groups-rings-fields

[crystal vale]

But IJ define as finite sum of ij

[mighty kiln]

Try to prove it

[crystal vale]

If it is correct then by co maximal (x) intersection (2) = (2x)

[mighty kiln]

?

Yes (2) ∩ (x) = (2x) in Z[x]

[crystal vale]

Yeah because of co maximal

[stuck fiber]

How do you write a polynomial in n variables as a finite sum? I've tried a few things but I don't see how to get it to work

[mighty kiln]

∑ a_ijk x^i y^j z^k

[dull marsh]

$\sum\limits_{\mathbf k \in { 0, \dots, N }^n}a_{\mathbf k}\prod_{i=1}^n x_i^{k_i}$ sotrue

[cloud walrus]

A Lonely Bean

[dull marsh]

Where k_i is the i-th entry of k and N is the degree

[stuck fiber]

Okay I see so you just use entries of a n-tuple whose entries are non negative integers as the exponents then sum over all such n-tuples?

[dull marsh]

Yes

[stuck fiber]

That makes sense I was trying to do some things with like subtraction and n sums to define the exponents but that was a nightmare

[cloud walrus]

A Lonely Bean

A Lonely Bean

[dull marsh]

Any hints?

[dull marsh]

Nvm I think I see

[rocky cloak]

||Replace b by its inverse||

[dull marsh]

It worked, thanks

[grizzled crane]

How do I write a permutation as cycles which aren’t disjoint?

I’m just getting started the permutation cycle notations

[rocky cloak]

Not sure what you mean. You have some permutation and you wish to write it as a product of cycles in just any arbitrary way or something specific?

[grizzled crane]

yes.

Just wanted to represent in a cycle notation which isn’t necessarily disjoint

Consider the permutation given there

in S6

Under disjoint cycles section

[rocky cloak]

Alright, will there a pretty standard way to write something as a product of transpositions.

Take some element, say 1 that is mapped to x1, then you want to write the permutation as (1, x1) times something. Say x1 maps to x2, then you can write this as
(1, x1)(x1, x2) times something.
Turn you continue until you get something that maps to 1

So for example the cycle
(1, 2, 3) can be written as
(1,2)(2,3)

This method can break up any cycle into transpositions, so if you just write your element as a product of disjoint cycles, you can make them not disjoint by doing something like this

[rocky cloak]

If you're fine with the cycles being disjoint that's easier.
Just consider some element and apply the permutation until it cycles back. Then that will be one of the cycles in the product

[grizzled crane]

so essentially write it as product of disjoint cycles then break it down

right?

[grizzled crane]

If I consider (1,3,6)(2,4)

[rocky cloak]

If you're just looking for a method to write it as a product of cycles, then writing it as a product of disjoint cycles is the easiest

And there's a unique way to do that

[grizzled crane]

I’m aware about writing it as product of disjoint cycles

[rocky cloak]

Then I don't understand your question

[grizzled crane]

I’m just trying to see how do I write is product of cycles which aren’t disjoint

like some numbers should appear again

[grizzled crane]

as in here

[rocky cloak]

Like you can always just add (1,2)(1,2) to your product if the only goal is for some cycles to not be disjoint

There's an infinite amount of ways to write it as a product of cycles that aren't disjoint

[grizzled crane]

makes sense

[languid trellis]

Im struggling with an exercise in Artin. "Suppose K and H are subgroups of G, where H is normal. Suppose also that K \cup H is a subgroup of G. Prove that it is normal."

I've tried to argue by contradiction but I haven't found one. I've also tried to construct a group homomorphism with K \cup H as its kernel but I can't figure out the details. Can someone please give me a hint?

[delicate orchid]

It is normal in which group? Also I think you mean cap instead of cup. Anyway, it’s not normal in G, pick G = S_5, H = A_5 and K any subgroup generated by an even permutation.

It is however normal in H

[languid trellis]

yeah i did mean cap

the exercise asks to prove that K \cap H is a normal subgroup of H

[delicate orchid]

Ah ok. If you want to go the route of finding a homomorphism with that intersection as it’s kernel id use the 2nd iso theorem?

Or you can just show that it’s normal directly. Up to you

[languid trellis]

haven't learnt 2nd iso theorem yet

I'm going to try again

am I stupid

Suppose $K \cap H$ is a subgroup of $H$, and $K$ is normal w.r.t to $G$. Then $K$ is certainly normal w.r.t $H$, as $H$ is a subgroup of $G$. Want to show that $K \cap H$ is a normal subgroup of $H$. This means that $\forall a \in K \cap H$, and $h \in H, hah^{-1} \in K \cap H$. Certainly $hah^{-1} \in K$, and closure of $H$ guarantees $hah^{-1} \in H$ Thus $K \cap H$ is a normal subgroup of $H$

[cloud walrus]

swifteeee

[languid trellis]

H is closed because it is a subgroup of G

i think that works. the reason I was having so much trouble earlier was because i was trying to prove K \cap H is a normal subgroup of G

im so stupid

[delicate orchid]

I presumed this was the problem

[languid trellis]

lol, thanks for the help wew

[naive whale]

Given a polynomial $p(x)$ such that $\mathbb{C}[x]\slash\left\langle p(x)\right\rangle \cong\mathbb{C}[x]\slash\left\langle x-\zeta_{0}\right\rangle \times\mathbb{C}[x]\slash\left\langle x-\zeta_{1}\right\rangle \times...\times\mathbb{C}[x]\slash\left\langle x-\zeta_{n-1}\right\rangle $, that $\zeta_0,\zeta_1$ is it necessarily the case that $\zeta_0,\zeta_1$ etc must be roots of $p(x)$? Also that p has degree $n$?

I feel like it makes intuitive sense that it should be true but I'm having a hard time putting my finger on why

[cloud walrus]

Nomad

[dull anvil]

@glad juniper Sorry for pinging you for this little issue

I was in a server with you along with biglawk and emme, but I cannot find that server again. Can you please again give me the link of that server?

[dim widget]

It must be true that p has degree n by counting dimensions. If the isomorphism is meant to take x to x in each factor then the zeta are roots. Otherwise no

[grizzled crane]

Can I say a permutation is a one to one function from A to itself rather than saying it’s a bijection?

i feel like onto is already guaranteed when I’m considering a one one function from A to A

[mighty kiln]

[Injective iff bijective] iff finite

[grizzled crane]

Are you saying, if two sets are finite and if there is a bijection then there is a injection?

[coral spindle]

if there is a bijection then there is a injection
every bijection is an injection; this is trivial

[grizzled crane]

yes

[coral spindle]

the nontrivial part is that a map from a finite set to itself (n.b., to itself!) is a bijection if it is an injection (or indeed a surjection).

[grizzled crane]

My question is if a function f is an injection and the cardinality of the both sets are equal. Can I say f is a bijection?

[coral spindle]

Absolutely not

Exercise: find a set A and an injection A → A that is not a bijection.

[grizzled crane]

Like here I have to map 3 to 3

[coral spindle]

That's not a function as you have not specified the direction nor have you decided where 3 lands.

[grizzled crane]

Yes, I can’t leave 3 and if I have to map, then only way I can map is 3 to 3

say from left to right

or say I map 1 to 3 and 2 to 2

Now I can only map 3 to 1,

like everytime I add an element in the codomain it gets added to the domain as well

[grizzled crane]

I just can’t see a injection between two finite equal sets which is not a bijection

[coral spindle]

Indeed, like I said here, no such thing exists.

Remember that equal cardinality means there is a bijection between the sets.

[grizzled crane]

Yes

so essentially i can just say a permutation is a one one function from A to A

I’ve only seen a permutation is a bijection in my lecture notes

[coral spindle]

For finite sets alone, as has already been said

[grizzled crane]

do we have permutations for infinite sets?

[coral spindle]

They are bijections

[grizzled crane]

makes sense

[naive whale]

alright, I think I have a feeling for why now, thanks for the sanity check

[feral timber]

How do I prove this in R? I cannot use Eisenstein or rational root test as this is in R, not Q

[tribal moss]

Cubics are never irreducible over R.

[feral timber]

They have not specified over what field are we supposed to show irreducability, so I assumed R

If it is Q, then let me try rational root test

[tribal moss]

It can't be R: there will be at least one real root r, so you can factor out (x-r).

[feral timber]

It's monic so it should have an integer solution. Constant is 2, so the only possible solution is +-2

[tribal moss]

Rational root test would work over Q, since if a cubic is reducible, at least one of the factors must be linear.

[feral timber]

That gives either 10-2x=0, so x=5 fails irreducability, or 2x-6=0, so 3 fails irreducability

Now +-1 can also be a root, so 3-x=0, gives 3 again. Plugging in -1 will give -1 as an objectionable solution.

So I guess this works?

I can apply the same method here as well, right? Because this is in Z, so if it splits into a cubic and a quadratic with integer factors, then I can apply rrt to the cubic (which must be monic), and any rational root for that must be a rational root for the quintic.

But what if the cubic we get is irreducible? It must have a real root, but what if it is not rational? The quintic is still reducible

[tribal moss]

Once you have a cubic factor, then by definition the quintic is reducible, no matter whether the cubic reduces further.

The trouble with the quintic case is that it can be reducible over Q even though it doesn't have any rational root.

[feral timber]

Yes, so how do I prove irreducability? Constant is -1, so cannot use Eisenstein

[tribal moss]

Beats me, at least right now. A technique that often seems to work is to substitute x = y+n for some small integer n and show by Eisenstein that the resulting polynomial in y is irreducible -- but if there's a way other than trial and error to find an n that works, I haven't come across an explanation of it...

[feral timber]

Hmm, that seems to be a nice technique

[dim widget]

5 = 4 + 1 or 3 + 2 or 1+… so you just have to understand what quadratic or linear factors this can have

You can do that by brute force

[glossy crag]

@dim widget since you're around, can I ask you something?

[dim widget]

Ok

[glossy crag]

In a couple of places I've seen a more general definition of an absolute value/place, namely something that satisfies the usual properties of an AV, except instead of the trinagle inequality it satisfies this

I don't understand how the open balls of such values make up a basis for a topology.

In order for that to be true, one needs to show that the interesection of any two contains a third, which can be reduced to showing that a ball contains a ball centered at its any point. This however I'm having trouble showing.

[dim widget]

This is stronger than the triangle inequality, except for the constant C

But the constant C won’t really make a difference as to what the topology looks like

[glossy crag]

Up-to hand-waving, I know this to be true, since every such value is a power of a proper absolute value and one could just say "take the topology of the equivalent absolute value", but I want to show directly such sets build a basis for a topology (namely, given a ball U_r(x) and a poiny y in it, that there is an s with U_s(y)<=U_r(x)).

[dim widget]

Okay well the distance between y and any z in that ball is bounded by 2Cr right

[glossy crag]

Yes

[dim widget]

So you can look at the maximum distance from y to any x in this ball and take half of that

[glossy crag]

In the original U_r(x) ball?

[dim widget]

Yeah

[glossy crag]

Let me double-check that

@dim widget I'm just not getting it for w/e reason, could you please spell it out for me?

[dim widget]

Okay you have to do something slightly different, take a ball of radius r/C

[glossy crag]

Lol that's exactly what I was trying this second

[dim widget]

Or no

[glossy crag]

I think take the minimum of 1/Cr and |x-y|

[dim widget]

That sounds right

Yeah

Anyway you just reverse engineer it

[glossy crag]

Or no, the maximum I think

Lemme check

[untold basalt]

If R is a ring and x \in R then the ideal generated by x can be expressed as xR? (R is commutative)

[dim widget]

Yes

[untold basalt]

thx

[glossy crag]

I just can't get seem to get this to work

Wlog the center of the original ball is 0, then I need to show that if |x|<r, then points close enough to it have length <r as well.

[dim widget]

Yeah I guess if |y - x| > r/C you can’t do this

Not this way anyway

[glossy crag]

I write |y|<=Cmax(|x-y|,|x|). Now if x is close to 0, then you can take the radius s to be larger than |x|, but <r/C, then you get what you want.

But if x is close to the boundary then there's no way to get a radius large enough and you just get the useless bound |y|<=C|x|. What do?

[dim widget]

Are you sure they allow C > 1?

[glossy crag]

They don't say anything whatsoever about C.

Except that it follows from the condition automatically that C>=1.

[dim widget]

Curious…

[glossy crag]

Not really, just stick 1 and 0 in. Or do you mean the sparcity of information.

[dim widget]

The spar city of info

What exactly is the setup?

[glossy crag]

Just generalities about valued fields. It's from Cassels' "Local Fields".

[dim widget]

Hmm but are there other conditions on the absolute value? Like how Q behaves?

[delicate bloom]

I think the trick is raise to the nth power then root and take the limit c^{1/n} -> 1

Something like that

[glossy crag]

If you mean the archimedean property, it's just a general definition.

[dim widget]

We’re trying to see if it can be done from the definition

[delicate bloom]

Yeah same

[dim widget]

No I meant like multiplicativity

I think that is pretty key

[glossy crag]

Nope, just a regular AV definition, except the C inequality instead of the triangle.

[delicate bloom]

I'm in a plane about to take off

[dim widget]

Because without that I had come up with some counter examples

[glossy crag]

Puff puff pass.

[chilly ocean]

blow up points... on the plane... algebraic geometry...

[glossy crag]

Really, what.

[steel pulsar]

[Group theory] Let n be an integer > 1, let d be a divisor of n and let G be the unique subgroup of cardinal d of Z/nZ. Show that (Z/nZ)/G is isomorphic to Z/(n/d)Z
I have the correction, but I don't understand it. I wanted to know when you have a statement like that
What should you start thinking about to resolve the problem?

[glossy crag]

Artin uses the same definition, but he doesn't speak of a topology base.

Although it's not at all clear to me how he straight-up defines a topology with these balls as a local base. Can one just do that for any collection of sets?

[crystal vale]

What your doubt?

[steel pulsar]

here is the correction

[crystal vale]

I think I do this by just showing G is my kernel of homomorphism mapping

Use First group isomorphism Theorem

[steel pulsar]

I don't have the courage to translate all this into English

[crystal vale]

Oh

[steel pulsar]

OK I'll do it, give me 5 minutes

[cloud solar]

G group of order 12k+10, f:G->G, f(x)=x^4 endomorphism. Show G abelian

If I can show f is surjective

Then x^3 in center and 3 coprime with 12k+10 so x in center so G abelian

Any idea

[crystal vale]

What is endomorphism?

[cloud solar]

homomorphism G->G

[steel pulsar]

We know that the subgroups of ℤ/ nℤ are the images of the subgroups of Z which contain nZ by the canonical projection ℤ -> ℤ / nℤ. They are therefore of the form aℤ/nℤ, with a | n. These groups are cyclic, generated by [a], and by Noether's first isomorphism theorem, they are isomorphic to ℤ/dℤ where d is the order of [a] in ℤ/nℤ. In other words, d = min(m, n | ma) = n / a.

This formula shows that there is no bijection between the set of subgroups of ℤ / nℤ and the divisors of d of n, given by the reciprocal bijections G -> | G | and d -> (n/d)ℤ / nℤ. For d | n, we note n' = n | d. There is a unique subgroup of order d of ℤ/nℤ which is n'ℤ / nℤ.

We consider the morphism of phi groups: ℤ / nℤ -> ℤ / nℤ, [x} -> [dx]. The image of this morphism is φ(ℤ / nℤ) = d ℤ / nℤ and its core is made up of classes [x] such that n | dx, that is to say such that n' | x. We therefore have ker(φ) = n'ℤ / nℤ. By the first isomorphism theorem of NOether we have: (ℤ / nℤ) / (n'ℤ / nℤ) ≃ d ℤ / nℤ

Furthermore, the kernel of the surjective morphism ℤ -> dℤ / nℤ given by x -> [dx] is the sosu-group n'ℤ, so we also have: ℤ / n'ℤ ≃ dℤ/nℤ

[crystal vale]

Is it question that If n=ds then we have to show that Zn-> Zs group homomorphism with kernel <a> which has order d

[steel pulsar]

finally :
(ℤ/nℤ)/(n′ℤ/nℤ) ≃ ℤ/n′ℤ

[crystal vale]

?

[steel pulsar]

Here's the correction, good luck making me understand that, it's a mess in my head

The question is : Let n be an integer > 1, let d be a divisor of n and let G be the unique subgroup of cardinal d of Z/nZ. Show that (Z/nZ)/G is isomorphic to Z/(n/d)Z

[crystal vale]

I think it's same

[steel pulsar]

Show that (Z/nZ)/G ≃ Z/(n/d)Z and d | n

[crystal vale]

Because you can make G as <a> where a has order d I think

What do you think about mapping?

Can I use Coset concept here?

Is <a> is <s> where n=ds

[cloud solar]

.

[crystal vale]

x^3 or x^4?

[cloud solar]

What

[crystal vale]

x^3 in centre or x^4 in centre?

[cloud solar]

x^3

But this not works for this problem

Cuz if f is surjective by G finite we get f bijective so f automorphism so f a permutation of G so 4 is coprime with 12+10 false

I think this is calculation problem

[crystal vale]

But how x^3 in centre if it is surjective?

[cloud solar]

Pick a in G. Since f surjective there x s.t. x^4=a. f(xy)=f(x)f(y) then (yx)^3=x^3y^3 (by simplifications) and now x^4y^3=x(x^3y^3)=x(yx)^3=xyxyxyx=(xy)^3x=y^3x^3x=y^3x^4. Cuz x^4=a we have ay^3=y^3a. Because a random we have y^3 in center

[crystal turtle]

It cannot be surjective. There must be some non-identity element of order 2, which must be mapped to the identity. So it's not injective, and since they are finite cannot be surjective either

[cloud solar]

Yeah

[cloud solar]

This is a lemma. If f(x)=x^(n+1) surjective endomorphism then x^n in center

But 12k+10 is coprime with 3

So if we can show x^3 in center in another way the problem is done

I tried to build a subgroup of order 4

By contradiction

But I dont see it

[crystal vale]

Okay

[crystal vale]

How we sure there must be an element of order 2?

[crystal turtle]

Cauchy's Theorem

[crystal vale]

I think I do not know

[cloud solar]

Sylow theorem>>😤

It is a pretty nice topic

Bruh I think this problems is a about a bunch of calculations

[crystal vale]

That question for Romania contest?

[cloud solar]

I hate this problems like we have this morphism1 and this morphism2 show G abelian

[crystal vale]

From

[cloud solar]

Idk I found in a book

[crystal vale]

Which book

[cloud solar]

Something like "matematica de excelenta"

[crystal vale]

Yesterday may be you told about Romania Contest so which Contest

[crystal vale]

Is not in English?

[cloud solar]

No

Abstract algebra problems from 12th grade math national olympiad are usually hard

And I dont like the polynomial problems

I feel like I dont have enough flexibility

[crystal vale]

Have you soft copy?

[cloud solar]

No

[crystal turtle]

Hint: ||gcd(4, 12k+10) = 2. Try composing f enough times that you get x |---> x^2 is a homomorphism. It should follow that G is abelian.||

[crystal vale]

I can not find on net

[cloud solar]

Thx

Let K be a finite field with q elements, q>=3. Let M be the set of polynomials with degree q-2 in K[X] with all non zero coefficients all distinct. Find the number of polynomials in M with q-2 distinct roots in K.

Bro what are these problems

I have 4 months left to do this kind of problems.

[long obsidian]

Are the generators of a product of ideals just the products of the generators?

[crystal vale]

If they are commutative then may be

[long obsidian]

If they have overlapping generators maybe some of the products of generators are redundant?

[crystal vale]

I have same doubt on (m)(n)=(mn) where m and n belongs to Z

[crystal vale]

But in that case they are co maximal

[crystal vale]

Like?

[long obsidian]

(x,y,z)^2=(xy,yz,xz)

[rocky cloak]

You're missing x^2 and so on

[crystal vale]

No idea

[rocky cloak]

But yeah, you don't need both xy and yx

[long obsidian]

Woops okay I am missing x^2 and things like that. Sorry for the confusion thanks!

[rustic crown]

do you mean all coefficients are non-zero and all of them are distinct?

or some weird thing like "all non-zero coefficients are distinct"

first one is easy, second idk

[cloud solar]

Yes

This problem is evil

[rustic crown]

not really

[rustic crown]

so wlog, you can assume you want monic polynomials, to get the total just multiply the shit by (q-1)

[summer path]

shit

[rustic crown]

now that monic polynomial of degree q-2 has q-2 distinct roots in K, so it looks like (prod_{a in K}(x-a))/(two factors)

the numerator from standard theory is x^q-x

you want the quotient to have all non-zero coeffs, so we have to divide by x at the very lest

so want number of a in K* such that (x^(q-1) - 1)/(x - a) has all distinct coefficients

since K* is a group, replace that 1 by a^(q-1) and now you can divide easily

something like sum of x^ia^j over i+j = q-2

[tribal moss]

Oooh.

[rustic crown]

when are a^0, ..., a^(q-2) all distinct?

that's precisely if a in K* is generates the cyclic group. there are phi(q-1) such thingies

[rustic crown]

yea so should be (q-1) * phi(q-1)

weird problem

[cloud solar]

Yes

[tribal moss]

Nice solution, but I'm not entirely behind calling it "easy".

[rustic crown]

<

i was thinking in head, and each step there was only one thing to do

[rustic crown]

if it was this, then i would be

[rustic crown]

cause this wont' work and stuff will be

[cloud solar]

In the original solution they use some lemma with a rank of cyclic matrix

[tribal moss]

Especially the "write 1 as a^(q-1)" step is more inspired than I was able to come up with.

[rustic crown]

(oh there i was thinking like what happens when a = 1, and then i realized "hmm...")

what's cyclic matrix

[white oxide]

can somebody explain to me how we would it makes sense to use the universal property here? because they're saying that the given map G1 * G2 --> H can be factored, but i thought it was the other way around

oh nvm

it's just this [\begin{tikzcd}
&& H \
& {G_1/N_1} && {G_2/N_2} \
{G_1} && {G_1 * G_2} && {G_2}
\arrow[from=3-1, to=3-3]
\arrow[from=3-5, to=3-3]
\arrow[from=3-1, to=2-2]
\arrow[from=2-2, to=1-3]
\arrow[from=3-5, to=2-4]
\arrow[from=2-4, to=1-3]
\arrow[from=3-3, to=1-3]
\end{tikzcd}]

[cloud walrus]

okeyokay

[white oxide]

can somebody explain to me how they incorporated the maps i1 and i2 here?

like where do they even use i_1 and i_2 in the second paragraph

isn't that a priori reasoning, assuming that (G_1 * G_2)/N is the external free product

[elder wave]

Im not sure what your question is

[white oxide]

yeah i probably need to read it again tbh

[elder wave]

The paragraph shows this by considering arbitrary homs from the factor groups and showing that (G1 * G2)/N extends them

And because this characterizes your group up to isomorphism you’re done

Okay this is just restating things so I don’t think it’s helpful

[white oxide]

yeah, it's all good

i'll just revisit later

thanks for the help tho

[chilly ocean]

Can anyone help with part 3 of this problem?

[inner needle]

Why does the composition of Mobiüs transformations correspond exactly to matrix multiplication? Is there any deeper underlying structure that is preserved?
As a rephrasing why is the group of Mobiüs transformations simply isomorphic to the quotient of the group of 2x2 complex matrices by its center?

What's special about composing Mobiüs transformations is that it somehow corresponds exactly to matrix multiplication, or vice versa?

I'm hard pressed to believe it is simply a coincidence.

[coral spindle]

1. Because matrix multiplication already corresponds to function composition, of linear functions.
2. You can see this very easily from the first isomorphism theorem, via the obvious map sending a matrix to a Möbius transformation, and recalling that the centre of GL_2 is the scalar matrices.
3. See above.

The fact is that actually, we see Möbius transformations as maps on projective complex space. This is a particular quotient of C^2. Matrices act on C^2, and once we quotient appropriately, Möbius transformations act on P^1(C)

[inner needle]

I'm not asking for a proof, that is very easy to check.

[coral spindle]

Keyword: projective linear group.

[inner needle]

Yes, this is what I was looking for.

Hmm, so are Mobiüs transformations indeed linear is some sense?

Because matrix multiplication corresponds to multiplication of linear functions, but these aren't linear when considered as functions of the plane

But perhaps they have a similar structure when considering the projective complex plane?

They aren't affine, that is for sure.

Oh, it is just a linear transformation of the projective space!

That's why it takes lines/circles to line/circles. That makes a lot of sense.

So it's actually a very geometric object in the projective space, and not just an analytic one.

[delicate bloom]

To be a bit more literal,

in P^1 you have any point represented by [x, y] up to scalar multiple, so as long as y != 0 you might as well scale down to represent any point as [x, 1], and worry about the special case y=0 later. (We really have just the regular affine line here, with [1,0] being our point at infinity.)

so now you can literally multiply the matrix [a, b; c, d] by [x; 1] to get [ax+b; cx+d] and since we're in projective space, so long as the y component is not 0, this is the same as [(ax+b)/(cx+d); 1] in projective space. The rest of it sort of just comes along for the ride, like modding out by the center, and now you can also see how to handle the point at infinity without restricting yourself, etc etc

[long obsidian]

Hey hungerford has an exercise should that f:B->C an epimorphism of R modules of a ring which is an integral domains might not have it's induced map on torsion submodules F:T(B)->T(C) be an epimorphism.

The example I came with is the integers Z and with Z itself and Z_6 modules over Z then f:Z->Z_6 the canonical epimorphism onto the quotient is a surjection. But T(Z)=0 and T(Z_6)=Z_6 since all elements have finite order.

Then F:0:->Z_6 cannot be an epimorphism on the torsion submodules. Does this sound about right?

[chilly radish]

Looks good

Any epi from a torsionfree module to a torsion one would work

Or just to a module with nontrivial torsion

[feral timber]

Just confirming, an algebraically closed field need not be the splitting field of every polynomial in the field, but rather must contain it, right?

[coral spindle]

It only needs to contain it.

E.g. C is algebraically closed, yet is most certainly not equal to the splitting field of the Q-polynomial x^2 + 1.

[feral timber]

Like I can take C. It is algebraically closed, but consider x-2. The splitting field would be Q ig

Yeah

[glossy crag]

@dim widget any further ideas on what we discussed? I sure didn't get any. Also you didn't mention what the potential counterexamples were.

[feral timber]

How should I go on proving that given a finite field extension K over F, if every irreducible polynomial in F which has a root in K splits completely in K, then K must be a splitting field?

[south patrol]

Consider the minimal polynomials of generators

[feral timber]

Okay, so say K=F(a1,a2,...,an). If I look at min poly of ai say, it contains a root in K, so it contains all roots in K, so the min poly splits in K. Now I take the product of all min polys. That splits in K, and since K is the smallest extension containing a1,a2,...,an, it automatically is the smallest extension containing all the roots of the product, hence would be the splitting field of the product

This works?

[white oxide]

Yeah, or you could just say it’s the splitting field for all of the minimal polynomials instead of their product

But that works too

[dire igloo]

How do I prove (1) implies (2)

[next obsidian]

How much ring theory do you know

[dire igloo]

Just basics definition of ring ,unit,zero divisors

[next obsidian]

Darn

[dire igloo]

Do I need to study more?

[next obsidian]

Okay well, I guess you just try playing with equations then

Probably not

But my solution requires a lot more theory

[dire igloo]

Okay

[tribal moss]

My first idea was to square both sides of ap+bq=1, but that led nowhere. ||Cubing, though ...||

[delicate bloom]

are we reading a,b in R to mean ALL pairs a,b in R or there's just some fixed two elements, a,b in R alone 😿

[tribal moss]

By its wording, it seems to be a claim about the triple (R,a,b).

Oh, I think I see your point. Its either a claim about "a ring R with [identity and elements a,b]" or a claim about "[a ring R with identity] and elements a, b".

(If it means "all pairs a,b", then both claims are false: consider a=b=0).

[delicate bloom]

I guess that's a relief

[cobalt heath]

Uh, in that case (1) still implies (2) tho

Since (1) is false then
(Welp I misread)

[tribal moss]

Yes, but boringly so.

[delicate bloom]

I'll allow it lol

[tribal moss]

(They are both true in the zero ring, even more boringly so).

[cobalt heath]

0 = 1 😱

[delicate bloom]

the fact that it's not necessarily euclidean is interesting

[delicate bloom]

do you know what an ideal is

[dire igloo]

Yeah

[tribal moss]

You don't even need ideals.

[cobalt heath]

Yea imo ideals might overcomplicate this

[delicate bloom]

you've already solved it?

[cobalt heath]

This indicates so; sorry for stealing words

[delicate bloom]

ah

[dire igloo]

So that statement is for some fixed a,b?

[tribal moss]

Yes.

[cobalt heath]

Oh were you guys interpreting it as
(1) for all a, b, there exists p, q ...
Now I see.
Imo it is about showing that
For all a, b \in R, (1) <=> (2).
Then you may fix a, b.

[tribal moss]

In other words the claim to be proved is that for every triple (R,a,b) that satisfies (I), the same (R,a,b) also satisfies (II).

[dire igloo]

Okay I figured out 2 implies 1

[tribal moss]

Right, that direction is immediate.

If a²·p + b²·q = 1, then a·ap + b·bq = 1.

[dire igloo]

Yeah exactly

That question looks so confusing they have used same p,q in both statement

[delicate bloom]

each line exists within its own scope, so they're not really the same p,q

[dire igloo]

Yeah I realized that later

[tribal moss]

It would be easier to speak about the argument if they had distinct names, for sure.

[cobalt heath]

Can you guess the solution of inverse direction? @dire igloo

[dire igloo]

It's ap and bq

[cobalt heath]

I mean (1) => (2) one

[dire igloo]

(a^-1)p and (b^-1)q

Oh but wait it's ring

Inverse may not exist

[cobalt heath]

Yep

[brittle ether]

I have a question:

An ordered field is defined as:

A field F that contains a subset P such that it is closed in terms of addition and multiplication, and all elements x in F are either in P, not in P, or zero.

I understand abstractly that in abstract algebra, we have some weirdly arbitrary looking definitions that are motivated by its ability to describe a wide range of seemingly disparate concepts through only their abstracted properties, but I'm curious what the motivation here is

I understand that we are basically saying that there exists a subset P that is essentially the set of all positive values, and that no matter how you add or multiply positive values, you only get other positive values. What I don't understand is why the existence of a positive subset implies the field is ordered. Why is this enough to qualify that the field is ordered? Can we not have addition and multiplication operations create a mixed bag of positive and negative values and still get something ordered?

I know the complex numbers are not ordered, and I can sort of get why but it's not clear in my head. Maybe some essential order properties clearly violated by the complex numbers might help me understand this motivation?

[tribal moss]

We can define an order relation on the field by saying that x "<" y means y-x in P.

I think the definition is slightly garbled, though. It should be

... closed under addition and multiplication, and for every x in F we have either x in P or -x in P or x=0, and 0 is not in P.

Otherwise C could become an ordered field by declaring P = {1,2,3,4,....} subset of C, and that wouldn't give us any ordering.

[tribal moss]

With this correction, most of the requirements on P map to desirable properties of "<":

• "closed under addition" guarantees that "<" is transitive.
• "0 is not in P" implies "<" is irreflexive.
• "either x in P or -x in P or x=0" guarantees that "<" is total.
  These three properies mean that "<" is a strict total order.
  The very definition of "<" above ensures that the order is compatible with addition in the sense that a < b implies a+c < b+c.
  It's more involved to state precisely how multiplication interacts with the ordering of the reals, but it turns out that "P is closed under multiplication" will allow us to derive the usual middle-school rules for it.

[brittle ether]

Thanks so much, what a beautiful answer!

[untold basalt]

Is the boolean ring $\mathcal{P}(n):=\mathcal{P}({1, \dots, n})$ with symmetric difference and intersection as sum and product isomorphic to $\mathbb{Z}_2 \times \dots \times \mathbb{Z}_2$ the product with $2^n$ factors?

[cloud walrus]

Seagull

[rocky cloak]

Yes, the bijection is given by sending a set X to the 'indicator function', that is 1 on X and 0 elsewhere

[untold basalt]

I see, thanks

[tribal moss]

There are only n factors in the product, though. The resulting ring has 2^n elements.

[dire igloo]

Can we consider group as a vector space over some field ?

[chilly ocean]

consider the size a finite vector space must have

[dire igloo]

What do you mean

[barren sierra]

Well if it's a finite vector space what fields can we use?

[dire igloo]

Z_p where p is prime it depends on group

I think

[barren sierra]

Yea or F_q where q is a power of a prime

[barren sierra]

Ok and now if this vector space is dimension d

[dire igloo]

I was reading this on stack exchange

[barren sierra]

Hold on there are easier examples

So consider some finite vector space of dimension d over F_p

What's the size of this vector space?

[dire igloo]

What do you mean by size are you talking about Cardinality?

[barren sierra]

Ya

[dire igloo]

Infinite

[barren sierra]

How?

[dire igloo]

Wait

P^d

[barren sierra]

cool

So for something to even have a chance of being a vector space it needs to be of size p^d for some prime p

But not all groups are of that order

[dire igloo]

Oh okay

[barren sierra]

Hence the answer to this is no

[lapis latch]

or.. the vector space is infinite dimensional

[rocky cloak]

An abelian group can be made into a vector space over Fp iff every element has order p.

And because every field has either Fp or Q as a subfield every vector space can be seen as a vector space over one of these fields

So for example Z/6Z cannot be made into a vector space

[dire igloo]

Okay I got it.but suppose there is some group and it's Cardinality can be expressed as p^d for some prime p and int d then can we consider it as a vector space over z_p or some other field ? And if yes what are operation on these vector space

[rocky cloak]

Z/p^2Z has order p^2, but cannot be made into a vector space.

The important property is every element having order p. In which case it becomes a vector space by n*x = x + x + ... + x (n times)

[dire igloo]

Okay thanks !

[south patrol]

||yes, F_1||

[rocky cloak]

No need for a similar proof, you can just apply the theorem

[median rock]

so how i start it

[rocky cloak]

Can you think of some ways a group can act on itself?

Not sure what you're thinking of there

The centralizer (of an element) is a subgroup

[median rock]

damn wrong chat

i have no idea

[rocky cloak]

Alright, so the formula looks a lot like Burnside's right. So you just need C(g) to be the fixed points of some action

Maybe think about how you can express the fact that g and x commute in terms of an equation

[median rock]

aight i am gonna try

so can i say that i count a random x as |C(x)| and use that?

[rocky cloak]

Not sure what you mean by 'count as' here, but what I was getting at was that x is in C(g) if g and x commute.

[wraith oak]

Did I write down this permutation correctly?

[white oxide]

yeah looks good

[wraith oak]

cool thx

[delicate bloom]

did you end up getting this? While I was lying in bed last night I realized this also has the fun extra that I implies II also implies that there exist p,q in R such that a^n p + b^n q = 1

[tribal moss]

I would be interested in seeing the fancy ring-theory solution, by the way.

[delicate bloom]

same

[white oxide]

how exactly do we know that the highlighted portion is true? because what if sigma restricted to K sends some k in K to some element in F (since sigma has codomain KF)

oh nvm i'm dumb embeddings don't have to be automorphisms lol

[wraith oak]

what algebra course is this?

[white oxide]

i'm in a graduate abstract algebra course rn, so i would say equivalent to about a second or third course in algebra

rn we're going over Galois theory

my beloved 😍

[wraith oak]

oh damn you have my respect

[white oxide]

well i'm not doing well in it so there's that 😭

lowk i should've taken the second course in algebra, the undergrad version

before jumping to grad

but for some reason i thought i was ready 🤦‍♂️

[broken quartz]

I h8 galois

even tho he was handsome

[white oxide]

bro what

galois theory is my favorite

[white oxide]

mf got too cocky and died cuz of it 😭

why is this part necessary? can't we just say that [Q(b): Q] = 3, and just use the line where Q(b)Q(a) = Q(a, \sqrt{-3}) to show that [Q(b)Q(a): Q(a)] = 2, so they can't divide each other?

[median rock]

yo if two conjugacy classes are equal then they have the same amount of elements?

[tribal moss]

If they're equal, then they are the same set.

[median rock]

if x,y belong to the same conjugacy class then can i say that |C(x)|=|C(y)| (centralizer)?

[delicate orchid]

see if you can prove it

[tribal moss]

Ideally your goal here should be to understand why "yes, because conjugation is an automorphism" answers that question.

[gusty moon]

what book is this from?

[white oxide]

Lang

[gusty moon]

ah

thanks

[broken quartz]

Yeah I also like it I just don’t like galois as a person

[broken quartz]

Fr lmao

[white oxide]

Oh what did you read a biography on him or smt

[broken quartz]

Yeah

[white oxide]

damn what was the name of it

i might have to check it out

[median rock]

(G is finite with order (2t)*k ,k odd) Consider an element g of order (2t)) why the following is true can someone explain? The image of g in the regular representation must consist of 2t-cycles, so k of them.

[opal osprey]

Let $V$ be a finite dimensional real vector space and consider $L(V)$ its lattice of linear subspaces. Notice that if $g$ is any inner product on $V$, then it defines an operation of orthocomplementation:
$$
\perp_{g} : L(V) \rightarrow L(V)
$$
satisfying the following properties $\forall W_{1},W_{2} \in L(V)$:
\begin{itemize}
\item ${0}^{\perp{g}} = V$
\item If $W_{1} \subseteq W_{2}$, then $W_{2}^{\perp_{g}} \subseteq W_{1}^{\perp_{g}}$
\item $(W_{1}^{\perp_{g}})^{\perp_{g}} = W_{1}$
\item $W_{1} \oplus W_{1}^{\perp_{g}} = V$
\end{itemize}
Now, consider a function $P : L(V) \rightarrow L(V)$ satisfying the same properties above. Is it necessarily true that there exists an inner product $g$ on $V$ where $\perp_{g} = P$? And if $g$ and $h$ are two inner products on $V$ inducing the same orthocomplementation operation, how are they related to each other?

[cloud walrus]

MisterSystem

[opal osprey]

Intuitively I'd say yes, there is always some inner product such that P is the induced orthocomplementation operation

and that two inner products induce the same orthocomplementation operation iff they are conformal (g = c^2*h for some constant c)

[opal osprey]

wonder how this fifth property comes into play

[rustic crown]

but these digonal scalings don't induce the same ortho-complements right

like if for a fixed vector (v_i) if sum(v_i w_i) = 0 iff sum(v_i x_i w_i) = 0, then i'm sure one can say nice things about x, (assuming v_i's weren't just zeros)

isn't asking every complement to be stable under diag(x1, ..., xn) too much?

like that would be saying every vector si an eigen-vector

[opal osprey]

two inner products are conformally equivalent iff they induce the same notion of angle between two vectors.

[next obsidian]

Chmonkey

[opal osprey]

this is an exercise from spivak's calculus on manifolds btw

[crystal turtle]

I might have been thinking a little to strictly in what I was picturing as orthogonal

lol

I will shut up now

[next obsidian]

No don’t shut up

[crystal turtle]

No I said something stupid. I will go an mald over it.

[next obsidian]

Loser

That’s such a weak mindset

You should just say

[chilly ocean]

chill bro. ur appa is over, u can relax

[next obsidian]

“Oops haha so tired”

That’s what I do

Anytime I say something dumb

Or “I should go to sleep lol”

Even if it’s 2pm

[rustic crown]

it's almost 4:30am for det, have an 8am lecture.

[next obsidian]

Oog

[opal osprey]

https://math.stackexchange.com/questions/177005/question-about-angle-preserving-operators?noredirect=1&lq=1

[rustic crown]

btw, i don't fully remember my linear algebra... but can't you also do complements with skew-symmetric forms? too tired to think about this ><

[opal osprey]

You can, but it doesn't satisfy axiom four nor the third one

the failure of axiom 3 has to do with the distinction between lagrangian, isotropic and coisotropic subspaces.

[rustic crown]

you sure? feels like they prolly work as long as we have non-deg

okie idk that stuff

[opal osprey]

The failure of axiom 4 has to do with the existence of non-symplectic subspaces.

[rustic crown]

(i thought we were doing some scary physics stuff, now i'm not so scared)

[opal osprey]

The terminology comes from physics (sadly lol)

u can also have the notion of complementation wrt a not necessarily positive definite symmetric bilinear form

But it also satisfies weaker properties

That's why I was thinking

That if you have those stronger properties

You may be able to recover an inner product up to its conformal class

[rustic crown]

what if you like start with any non-zero e1, then pick non-zero e2 in P(e1), then non-zero e3 in P(e1, e2) and so on. Now define an inner product where we declare these to be orthonormal. feels like we might recover P from this

also feels like i'm falling into a simlar trap like ryx

might need to carefully choose the length of each e_i, and not just declare all of them to be 1

[rocky cloak]

Seems to me that property 2 and 3 imply property 1 and 5.

Property 2 and 3 together imply that the complement is an order preserving bijection.

W1+W2 contains both W1 and W2, hence property 2 gives that the complement is contained in the intersection. Applying the complement again and using property 3 we see that they are in fact equal.

Similarly 0 is contained in everything, so the complement must contain everything.

[rustic crown]

me slept 2 hours

[rocky cloak]

I think the answer to the inner product question is ||no.||

Consider ||V = R^2 and let e1, e2 be the standard basis. Define e1 and e2 to be orthogonal and for x nonzero let the orthogonal complement to (x, 1) be (x, -1). This should satisfy all the requirements. Now assume we have an inner product where <e1,e1> = a and <e2,e2> = b. Then the inner product of (x, 1) with (x, -1) is x^2a - b. This can't be 0 for all nonzero x.||

[rustic crown]

|| this orthocomp is obtained from the bilinear form with 1s in the off-diagonals right. Its non deg so does have all the properties you said. ||

[rocky cloak]

||not quite I don't think, since with respect to that bilinear form, e1 and e2 are not orthogonal right?||

[opal osprey]

oooooh

[rocky cloak]

More generally I guess any orthogonal decomposition coming from a bilinear form should be continuous, but this one isn't

[opal osprey]

In the case of R², where we think of L(V) as P¹(R) union two points right?

[rocky cloak]

Yeah

[opal osprey]

Or more generally, L(V) as being the disjoint union of the k-th grassmanians up to the dim of V.

hm

makes sense

[feral timber]

How do I show that x^4-x-1 is irreducible over Q?

One method would be using x->y+n transformation and checking using Eisenstein then, but if it is reducible, then none of the transformation will yield an irreducible

So unless I find a transformation that works, I cannot conclude at all whether it is irreducible or not

[rocky cloak]

You can show that it's irreducible mod 2, to conclude that it's irreducible over Z

[feral timber]

And why does that work?

Also, that only shows irreducability over Z, right?

[rocky cloak]

Then monic polynomial irreducible over Z implies irreducible over Q

[rocky cloak]

Imagine f = gh, then in particular f = gh (mod 2)

[feral timber]

Both g and h will be quads, as I can rule out linear factors using rational root test

f(0)=f(1)=1 mod 2, hence same must be true for g and h. Then?

[rocky cloak]

Yeah, so f doesn't have any roots mod 2, so if it factors it must be as a product of two irreducible quadratic polynomials

So then you just check all the irreducible quadratics modulo 2 (there is only one)

[feral timber]

Ah yes, x^2+x+1?

So this sort of test I can use for any prime p, right?

[delicate orchid]

Yeah, if it’s never 0 mod p then it can’t possibly ever be 0 in Z - hopefully it’s clear to see why

[feral timber]

Ah, thanks.

[cloud solar]

If A finite ring show there is m>p>=1 s.t. x^m = x^p for every x in A

I said A={x1,x2,..,xn}

And the set {(x1^m,x2^m,...,xn^m) with m non zero natural} is included in A×A×...A

But A×A×..×A finite so there is m>p s.t. x^m=x^p for every x in A

It is right?

[dire siren]

yes

[cloud solar]

Thx

[elder lance]

hey guys, can anyone help me with this?

The game of Chomp starts with a half-infinite checkboard with squares labeled (n,m)(n,m) for integers n,m≥0n,m≥0, with a cookie on each square. Two players take turns choosing a cookie (i.e., not an empty square) and eating all the cookies on that square and the ones that are not to the left or below it; that is, if a player chooses the cookie on (n,m)(n,m), then they eat all the cookies on (n′,m′)(n′,m′) for n′≥nn′≥n and m′≥mm′≥m. The cookie on (0,0)(0,0) is poisoned, and the player who eats it immediately loses. Prove that any game of Chomp ends after finitely many moves.

I'm thinking about using the fact that Z^2 is well ordered with <

[untold basalt]

Suppose $R$ is a principal ideal domain. Show that if $ \langle a \rangle$ is a prime ideal of $R$ then $a$ is irreducible.
Proof: If $a$ is not irreducible then there exist two non-invertible elements $x,y \in R$ such that $a=xy$. Since $a \in \langle a \rangle$ we deduce, without loss of generality, that $x \in \langle a \rangle$, so $x=ab$ for some $b \in R$, so $a=aby$, and because $R$ is a domain we conlclude $by=1$, a contradiction.

[cloud walrus]

Seagull

[untold basalt]

Is this true in principal ideal rings too? I was initially trying to prove it in the general case but couldn't, and I can't think of a counter example rn.

[tribal moss]

You should cite your source: https://math.stackexchange.com/a/3542753 (and also fix up the copy-pasting so you don't post every formula twice).

[elder lance]

sorry, i pasted it on a rush

[tribal moss]

Since the context explicitly mentions Hilbert's basis theorem, it would be natural to consider the board to be a visualization of R[x,y] (for your favorite Noetherian ring R), and then each bite corresponds to adding a monomial as a generator to an ideal you're building.

[teal vessel]

so I'm proving that the special linear group of size 2 over F_3 is both of the same order and non-isomorphic to S_4. I've successfully shown that, because the special linear group is a subgroup of the general linear group, I can safely say that 8=<|SL2(F3)|=<48, and because there exists a subgroup of the SL group isomorphic to the quaternion unit group (order 8), the order of this SL group must be a multiple of 8. It would be nice if I could find a subgroup of order 3 to force it to be a multiple of 24 (which cannot be 48 because there exist members of the GL with determinant -1), but I'm sure there's a better way.

[delicate orchid]

if you're in F_3 there's a pretty obvious subgroup of order 3

but that wouldn't rule out everything - there could be one of order 9 for all we know

[teal vessel]

it's just early in the morning and I haven't done this in a minute.

[teal vessel]

not possible, it's a subgroup of a group of order 48

[tribal moss]

Can you count the size of GL(2,F3)? If so, SL(2,F3) contains exactly half of its elements - it's the kernel of a homomorphism (det) whose range has 2 elements.

[delicate orchid]

oh ok sure

ngl you identified the important part - that the sylow 2-subgroup is Q_8

[teal vessel]

https://tenor.com/view/jfk-clone-high-i-like-your-funny-words-magic-man-jack-black-gif-18659433

[delicate orchid]

[teal vessel]

haven't gotten around to these notions yet

oh, wait, hold on. the determinant for a size 2 matrix is antisymmetric, so it would have to cut the space in half between 1 and -1 in this field.

[delicate orchid]

that's basically what tropo said but it has nothing to do with det being antisymmetric

[tribal moss]

The antisymmetry agument is that you can pair each det +1 matrix up to a det -1 matrix by swapping the two columns of each matrix.

[teal vessel]

yeh

[velvet steeple]

Btw, would what I asked in #help-12 fit here?
It's about groups, but it's first-year uni

[delicate orchid]

the possible determinants of matrices in GL(2, 3) (or hell, GL(n, 3)) are -1, 1. SL are exactly those matrices with det = 1, so GL/SL has order 2

[delicate orchid]

yeah this channel is for early group/ring theory stuff

I'll go help you in there though

[velvet steeple]

Oh, ok, weird that it's not in the early uni category then

[velvet steeple]

Thanks!!

[delicate orchid]

it's (unfortunately) rather rare for people to be introduced to groups in their first year of uni

[velvet steeple]

We do this in linear algebra

[delicate orchid]

based but strange

[tribal moss]

There was a long discussion about the categorization when this channel was created -- the view that won out is that topics that are taught "proof based" belong in "Advanced mathematics" no mater that they might be taught early at some places.

[velvet steeple]

Let $(R, +, \cdot)$ be a ring. \ Prove that $-(a \cdot b) = (-a) \cdot b = a \cdot (-b) \quad \forall a, b \in R$.

[elder wave]

#prealg-and-algebra

[velvet steeple]

So from $(R, +, \cdot)$ being a ring, we have that $\cdot$ is associative and the distributive laws hold.

[delicate orchid]

do you have that -a = -1*a

[tribal moss]

You don't even need -1; it's also true in rngs.
You can compute using distributive law that (-a)b + ab = 0, and since inverses are unique in the additive group this means that (-a)b must be -(ab).

[velvet steeple]

It's not really given in our lecture notes, but I guess yeah

[delicate orchid]

yeah but the proof is SO much faster

[delicate orchid]

well then you don't have it

[tribal moss]

||(-a)b + ab = (-a + a)b = 0b = 0|| is pretty fast, I think.

[velvet steeple]

Oh

[rocky cloak]

Not really, it's not any faster to prove that -a = (-1)a than that -(ab) = (-a)b

[delicate orchid]

grrrrrrrrrrrrrrrrrrrrrrrrr

anyway yeah tropo don't spoil the entire thing lol ty

[rocky cloak]

In fact I think it would make more sense to use the latter to prove the former

[tribal moss]

Okay, spoliered.

[velvet steeple]

(Woah, so many Graduate+ people in chat, lol)

[velvet steeple]

Hm

[tribal moss]

*(Actually 0x = 0 might need a separate proof if you haven't established that already).

[velvet steeple]

My lectures notes did establish that

[velvet steeple]

So something like this: \ We first study the inverse of $a \cdot b$. \ Since $a(-b + b) = a \cdot 0 = 0 \iff a \cdot (-b) + ab = 0$. \ Similarly, $-a(b - b) = 0$, thus \dots \ Thus, $(-a) \cdot b = a \cdot (-b)$.

Wait, forgot the second

[delicate orchid]

strange phraseology and the correct conclusion is (-a)b = -(ab)

[velvet steeple]

Well I conclude (-a) * b = a * (-b)

I need to get the third in there, somehow. I guess just distributive law too

ab + -(ab) = ab(1 - 1) = 0

So -(ab) is an inverse too

[delicate orchid]

I don't know what you mean by "the second" and "the third" but if you have (-a)b = -(ab) = a(-b) then you automatically have (-a)b = a(-b)

[tribal moss]

-(ab) is an additive inverse of ab because that's what the - sign means.

[velvet steeple]

Ah, so we don't really need to show that one

[velvet steeple]

We have that $(-a)b$ is an inverse of $ab$ because $(-a)b + ab = b(-a + a) = b(0) = 0$. \ $a(-b)$ is an inverse of $ab$ because $a(-b) + ab = a(-b + b) = a(0) = 0$.

And -(ab) follows because of what Tropo said

Does that justification look fine?

[delicate orchid]

yeah this is much better written

maybe mention that inverses in a group are unique at the end hence why we can conclude everything is equal

[velvet steeple]

Alright, thank you Wew, Tropo and jagr

[velvet steeple]

Btw, would we also need to show that 1 in this case is the neutral element?

Or is that not needed

[delicate orchid]

no because that's not true? a+1 isn't a

and of course 0 is the neutral element under addition that's the definition of 0

[velvet steeple]

For multiplication, for addition, it's 0

Yeah

[delicate orchid]

we're not dealing with multiplicative inverses here so I don't understand the relevance

[tribal moss]

No, because that (multiplicative identity) is what the the "1" sign means.

[velvet steeple]

Oh, so it's basically just by definition

Thanks

[velvet steeple]

Well I thought maybe to make this rigorous we need to state that 0 is the neutral element, for our argument with the inverse to work

[tribal moss]

There's no "1" in the proof you posted, though ....

[velvet steeple]

We need to get the neutral element when adding up x and its inverse

And we are getting 0

But it's just by definition, then, that 0 is the neutral element

[velvet steeple]

Yep, 1 doesn't have any relevance, true. I just meant the 0

[tribal moss]

That's what "0" means in a ring, yes.

[velvet steeple]

For $(A, \star_1, \star_2)$ to be a ring, does $(A, \star_1)$ need to be a group?

[south patrol]

Well this is unclear notation imo lol

If, more conventionally, you mean (A, + , x) then yes (A,+) must be an (abelian!) group

indeed you can view a ring as an abelian group equipped with a (in some sense) compatible multiplication

[untold basalt]

Suppose $R$ is a principal ideal domain. Show that if $ \langle a \rangle$ is a prime ideal of $R$ then $a$ is irreducible.
Proof: If $a$ is not irreducible then there exist two non-invertible elements $x,y \in R$ such that $a=xy$. Since $a \in \langle a \rangle$ we deduce, without loss of generality, that $x \in \langle a \rangle$, so $x=ab$ for some $b \in R$, so $a=aby$, and because $R$ is a domain we conlclude $by=1$, a contradiction.

[cloud walrus]

Seagull

[untold basalt]

^

[delicate orchid]

you can't deduce that x is in <a>

infact <x> properly contains <a>, which is where the contradiction comes from

[crystal turtle]

[delicate orchid]

I think this works: assume <a> is prime, if a is not irreducible then a = xy for x,y as you said. Then since y is non-invertible, <x> properly contains <a>. Identically because x is non-invertible, <y> properly contains <a>. Hence x,y are not in <a> but xy is, so <a> isn't prime which is a contradiction (a is prime iff <a> is prime)

get someone to double check this though I'm not confident

[delicate orchid]

actually no, sorry - I see it now. Your proof is fine

[velvet steeple]

Yeah, but it doesn't have to be + and *, it could be some other random operations, right?

[south patrol]

I mean those are the typical notations

It's just cause the order matters

[velvet steeple]

So even if it wasn't conventional addition, you would denote it by +?

[south patrol]

Basically always yes

[velvet steeple]

Oh

[south patrol]

I mean, in a sense what is convenntional addition? Just addition in reals? So yeah

[velvet steeple]

Well we could define a + b as 2a + b, where the first is some non-conventional addition and the second is the conventional one

And we could create a group with that, no?

(A, (unconventional) +)

[crystal turtle]

It's still standard to just use + and *

[velvet steeple]

Alright, thank you

[teal vessel]

I have found the subgroup of order 3 for SL(2,F3). Now I need to figure out how to show that it's not isomorphic to S4

[median rock]

yo any idea?

i have something on my mind but i cant continue

[delicate orchid]

hint: what is the order 8 subgroup in S_4

[teal vessel]

I'll track it down, dw

[delicate orchid]

lets call this element of order 2^m x. Then the permutation representation of x given via Cayley's theorem consists of a product of 2^m-cycles and acts on 2^mk elements, so there are k 2^m-cycles in x. So x is an odd permutation.

so the permutation representation of G is not contained in the alternating subgroup of whatever symmetric group we're mapping into, see if you can go from there

[median rock]

can you explain me this :'Then the permutation representation of x given via Cayley's theorem consists of a product of 2^m-cycles and acts on 2^mk elements, so there are k 2^m-cycles in x'

[delicate orchid]

x is order 2^m

[median rock]

ok

[delicate orchid]

and we're embedding it inside S_(2^mk)

[median rock]

via regular representation?

[delicate orchid]

yeah via the regular representation

[median rock]

can you explaion to me what the regular representaion does i think that i have lost this part

[delicate orchid]

wait if you know about regular representations do you know about normal p-complements, because there's a theorem of Burnside's that proves this super quick

ah nevermind

[median rock]

not yet

[delicate orchid]

multiplication by an element of a group is a permutation of elements of that group, so we can map that element to that permutation inside of S_|G|

this map is an injective group homomorphism and is called the regular representation

[median rock]

ok

if i prove it can i dm you just to check?

[delicate orchid]

just ping me in here I don't mind

[median rock]

thank you m8

if i knew sylow would it be easier?

[delicate orchid]

it would be

Cyclic sylow subgroups are nice

[median rock]

next week i think i am going to see sylow

[delicate orchid]

well you can't use them yet then!

[median rock]

if i take G goes to S|G| and S|G| goes to sgn function and take the ker of this composition would it be useful?

[delicate orchid]

that shows that the elements of G which map to even permutations is a normal subgroup of G - which is indeed useful!

in fact, it's an index 2 subgroup

[median rock]

so if i show that the ker is the set of elements with odd order then i am done

[delicate orchid]

yup

[median rock]

so here comes the hard part

[delicate orchid]

a good start is to show that the elements of odd order are in ker

yup exactly

[median rock]

if i show that then its over i mean i know that ker is subgroup so i dont care what other elements are in right?

[delicate orchid]

honestly this problem is really, really hard (without any machinery)

[delicate orchid]

oh wait this is a good point

but no, because you just know that the odd elements are a subset rather than a subgroup

[median rock]

yeah but if i show that they belong to the ker?

or its not right

i mean that we know ker is a subgroup which contains some elements if i show that some them are those with odd order then i finished

[delicate orchid]

no, you don't

[median rock]

why?

[delicate orchid]

_ _

not every subset of a group is a subgroup

[median rock]

so i have to show that the whole ker has elements with odd order

damn

too good too be true

[delicate orchid]

this also isn't true, since the permutation corrisponding to x is odd the permutation corrisponding to x^2 is even, and so x^2 is both in the kernel and has even order

just see if you can show that all of the odd elements are in the kernel first

[median rock]

kk

[median rock]

nah m8 i cant i m so bad

this fu**** regular representation is destroying me

[delicate orchid]

take an element g of odd order (say, o(g) = j) in G, then the permutation corresponding to g will consist of 2^mk/j j-cycles. Since j is odd, j-cycles are even and so the entire thing is even

thus g is in the kernel

because the kernel of sgn intersected with the image of G in S_|G| is index 2, and the regular rep is injective, the kernel in G must also be an index 2 subgroup. So there is a subgroup H_1 of G of order 2^**(m-1)**k that contains all of the odd elements. See if you can now use an inductive argument to find a subgroup of G of order k that contains all of the odd elements

[median rock]

if i repeat the procedure and i can gain a 'chain' of subgroups each of index 2 of the previous one until i get one with index 2**m (lets say it R) thus order k which contains all the elements of odd order in G(Since is odd, by Lagrange R consists of exactly the elements of odd order.)

am i right?

@delicate orchid

[delicate orchid]

what consists of the elements of odd order

I think you forgot some words there

but yes that's correct

[median rock]

is it correct now?

[delicate orchid]

yeah - but can you see why this subgroup has to be normal?

you just said the index is 2**m

[median rock]

w8

bcs k is odd?

[delicate orchid]

take some g in G and some r in your R, what can you say about the order of grg^-1

[median rock]

i have a chain of subgroups with ine

index 2

[delicate orchid]

ok so each one is normal in the next one but that doesn't mean they're normal in the whole group

[median rock]

so i crate a chain of normal subgroups

yeah yeah

[delicate orchid]

think about this

[median rock]

give me a minute

[median rock]

it has the same order as r?

[delicate orchid]

yes! So in particular grg^-1 also has odd order (because r is in R), so R is normal

[median rock]

damn so we could recreate the exercise and say that the elements of odd order create a normal subgroup?

[delicate orchid]

yup, now we're actually honestly truly done

[median rock]

damn man thank you so much

are you graduate or what?

[delicate orchid]

I'm a PhD student

[median rock]

damn

[delicate orchid]

this was a very tough problem

[median rock]

in which country

[delicate orchid]

UK

[median rock]

rly?

[chilly ocean]

doxxed

[delicate orchid]

there's like 4 different annoying steps

[median rock]

i am an undergraduate

3rd year

so its alittle tough for me

[delicate orchid]

feel free to ask about any other problems you're stuck on

[median rock]

thank you

[delicate orchid]

east midlands

nowhere particularly exciting/prestigious

leeds is like

6 train stops if I've counted correctly

[mint granite]

Hello, can someone explain the different steps to answer these questions? especially (iii) (v) and (vi)
for (i) and (ii), in my opinion, you just need to show "the implications".
I suppose that for (iv), you just need to use the definition of equivalence relation, but for the rest, I'm struggling.
Since my course is on group theory, I really can't see how to solve (iii), (v) and (vi). If anyone can give me "the beginning of a solution", that would be great!

[delicate orchid]

for iii the quickest way is to think about elements of H as 2x2 matrices acting on the plane that preserve the unit circle, I believe that you yourself can think of the action isometries of R^2 might have on lines in R^2 for v, and for vi just follow the hint

[mint granite]

okay, thank you! 😊

[velvet steeple]

It seems though we aren't allowed to use distributivity

Only associativity is given

[velvet steeple]

(according to the solutions of my prof)

[uneven jackal]

I'm quite confused by a Lie algebra fact ; in our course there's a proof that if we have a complex representation of a solvable group, it is upper triangular in some basis. it then follows by saying that "nilpotent lie algebras are like strict upper triangular matrices"

but I can find abelian lie subalgebras of gl_n that are diagonal ; is the subtlety that nilpotent algebras in fact don't have a strict upper triangular matrices, or is it mixing up between the matrices in the reps and their action by bracket ?

[delicate orchid]

you're working in a ring...

[coral spindle]

Haha well diagonal matrices are also upper diagonal!

whoopsie I mean

upper triangular

[uneven jackal]

not strictly. I edited

[coral spindle]

OK what do you mean ‘strictly’ then – with zeroes on the diagonal?

[uneven jackal]

yes

[velvet steeple]

Hm, true, so it should be allowed

It's an axiom of rings

[coral spindle]

Well yes it boils down to this I suppose

The difference is in the representations

I forget my Lie theory but there is a difference in the Lie algebra itself and its representations, so there you go.

[uneven jackal]

that doen't tell me much lol

[delicate orchid]

they're like strict upper triangular matrices in the sense that their both their bracket and strict upper triangular matrices are nilpotent

I have no idea what this statement is supposed to mean and I'd be surprised if it was anymore rigorous passed this point

[uneven jackal]

I'm just trying to find if I have a gross misconception somewhere or not

[coral spindle]

It's not as if you can force the Lie algebra itself to be strictly upper triangular iirc, so there's not much more to say.

Well you haven't made it clear what misconception you think you have...

[delicate orchid]

well if it's any consolation I completely agree with you on the abelian lie subalgebra thing

[delicate orchid]

"how on earth can reps of nilpotent matrices be strictly upper triangular when reps are maps into invertible matrices and I have abelian counter examples"

[coral spindle]

Well we've answered that: because there's a difference between the reps and the actual algebra

[delicate orchid]

yur

[coral spindle]

They represent like upper triangular matrices

Job done

[delicate orchid]

hmmm

[uneven jackal]

boyt I think you're minsunderstanding

[coral spindle]

OK

[uneven jackal]

this is not the problem

[delicate orchid]

say the algebra is finitely generated by X_1 ... X_n. If we construct the matrix (M_ij) = [X_i, X_j] then is this strictly upper triangular when the algebra is nilpotent? It's obviously diagonal when the algebra is abelian right so hmm perhaps not
oh it's like a markov chain. The matrix has to be nilpotent so it just has have 0 trace, does this mean it's similar to a strictly upper triangular one? Perhaps not
this isn't relevant I'm just sludgeposting

[uneven jackal]

roughly the thing was :

• reps of solvable groups are upper triangular up to basis
• they say nilpotent lie algebras are "like strict upper triangular matrices" (morally)
• I have an example where a rep of nilpotent (even abelian) lie algebras isn't strictly triangular, seeming to contradict the "intuition" he threw at me lol

[coral spindle]

OK fixed then

[uneven jackal]

anyways

[coral spindle]

Oh wait no I see

It's about this strictness, oh this is so obvious I'm so silly

Yes that is just not the statement of the theorem!

The theorem doesn't guarantee strictness.

[delicate orchid]

oh right so it's the author not knowing how to write stuff. Great

this is why "vague" moral statements should be avoided

[coral spindle]

Lie's theorem only guarantees a stable flag. It doesn't allow us to say that we're gonna have strictness.

[uneven jackal]

I think what he means is that (maybe?) the endomorphism griven by bracketing with something is strict upper triangular

[delicate orchid]

surely your abelian example disproves this as well

[coral spindle]

The adjoint representation? That would kill off the centre, so I imagine so

[uneven jackal]

hm

[delicate orchid]

this is what I've been sludge babbling about up here #groups-rings-fields message

wait isn't there a thing

I remember there being a thing

[coral spindle]

ngl Wew I do not understand what you were babbling about

[delicate orchid]

trying to see if what PearlSek was saying about the endomorphism holds any weight

[delicate orchid]

Schur decomposition, that's the one

every matrix over the complex numbers is conjugate to an upper triangular matrix via a unitary matrix, so viewing the Endomorphism as a matrix (I'm assuming f.g.) this matrix is nilpotent and thus has 0 trace - hence it is conjugate to an upper triangular matrix that has 0 trace

which is a strictly upper triangular matrix

yeah I'm happy with that

[uneven jackal]

uhhh

upper triangular with 1 and -1 has zero trace

[delicate orchid]

that's also diagonal

I said I was happy with it so I don't really care if I was wrong

[wet zodiac]

is there a nice way to show that a finite non-cyclic abelian group always contians a subgroup isomorphic to the direct product of Z_pxZ_p for some prime p without relying on classification of finite abelian groups

[coral spindle]

Well sure. What you're asking for is to find two commuting elements of order p that aren't in each other's cyclic subgroups.

Let G be a finite non-cyclic Abelian group of minimal size such that there exists no such pair. I will assume it's obvious to you that there must be some prime p such that p^2 | |G| and there is some element g of prime-squared order.

Now consider the group G/<g>. By minimality, this quotient must be cyclic, say G/<g> = <h + <g>>, and we know that <h> =/= G by assumption. The order of h + <g> divides the order of h, so we conclude that the order of h is equal to the order of h + <g>, i.e., |G|/p. [Elaboration: since the order is at least the order of h+<g>, and at most a prime multiple of it, there is only one possibility for the divisor.]

Now H = <h> is a subgroup of order |G|/p, and H n <g> = 0 necessarily. There is an element of order p in H, and we are done, contradicting minimality.

That should do it.

[wet zodiac]

doesnt "some prime p such that p^2 | |G| and there is some element g of prime-squared order" rely on classification of finite abelian groups

[coral spindle]

No.

Since it's not obvious I will explain

There's a theorem – I believe it's called Cauchy's theorem? – that states that if a prime p | |G| then there is an element of order p in G.

[wet zodiac]

yes that is cauchy

oh wait

i see now

[coral spindle]

Yup you should be done then

[wet zodiac]

if we can repeatedly do this such that all the prime factors are distinct

then we just end up with a cyclic group of order G

a contradiction

[coral spindle]

If |G| is squarefree then you get the product. Job done. You only need CRT.

[wet zodiac]

based on the fact that Z_mn is isomorphic to Z_m x Z_n if gcd(m,n)=1

[coral spindle]

...aka the CRT.

[wet zodiac]

my brain is not functioning

i proved CRT for rings and how did i not recognize that this was just that

[cloud solar]

Let n>=2 natural. Find the greatest number A s.t. there is P monic polynomial in Z[X] with deg P = n s.t. P(1),P(2),..,P(A) are divisible by A and P(0)=0

P is not constant cuz P monic and P(0)=0. P(x)=xG(x) x doesnt divide G(x).

Idk how to find the max of A. Any idea.

[rotund aurora]

idk if I'm missing something or you are

ah yeah

I am sorry

I didn't read the commas xD

[cloud solar]

Since P not constant by Schur lemma there is p prime s.t. p divides P(k)=kG(k) with k integer

[rotund aurora]

if A is prime, then P(x) has at most n roots mod A

[cloud solar]

Yes

Hmm

[delicate orchid]

ermmmm aktually... it's CRT passed through the uhh forgetful functor 🤓 it just so happesn to ermmm commute with limits

[cloud solar]

Any idea guys

[rotund aurora]

I gave you one

actually, I have no idea why they require you to have P(0)=0

[rotund aurora]

by this I meant that it's irrelevant. If P(A)=0 mod A then P(0)=0 mod A and if P works then so does P+Ak for any constant k

[celest furnace]

🤓 Erm that was super obvious (I don't know what 75% of those words mean)

[delicate orchid]

you and me both lil buddy

[celest furnace]

Lmao

[cloud solar]

If A prime p:=A. P(mod p) has p distinct roots in Z/pZ.

[rotund aurora]

there is something somewhat annoying

well

you can write the answer, but I think it is not very nice

[rotund aurora]

I mean the point is, if p|A then you know p<=n. If p=n then P(x)=(x-1)(x-2)...(x-p) mod p and mod p^2 there are no more than p roots.

If p<n, say n-p=m then P(x)=(x-1)(x-2)...(x-p)*G(x) mod p where G(x) has degree m. Then mod p^2, p^3,... you could have more roots. In fact, I think the best you could do is set G(x)=(x-a)^m for some a mod p

like in (x-1)^3 mod 27, this has 9 roots

wait but it's irrelevant

G(x)=(x-a)^m will give several roots, but since you want to have roots for 1,2,3,....,p^k you need to be a little more careful

like if P(x)=(x-a)G(x) mod p with G(a)!=0 mod p then P(a+p)!=0 mod p^2

[cloud solar]

Oh I think I get it

[rotund aurora]

whats ur answer

[median rock]

@delicate orchid yoo sry to bother you AGAIN but can you explain in the first line why is the index 2?

[delicate orchid]

the kernel of sgn is A_|G| which is index 2 in S_|G|

if you want to be really explicit, multiplying by a random 2-cycle is a bijection between the set of even permutations and odd permutations

so there are two cosets of A_|G| in S_|G|

[median rock]

it was so obvious ,sry to bother you ,i have been working on this problem for 3 days and my brain is a mess

[delicate orchid]

it's alright - I know the feeling all too well

[white oxide]

how does corollary 1.3 imply what's highlighted if the groups they belong to are compositums, not intersections?

[wraith cargo]

You intersect the groups but you compose the fields
I'm not sure what you're asking tho?

[white oxide]

I guess I was asking how corollary 1.3 implied what was highlighted but I didn’t even articulate my question properly

never mind lol

i think it makes sense

just take the Kis to be your fixed field

wait bruh he said that

reading comprehension

well technically he phrased it wrong, i think it should be that the intersection of the corresponding groups of the Ki belongs to the compositum of all Ki

but whatever

[white oxide]

i'm not really sure how theorem 1.14 implies the first result - how exactly do we know that the intersection of K and L is k?

i looked at theorem 1.12, and it doesn't seem relevant, so possibly it's used for the other propositions, but i'll post it anyways

[chilly radish]

It needn't be. It injects into an abelian group hence it is abelian

That map in 1.14 is always injectice

[olive jewel]

i am the human railgun slug, if x1+2a=4a - 2x + 4a x2+2x = 4ax2 then what is y if y equals 1 and 1 is equal to zero minus 4ax2squared plus 4a times 2xsquared minus 4a2x2

[lusty marlin]

Care to elaborate what this has to do with groups, rings or fields? After you've typed it in a way that is less grating on the eyes, that is.

[summer path]

looks like it belongs in #prealg-and-algebra or something tbh

although I can't really tell because it's not very readable

det

[rustic crown]

tubu

[cloud solar]

Let a>1 natural number and P a polynomial in Z[X] and P(q) divides a^q - 1 for every q prime. Show that P is constant

Suppose P is not constant. If P(0)=0 then q divides P(q) so q divides a^q - 1 for every q prime. From here a=1 false. So P(0)≠0.

[olive jewel]

yes pretty much but it is not a constant it represents uncertainty as 1 and 0 in a ring traveling through time

[dim widget]

Do you know how to factor the polynomial a^q - 1? How much do you know?

[olive jewel]

i know nothing, you would have to consult the super collidgerscope in question as the reality is uncertain you would verily witness the "god particle" for reasonable data but science would be an excuse to examine a mathematical problem on a scale that is dealt with more effectivly by a more motivated ai with data updated which it will tell u is but a lie anyway

[coral spindle]

So wise.

[olive jewel]

wise enough i could crack the earth in half, but if u look to space you will find your time in the universe is limited and your uncertain at all times

[dim widget]

I feel like you might get along better with @split cipher

[olive jewel]

well if u seek knowledge u shall find it but i dont see how it would do any good when one should choose the lesser of two evils i cant imagine what you might discover with machines calculating as you would a problem

[cloud solar]

(a-1)(a^(q-1)+a^(q-2)+...+a+1)

[olive jewel]

its just a simple 1+1 and 2x2 = +1

[dim widget]

have you considered visiting a grippy sock hotel?

[coral spindle]

woah, taking no prisoners

[velvet steeple]

Does anyone have some introductory exercise on groups and rings (to practice their definition, construction, ... after you first learn about it)?

[dull marsh]

Did you learn about normal subgroups yet?

[velvet steeple]

No

[dull marsh]

Prove or disprove: There is only one ring up to isomorphism with 3 elements

[velvet steeple]

My lecture notes defined groups and rings and I wanted some exercise on it to practice their definition

[velvet steeple]

We haven't had isomorphisms yet, lol

[dull marsh]

Draw the multiplication table for S_3

[velvet steeple]

Thanks

[velvet steeple]

By the way, what's the difference between a relation and an operation (A x A --> A, (a, b) |-> a * b)? Couldn't we define them similarly

Oh, nvm

if two elements are in relation, that's like a property of them

It's completely different with operations

[dull marsh]

I guess you could implement a binary operation via a relation

But it will have to take 3 inputs

And right-unique

Something like (a, b, c) in R iff ab = c

[velvet steeple]

We can prove that the inverse is uniquely defined by letting the inverse of $a \in A$ be $b \in A$, and if $c \in A$ is the inverse of $a \in A$ too, then we have [c = c \star e = c \star (a \star b) = (c \star a) \star b = e \star b = b,] {\bf assuming that $\star$ is associative}. \[5pt] But it could be the case that $\star$ is not associative?

Then would the inverse not be unique?

[dull marsh]

Yes, but you won't be able to prove uniqueness

Or would need something else

[velvet steeple]

But would it still be unique

[dull marsh]

Yeah, inverse may not be unique if you just have a binary operation

[velvet steeple]

Oh

[velvet steeple]

Do you have an example?

[dull marsh]

Let e, a, b, c be distinct. Define a commutative multiplication on {e, a, b, c} with neutral element e and ab = e = cb

And the rest as you wish

b has more than 1 inverses

Magmas (sets with just a binary operation) are never interesting

[velvet steeple]

Where did you get that equation from?

Why ab = e = cb

[dull marsh]

I defined it to be so

[velvet steeple]

But then how do you know e, a, b, c are really distinct

Perhaps these two definitions don't match up?

[dull marsh]

Let them be

I can't see any issues in the definition

[velvet steeple]

Like doesn't ab = cb mean that a = c here?

[dull marsh]

No, I never claimed the cancellation law to hold

[velvet steeple]

Oh

[velvet steeple]

What'd the inverse of b be?

[dull marsh]

a and c

Perhaps b itself too if you let bb = e

[velvet steeple]

Ah

[velvet steeple]

And your set A here is just {e, a, b, c}, right?

[dull marsh]

Yes

[velvet steeple]

Thanks

[dull marsh]

Again, if the only requirement is closure, then there are pretty much no results you can derive

[rustic crown]

my first thought was to show that if P is non-constant, then there are infinitely many primes p such that P has a root mod p (this would be a euclid type argument). say P(k) = 0 mod p, we may assume p didn't divide the constant term, so k and p are coprime, and so there is a prime q > p such that q = k mod p. then P(q) is divisible by p and so a^q = 1 (mod p). as q is prime, a = 1 (mod p) and as that's true for infinitely many p, i get a = 1.

but yea, its not nice if i'm using there are infinitely many primes in arithmetic progressions ><

[crystal vale]

I am stuck at conceptual doubt of splitting field ....I am not getting why we want some subfield of E which contain F and some elements of E (and if I am right then contain means Isomorphic copy)

[cloud walrus]

Seagull

[untold basalt]

I only used the fact that R is a domain for the last implication and I was initially trying to prove this for principal ideal rings in general, but I can't find a counter example to show the domain condition is required. For example in Z_n I think this is true too?

[crystal vale]

Yes for integral domain it's true every prime element is irreducible element

[uneven fossil]

wanted to ask a notation question in dummit foote. for the bottom half of exercise 6 does σ(h) mean to conjugate h by σ?

or am i just overthinking this and it's literally σh

[uneven fossil]

doing it this way i can't seem to get the homomorphism so i feel i need to ask

[crystal turtle]

(4 => 1) <x> =/= R properly contains <a> since a is in <x> as you should take a = xy (you are showing a is prime, not <a>)

[crystal turtle]

As for whether this holds in non-domain PIRs, you only need a to not be a zero divisor

[cloud solar]

From where a = 1 mod p

[rustic crown]

order of a mod p divides p-1 by lagrange and q by the assumption. so must be 1.

(as we choose q > p-1)

[untold basalt]

thx