groups-rings-fields

[quartz thistle]

why did you take (a-i)(a+i) to disprove it for integral domain

[tribal moss]

Um, because that works?

[quartz thistle]

is a an element from set of Zp

[tribal moss]

My assumption was that Zp contains some element whose square is -1, and I chose to call that element a.

[quartz thistle]

got it

can you explain about Z2[i]

i^2 = -1, is it not an integral domain then?

[tribal moss]

The trouble is not i^2 itself.

The trouble is that the element 1, which is already in Z/2Z satisfies 1² = -1.

[quartz thistle]

okay

ill have to study more about this

[tribal moss]

Another way to look at it is (which unfortunately fails to quite work in the p=2 case):
In an integral domain, a polynomial equation cannot have more solutions than its degree.
If Zp already contains a solution to the equation x² = -1, then going to Zp[i] would add two new solutions (namely i and -i), so suddenly x² = -1 has more than the two solutions that it's possible for it to have in an integral domain. The only way out is to conclude that Zp[i] cannot then be an integral domain.

(This argument breaks down for p=2, because then (but only then!) i and -i are the same element).

[quartz thistle]

this does not imply on Z3? As Z3[i] is also a field

[tribal moss]

Does Z/3Z already contain an element that squares to -1?

I'm somehow getting the impression that you have completely ignored the many times I've mentioned that condition.

[quartz thistle]

no it does not

[tribal moss]

Then my argument does not apply to Z/3Z.

[quartz thistle]

i am trying to figure it out with the existing knowledge, the textbook im following does not exclusively mention this condition
it only says Zn[i] is an int domain if n is prime but now that youve mentioned it i will use this condition to solve for sets of complex numbers

[tribal moss]

it only says Zn[i] is an int domain if n is prime
If it says that, it's lying.
Or possibly it means something different with the notation "Zn[i]" than we assume it means.

[quartz thistle]

i was in the idea that all Zn[i] with n being prime forms an integral domain as Zp forms integral domain which is the generalised theorem written in book

[quartz thistle]

set of gaussian integers

[tribal moss]

The Gaussian integers are usually notated simply Z[i]; I'm not sure what the n in your book's notation would denote.

[quartz thistle]

modulo

[tribal moss]

I mean, it is reasonable that you would believe your textbook over a random stranger on the internet who says it is wrong, but if even our concrete examples of cases where Zp[i] is not an integral domain will not convince you, then I don't really think I can be of any further help. ¯_(ツ)_/¯

[quartz thistle]

Like here in Z3[i]

[quartz thistle]

I'm not saying that youre wrong, but at undergrad level maybe they're not teaching us the exceptions before we grasp onto the concept first idk i just want score good in semester exam😭

[tribal moss]

Hmm, it looks like we've actually given only one concrete example. A few more, then I'll shut up:

In Z5[i] we have (1+2i)(1+3i) = 0.

[quartz thistle]

so its a zero divisor

[tribal moss]

In Z13[i] we have (1+5i)(1+8i) = 0.

[quartz thistle]

wow

im getting some clarity now

we can not generalise for gaussian integers with prime modulo to be integral domain

Does example 5 hold true in all cases?

[tribal moss]

That is true for all p, yes.

[amber cradle]

No, it’s elementary number theory to show gaussian primes are primes congruent 3 mod 4, then modding by a prime gives a field (the justification is more nuanced, but wtv)

[quartz thistle]

Also for example 4, if we generalise it as Z[root B]?

[tribal moss]

Yes, because all elements of that ring are real numbers, and a subring of R (or C, or any field) is automatically an integral domain.

[amber cradle]

What is the norm of ring?

And when is the norm zero

[quartz thistle]

Got it got it

[quartz thistle]

What is norm of ring

I havent studied that yet

[amber cradle]

Just the euclidean norm

Or norm or of a complex number

[quartz thistle]

Not studied

[amber cradle]

a^2 + b^2?

pythagorean theorem?

[quartz thistle]

I dont know what norm is, or maybe it is referred as something else in my curriculum

[amber cradle]

Absolute value

Magnitude

Length

[quartz thistle]

oh right okay okay

[amber cradle]

So if sqrt(D) is real it’s just the number squared

[tribal moss]

Your question does not make sense. Please show the problem you're trying to solve exactly as it was given to you, with a picture if necessary.

[amber cradle]

I’m not asking a question I’m giving the student a hint instead of just giving him the answer

[tribal moss]

Your hint does not make any sense, then..

[amber cradle]

Yea it does, sincr we have the norm is zero if and only if there are zero divisors

And norm is homomorphic

[tribal moss]

Define "norm"?

[amber cradle]

So the question becomes are there zero divisors in the reals

[amber cradle]

I assumed, mistakenly, that was already taught

[tribal moss]

You can sometimes equip a ring with a norm, but it's not something rings come with intrinsically.

[amber cradle]

Right but for these rings we have it

And I assumed that was covered

Since in D&F it is

[quartz thistle]

should i know about norms to be able to answer such questions? as it hasnt been mentioned in the chapters so far

[tribal moss]

No.

[amber cradle]

No, but that’s the most natural way imo

[quartz thistle]

okay

i only studied norms for vector spaces

[amber cradle]

The magical part is that Z[\sqrt{d}] is a vector space

Or, to be precise, a Z-module

Since Z is not a field

A module is basically a vector space but over a ring instead of over a field

[quartz thistle]

im having a hard time understanding this

[tribal moss]

This seems to be an extremely complex way of thinking compared to simply "a subring of a field is always an integral domain".

[dull ginkgo]

I still call vector spaces modules a lot out of laziness

Which is bad

[amber cradle]

Well, can you see how Q(i) is a vector space

[quartz thistle]

a field always has integral domains but the converse might not be true?

[dull ginkgo]

This made me think for a moment but Q[i] = Q(i) lmao

[tribal moss]

I'm not sure what you mean by "has integral domains".

[dull ginkgo]

Subrings of a field are integral domains :3

[quartz thistle]

i mean field is an integral domain

i phrased it wrong

[dull ginkgo]

Yep

Integral domain basically means no zero divisors

No xy = 0 funny business

[quartz thistle]

yess yes got it

[tribal moss]

And if you have a ring that sits inside a field, then clearly you can't have xy=0 in the ring, because then you would also have xy=0 in the field (the multiplication operation is the same; that's what makes it a subring!) and we know we don't have that.

[dull ginkgo]

if all rings between two fields are fields, then it’s an algebraic extension. That’s an iff :3

[quartz thistle]

subrings should have the same operation

[tribal moss]

Exactly.

[crystal vale]

Any hint , and I don't want to use group action results here

[quartz thistle]

thankyou🫡

[rocky cloak]

What's the book?

And where in the book does it say that? Maybe they mean something else by Zn[i] in that case

[tribal moss]

It looks like Seaturtle generalized too hastily from "Zp is an integral domain" and an example showing "Z3[i] is an integral domain", and the book didn't actually make a general claim about Zn[i].

[quartz thistle]

yes that is exactly what happened until Z2[i] didnt fit the generalisation i formed based off an example

that was my bad

the book states for Z[i] to be an integral domain not with the modulo

[south patrol]

I kinda wanna say like

Pretty sure Z_n[i] or (Z/nZ)[i] is very nonstandard notation

I for one have like never seen it

[wraith swan]

I made a walk and thought about the amalgamation of the free product. Is this a quotient set (group?) on the free product?

What structure is normally used to glue those elements?

[dull ginkgo]

I think it’s a push out?

[wraith swan]

yes a pushout, my book calls it a fibered coproduct

[dull ginkgo]

I don’t know much about them since I worked on them by accident

[tribal moss]

Here's how Wikipedia describes it, which seems to make sense.
(https://en.wikipedia.org/wiki/Free_product#Generalization:_Free_product_with_amalgamation)

[dull ginkgo]

Oh I see

[tribal moss]

(Wikipedia at its best, thinking a reader who has any chance of making sense of this paragraph will benefit from having a link that explains what "left-hand side" means)..

[dull ginkgo]

Better than ncatlab

[tribal moss]

Low bar.

[wraith swan]

I dont understand this yet, i guess the next few chapters mention normal and quotiënt groups

[dull ginkgo]

It’s funny how like some articles on ncatlab are fucking gold

Like the salamander lemma one

[tribal moss]

Hmm, bit of a bold pedagogical choice of your book to expect you to talk about category-theoretic properties of Ab before it has even explained quotient groups. But I suppose it can be done.

[teal vessel]

dumb thought that might help others new in their journey:
Center (n):

the normal-est normal subgroup

as a joke definition. it's the normal subgroup that is not only invariant under conjugation, but also unpermuted by conjugation.

[wraith swan]

it explains CT and AA in parallel, since it had an exercise about pushouts and on Set I guess it isn't that strange to do one on Ab. It isn't a recommended exercise

Sometimes exercises also mentioned it may be better to come back to it later

[dull ginkgo]

The normalizer is the stabilizer of G acting on P(G) (powerset) by conjugation

[runic lava]

hello, i have a question please 🙏 to prove that H is a Normal subgroup of a group G, the "classic" method is to take h in H and show that for all g in G, g h g^1 is in H, yes ?

[dull ginkgo]

that is a method yes

[runic lava]

thank you very much

[teal vessel]

that's the definition, yeah. One thing to notice is that if g works, then so does g inverse, so that can cut your time in nearly half

(assuming the majority of elements in G are not of order 2)

[runic lava]

gotcha

the definition we were given is for all g in G, g H g^-1 = H

[runic lava]

just wanted to ease my mind in knowing that this is <=>

[tribal moss]

Often it is easier to prove it by finding a homomorphism that the subgroup is the kernel of.

[teal vessel]

whenever you have a definition of a term, the name of the term and its description are logically equivalent, i.e. <=>, there's no way to be normal without satisfying those conditions, and there's no subgroup which satisfies those conditions without being normal.

there are often other ways to prove that something fits that definition, such as what Troposphere said, but definitions are, by their very nature, slapping a name on all things with these properties, thus rendering the name equivalent to the collection of properties.

[tribal moss]

I think the question was slightly more substantive than that. It can be read as roughly: "Instead of proving gHg^-1=H for all g, is it enough to prove that gHg^-1 subseteq H for all g?"

[runic lava]

english isn't my native language, this is precisely what i had in mind. thank you !

[tribal moss]

And the answer is yes, but it is not trivial that it is true.
For example, it's possible to find an example of a group G and an element g and a subgroup H such that gHg^-1 is a proper subset of H.
That situation just cannot appear for all g simultaneously.

[runic lava]

proper subset meaning a none normal subset of G ?

[tribal moss]

"A is a proper subset of B" == "A is a subset of B, but A is not equal to B". That is, there's at least one element of B that is not an element of A.

[runic lava]

i see

thank you very much !!

[chilly ocean]

hey guys, i'm trying to understand ring adjunctions (constructing the smallest subring that contains bla), I understand the 1 element case and I think I understand the finite elements case in which you recursively adjoin 1 element at a time. But how does it work if I need to adjoin infinitely many elements? even possibly uncountably many?

[tribal moss]

Generally you can start by constructing a polynomial ring with infinitely many variables, and then in one fell swoop quotient out the ideal generated by the relations you want your new elements to satisfy.

Proving nice facts about the resulting ring can get complicated, though.

[dull ginkgo]

this is used to a horrifying extent to construct algebraic closures

I forget the proof of that, I think you take F[X], then define F[|F[X]|] and use Zorn's Lemma to get a maximal F-disjoint ideal to quotient out by and have a field of,

Yeah because you can kind of pass through F[|F[X]|]

[tribal moss]

I think you need to be a little more careful than that when you construct an algebraic closure -- you don't want to accidentally adjoin different independent roots to the same polynomial, or you'd end up with zero divisors.

[dull ginkgo]

Let me clarify

Monic polynomials

and the maximal ideal needs to contain an ideal with a "representation" you want

[tribal moss]

The alternative is to adjoin elements one by one, giving an infinite sequence of rings, and then take the direct limits of those rings. That is more flexible, but also more abstract and conceptually demanding.

(That was an answer to Matias' original question, not specifically for algebraic closures).

[dull ginkgo]

oh

[tribal moss]

Even for monic polynomials, you had better not try to adjoin roots of x²+x+1 and x²-x+1 separately, for example.

[dull ginkgo]

Would this be a way to construct algebraic closures?

Take a field F, and consider F[X]. Let K be the set of irreducible monic polynomials in F[X]. We want to append each monic polynomial's roots to the field, and so we need to adjoin deg(f) roots for each f in K, so we can identify each root with the pair (f, n) for n < deg(f).

and namely we want $f(X) = \prod_{n = 1}^{\mathrm{deg}(f)}{(X - x_{(f,n)})}$

[cloud walrus]

THE TUBE

[tribal moss]

Generally, when you adjoin just one root of an irreducible polynomial, you'll get all of the oher roots as linear combinations of its powers, too.

[dull ginkgo]

oh that is a good point

[frank cosmos]

shouldnt the coeffieints be quotients of units in A?

[chilly ocean]

then I would have to evaluate all polynomials in elements of my set of elements to be adjoined correct? and that would be my final ring? (say I don't have any relations to be satisfied)

[dull ginkgo]

because the quotient of it's principal ideal would be a field hmm

[tribal moss]

And as the x²+x+1 versus x²-x+1 example shows, you also get roots of many other irreducible polynomials for free.

[dull ginkgo]

I wonder if you can construct it as a direct limit

Over the set of irreducibles given some relation

or less so set, more like lattice or pset

[tribal moss]

Yes, I think a direct limit is what I would go for. In that case, we can choose in each step to either adjoin a new root or to do nothing, depending on whether the polynomial we're looking at already has a root from one of the earlier steps.

[dull ginkgo]

transfinite induction here though?

or regular by degree?

[tribal moss]

If the original field was uncountable, it would need transfinite induction, yes.

[dull ginkgo]

I was just going to immediately partition K by degree, well order the equivalence classes and use lexicographic order

[tribal moss]

But each of those individual well-orders will be complex enough to require transfinite induction on its own.

[dull ginkgo]

ah yeah

[tribal moss]

There's probably a way to get around doing it manually with Zorn...

[dull ginkgo]

I was originally going to try instead just using monics in F[X] here and appending x_{f,n}

[tribal moss]

Hmm, I don't think I get what you mean by "have to evaluate all polynomials" there.

[chilly ocean]

Sorry I mean similar to the 1 element case in which the resulting ring for adjoining an element $\alpha$ is $R[\alpha] = { f(\alpha) : f(X) \in R[X] }$ i.e all evaluations of polynomials on alpha

[cloud walrus]

Matias

[dull ginkgo]

Either you need to append "alpha" by quotienting out by an ideal that represents its relations (much like how you construct a free group with relations), or it already exists in some ring with the relations you want, so you map F[X] into the ring such that f(X) = \lapha. Its image is isomorphic to F[alpha]

[tribal moss]

Oh, so your setting is that the ring you're starting with is already a subring of some larger ring, and you're adjoining a particular already existing element of the larger ring?

[rocky cloak]

Sort of a similar way to this to construct the algebraic closure:

Let {f_i} be the set of irreducible monic polynomials, and consider the ring F[X_i] of polynomials with one indeterminate for each f. Let I be the ideal generated by f_i(X_i) and let M be a maximal ideal containing I. Then F[X_i]/M is the algebraic closure of F.

[dull ginkgo]

Like for example, lets say you wanted to take Z and add the inverse of 2?

well this new element, call it a, satisfies 2a = 1. then we know that f(X) = 2X - 1 is 0 at a. So we take Z[X] {ring of polynomials with integer coefficients} and quotient out by the principal ideal (2X - 1). so F[X]/(2X - 1) is isomorphic to Z[1/2].

OR we know that the rationals, Q contains 1/2, so we can just map F[X] into Q sending X to 1/2, then the image is Z[1/2]

:3

One is a constructive approach

it's incredibly similar to how you'd like, append elements to a group

[hollow mica]

I have seen the statement "In PIDs, (p) is maximal iff p is irreducible" before. How does that relate to what you're doing

[chilly ocean]

Yeah, maybe it'll be best if I give full context. My problem was:

Given a large ring $R$, a subring $S$, and a subset $A \subseteq R$, prove there exists a subring "$S[A]$" of $R$ that cointains both $S$ and $A$, and is the smallest subring with this property (i.e any other subring that contains $S$ and $A$ also contains $S[A]$).

I think I got it in the case $A$ is a finite set, by the ring adjunction construction I mentioned first, but I got stuck in the case of $A$ being an infinite set, thus my original question. I'm trying to self-learn this but all of my books/material I found just gloss over this topic

[cloud walrus]

Matias

[tribal moss]

Oh, neat idea just to say "maximal ideal" to get rid of the supernumerary roots.

[dull ginkgo]

I was going to consider the maximal ideal containing the "relators" lmao

[dull ginkgo]

Do you know a priori that the polynomial ring F[S] exists for any indexing set S?

[rocky cloak]

Yes, we delegate the axiom of choice to work in the shadows

[tribal moss]

You need it to be "a" maximal ideal rather than "the" maximal ideal.

[dull ginkgo]

Just realized $R[S] \cong R\left[\bigoplus_{s \in S}{\mathbb{N}}\right]$ as a monoid ring htf did I not notice this before

[cloud walrus]

THE TUBE

[tribal moss]

Oh, in that case there's a standard shortcut: Just consider the set of all possible subrings of R that contain S ∪ A. (There's at least one of these since R itself qualifies). Take the set-theoretic intersection of all of them. Prove this gives a subring, and that it is a solution to your problem.

[hollow mica]

Like if I have a a polynomial f in r, I want to find a polynomial g such that fg = 1. Why does finding a g such that fg = 1 + qp for some q satisfy my needs?

To catch others up to speed:

• E/F is a finite extension
• r is an element of E
• p is the minimal polynomial of r
• n = [F(r):F]

and I'm trying to show that S = {1, r, ... r^{n-1}} is a basis for F(r) over F.
To do this I need to show every ratio of polynomials in r is in Span(S), or that every f in F[r] has a g in F[r] such that fg = 1

[dull ginkgo]

i like this problem

So f and g are in F[X] i take it.

[tardy hedge]

When looking at third isomorphism theorem, (G/N)/(H/N) = G/H, do you guys have any intuition or concrete visualization for it? Like, sure i have a vague idea of why its true, but especially when i look at an example of it being used on say, a dihedral group, im just lost in the intuition, besides being able to follow along the computation

[hollow mica]

F[r] but basically

[dull ginkgo]

do you want me to open up a brief thread

[hollow mica]

sure

[dull ginkgo]

so kiand can get help too

Like if I have a a polynomial f in r

[chilly ocean]

got it, I'll just go with the intersection definition, thanks tropos and tube

[sonic coral]

it’s just fraction cancelation!

[rocky cloak]

Well, the intuition is basically if you take G, then remove N, then remove H, you get the same thing as if you just removed H to begin with.

[frank cosmos]

do we call associates of the content also a content?

namely, when all coefieints of f are quotients of units in A

since then every prime divides f 0 times so by this definition the content is 0, no matter which set of non-associate primes we pick

but then f=c*f_1 where f_1 has coeffients only in A does not work if f is of the form above and the content has to be one

[tardy hedge]

Ok yeah youre “removing” H by passing through N

As like a bridge

Thats fine like yeah thats how I was thinking about it but with normal subgroups, cosets, quotients i tend to get so hung up on all the mechanics of it and what every piece means intuitively

I think i might be trying too hard on it and would be better off just pushing forward

[tough raven]

It's a direct application of that, because the polynomial ring F[X] is a PID.

[frank cosmos]

im thinking more, my question boils down to "why is the content of cx \in K[x], where c!=1 is a unit, one"

[dull ginkgo]

I want to show that |F[s_1... s_r] = q^q^r. Does this work?
Let F_n = F[s_1... s_n]
Inductive Hypothesis: |F_n| = q^q^n.
Base (0): |F_0| = |F| = q = q^1 = q^q^0
Step:
F_(n+1) = F_n[X]/(X^q - X), thus is spanned by s_(n+1)^m for m in {0... q - 1}. This is a basis, as the difference of two polynomials of degree less than q who are mapped to 0 must be divisible by X^q - X of degree q, which is impossible.

Therefore every element can be uniquely indexed by a q-tuple of F_n, thus |F_(n+1)| = |F_n|^q = (q^q^n)^q = q^q^(n+1)

[hidden wind]

this article i’m reading seem to use “G covers H” to mean H is isomorphic to a quotient of G, is this standard terminology?

[dull ginkgo]

That means there is a surjective homomorphism (epimorphism) from G to H

Thus im(f) = H, and by first isomorphism theorem

H is isomorphic to G/ker(f)

but that's very dumb and out of the way

or not ig I guess it's nice for shategory theorists

[lusty marlin]

What is the article about?

[hidden wind]

it is also very possible i'm misunderstanding something

[hidden wind]

https://www.mat.ucm.es/serv/revista/vol1-123/vol1-123h.pdf

[hidden wind]

the distinction between SL and PSL is clear enough... now they just had to bring SL^+- and PSL^+- into the mix as well huh

[kind temple]

what book is this from?

[dull ginkgo]

Jacobson

[untold garnet]

Let U, V, W be finite-dimensional representations of a finite group G over some base field. Assume that for every cyclic subgroup H \leq G, U_{|H} is isomorphic to V_{|H} or W_{|H}. Is it true that U is isomorphic to V or W?

(I don’t know if that’s helpful, but in my situation, U, V and W have the same dimension and U embeds in V \oplus W).

[rocky cloak]

I don't know if I understand your question correctly, but

Let G be S3, and let X be the standard representation, S the sign representation and T the trivial representation.

Let U = X (+) S, V = T (+) S^2, W = X (+) T.

Then U embeds into V(+)W, the restriction of U to an order 2 subgroup is isomorphic to the restriction of V, and the restriction of U to a order 3 subgroup is isomorphic to the restriction of W.

[fast heart]

Can gHg^-1 be a proper subset of H for a subgroup H of a group G?

[barren sierra]

yes

the example that comes to mind is some matrix stuff

but maybe there's something even easier

[rocky cloak]

As a medium difficulty exercise see if you can come up with one yourself

[barren sierra]

yea it's not entirely obvious, it's a good question

[rocky cloak]

I think ||semidirect products|| might yield the easiest examples

[barren sierra]

maybe yea

[rocky cloak]

A mix of the two might be ||let G be the group of functions f(x) = ax + b with a nonzero. Let H be the subgroup generated by h(x) = x+1.||

||Let g be g(x) = 2x. Then ghg^- = x + 2||

[untold garnet]

Okay, I guess that makes sense. Thank you!

[rocky cloak]

So you might think of this as ||the semidirect product of Q and Q^||
or as
||the matrix group [Q^, Q; 0, 1]||

[hidden wind]

[round jay]

The exact sequence at the bottom is not split because Z is not the product of Z and Z/2Z right?

[agile burrow]

Yes, that's right

[next obsidian]

hi walter

[fast heart]

thanks

[fast heart]

thanks, this was very instructive

[tardy hedge]

Automorphisms are like global symmetries of the group?

[agile burrow]

Hi chmonkey. Sorry for delayed response, I was putting my toaster away

[next obsidian]

Dear Walter,

Please do not fret, I understand that one has priorities, and the prompt return of a toaster to its dutiful position ranks high on this list

[barren sierra]

Not a bad way to put it

[round jay]

If f has a left inverse h, is coker(f) isomorphic to ker(h)

?

[rocky cloak]

Yes, if hf = 1, then f is mono and h is epi and fh is idempotent. So the codomain of f equals
Im(fh) (+) ker(fh) = Im(f) (+) ker(h)
Hence ker(h) = cok(f)

[rotund aurora]

Let G be a topological group (assume it is Hausdorff, although idk how relevant this is). Let F, U subseteq G with F compact and U open. Is bigcap_{f in F} fU necessarily an open subset of G?

I would say this is false. But idk counter-examples

[rocky cloak]

So if C is the complent of U, then your question amounts to asking if FC is still closed.

Which seems to be true
https://math.stackexchange.com/q/71983/306319

[rotund aurora]

are you claiming that $\left(\bigcap_{f\in F} fU\right)^c=F U^c$?

[cloud walrus]

croqueta3385

[coral steeple]

Anyone know any rings where for all subsets X, Y, we have (X)=(Y) implies X=Y?

[rocky cloak]

Yes, complent of intersection is union and
(fU)^c = f(U^c)

[coral spindle]

If the ring has any nontrivial units, then this does not hold. So certainly at least this ring has characteristic 2.

Now on the other hand

If this ring is not the trivial ring, note (1) = R = (R), so

So in fact the only ring for which this holds is the ring with one element.

[coral steeple]

Shoot

[rocky cloak]

If we replace subset with element there can be more interesting answers I guess. Like Z/2 or Z/2[x]

[rotund aurora]

I don't know why I find many "simple" problems regarding topological groups somewhat hard...

[coral steeple]

My friend is interested in taking a subset X and asking what the "smallest" set Y for which (X)=(Y) is, I was thinking this might fail in that way but ig not

[rocky cloak]

I'm not convinced this problem was particularly simple

[coral spindle]

There is typically not a unique smallest subset, and often there is not even a minimal generating set for an ideal. Nonetheless such things sometimes exist.

Rings are not like vector spaces, they don't have nice bases

[rocky cloak]

I guess Z/2 and Z/2[x] are examples where its always possible to find such a unique smallest set

[coral spindle]

I wonder if Boolean rings are also examples?

Well. Finitely generated ones are, if I'm remembering the structure of Boolean rings correctly

No, scratch that.

[rocky cloak]

Yeah, I guess any ideal of (Z/2)^n is just generated by a single element. And that should be unique

[coral steeple]

Wouldn't this property be had by Z though?

[rocky cloak]

(1) = (-1) for example in Z

[coral steeple]

Ah shoot

Thank you both

[worn spindle]

if u have any n in {2, 4, p^k, 2p^k | p > 2, p is prime} then (Z/nZ)^x = <m> for which m?

hint: (Z/nZ)^x is cyclic due to the conditions and is the ring of units

aka multiplicative group mod n

/nick ping when responding

ik it's a primitive root modulo n

[tribal moss]

You've read https://en.wikipedia.org/wiki/Primitive_root_modulo_n#Finding_primitive_roots ?

[worn spindle]

ya

what would be a general form for a primitive root mod n?

[tribal moss]

The section I linked to explains that the way you find a primitive root is generally by trying small (coprime) numbers one by one until you hit one, with various tricks to speed up the test of whether the one you're looking at has the right order.

[opal plover]

what does it mean to take a quotient group to a power

[tribal moss]

Context?

[worn spindle]

No simple general formula to compute primitive roots modulo n is known
ah ic ty

[lusty marlin]

I think they confused the ^x in (Z/nZ)^x with an exponent

[tardy hedge]

Wow i just got new intuition for the gN = Ng criteria for N being normal subgroup

And how it relates to being able to quotient by N

That made it click more for me rather than thinking of it as gNg^-1 = N somehow

[barren sierra]

🔥

that's IMO one of the big hurdles of learning algebra for the first time

[tardy hedge]

Yeah. But like ive already studied this before yet im STILL trying to think about it 😂

Going back to review things rn

[barren sierra]

of course of course

[tardy hedge]

Yeah thats crazy how that works hahaha

[barren sierra]

sometimes things only clear up on the 2nd look

or the 3rd look

or the 10th

[tardy hedge]

Like it does take a long time to really internalize concepts and understand them better

It *can at least

For some things

I been hung up on normal subgroups and how it relates to kernel, quotients, cosets etc etc

For a while

I understood it somewhat but it didnt feel satisfying

Cause ppl ALWAYS just say something like “normal subgroups are somehow compatible with the overall group structure”… but like, what does that actually MEAN?

Why arent exactly any old subgroup compatible with “overall group structure”

Etc

[next obsidian]

Cuz you can show that if G satisfies P, N satisfies P only when N is normal

Cuz of weird stuff, it just is how it is

[tardy hedge]

Yes but im trying to think in terms of intuition so that the results seem to make more intuitive sende

Sense

If i just see algebra symbols and im not “understanding”, thats not satisfying

If you get what i mean

[next obsidian]

Yeah

[tardy hedge]

Maybe some ppl “see” it quicker ofc

[next obsidian]

But it’s just something that you justify by observing how it’s different

[tardy hedge]

Good point!

By working with it directly more

[next obsidian]

You see enough theorems that justify the statement that you need normality to pass down properties from G to N

[tardy hedge]

You would gain better intuition based on what u said

Yea thats interesting and true!

[next obsidian]

And then somehow the statement that you need normality becomes the intuition

[tardy hedge]

Yeah so true! Im noticing that myself rn!

[next obsidian]

Because you look at a new situation and think “hmm I think I need this to be normal”

[tardy hedge]

Im learning new material i havent before and its cleared up things

A bit more at least

Providing new perspectives on i

It

[next obsidian]

It’s unsatisfying that sometimes it’s this way but it’s just how it is sometimes

[tardy hedge]

Yeah

I think i like algebra a hell of a lot more than analysis thats for sure hahaha

I was studying analysis before this and didnt get nearly as hyped

Maybe im not at the more interesting parts though

Im not that far in it admittedly so

[next obsidian]

I decided like 1 week into algebra that "it just doesn't make as much sense as analysis :("

and that I wouldn't like it

[barren sierra]

God had to nerf me somehow

so he made me bad at analysis

[tardy hedge]

Yeah i dont know i just dont find convergence that interesting

The kinds of questions that “come up” in analysis i just dont really care that much about

[sonic coral]

my favorite part about algebra is having access to a lot of tools and being able to use them

analysis is just trying different inequalities

[tardy hedge]

Yeahhh

[tardy hedge]

I guess studying universal algebra provides some intuition too!

[muted comet]

taking the rotations in D_4 as a subgroup. why is it not eneough to see that it is abelian to know it is normal?

[dull marsh]

Why would it in general be true that an abelian subgroup is necessarily normal?

[muted comet]

well every subgroup of an abelian group is normal. the rotations are an abelian group. and is a subgroup of itself, thus it is normal?

[dull marsh]

Every subgroup of abelian group is normal in that group, yes

Is D_4 abelian?

Ah

Is your conclusion that <r> is normal in <r>?

[muted comet]

well yes but i guess i just dont see the difference i guess. gotta get more familiar with the definition.

[dull marsh]

A subgroup H of a group G is normal iff is it closed under conjugation by every element of G

It is trivial that every group is normal in itself, you don't need it to be abelian

<r> is normal in <r>

<r> is normal in D_4 as well but <r> being abelian is not enough to show that

[languid trellis]

,rotate

[languid trellis]

I'm struggling to make progress on 8

I have that p | a^2 + 1 for some a

so bp = a^2 + 1

And i am completely stuck. Can anyone give a pointer please?

[rotund aurora]

Use that it is not a prime in Z[i]

[languid trellis]

So there exist a, b in Z[i] such that p | ab but p doesn't divide a or b. Should I then claim that a^2 + b^2 satisfies this property?

[tranquil musk]

Hey guys I have a question that I am not sure how to solve and any help would be appreciated greatly.

Let F be a field. Suppose that f = x^2 + x + 1 ∈ F [x] is irreducible, so F [x]/(f ) is a field.
(i) Find q ∈ F [x] so that deg(q) ≤ 1 and q+(f ) = ((x+3)+(f ))·((2x+1)+(f ))

[rotund aurora]

No

[rocky cloak]

Well, there's not so much to it. You just multiply out the expression, then use that
x^2 + (f) = -x - 1 + (f)

[tranquil musk]

yeah so im sure that im not thinking about the question right, so if i multiply that expression out it gives a quartic polynomial and then I have: q + (x^2 + x + 1) = some quartic polynomial and q cant be higher than degree one so it doesnt make sense to me. do you know what im saying?

[languid trellis]

Irreducible implies prime, so not prime implies not irreducible, so p = p1^e1 ... pn^en, pi primes in Z[i], as p has complex part 0, it follows that p =zz* so p = a^2 + b^2?

[rocky cloak]

I'm not sure I do follow what you're saying.

But if you multiply (x+3) and (2x + 1) you should get a quadratic polynomial. Then you can remove the x^2 term using that x^2 = -x - 1

That's all there is to it

[tranquil musk]

I dont understand why I would multiply (x+3) and (2x+1) because it is ((x+3) + f) ((2x+1)+f)

and f is a quadratic

so it becomes a quadratic times a quadratic

[rocky cloak]

It's not (x+3) + f, it's (x+3) + (f) where (f) is the ideal generated by (f)

Do you know how multiplication in F[x]/(f) works?

[rotund aurora]

ye

[tranquil musk]

ahh ok yeah well I am confused but I know what you are getting at. No i am not totally sure how that multiplication works, I dont get where the question specifies that we are working in F[x]/f

[rocky cloak]

Well the notation (x+3) + (f) is how you would notate an element of F[x]/(f). So I guess that's the main reason. And then the start of the exercise also hints at it a bit

[tranquil musk]

Ok, forgive me I'm not well versed with this, but does it go this way? ((x+3) + f) ((2x+1) + f) = (x+3)(2x+1) + (f)(x+3) + (f)(2x+1) + (f)^2, then do we divide out the (f)?

[rocky cloak]

So when you multiply an ideal by something you just land in the ideal again. So this whole thing actually just reduces to
(x+3)(2x+1) + (f) In the quotient ring.

[tranquil musk]

sorry bro Im so confused im trying to read up on this stuff but I cant find much. ChatGPT which is terrible at math is telling me its equal to the remainder of (2x+1)(x+3) divided by f.

Im trying to find the operations in F[x]/(f) online

[rocky cloak]

Well, they're pretty much what you've shown already.

If you have p + (f) and q + (f) then their product is pq + (f) and their sum is p+q + (f).

And for any polynomial r that is contained in (f) (i.e. is a multiple of f) you have p+r + (f) = p + (f)

So by choosing r appropriately you might be able to reduce the degree of p, and this is exactly the division algorithm

[dire siren]

@languid trellis alternative ending: p=xy, with x and y non-units, so N(x)N(y)=p^2, so N(x)=p

[tranquil musk]

Ok first up thankyou heaps for your help. I did everything you said and I got q = 5x+1. I just dont really understand the last part where you said x^2 = -x -1. ChatGPT is telling me some other way where it divides the expression 'pq + f' by the ideal x^2 + x +1 and takes the remainder and it gives 5x+1 as well. I dont really get what weve done here and why both methods work.

[rocky cloak]

Yeah, so since
x^2 + x + 1 is in (f) we would have
x^2 + x + 1 + (f) = (f)
which we can rearrange to
x^2 + (f) = -x - 1 + (f)

This means that every time we have an expression of the form x^2 + (f) we can replace it by the right hand side.

Another way to think about this is that were just subtracting by x^2 + x + 1, cancelling the x^2 and leaving -x - 1.

The division algorithm just determines which multiple of x^2 + x + 1 we should subtract to get the polynomial to lowest possible degree.

For example (x+3)(2x+1) = 2x^2 + 7x + 3
So we want to subtract 2 times x^2 + x + 1, to get it down to degree 1

[obsidian cedar]

Im trying to show a group G of order 36 is simple. i first showed G can have 1 or 4 sylow 3-subgroups and 1,3 or 9 sylow 2-subgroups.

i then assumed G has 4 sylow 3-subgroups and 3 sylow 2-subgroups. each sylow 3-subgroup has order 9 and each sylow 2-subgroup has order 4. the only common element in each subgroup is the identity.

i then showed the 4 sylow 3-subgroups contain 4 * 9-3=33 unique elements, and the 3 sylow 2-subgroups contain 3 * 4-2=10 unique elements, which exceeds the order of G. Therefore G has either 1 sylow 3-subgroup or 1 sylow 2-subgroup => not simple

is this valid?

[cobalt heath]

Uh, if you have 1 sylow subgroup, isn't that group a normal subgrouo

[obsidian cedar]

yeah, hence => not simple

[rocky cloak]

I think you mean hence not simple

[obsidian cedar]

oh yeah, my bad

[rocky cloak]

Anyway, there is some issue with assuming the only common element between sylow groups is the identity.

It could for example happen that the intersection of two subgroups of order 9 has order 3

[obsidian cedar]

can i assume the intersection has at most 3 elements, and then show that it still exceeds the order of G?

[cobalt heath]

Yeah, that's possible, I guess

Hmm but I think you need to be careful in counting

[obsidian cedar]

how so

[tranquil musk]

Thanks heaps, makes sense to me especially the example at the end. Btw, does the same method work to do (q + (f)) ((x+1)+(f)) = 1 + (f)? Ive done multiplication using the rule for multiplication and I get 'qx + q + (f) = 1 + (f)'

[cobalt heath]

If H = {a1, .., a9}, potentially {a1, a2, a3} is in another sylow 3-gp, {a4, a5, a6} is in yet another sylow 3-gp, and similar for {a7, a8, a9}.

[rocky cloak]

Well, the identity element is in all the sylow subgroups, so you can do a little bit better I guess

[obsidian cedar]

i see, but wont assuming that the total intersection has at most 3 elements cover all those cases?

[cobalt heath]

Ah, yeah

[cobalt heath]

Yep but in this case, you can't like (9-3) * (number of Sylow subgroups)

[obsidian cedar]

yeah that makes sense

counting gave me 27 unique elements in the sylow 3-subgroups and 8 unique elements in the sylow 2-subgroups, which gives me 34 total elements.. so i guess this method wont work

[languid trellis]

How does this get us to a^2 + b^2 = p sorry

[dire siren]

the norm of x is a sum of two squares

[rocky cloak]

If you say every group intersects every other group in an order 3 group, and no three groups have nontrivial intersection. That means you have 2 unique elements for each group, 2 unique elements for each pair of groups and the identity.

Totaling 4*2 + 6*2 + 1 = 21 elements. Now, I don't know if there's an argument for why such a configuration can't ocour, but naively you can get it down that low.

[obsidian cedar]

Oh wow, ill try a different approach then😅

[languid trellis]

Oh right of course, definition of norm in Z[i]

[rocky cloak]

Here's a hint for a different approach
||If you have four 3-sylow subgroups, then G acts on them by conjugation. This gives you a homomorphism G -> S4||

||Think about what such a map could be||

[obsidian cedar]

ah okay, ty

[rocky cloak]

I guess from this you get the more general statement that ||anytime you have four 3-sylow subgroups bigger than C3, they must have a common intersection.||

[obsidian cedar]

kernel cannot be trivial as |G|>|S4|, so we get a nontrivial normal subgroup of G (?)

[rocky cloak]

Indeed

[obsidian cedar]

perfect, ty

[worn spindle]

can i strengthen this to iff?

[obsidian cedar]

yes, theyre equivalent

[worn spindle]

does naturally identified mean isomorphic?

or equal?

or sth else?

[dire siren]

but what do you even mean? that's a definition; we don't use iff in definitions

[worn spindle]

well all definitions are iff

i got that

any idea abt what naturally identified means

im guessing "identification" means isomorphism?

[dire siren]

it means that the class of an element is identified with that element

[worn spindle]

oh yeah https://www.reddit.com/r/math/comments/13uq6pz/comment/jm1y3wm/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button

[hidden wind]

ok it does NOT feel right that the tiling of hyperbolic plane by ideal (∞,∞,∞)-triangles is a quotient of the tiling by (2,3,∞)-triangles

totally unrelated but real

[obsidian cedar]

i like my unique factorization domains chinese

[rocky cloak]

What's a Chinese domain?

[hidden wind]

https://www.sciencedirect.com/science/article/pii/0022404982900391?via%3Dihub

i have no idea, i just have a (bad) habit of just clicking on and glancing through random things

[mighty kiln]

Chinese rings now 😭

[hidden wind]

wait can this possibly be true

the automormphism group of the (∞,∞,∞) tiling is isomorphic to the group presented < x,y,z | x^2 = y^2 = z^2 >

while the group for the (2,3,∞) tiling is < x,y,z | x^2 = y^3 = xyz >

[wraith swan]

Sorry another question about the fibred direct sum, pushout in Ab. Does A +_C B contain elements from A and B that are not in the image of alpha and beta?

[dawn quest]

guy who named them is european lol

[rocky cloak]

I mean it's named after the Chinese remainder theorem, so I guess where the author is from is not that relevant.

[rocky cloak]

You mean in the image of iota_1 and iota_2? You will also have sums of things from A and things from B, but otherwise that's everything

[wraith swan]

Right so there does not exist a c for every element in A + B

That's only for elements in A and B that are equivalent

[rocky cloak]

You have a surjective homomorphism from A(+)B onto A (+)_C B, so every element of A(+)B gives you something in the pushout

Really, the pushout is just A(+)B where we identify elements that come from the same element in C

[worn spindle]

why is uniqueness needed here?

[wraith swan]

Ok thank you

[rocky cloak]

You don't actually need to assume it. Identities are always unique (try proving it yourself!)

[tardy hedge]

in showing that Inn(G) is a normal subgroup of Aut(G), they played with B(a(B^-1(x))a^-1) = B(a)B(B^-1(x))B(a^-1)

idk why im confused on how they split the function composition like that

ohh because they are all isomorphisms

so its just using homomorphism property

[worn spindle]

wb in a boolean ring? there -1 and 1 are both

also we use d&f

The definition of “group” on page 16 of Abstract Algebra by Dummitt and Foote (2004) says “(ii) there exists an element 𝑒 in 𝐺, called an identity of 𝐺, such that for all 𝑎∈𝐺 we have 𝑎⋆𝑒=𝑒⋆𝑎=𝑎.” Note that they say there exists “an element” not “a unique element”, and that say that 𝑒 is called “an identity”, not “the identity”. The axioms do not require that there is only one identity, only that there is at least one

[rocky cloak]

Yes, but 1 = -1, so that's still just one element

[rocky cloak]

Exactly, you don't need to require it, because you can prove it

[tardy hedge]

True?

[worn spindle]

how is a subring a ring itself?

when you dont require it to be an abelian group under addition

[hollow mica]

If you require it to be a ring, then by definition it is a (sub)group too

[dull marsh]

1. Any subgroup of an abelian group is also abelian
2. Read the definition

[worn spindle]

i didnt know 1 😮 ty, makes sense since if they were abelian before then theyll still commute since they were in G

If R has zero divisors then so does R[x]
does the converse hold?

[hidden wind]

an american would never honor the chinese like that even if due

[teal wind]

Can anyone help me with part b and c?

[lusty marlin]

For part b, consider the subgroup generated by 3

[hidden wind]

OK yeah i am definitely off... i have a mostly clear geometric image in my head (each (∞,∞,∞) triangle consists of six (2,3,∞) triangles) but i struggle to translate it into algebra

i'd imagine that gluing six of these (2,3,∞) triangle corresponds to simply quotienting out the generator x of order two and y of order 3 but i may have that wrong

[teal wind]

I did, but I was told this:

[hidden wind]

it seems rather like (2,3, ∞) ought to be a quotient of (∞,∞,∞)

[teal wind]

"I actually meant u_10 so you could use previous part..."

[terse crystal]

U(Z/nZ) is direct sum of U(Z/p^e Z) where n=Πp^e

So you can start factorizing 10

Or 12

[worn spindle]

If S is a subring of R then S[x] is a subring of R[x],
does the converse hold?

[worn spindle]

also bump on similar but unrelated q

[dire siren]

yes, hint: look at the leading coefficients

[teal wind]

Could someone help me with this problem?

[dire siren]

hint: consider constant polynomials in S[x]

[teal wind]

Right now, I have this, what are your thoughts guys?

[south patrol]

What is R_n

[teal wind]

not defined

[south patrol]

Oop so how are we supposed to help you

[rapid junco]

is there a straightforward way to prove that (x+1)^n - x^n has no roots in Q other than -1/2.

for even n

and then for odd n prove no such roots exist.

[south patrol]

Shift it to $(x + 1/2)^n - (x - 1/2)^n = 0$. since $x + 1/2 \ne 0$ we have $1 = \Big(\frac{x+1/2}{x-1/2}\Big)^n$

[cloud walrus]

Crystalline Potato

[south patrol]

so (x + 1/2)/(x-1/2) = 1 (or -1 if n is even)

So x + 1/2 = x - 1/2 [impossible], or n is even and x + 1/2 = 1/2 - x so x = 0

[rapid junco]

ah well done

[south patrol]

But yeah bit of a trick i suppose

[rapid junco]

indeed

[south patrol]

I wonder if there is a nicer way lol

Yeah so uh

If $f_n(x) = (x + 1/2)^n - (x-1/2)^n$ then $f_{n}'= nf_{n-1}$

[cloud walrus]

Crystalline Potato

[south patrol]

f_1 has no roots [it's just costantly 1]

[rapid junco]

how does the formal derivative relate roots again?

[teal wind]

im sorry, this is it:

[south patrol]

No need to even use the formal derivative since we are over Q lol

But yeah it's more an analytic method rather than algebraic

[rapid junco]

but how does f_'n not having any roots imply that f_n doesnt have any.

[worn spindle]

ah tahst what i thought -- just to confirm in writing (haha) it is true, right?

[south patrol]

i didn't finish the argument

[dire siren]

yes

[south patrol]

So uh suppose n is odd and that f_n(x) > 0 for all x

Then f_{n+1} is monotonically increasing since its derivative is nf_n, and f_{n+1}(0)=0 [so it has exactly one root]

Since f_{n+1} is monotonically increasing with f_{n+1}(0)=0, it follows that f_{n+2}' > 0 for x > 0 and f_{n+2}' < 0 for x < 0

So f_{n+2}(x) >= f_{n+2}(0) > 0

Yeah that's a funny proof lol

[rapid junco]

lol nice argument

[vapid vale]

isomorphisms map generators to generators

[teal wind]

could we go over this?

[vapid vale]

what have you gotten so far

[strange grove]

How do you determine all isomorphisms from Zn to itself?
I used the fact that if f(1) = a is an isomorphism then g(a) = 1 is inverse isomorphism of f. Then |a| divides |1| and |1| divides |a| so |a| = |1| so |a| = n, and in Zn order of generators is n so all those homomorphisms that map 1 to a generator are isomorphisms. So there are phi(n) isomorphisms in total

Is this valid proof?

[teal wind]

anyone?

[south patrol]

Yes, nice (though i wouldn't just say "g(a)=1")

One way to look at this is as follows: every map $\mathbf Z/n \to \mathbf Z/n$ is given by multiplication by something. It's invertible iff you multiply by something in $(\mathbf Z/n)^\times$

[cloud walrus]

Crystalline Potato

[clear ingot]

for this problem, can i just say that by bazouts, there exists a, b in Z for which 2a+3b=gcd(2, 3)=1, which is a generator of Z. qed

[terse crystal]

Yeah. Better give one specific pair of a,b

[clear ingot]

u just need any bijection from Zn to Zn such that the isomorphism is also a bijection between the set of generators of Zn and the set of generators of Zn right, so varphi^2 of them?

[terse crystal]

Just φ(n)

specify just one generator of the domain

[pure arrow]

This is a proof from Artin's Algebra textbook.

What I don't understand is, if a is the smallest postive integer in S, then how does that imply r=0?

[terse crystal]

“When a is the smallest positive integer in S”

r=n-qa is in S, and it’s non-negative

[hollow mica]

by construction 0 < r < a

[clear ingot]

ye thats true

[warm ember]

is it true that every irreducible polynomial in Fq divides x^p^r-x for some r

where q is a power of p

i think it should be true cuz the splitting field for f is an extension over Fp

so it divides the characteristic polynomial?

[strange grove]

I mean f^-1(a) = 1?

[hollow mica]

In Z[x], the ideal (x - 2) is prime but not maximal, correct?

[celest furnace]

Yep, looks good to me. With this one in particular you can verify that by saying that (x-2) is the kernel of the map evaluating a polynomial at x = 2, and Z is an integral domain but certainly not a field so the ideal is prime but not maximal

[hollow mica]

Lmao I was asked to come up with a prime ideal in an integral domain that isn't maximal

When I could've just done (0) in Z

I also had brain fog and showed (x - 2) was not maximal by showing that 1 isn't inside (x, x-2)

quotients is way cleaner

[dawn quest]

To be clear, the surjectivity part is only necessary for the diagram to commute; an isomorphism phi^bar always exists regardless?

[languid trellis]

Nope. If phi is not surjective them phi bar isn't an iso, it's just an injection

Its always an iso between G bar and the image of G under phi though

[dawn quest]

sorry, I also meant to add that we change the range of phi^bar to be im(phi)

Wait in that case, if we change G' in the diagram to im(phi) as well, does the diagram also commute then?

[wraith swan]

Wait, isn't G bar initial and therefore there is a unique morphism phi bar whatever phi is?

[dawn quest]

Bro I don't know what initial is

[languid trellis]

Yep and yep

[wraith swan]

Oh sorry, unique not a unique isomorphism

[cobalt heath]

Initial in which category

[wraith swan]

testing my knowledge 😛
Uh, coslice category Grp/G ?

[cobalt heath]

But G bar shall depend on both data of G and data of phi

[wraith swan]

I only know about Set/X so I guess you need extra restrictions on the morphisms

[cobalt heath]

Difficult stuff

[wraith swan]

Can someone give some feedback on the first part of my proof that Ab has fibred coproducts? The first part only constructs it as a group on the quotient set A x_C B.

The next chapter is about quotient groups, but this exercise let's you optionally try it out before.

I have implicitly used that all morphisms are homo and all elements are commutative.

Are there some things straight-up wrong? Have I done some things I can't just conclude and need additional lemmas first? Maybe some critque on my notation and way of writing it down?

With defining this operation on A x B/~ have I shown A x B/~ is a group and am i now only left to show it is a pushout?

The diagram

I am also not really sure if my notation and usage of the equivalent relation is that correct... Like i_1 a = i_2 b, but that does not define an equivalence relation explicitly, do I need to define this first? Can I just say that from i_1(a - a') = i_2(b' - b) that there exists a c by the equivalence relation?

[languid trellis]

We have already shown that primes of the form 4n+1 in Z aren't prime in Z[i]. Moreover, it's clear that if gcd(a,b) = 1 in Z, then a + bi is prime in Z[i]. Now, I suspect that primes of the form 4n+3 in Z remain prime in Z[i] just from working with a few small examples, but I don't know how to prove this. Does anyone have any hints?

[tribal moss]

gcd(11,3) = 1, yet 11+3i = (1-2i)(1+5i)

[languid trellis]

ohhh that complicates things

also 2 = (1+i)(1-i)

thats a prime not of the form 4n+1, 4n+3

the only one

but there we are

[dull ginkgo]

the whole 4n + 3 thing is because 3 is the only non residue mod 4 right

[languid trellis]

it may be

[dull ginkgo]

Let me check

[languid trellis]

its not a qr mod 4 if thats what you're asking tho

wait

[dull ginkgo]

it’s the only non qr mod 4

[languid trellis]

i don't know anywhere near enough numebr theory. fuck this book

[dull ginkgo]

That’s more ring theory

A lot of basic number theory is flat out just basic ring theory in a way the integers are the simplest domain

[languid trellis]

now I'm not convinced anything of the form a+bi, a,b neq 0 should be prime tbh

[dull ginkgo]

check it’s norm

[tribal moss]

3 is also the only residue mod 4 that is not the sum of two squares.

[dull ginkgo]

Which is multiplicative

[hidden wind]

why can i never speel gruop correvtly

[dull ginkgo]

Two squares within Z/(4)Z or in general

[tribal moss]

Squares in Z/4Z (but those are also the images of squares in Z, so it doesn't really matter).

[dull ginkgo]

x^2 = 1 (mod 4) iff x = 1 (mod 2)
x^2 = 0 (mod 4) iff x = 0 (mod 2)

Do analogies like this hold for other Z/p^2Z rings?

[tribal moss]

The point is that if p is prime in Z, then its norm is p², so it if factored, each of the factors would have norm p. However if p = 4n+3, it is impossible for it to be a²+b² for integer a,b, because 3 is not a sum of two squares mod 4.

[dull ginkgo]

Oh shit this implies the sum of two squares can only be 0, 1, or 2

[languid trellis]

Can you explain to me in full detail how we get to "However if p = 4n+3, it is impossible for it to be a²+b² for integer a,b, because 3 is not a sum of two squares mod 4.", in particular from "because"

I don't see why we start to work mod 4

[dull ginkgo]

I by default if I see a square i take shit mod 2

And then take shit mod 4

[tribal moss]

If a²+b² = p = 4n+3 in Z, then (a mod 4)² + (b mod 4)² = 3. But we have just seen that isn't possible.

[dull ginkgo]

Hey Tropo

[languid trellis]

"we have just seen that isn't possible", where?

[tribal moss]

Because the only squares in Z/4Z are 0 and 1, and no combination of those add up to 3.

[languid trellis]

oh of course

[dull ginkgo]

Let’s say we have a monic integer polynomial p(x) with root u

Then does Z[u] necessarily have a norm?

[tribal moss]

https://en.wikipedia.org/wiki/Sum_of_two_squares_theorem gives a useful converse.

[dull ginkgo]

I assume for each element x in Z[u] you can take the product of all its conjugates to get an integer?

[wraith swan]

<@&268886789983436800> ^ 18+

[languid trellis]

and to check whether e+fi is prime, we need to show that there is no (a+bi)(c+di) = e+fi. in other words, e = ac-bd f = ad +bc. Intuitively, this seems like the result that A: V->W a linear map is surjective if dim V >= dim W

except 1+i

maybe that's prime

[dull ginkgo]

can xi still be prime

even though it’s x * i

I forget how prime elements are defined in general

[languid trellis]

yes because i is a unit

[dull ginkgo]

Assume a + bi is prime

Assume both are nonzero too

then a and b must be coprime as otherwise d | (a + bi)

[languid trellis]

tropo gave a counteraxample

"gcd(11,3) = 1, yet 11+3i = (1-2i)(1+5i)"

[dull ginkgo]

11 + 3i isn’t prime though?

Is it

[languid trellis]

oh I see what you're saying

one way implication

[dull ginkgo]

I’m working my way to a condition and then trying to find an if

[languid trellis]

yeah i agree they should be coprime

[dull ginkgo]

well then the norm is multiplicative

N(a + bi) = a^2 + b^2

hmm

N(a + bi) must be coprime to a and b

[languid trellis]

sps a+bi is prime. then if a+bi | cd, then a+bi |c or a+bi | d. also a^2 + b^2 | N(cd) = N(c)N(d)

also a^2 + b^2 | N(c) or N(d)

[dull ginkgo]

Hm

[languid trellis]

sps wlog a+bi | c

actually

actually idk

doesn't this imply a^2 + b^2 is prime in Z ? ( i dont think so)

by the norm argument

[dull ginkgo]

The fuck

I have no idea what you did here

[languid trellis]

a+bi | cd, so cd = u(a+bi), so N(cd) = N(u)N(a+bi) so N(a+bi) | N(cd)

no?

[dull ginkgo]

What is cd here?

[languid trellis]

c, d gaussian integers

[dull ginkgo]

just any two Gaussian integers?

[languid trellis]

yep

any two such that a+bi | cd

[dull ginkgo]

You’d need to show there is a Gaussian integer x with norm N(x) = n for any n to show primality, no?

But no 4n + 3 can?

[languid trellis]

what does this sentence mean?

[dull ginkgo]

Why are you doing this?

[languid trellis]

I'm trying any idea

[dull ginkgo]

Are you trying to show N(a + bi) | N(c)N(d) => N(a + bi)|N(c) or N(d)

[languid trellis]

oh shit I see. I assumed a+bi is prime to conclude that

I'm not sure what I'm trying to do

wiat wait wait wait

[dull ginkgo]

Our goal is to show N(a + bi) is prime

[languid trellis]

consider 1+i. suppose it is not prime. then it is reducible, so there are xy s.t. 1+i = p, so 2 = N(x)N(y), but 2 is prime, so x or y is a unit and hence 1+i is irreducible hence prime

I think the norm being prime might be what we need (unless the norm is a prime of the form 4n+1)

[tribal moss]

Obviously if a²+b² is prime in Z, then a+bi has to be prime (or at least irreucible) in Z[i].

On the other hand, 3 is prime in Z[i], but its norm is 9, which isn't

[languid trellis]

jeebus christ

[languid trellis]

i though irreducible -> prime (in UFD's at least), isn't irreducibility a stronger condition?

actually I think they're equiv in ufd

[tribal moss]

In UFDs primes and irreducibles are the same, yes.

And primes are always irreducible anywhere.

But there are non-UFD rings where some elements are irreducible without being prime.

[languid trellis]

lovely

[dull ginkgo]

How does a + bi being prime imply N(a + bi) being prime

[languid trellis]

other way around

[dull ginkgo]

I see

No i mean if a + bi and both a and b are nonzero then does N(a + bi) have to be prime

[languid trellis]

we're assuming a+bi prime right

[dull ginkgo]

Yes

[languid trellis]

a+bi reducible implies N(a+bi) reducible as a+bi = xy (x, y non-units), N(a+bi) = N(x)N(y). so N(a+bi) irred implies a+bi irred by contrapositive and hence prime?

[tribal moss]

If a+bi is prime, I think that is true, but I don't recall how to prove it.

[dull ginkgo]

There need not exist an x such that N(x) = 4n + 3

[languid trellis]

this isn't saying anything beyond a^2+b^2 prime in Z then a+bi prime in Z[i]

which was said earlier

[tribal moss]

"And hence prime" if you already know Z[i] is UFD.

[languid trellis]

we've already shown the norm makes Z[i] into a euclidean domain and hence ufd

[tribal moss]

Okay then.

[languid trellis]

okay to summarise. primes of the form 4n+3 in Z are prime in Z[i]. primes of the form 4n+1 aren't. also, if N(a+bi) irred in Z implies a+bi is prime in Z[i], so anything of the form 4n + i is prime in Z[i] if 4n+1 is prime in Z (as is 2n + (2n + 1) i and other combinations of the sort)

what is missing to complete this "classification"?

[dull ginkgo]

Also note that xy = a in Z iff x and y are conjugates up to scaling

[languid trellis]

yeah i used that a few days ago when showing primes of the form 4n+1 can be written as a^2 + b^2 in Z

[dull ginkgo]

Thus a + bi | x iff a^2 + b^2 | x I’m pretty sure

[tribal moss]

so anything of the form 4n + i is prime in Z[i] if 4n+1 is prime in Z
That doesn't sound right.

(3-2i)(i+1) = 5+i

[languid trellis]

neq 4+ i

[tribal moss]

Oops, I misread.

[languid trellis]

I feel like the condition may still be wrong but the sentiment may be there xp

[dull ginkgo]

So assume p > 0 is prime in Z[i]. Then there doesn’t exist a, b such that a^2 + b^2 | x. Assume p ≠ 3 mod 4, then there exists a, b such that a^2 + b^2 = p

thus (a + bi)(a - bi) = a^2 + b^2 = p, contradicting p’s primality

So p = 3 (mod 4)

[tribal moss]

(5-2i)(2+i) = 12+i

[languid trellis]

4+i works though for sure

because 17 is prime

[dull ginkgo]

p is a positive integer

Now assume a + bi (a nor b is nonzero) is prime in Z[i]

[languid trellis]

god damn it

[dull ginkgo]

(a + bi)(a - bi) = N(a + bi)

(a + bi) | N(a + bi)

[languid trellis]

im getting very close to giving up and going to play video games

[dull ginkgo]

now assume N(a + bi) is not prime in Z, equal to xy

Thus (a + bi) | xy so (a + bi) | x or (a + bi) | y

Oh that’s fucking smart

@languid trellis i have an idea how to show a + bi’s norm is prime if neither a nor b is zero

Take a + bi right, then (a + bi)(a - bi) = a^2 + b^2 = N(a + bi)

Thus (a + bi) | N(a + bi)

Because a + bi is prime, N(a + bi) must divide one of the prime factors of N(a + bi)

Which can only happen if N(a + bi) is prime

so a^2 + b^2 is prime

[languid trellis]

i dont see what you're trying to say here

[dull ginkgo]

If prime a + bi is not strictly real or imaginary, then it’s norm is a prime number

And you can prove the reverse to get an iff

And the other case is where it’s just a or ai

Which you can prove is an iff

Providing a classification

[languid trellis]

pause. "Because a+bi is prime, N(a+bi) must divide one of the prime factors of N(a+bi)"

explain that sentence

[dull ginkgo]

a + bi is prime

So for any x, y in Z[i]

If a + bi divides xy, then it must divide x or y

And a + bi | N(x + yi)

so a + bi | p for some prime factor p of N(x + yi)

thus N(x + yi) | N(p) = p^2

[tribal moss]

Why?

[dull ginkgo]

Mistyped i meant a + bi

a + bi | N(a + bi)

For any x in Z[i], x | N(x)

Since N(x) is the product of x and its conjugate

[languid trellis]

why can't the prime factor p just be a+bi

[tribal moss]

Yeah, it looks like there's some confusion between Z-prime factors and Z[i]-prime factors going on.

[dull ginkgo]

no,

a + bi is prime in Z[i], and divides N(a + bi) in Z

N(a + bi) has prime factorization in Z, subring of Z[i]

So a + bi must divide some prime factor of N(a + bi)

[tribal moss]

That doesn't make sense; a+bi is not even in Z.

[dull ginkgo]

a + bi divides N(a + bi) = (a + bi)(a - bi) in Z[i]

and N(a + bi) has prime factorization in the subring Z, so a + bi must divide one of those prime Z-factors because it’s prime in Z[i]

[tribal moss]

Okay, with you so far.

[dull ginkgo]

a + bi | p for Z-prime p

I need to show this implies N(a + bi) is prime

Take any x + yi, assume x + yi | n for n in Z

This can only happen if N(x + yi)^2 | n^2

As that’s the only way for the imaginary part to vanish

So N(a + bi)^2 | p^2

[tribal moss]

Stop a moment.

[dull ginkgo]

Ergo either N(a + bi) = 1 or p

[tribal moss]

What does x,y,n have to do with the a,b,p you were talking about?

[dull ginkgo]

It’s a lemma

I probably should’ve stated that

[tribal moss]

What does the lemma state?

[dull ginkgo]

In Z[i]

If z in Z[i] divides x in Z

then N(z)^2 divide x^2 in Z

[tribal moss]

What does z now have to do with a, b, p, n, x, y?

[dull ginkgo]

so using this lemma on z = a + bi, x = p

N(a + bi)^2 | p^2

[tribal moss]

Which lemma?

[dull ginkgo]

The one stated above, that z | x => N(z)^2 | x^2

[tribal moss]

It looks like you keep just piling on new letters.

[dull ginkgo]

Just a lemma i was using

I used it for both parts

[tribal moss]

State. The. Lemma.

[dull ginkgo]

Forall z in Z[i], x in Z
z | x implies N(z)^2 | x^2

[tribal moss]

You need to explain how all of the equations containing totally different letters you keep throwing around relate to each other.

[dull ginkgo]

Okay sorry,

[languid trellis]

3 | 6 but 81 doesn't divide 36

[dull ginkgo]

sorry z in Z[i]\Z

[languid trellis]

1 +i | 2 but 16 doesn't divide 4

[dull ginkgo]

Oh shit i meant N(z) | x^2

[languid trellis]

oh wait

[tribal moss]

The last time I had any idea what you were doing was here.

[languid trellis]

I'll let you guys have fun, im done with this problem for today

[dull ginkgo]

What I’m doing is using a + bi, a ≠ 0 ≠ b, primality in Z[i] to show it’s norm is prime in Z, since a + bi must divide some Z-prime factor of the norm, which then forces the norm to divide a Z-prime factor of itself

[tribal moss]

I don't follow that reasoning.

[dull ginkgo]

What part are you confused about

[tribal moss]

And you seem to have stopped distinguishing betwee Z-prime factors and Z[i]-prime factors again.

[dull ginkgo]

I said Z-prime factor

N(a + bi) factorizes over Z

[tribal moss]

Which kind of prime factor do you mean when you say "forces the norm to divide a prime factor of itself"?

[dull ginkgo]

Thus over Z[i]

[dull ginkgo]

A Z-prime factor

[tribal moss]

Then say that.

[dull ginkgo]

Sorry i corrected it

I did the work symbolically and I’m trying to translate it based off of that

[tribal moss]

Okay, I accepted your argument at a+bi divides a Z-prime factor of a²+b².
How does that imply that a²+b² "divides a Z-prime factor of itself"?

[dull ginkgo]

a + bi can divide a positive natural number p if and only if it’s conjugate does aswell

[tribal moss]

Okay.

[dull ginkgo]

So thus (a + bi)(a - bi) = N(a + bi) | p

[tribal moss]

By which I mean, I agree that is true, but I don't see how it relates to my question.

[dull ginkgo]

p is a Z-prime factor of N(a + bi)

[tribal moss]

Yes.

[dull ginkgo]

so p | N(a + bi) and N(a + bi) | p

[tribal moss]

Huh.

Where does that "so" come from?

[dull ginkgo]

p is a Z-prime factor of N(a + bi) thus by definition p divides N(a + bi) in Z

[tribal moss]

Yes.

[dull ginkgo]

but we showed N(a + bi) divides p too

[tribal moss]

I didn't follow that.

It must have been one of the things you posted while I was typing clarifying questions to something you posted earlier.

[dull ginkgo]

Here

[tribal moss]

That asserts that N(a+bi) | p, but what is the argument that this would be true?

[dull ginkgo]

Assume p is in Z, and (a + bi) divides p

Therefore there exists (c + di) such that
(a + bi)(c + di) = p + 0i

p = ac - db
0 = ad + bc

-ad = bc

because a + bi is prime in Z[i], no n in Z divides both a and b, so a and b are coprime

Thus d = b, c = -a, or d = -b, c = a

[tribal moss]

They could be multiples of that.

[dull ginkgo]

Or multiples yes

Sorry, thanks for clarifying

d = -mb, c = ma

so (c + di) = m(a - bi)

m(a + bi)(a - bi) = m(a^2 + b^2) = mN(a + bi) = p all in Z

[tribal moss]

Okay.

[dull ginkgo]

Ergo, N(a + bi) divides p, which is a prime factor of N(a + bi)

Thus N(a + bi) = p prime

[tribal moss]

Okay. I see the structure of your argument now.

[dull ginkgo]

My other attempt uses that same Z[i] element dividing Z element argument

[tribal moss]

I think I was confused by speaking about (a+bi)(a-bi) as the norm in this context, since its multiplicative property never was used. All you really needed is that a+bi divides some integer.

[dull ginkgo]

Yeah

Sorry

So for prime a + bi, we have two cases:
a ≠ 0 ≠ b : N(a + bi) is prime
else: a = 4n + 3 prime or b = 4n + 3 prime

[tribal moss]

Thanks for the patient explanation. :-)

[dull ginkgo]

No worries, thanks for helping me get better at explaining things

[tribal moss]

Have we excluded the possibility that a Z-prime of the form 4n+1 could still be prime in Z[i]?
(The sum of squares theorem does do that, though).

[dull ginkgo]

The converse of these two cases is way easier

[dull ginkgo]

No, can’t be

Because a^2 + b^2 = 4n + 1

(a + bi)(a - bi) = 4n - 1 = p

so it factorizes over Z[i]

[tribal moss]

Yes, I wasn't sure if we officially know that a prime of the form 4n+1 is the sum of two squares.

(Or rather whether Swifteee oficially knows it, but he has left the room by now ...)

[dull ginkgo]

I’d need to think of a way to prove that there is a pair (a,b) such that a^2 + b^2 = c for c not congruent to 3 mod 4

Might be an inductive approach

(a - b)^2 + (a + b)^2 = 2(a^2 + b^2)

Or the Pythagorean route

Gonna ponder for a bit

Btw we are both working from the same textbook

Swifteee and I

[languid trellis]

lurked back in here. that was the previous exercise : )

[dull ginkgo]

Was this a problem at the end of the section lol

idk where my textbook is rn

[wraith swan]

After contemplating I think understand the first part of my proof. Just some small things missing, I think. THanks!

[languid trellis]

its 2.15 ex 7

[dull ginkgo]

ooooh number 20 looks fun

Wait FUCK I can’t use strong field theory stuff yet

Shit.

[languid trellis]

all of these exercises look hard

[dull ginkgo]

Some of them just look like annoying induction arguments

Particularly the Euclidean algorithm one

Some parts of my life I will not get back

[languid trellis]

I've been doing about one a day lmfao

and I've needed a ton of help

[dull ginkgo]

Eg proving generators of SL(2,D) for Euclidean domain D

[wraith swan]

,rotate should work

[dull ginkgo]

F_q[X]/(p(x)) ~= F_(q^deg(p)) if p is irreducible

we’ve done something similar to this before

[tribal moss]

Finite field theory is probably a detour there -- just count the total number of monic polynomials, and use combinatorics to count how many reducible ones there are.

[dull ginkgo]

There’s q^(n - 1) monic polynomials of degree n hmm

Oh I see

wait

[dull ginkgo]

Let $N(d)$ be the number of irreducible monic polys in $F_q$ of degree $d$ then wouldn’t $N(x) = \sum_{d | n}{N(d) N\left(\frac{x}{d}\right)}$

[cloud walrus]

THE TUBE

[dull ginkgo]

wait no

[tribal moss]

No, for two reasons. First, you add degrees rather than multiply them.

[dull ginkgo]

Not dirichlet convolution it’s regular adding

Whoopsie on my end

Let me rewrite it rq

$N(x) = q^{x- 1}- \sum_{n = 1}^{x - 1}{N(n)N(x - n)}$

[tribal moss]

Also make sure that e.g. cubics that split into 3 linear factor are not overcounted or undercounted.

I think it's deliberate that the exercise stops at degree 3, because the combo gets wilder the higher the degree.

[dull ginkgo]

Ah yeah true

It would be over partitions of x

[tribal moss]

But as long as we're only looking at cubics, a reducible onea are either an irreducible quadratic times a linear factor, or a product of three linear factors (and the latter can be counted by stars-and-bars, without needing to split into whether some of the factors are identical).

[dull ginkgo]

So… multindicies?

[tribal moss]

That would be a way to phrase it.

[dull ginkgo]

$N(d) = q^{d - 1} - \sum_{|i| = d}{\prod_{n = 1}^{d}{N(i_n)}}}$

[cloud walrus]

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

[dull ginkgo]

The combo psycho in my head wants to try fucking with generating functions but this seems like a very very bad idea

That is incorrect but idc

[glad osprey]

Is $(x^{-1})^n = (x^n)^{-1}$ for x in a group G only true when G is abelian?

[cloud walrus]

sheddow

[south patrol]

No, since the proof doesn't refer to any elements of G other than x

[glad osprey]

ah, of course

x and x^(-1) commute even in a non-abelian group

[hidden wind]

you can think of this as being contained in the the subgroup of G generated by x, which is abelian

[glad osprey]

Yeah, good point 👍

[delicate orchid]

commutivity isn't even needed
$(x^{-1})^nx^n = x^{-1}...x^{-1}x^{-1}xx...x = x^{-1}...x^{-1}x...x = ... = 1$

[cloud walrus]

Wew Lads Tbh

[cobalt heath]

x^(-1) and x commutes

[tribal moss]

Prof. First I will show that x^-1 commutes. [...] It remains to show that x also commutes. [...]

[wraith swan]

Anyone have an idea on how to factorise the elements of A x_C B (abelian pushout) in terms of element in the image of \alpha and \beta?

I have a problem with constructing a unique morphism from the equivalence classes of A x_C B. If the class is just {(a, e)} or {(e, b)}, it is easy. But what if it is {(a + alpha(c), b + beta(d)), (a + alpha(c-d), b), etc..}, I want to deconstruct all the elements in the class into (a, e) + (e, b) + ...

[south patrol]

Hm well for a pushout, just use the universal property

(Or first isomorphism theorem)

[wraith swan]

thats what i want to prove

[south patrol]

Ah

Well then use the first iso theorem

Like you can view the pushout as A (+) B / (subgroup generated by f(c) - g(c))

Where f: C -> A and g : C -> B are the maps

You can let stuff with quotients do the heavy lifting rather than worrying about finding unique represenatives

(Uniqueness is the easy bit - existence is what you need this for)

[cobalt heath]

I have forever wondered why this kind of stuff takes effort to prove

[wraith swan]

Haven't yet read about quotient groups, iso theorem, etc. This is an exercise that mentions that it is easier later on

[south patrol]

So how are you defining pushouts...

By an equivalence relation?

[wraith swan]

Quotient set and then proving it is a group, etc.

[south patrol]

Then just use that instead of quotients yeah

So like

Say you have maps f:C -> A and g: C -> B

And you gave maps A -> D and B -> D compatible with f and g

[wraith swan]

right

[south patrol]

Well you can define a map A (+) B -> D, let's call it h

And your task is to check you can "factor that through"

As in, show that if (a, b) ~ (a',b') (where ~ is the equivalence relation defining the quotient set) then h(a,b) = h(a',b')

Does that make sense

[wraith swan]

oh I can do that as well of course

[south patrol]

Yes

Nice

Yeah this is how mapping out of quotients is essentially always done

[wraith swan]

I wanted to create an algorithm

[south patrol]

Idk what you mean

The point is you can just pick any representative and send it according to h

[wraith swan]

I wanted to construct an algorithm that deconstructed all elements in the quotient set into (a, e) + (e, b) + alpha(c) + beta(d). Then I could just map (a, e) and (e, b) with the standard morphism from the coproduct/direct sum and use the commutativity of the diagram to map alpha(c) + beta(d)

this is difficult to explain 😛

[south patrol]

Okay yeah

Well the point is you can always write as (a,b) anyway and map that according to ur things

And the representative doesn't matter at all by how it is set up

I guess maybe what you mean is like you want to find subsets of A , B, C such that you can uniquely write stuff as combinations?

[wraith swan]

Right but the problem i had is that the quotient set contains multiple elements and I wouldn't know if mapping one of them with the unique morphism from the coproduct would also work for the other elements in the set

[south patrol]

That's just what you check using the various assumptions on ur maps

[wraith swan]

But just like I checked if the operation on A x_C B is well-defined, and can just check if the unique morphism is well defined on the equivalence class

Altough how do I then show it is unique

Thank you Crstalline, I think I just need to think about it a little

[wraith swan]

Ah I get it now, the problem I had with this was how to show this is the unique morphism by just looking at one representative. Because obviously A x_C B is not isomorphic to A x B. But I guess the uniqueness follows from 1 representative and proving it is well-defined show that it actually exists

[winged void]

I have a question

what does it mean for a group to be transitive

im suppose to solve the following question

Let G act transitively on X, and let N be a normal subgroup of $G$. Prove that the orbits of in other words: for all x and y in X, $#Nx$ = $#Ny$

[cloud walrus]

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

[hidden wind]

G acts transitively on X, means for any pair x,y in X there is a g i G with y = gx

which one can show is equivalent to there being only one orbit

[flint cave]

Im working in SLn(Z) (n >= 3), If I have two unipotent elements u,v such that $u^r = v^s$ and rank(u - Id) = rank(v-Id) = 1 for some integers r,s, why does Im(u-Id) = Im(v-Id) and Ker(u-Id) = Ker(v-Id)

[cloud walrus]

Symmetry enjoyer

[flint cave]

Unipotent means difference with identity is nilpotent

In fact this is iff apparently

[hidden wind]

@winged void

[flint cave]

Not able to see either direction : (

[rocky cloak]

So let N = u-1. Then N is nilpotent of rank 1, so must have N^2 = 0.

Hence u^r = 1 + rN.

Similarly v = 1 + M and v^s = 1 + sM, so M = (r/s)N, so they have the same image and kernel.

[flint cave]

Omg

It was easy

I jhst needed to see N^2 = 0. It was staring me in my face, especially since i was thinking if rank(N) = 1 it has to act by scaling on that one line if N^2 != 0 but this can't happen cus it's nilpotent... but I kept brushing it aside

[hidden wind]

unipotent means a unit away from being nilpotent

[winter fog]

hello, If $A = {a + ib\sqrt{2} ; a, b \in \mathbb{Z}}$ and $N(z) = z\bar{z}$. $\exists x, y \in \mathbb{Q}$ such that $\frac{z_1}{z_2} = x + iy\sqrt{2}$ and u, v $\in \mathbb{Z}$ such that :

[cloud walrus]

abyssworld

[winter fog]

if $q = u + iv\sqrt{2}$ and $r = z_1 - qz_2$ then why $N(r) < N(z_2)$?

[cloud walrus]

abyssworld

[winter fog]

can someone help me pls

[rocky cloak]

Notice that by dividing by z2 you have
r/z2 = z1/z2 - q

So you just need to show that the norm of this is less than 1.

[winter fog]

ok

[winged void]

i see thanks alot

then its easy to show that

[hidden wind]

be careful to distinguish groups and their elements from group actions!

the archetypical example is the group $S_X$ of permutations on a set $X$

[cloud walrus]

rødbet

[hidden wind]

the group consists of all bijections of $X$ with $X$ under composition, and the obvious grouo action is to map each pair $(\alpha, x)$ in $S_X \times X$ to $\alpha(x)$ in $X$

[cloud walrus]

rødbet

[wraith swan]

My book defines an action of a group G as a homomorphism G -> Aut_C(A). That is in Grp, right? since Hom(A, A) is a group?

[glossy crag]

Hom(A,A) isn't a group (not all homs A->A are invertible), Aut(A) is a group.

[wraith swan]

Oh right, Aut are the automorphisms. Thank you