bitter locust]

hmm

in R[x] theres (x) which is a prime ideal thats not maximal

yet we talk about factoring polynomials into linear terms

[coral spindle]

No I’m pretty sure that is maximal?

[bitter locust]

(2, x) would be a proper maximal ideal

[coral spindle]

You mean R as in the real numbers, right?

[bitter locust]

yeah

[coral spindle]

So (2, x) = (2) = R[x].

Because 2 is a unit.

In fact, any element of R[x] differs from some element of (x) by a unit. I.e., (x) is maximal in R[x].

[bitter locust]

ohhh

[coral spindle]

On the other hand if you meant Z[x] I would agree with you

[bitter locust]

yeah ok i think i got confused

Z[x] makes more sense

for (x) being prime but not maximal

[coral spindle]

Yeah

[hasty lynx]

Probably rings -> fields -> complex numbers. A sufficient condition for A to come before B is if you can only define B if you already have A.

[rocky cloak]

Well, then anything should motivate group theory right, since it's so low on the tech tree

[hasty lynx]

I'm looking for the kind of motivation that proceeds from stuff lower on the tech tree

[slim kayak]

Good to remember that R being a field iff R[x] is a PID, all ideals look like (f) there

[rocky cloak]

Right, but is there really anything below? Like something as simple as addition of integers would be higher in the tree right?

[dull ginkgo]

It is maximal amongst ideals disjoint from R

[hasty lynx]

sets are below. quivers arguably don't need groups

[slim kayak]

Wouldn't this tech tree need to be continuously updated on the left/top

[hasty lynx]

and "the structure of the set of automorphisms of a quiver" is enough to get all the groups, right?

[slim kayak]

I don't think we've reached the mathematical object which has every other as a special case yet and probably never will

[hasty lynx]

skill issue

[bitter locust]

so we only allow ideals that don't have any constant terms as generators?

[slim kayak]

Yeah, sorta like that

The actual idea behind it was probably looking at the prime ideals of Frac(Z)[x]=Q[x], where (x) is again maximal

[dull ginkgo]

Neat thing about localization

It’s like “opposite” of fourth isomorphism theorem

In terms of like how the ideal lattice is affected

[slim kayak]

And it's even nicely compatible with zariski

[dull ginkgo]

Zariski topology’s funny little topology thing comes from the product def lmao

[dull ginkgo]

But yeah

I like to view units as non-ideal elements

[slim kayak]

Hm?

[dull ginkgo]

Well inverse-topology

[slim kayak]

Oh, right. Final topology unless I mixed up my lefts and rights again

[dull ginkgo]

Well the chains of prime ideals containing a given ideal form a topology with their complements

And that should come rather intuitively

Like consider binary unions

Instead through products

[dull ginkgo]

But yeah my intuition for localization is almost like the opposite of a quotient map

Instead of sending your designated ideal to the additive identity, you are sending your multiplicative set to units

Which are opposite in the fact that 0 is in all ideals, but an element is a unit iff it doesn’t lie in any proper ideal

And how the ideals are preserved or not is from that fact

[dull ginkgo]

For proof of this fact, consider that x is a (two-sided) unit, and consider the ideal (x) lol, or imagine x is in a proper ideal

[slim kayak]

This is a bit of a non-standard way of looking at things to my eyes, but yeah

[limpid ferry]

This is my teacher's proof for a theorem. For the (=>) part, I don't understand why "(1,1) is a generator" according to the preceding clause

[slim kayak]

That gcd(n,m)=1?

[limpid ferry]

yes with gcd(n,m) = 1, then that would be obvious, then may I ask how gcd(n,m)=1 by the isomorphism?

[rocky cloak]

I agree there is a little bit missing here.

It would be better if the statement was that "the map Z_nm -> Z_n x Z_m, x |-> (x, x) is an isomorphism"

Alternatively you can show that (1,1) is the element with the largest order in Z_n x Z_m

[limpid ferry]

Yikes, if the mapping was explicitly defined, then I should have been good.

[dull ginkgo]

It works lo

[ornate tiger]

are they cosets becuase they keep that the differance is 3 ?

[dull ginkgo]

Yep!

Do you know why?

[ornate tiger]

why

[dull ginkgo]

Do you know the equivalence relation for group quotients

[cinder fox]

If $p$ is prime, then $\mathbb{Z}/p\mathbb{Z}$ is a field, i.e. $(\mathbb{Z}/p\mathbb{Z}\setminus{\bar{0}},\times)$ is a group of order $p$.
Then shouldn't $\bar{a}^p=\bar{1},\forall \bar{a}\in\mathbb{Z}/p\mathbb{Z}\setminus{\bar{0}}$, conflicting with what Fermat's little theorem states?

[cloud walrus]

VirtualCode

[dull ginkgo]

It’s a group of order p - 1

[cinder fox]

oh right

[dull ginkgo]

In fact any finite field has a cyclic multiplicative group of order one less than the field’s order

[cinder fox]

0 excluded

bruh

🗿

thanks man

[dull ginkgo]

You’re welcome lo

[south patrol]

there's also the mild generalisation that this works for any finite subgroup of the group of units of a field

[dull ginkgo]

me when polynomials… finite roots of polynomial … order of subgroup gives polynomial…

[south patrol]

Lol

Well you need an extra ingredient right lol

[dull ginkgo]

Isn’t this the lemma used to prove the finite field case lmao

[south patrol]

There are two ways I've aware of actually

[dull ginkgo]

Yeah afaik

[south patrol]

I mean, this includes the finite field case so idk what you mean

[dull ginkgo]

The way it is presented in Jacobson presents the finite field case as a corollary of what you said

[south patrol]

oh lol

Well yeah the proof for finite fields doesn't need anything beynod what you get from being in a field ye

[dull ginkgo]

Tldr |G| is group in F^x, then x^|G| = x for each x in |G| so each is a root of the polynomial x^|G| - x then there’s a small argument I need to remember to further prove it’s generated by a single element

OH YEAH, it’s by an argument with the powers still being roots

[south patrol]

ig the proofs i'm aware of are 1) if G is the subgroup and has order n, show that there is at most one subgroup of order d for each d | n, then compare that that to Z/nZ where we have exactly one subgroup of order d but the same number of elements overall

2. classification of finite abelian groups lol

[dull ginkgo]

Oh

I think 1 is how Jacobson did it

[south patrol]

yeah i have rephrased it slightly

[dull ginkgo]

I think I tried it myself and got a contradiction in the minimality of the order of |G|

[south patrol]

basically uses fact sum of phi(d) over d|n is n

[dull ginkgo]

I.e assuming the maximum order of an element is d | |G|

[south patrol]

but you can prove that by nonsidering Z/n, say

[south patrol]

ah

[dull ginkgo]

If there is an element of order |G| then it must generate the whole group

otherwise there would be a smaller order if it pigeonhole’d

I tend to immediately go for contradiction proofs with anything order related

Try to prove there is some smaller order than the actual order contradicting minimality

[dull ginkgo]

Here’s what I did explicitly.

Assume G in F^x has order N, then for each x in G, we have x^N = 1 thus each x in G is a root of x^N - 1.

Due to euclid’s algorithm and degree inequalities, there can only be N many roots of x^N - 1, thus the only roots of x^N - 1 are elements of G.

However assume G isn’t cyclic, then assume the maximum order that an element of G can have is d | N. If d = N, then an element of order N must generate the whole group, as otherwise it would have a smaller order.

But if d < N, then for each x in G, x^d = 1, so all of G would be roots of x^d - 1, which can only have d many roots, a contradiction.

Thus G is cyclic

[south patrol]

ig you have to justify that d being the largest order means they all have order dividing that

[dull ginkgo]

Yep

[cinder fox]

what exactly is being generalized here?

[south patrol]

this

[cinder fox]

you mean that the finite subgroup is cyclic?

[dull ginkgo]

This is a bit more group theoretic yeah, but now that I think about it I don’t think it’s true

Like consider Z/3Z x Z/2Z

[south patrol]

that is just Z/6 though

[dull ginkgo]

Wait good point

[south patrol]

for finite abelian groups it is true

but requires proof

[dull ginkgo]

Well if there is an element of order n not dividing d

Then consider the product of xy, x of order n, y of order d or use Bezout yeah

[cinder fox]

wait is G an additive or multiplicative group here?

[dull ginkgo]

Multiplicative

[cinder fox]

then why is x^|G| not 1?

[dull ginkgo]

Mistype on my end

[cinder fox]

oh alright

[dull ginkgo]

I was thinking of if G was a finite field

Because then we’d have the 0

So it would be x^|F| - x

[long geyser]

nvm, let me think about this a bit more

ok, thought about it for 1 more minute and decided to ask anyway

how would one prove this without bases?

this is being split just in case terminology isn't uniform

[void cosmos]

ig you can just

any vector space is both injective and projective

so meh

[long geyser]

idk what those words mean

(in this context)

[void cosmos]

oh okay

[long geyser]

maybe you need bases

[void cosmos]

no you don't

you just need

the existence

[long geyser]

well, with the knowledge I have available at least

[void cosmos]

of basis

[long geyser]

so, the thing is that the book covers bases in infinite dimensions in the next section

so I'd asssume you wouldn't need their existence for this, but could be an error

(book has a lot of errors)

[rocky cloak]

The statement is equivalent (in ZF) to every vector space having a basis. So you probably want to use bases to prove it

[long geyser]

lol

alright

another tiny mishap by the author then

[rocky cloak]

I mean you can prove it with Zorn's lemma or something if you prefer. So I guess it's not necessarily a mishap by the author

[long geyser]

no it is, he introduces zorn's lemma literally in the next page

[rocky cloak]

I see

[long geyser]

and does a bunch of zorn's lemma related stuff in the section, which includes proving vector spaces have bases

and other stuff

[serene ether]

Hi A a ring and I subring of A what is AI subset of I rule for ideal? Wdym on about AI?

[long geyser]

AI = {ai | a in A, i in I}

so it is saying, for all a in A and i in I, the product ai belongs to I

[serene ether]

Oh okayyyy thanks

[slim kayak]

"Some vector spaces might not be projective"

Statements dreamt up by the utterly deranged

[long geyser]

damn...

not asking a question

just, damn

time for torture