groups-rings-fields

[iron bolt]

in odd cubes is easy to solve them if u r doin the last two edges but in even cubes u cant notice them

until u get to the last layer

[eager willow]

I solve it the way I read in a book published in 1981... Not the standard way that gets taught nowadays, and not the fastest way either.

[chilly ocean]

Can I have a group that is only the identity{e}

[eager willow]

When you think about it, moves that change parity is exactly the same kind of consideration as introducing illegal moves. There is a subgroup of moves on a big cube that doesn't change parity at all. If you stick to these moves while mixing it up, you never have to worry about it when you're solving, unless your solving algorithm also uses moves that change parity. The tricky part of finding the parity only later is really an illusion, a failure of the human brain. Your algorithm is sticking with simpler parity preserving moves while solving the first few layers and diverting the more complicated algorithms for later, because this allows you not to have a huge arsenal of algorithms that are difficult to memorize.

Imagine if twisting the corners was just a standard technique for solving the cube. You could solve the first layer with incorrect orientations and then just twist the corners into place. Then oops once you get to the last layer you may find that it is impossible to solve without twisting corners again. No matter, you can just use your parity switching algorithm (twisting) and you're good.

[iron bolt]

mmake sense

ok thank uu

[midnight summit]

Hello

I am currently reading about ladder operators

and I can not find the explanation (physics pov)

what does increasing (or lowering) an eigenvalue do?

[chilly ocean]

can't help you there, sorry

I have a problem "If G is non-abelian group, prove G has an automorphism that is not the identity. I tried saying $\phi =axa^-1$ Where a is some element in G. I have proved this is one-to-one, and order preserving, but am stuck trying to prove it is onto. I have to prove that $\forall y, \exists x such that ax=ya$

is this possible to prove?

[cloud walrus]

Avina

[paper flint]

phi(a^(-1)xa)=x

[delicate bloom]

since you want to find an $x$ such that $y=axa^{-1}$, I would pick $x=a^{-1}ya$

[cloud walrus]

Merosity

[paper flint]

But you wouldn't be done; you also need to show that at least one such automorphism is also not the identity

Note that the automorphism you defined always reduces to the identity when your group is Abelian. Try to see why a pair of elements failing to commute will give rise to at least one such automorphism that is not the identity.

[young fiber]

You want to find a,x in G such that axa^(-1) =/= x

axa^(-1) = x if and only if ax = xa

i.e. a and x commute

[south patrol]

@ shin

[tribal moss]

Pick two colors that are opposite each other on the cube, such as yellow and white. Assign to each configuration of a cube a "score", which is the smallest number of in-place 120° twists of corners necessary to make each corner cubie have its yellow-or-white side on a side of the cube whose center is yellow-or-white. Obviously the score is 0 in the solved state. Prove that each quarter-turn changes the total score by some multiple of 3, no matter what the initial configuration.

[oblique hinge]

Can someone help me?

I have a question: what's the difference between a field isomorphism and a vector space isomorphism?

I'm working with Q[√2] and Q[√3].

[chilly ocean]

A field isomorphism is an isomorphism of fields. A vector space isomorphism is an isomorphism of vector spaces.

[iron bolt]

wow such a great idea thank uu

[chilly ocean]

These are isomorphic as vector spaces over Q because they both have dimension 2.

They are not isomorphic as fields.

Try to find a field-theoretic property of one that fails in the other.

Something explicitly involving the field structure (multiplying elements).

For example, Q[sqrt 2] has an element which squares to 2. Does the other?

[oblique hinge]

Thanks for the useful hints.

[white yoke]

Can anyone give me a reference for the study of non-commutative rings? Specifically, I want a book that will give the definitions of Euclidean domains and PID's in the non-commutative case. And also prove that Euclidean 'rings' are Principal ideal 'rings'. Thank you 🙂

[chilly ocean]

If a problem mentions V in the context of it being an element of Dihedral group, what is V?

[delicate bloom]

possibly vertical reflection

[tardy yacht]

I have seen people use "A first course in Non-commutative Rings" by Tsit Yuen Lam

[tall mountain]

working with field extensions, I have seen extensions, for example the field Q(√2) variously defined as "the smallest field containing both Q and √2." I'm satisfied with that if you define a field F containing elements of the form a + b√2, with a,b in Q, then this is in fact a field, contains Q and √2 is obvious, but how would you go about proving that this is the smallest such field?

in some resources I have also seen Q(√2) simply defined as a set of elements of the form a + b√2 which makes this more confusing as well

[tardy yacht]

If a field K contains both Q and sqrt (2), surely it contains all linear combinations of the form a+b(sqrt(2)), where a and b are in Q, so it contains Q(sqrt(2))

[tall mountain]

ok, that makes sense

well I feel stupid now

[tardy yacht]

nah, I am sure I wondered the same a couple of years ago

[north parrot]

If $y\in \text{Jac(R)}$ (where the Jacobson Radical is the intersection of all maximal left ideals) for a non commutative ring R, can we show that 1-xy is a unit (not only left invertible) without using its Annihilator of simple modules definition?

[cloud walrus]

loliiipop

[sweet sundial]

can someone please take a look at this and see if i did it right

feel free to at me cause im going to be away from my laptop and phone for the next 15-20 minutes

[ember field]

if a polynomial is zero then is each coefficient is 0 because the x's of different powers cannot cancel out ??

[sweet sundial]

what

are you referring to my problem?

[delicate bloom]

are you asking if a polynomial is zero for all x in a field, is it necessarily the zero polynomial? the answer is no, x^q-x in the field with q elements is nonzero but zero for all elements in the field.

[ember field]

no i just mean for a polynomial over a ring without evaluating it, if its equal to 0 then every coefficient must be 0 because there is no relation between the powers of x

[vagrant zinc]

Hi guys, could someone be so kind as to help me to solve this exercise?

[tardy yacht]

sabes las leyes de exponentes y logaritmos?

[vagrant zinc]

Si

tienes alguna idea de como comprobarlo xde

[tardy yacht]

las reglas o que sean homomorfismos?

[vagrant zinc]

de que sean homomorfismo

[tardy yacht]

por cierto, log y e no te llevan a Z porque por ejemplo e^1 no es un entero

igualmente log de un entero no es un entero

[vagrant zinc]

mmmmm

[tardy yacht]

El segundo puede ser R en vez de Z?

[vagrant zinc]

Es de R a R perdon amigo

si tienes razon

[tardy yacht]

de ahi, para ser un homomorfismo necesitas log(a+b)=log(a)+log(b) y log(ab)=log(a)log(b). Es esto cierto?

[vagrant zinc]

falso

[tardy yacht]

entonces no es un homomorfismo de anillos

[vagrant zinc]

Mi amigo dice que es verdadero

pero no veo ninguna propiedad que cumple

ademas de que las propiedades de logaritmo no es igual a lo que me dices

Mi profesor dijo que si cumplia ser homomorfismo

Bro ? 😦

[tardy yacht]

un homomorfismo de anillos cumple con f(xy)=f(x)f(y), f(x+y)=f(x)+f(y), y f(1)=1

sin embargo, log(xy) no es log(x)log(y), si no que log(xy)=log(x)+log(y)

pero aunq quisieramos que log sea un homomorfismo entre (R,+,) y (R,,+), necesitariamos que log(x+y)=log(x)log(y) lo cual no es verdad

algo que si es verdad es que log es un homomorfismo de grupos entre los reales positivos con multiplicacion y los reales con suma

[vagrant zinc]

Ok gracias amigo te lo agrezco me ire a dormir, ten un bonita noche o dia 😄

[woeful sage]

I just opened up abstract-algebra and saw everything in spanish lmao I was so confused for a second

[ocean jolt]

I'm surprised I understand what I'm reading

[white yoke]

Thank you, I had a look but couldn't find exactly what I wanted

[median karma]

Can someone explain to me why there is an isomorphism between these two groups?

[quiet pelican]

An automorphism of Z/qZ is determined by the image of 1, and since it must have an inverse, we must have that f(g(1)) = 1, so that if f maps 1 to a, and g maps 1 to b then f(b) = bf(1) = ab, so that 1 = ab and we must have that the image of 1 is a unit mod q, and composition of these maps is clearly just multiplying the images of 1, as calculated

[median karma]

Thank you very much for the explanation, I understand now 🙂

[dim plinth]

I have to find a polynomial ideal J in F[x,y] such that members of J are 0 along either the x or y axis. I'm having trouble constructing such an ideal, and I'm wondering if I understand ideal generators properly. I was thinking the ideal could be generated by x and y, but then f = xp + yq would be a member of the ideal, and it's easy to construct a function of that form that isn't 0 on either axis. Could someone point me in the right direction?

[quiet pelican]

You’ve proven that it can’t be all functions of that form
So it must be a subset
So ||how about we ignore the “or y” part of the question||
Or am I misunderstanding what you’re asking?

[dim plinth]

Ok, so ignoring the "or y" part, we can get the set of functions of the form f = yq, which do indeed all vanish on the x axis

[quiet pelican]

xq, no?

[dim plinth]

But then how do I build an ideal that includes both those and the ones that vanish on the y axis, without also including ones that vanish on neither?

yq because we want y = 0 => f = 0

[quiet pelican]

Oh right then I think ||xyq|| works

And I was misunderstanding the question

[dim plinth]

But that has the issue that it doesn't include functions that vanish on one axis but not the other.

The first part of the homework question was to find that ideal, which was easy.

[quiet pelican]

But as you’ve proven, you can’t have all those functions

As that must include x, and y, but then x + y doesn’t vanish on either axis

[dim plinth]

I think my issue is that I don't quite understand if an ideal generated by f and g is the set of sums of elements from the ideal generated by f and the ideal generated by g, or if it's the union of those individual ideals

[quiet pelican]

The second wouldn’t be an ideal in general

[dim plinth]

Right, because it would have to have a single generator right

[quiet pelican]

No, unless you’ve already restricted to PIDs

[dim plinth]

oml

the problem is poorly written

J is indeed not an ideal

this 100% implies it's an ideal

(╯°□°）╯︵ ┻━┻

[quiet pelican]

Unless you’re using a very very non standard def that doesn’t require that an ideal is additively closed

Then it works I think

[dim plinth]

wait ok apparently he meant an inclusive or

in which case it is just the ideal generated by xy right?

[quiet pelican]

That doesn’t include x, which vanishes on the y axis

[next obsidian]

It’s the ideal generated by x and y

Unions of algebraic sets = adding the ideals

x-axis = x, y-axis = y

Wait

Lmfao oopsie

No it is xy

[quiet pelican]

That would include x + y

[next obsidian]

You intersect

Yeah so

[quiet pelican]

That’s the I ideal, not J, no?

[next obsidian]

Now I am confused, because if you say vanish on the x or y axis that’s not gonna be closed

[quiet pelican]

That’s our confusion

[dim plinth]

I think it is not an ideal

So I should just show it's not

[quiet pelican]

I would probably

Or email whoever’s responsible for that sort of stuff and ask

[dim plinth]

yeah the prof made a discord because he's super cool

fuck canvas

gamer time

[next obsidian]

Wait no it is (x,y)

Wait but

I think it’s supposed to end up being (x,y)

[quiet pelican]

What do you mean by that?
If it’s what I think it is, it includes x+y, which shouldn’t be in J

[next obsidian]

But the problem got said wrong

Like this is definitely supposed to be taking the sum of the ideals for the x and y-axis

Or intersecting them

[dim plinth]

Ok so showing non-closure should be easy because I just take a function that vanishes on the x axis, and one that vanishes on the y axis, and show that when I add them it vanishes on neither

[next obsidian]

That corresponds to doing the intersection resp union of the x and y-axis

[quiet pelican]

I think this is meant to be the ideal generated by that set, but the question worded it wrong

[next obsidian]

Yeah so like if you generate

[quiet pelican]

Yes, for most pairs of such functions

[next obsidian]

Then it contains x and it contains y

Then they just have to generate (x,y)

But as written it’s asking something different

[dim plinth]

Asking my prof if J isn't an ideal for this reason

I think that's right

Yeah, I'm happy with that

Consider p(x,y) = x, q(x,y) = y, p and q are in J, but p + q is not in J, thus J is not closed under addition, so it's not an ideal

real analysis time 😎

[astral galleon]

I'm writing some code to generate the even permutations of {1,2,3,4} and I get the following output

[[4, 1, 2, 3], [1, 2, 3, 4], [4, 1, 3, 2], [1, 3, 2, 4], [4, 2, 1, 3], [2, 1, 3, 4], [4, 2, 3, 1], [2, 3, 1, 4], [4, 3, 1, 2], [3, 1, 2, 4], [4, 3, 2, 1], [3, 2, 1, 4]]

Are all those even permutations ? I made sure that at most 2 swaps were being done in code

[quiet pelican]

The first one looks like a four cycle (so not even) to me

[astral galleon]

Yeah figured :\

[delicate bloom]

I think you can generate them in sagemath if you just want them for something else, but maybe I'm mistaken

[astral galleon]

I'll have to rewrite project i'm doing one has to do a bit of manual work external libares don't work

[white yoke]

How can I prove that (R^m+T_p) + (R^n + T_q) = R^(m+n) + (T_p+T_q) , where R is a ring (PID but i dont think it has to be) and T_p,T_q are torsion modules?

[mighty spade]

hint, ||use the fact that K is a unique factorization domain||

[celest cairn]

Who discovered the formula
$\Phi_n(x) = \prod_{k=1}^{\Phi(n)}(x-\zeta_k)$?

[cloud walrus]

Sapphire Gaming

[celest cairn]

Anyone know?

[sand cradle]

chmonkey

[celest cairn]

Your joking right? lmao

[hollow parrot]

Is it true that for $n \geq 2$, 
\[S_n = \{\sigma \tau \; | \; \sigma \in A_n, \tau \in S_2\}\]
?

[cloud walrus]

Skid

[hollow parrot]

I've stumbled across this as a consequence of second isomorphism theorem 'I think'

[next obsidian]

What is S_2?

There’s tons of copies of S_2 inside of S_n

If you mean everything is a product or something in A_n and a fixed transposition (or identity), this is true

Let x in S_n be arbitrary

If x is even, then it’s in A_n, so it’s just x•e

If x is odd, then x•tau is even where tau is any transposition

Then x = (x•tau)•tau is of the form you are looking for

[hollow parrot]

Yes this is what I mean

[hollow parrot]

I didn't even notice this...pretty obvious now 😆 thank you

[next obsidian]

There’s kinda an easier way to see it too

Eh

Basically the size of the subgroup generated by this

Has to be |A_n|•|S_2|/|their intersevtion|

But you know A_n has size n!/2

So just by a size reason as long as the S_2 isn’t contained in A_n it has to be everything

[hollow parrot]

This is how I proved it

[next obsidian]

Ah

I think that’s more clear

Or like

[hollow parrot]

But it wasn't obvious to me

[next obsidian]

Easier

Idk

Both are fine

And not hard

[hollow parrot]

I like your element wise version

[tribal niche]

i keep getting confused is a representation a vector space or a linear transformation instructor keeps referring to vector spaces as representations and i'm thinking that he is referring to the space on which the group acts

e.g. he refers to things like dim V where V is a representation of G

[barren sierra]

Can you give an example where they refer to a representation as a vector space?

[tribal niche]

here he called C the trivial representation and called V a representation. $\chi$ is a character, and here he is referring to the character of the symmetric and alternating squares of an arbitrary "representation" V

[cloud walrus]

monkeman

[tribal niche]

(ignore the mistake here, he made a sign mistake on the character of the symmetric square)

[chilly ocean]

Let $S$ be a subset of the primes. Define $F_S \coloneqq \mathbb Q(\sqrt p : p \in S)$. So for example, if I consider $S = {2, 3, 5}$, then $F_S = \mathbb Q(\sqrt 2, \sqrt 3, \sqrt 5)$. Defining $F \coloneqq \bigcup_S F_S$, where the union is defined over all finite subsets of the primes, it is intuitively clear that $F \subseteq \mathbb R$. To formally show it, would it be enough to say that, since each $F_S \subseteq \mathbb R$, then taking the (uncountably infinite) union of subsets is also a subset of $\mathbb R$?

[cloud walrus]

logical_mind

[chilly ocean]

It would be hard to consider any arbitrary element of F and show that it is in R I think

[ripe basalt]

"Show there is a nonabelian subgroup T of S_3 x Z_4 of order 12 generated by elements a, b such that |a| = 6, a^3 = b^2, and ba = a^{-1}b. I know i probably want a semidirect product of two things, but im not sure which things to look at and how to define the required homomorphism

[terse crystal]

You can let a=((123),2),b=((12),1)

[ripe basalt]

yeah i found it's the semidirect product $C_6 \rtimes C_2$

[cloud walrus]

*-algebra

[terse crystal]

Yeah, D_12

[median karma]

Does someone understand why G_i is a normal subgroup of G (2nd page)? Sorry it's in french

(it deals with solvable groups)

[hidden haven]

Yes this is common abuse of notation

[formal ermine]

in what context does it say it

like can you translate the text around it?

[ripe basalt]

ok

[median karma]

The theorem says that G is solvable if and only if there exists an integer n so that D^n(G) = {1}. And apparently a consequence of the theorem and of its demonstration is that if G is solvable and if G_i are the subgroups of the sequence, then G_i is a normal subgroup of G.

[ripe basalt]

what's D^n here?

[median karma]

By definition of solvability, G_i is a normal subgroup of G_(i+1) but why would it be a normal subgroup of G?

[ripe basalt]

oh is that like the nth factor of the derived series?

[median karma]

D^0(G)=G and D^(i+1)(G)=D(D^i(G)) (D(H) is the commutator subgroup of H)

[ripe basalt]

yeah this is a derived series terminates proof

[median karma]

Yepp

[ripe basalt]

so if G^n is 1

then G = G^0 > G^1 > ... > G^n is an abelian series ending with 1

which you can show meets your definition in a pretty straightforward way

got that part?

[median karma]

Yes okay

[ripe basalt]

ok

the other way is the hard part

it's proved using induction usually

So solvable implies it has an abelian series

thats a series like G_0 >= G_1 >= ... >= G_m = 1 where those inclusions are normal inclusions

and $G_i/G_{i+1}$ is abelian

[cloud walrus]

*-algebra

[median karma]

Yes

[ripe basalt]

sorry im working on another problem rn

preparing for a midterm

ok so

I think you can show G^i <= G_i for all i

[median karma]

No pb, good luck for your exams

[median karma]

I think it's G^i <= G_(n-i)

[ripe basalt]

no

er

no i think G^i <= G_i

so induction is easy when i = 0

then suppose for all i < k

then $G^k = [G^{k-1}, G^{k-1}] \le [G_{k-1}, G_{k-1}] \le G_k$

[cloud walrus]

*-algebra

[median karma]

Oh yes sorry it's because on the picture G_n=G or here we took G_0=G

[ripe basalt]

yes

anyways you get this since G_{k-1} / G_k is abelian

[median karma]

Yes

[ripe basalt]

so then G^m <= G_m = 1

means that G^m = 1

[median karma]

Yes

Then why G_i is normal in G?

[ripe basalt]

I think you can look at the series for like G/G_{i}

[median karma]

What do you mean?

[ripe basalt]

like the fact you have an abelian series for G

means you have an abelian series for G/G_i no?

[median karma]

But to define the group G/G_i you must already know that G_i is normal in G...

[ripe basalt]

oh right

not sure!

[next obsidian]

This is totally not true.

Take D_8, you get a counterexample

If you write D_8 as <r,f|rfrf = 1> then you get

{e} < {e,f} < {e,f,r^2,r^2f}< D_8

And I guess I didn’t spell it out, but {e,f} is not a normal subgroup

[chilly radish]

Does anyone know if historically H^1 and H^2 were studied before general group cohomology was understood? (Maybe not by that name)

I'm doing a talk and i'm trying to understood if first we discovered group cohomology and then found an explicit construction of H^1,H^2 or the other way

[agile burrow]

Schur did work in the early 1900s on projective representations. In his work he essentially described H^2(G, C*)

I would assume crossed homomorphisms were also studied prior to the unified definition of group cohomology, but I don't know of anything explicitly

[chilly radish]

Hilbert's theorem 90 came about before the language of group cohomo I think

[agile burrow]

Ah yeah, that's a good example

If you also want to add in remarks about group homology, the introduction to Brown's book seems to imply that it was already clear to Hurewicz when he was studying aspherical spaces that H_1(G) should be the abelianization of G. Hopf later gave an explicit formula for H_2, still before we had the unified approach to group cohomology

[gusty thistle]

Is it true that every Galois extension K/Q of odd degree is contained in R? The reasoning is that if K is not contained in R, then the compex conjugation map is an element of Gal(K/Q), and thus 2 | |Gal(K/Q)|.

Asking because it sounds not true

[uncut girder]

Not true

[gusty thistle]

Thanks

Do you know why my reasonings is wrong?

[uncut girder]

You can find a counter example

Wait

I think you're right

I don't see a flaw in your argument

💀

I guess it's because you're requiring Galois

[gusty thistle]

yeah

[uncut girder]

I think there are def odd degrree extensions of Q which are not contained in R

[gusty thistle]

otherwise K might not be closed under complex conjugation

[uncut girder]

But they're not galois

[coarse forge]

is the separability condition required?

it might be enough for the extension to be normal and finite

haven't thought it fully though yet

[gusty thistle]

For my purposes I was only interested in Galois extensions

[gusty thistle]

if u come up with something lmk though

it seems reasonable to me too

[coarse forge]

Let $x \in K$. Since the field extension $K / \mathbb Q$ is Galois (well, normality is enough here), $\bar x \in K$, where $\bar x$ is the (complex) conjugate of $x$. Defining $\sigma$ to be the complex conjugate mapping, looking at the mapping restricted to $K$ is an automorphism of $K$. In other words, $\sigma \in Gal(K / \mathbb Q)$. But since $K / \mathbb Q$ is Galois, note that $\lvert Gal(K / \mathbb Q) \rvert = \lvert K : \mathbb Q \rvert$. And indeed, by Lagrange's theorem, this would imply that $2 \mid \lvert K : \mathbb Q \rvert$ unless $\sigma = \operatorname{id}_K$

You can probably fill in the dots here but the idea of the proof is what you provided; indeed, if K wasn't contained in R, then the complex conjugate mapping (restricted to K) is an automorphism of K => [K : Q] is even

[zenith trellis]

Is it ax³ = xa³ or (ax)³ = (xa)³ ?

[gusty thistle]

Suppose $p,q$ prime with $p \neq q$. Can $\mathbb{Q}(\zeta_p)$ and $\mathbb{Q}(\zeta_q)$ share a subfield not contained in $\mathbb{Q}$?

[cloud walrus]

bwpvbzz

[gusty thistle]

I have a hunch that this should not be possible. But I am not sure and I really have no clue how to prove it.

[chilly ocean]

isn't Q(i) in both perhaps? I dont remember this well so just asking

or like Q(zeta^(gcd(p-1,q-1)))?

[gusty thistle]

$\mathbb{Q}(i)$ is not contained in $\mathbb{Q}(\zeta_3)$

[cloud walrus]

bwpvbzz

[zenith trellis]

Ok. I feel like I've done a mistake in what follows but here are my thoughts : Take x = e and get a² = e which gives for any x, ax³ = xa. Then, x³ = axa. Phi = x -> x³ is an automorphism with Phi^-1 = Phi so Phi² = Id. Then, for any x, x⁶ = x or x⁵ = e. Take back x³ = axa and notice ax = x³a = x³a³ = ax⁹ = ax⁴ which makes x³ = axa = e or x = e. G = {e} is obviously abelian.

[chilly radish]

cool. I'm not sure if i'll have time but we'll see

[zenith trellis]

Yup, that counts as defining the function Phi such that for any x in G Phi(x) = x³. I could even have said "x -> x³ is an automorphism"

Because for any x, Phi(x) = axa and a² = e

That's what I prove at the end only

Phi²(x) = Phi(Phi(x)) = aaxaa = exe = x

Oh ok

I messed up

No

x⁶ is not x

x⁶ is not Phi²(x)

Phi²(x) is x⁹

So I only have x⁸ = e

[long nebula]

why do we say "char 0" instead of "char ∞"?

[tardy yacht]

once I read the following on math stack exchange: since there is a unique homomorphism from phi: Z to R (R a ring), we can define characteristic as the canonical (non-negative) generator of ker(phi)

[zenith trellis]

(x³)³ = x⁹

Phi(Phi(x)) = Phi(x³) = (x³)³

[coral spindle]

char R is the unique nonnegative integer such that the kernel of the unique map Z -> R is (char R)Z.

This is the natural way to state the definition of the character

[long nebula]

ahhh

okay ty

[zenith trellis]

Doesnt seem like it

If G was abelian that would be 2

[simple mulch]

Hey, suppose $\phi : G_1 \rightarrow G_2$ is an epimorphism $k : 1$ between two groups. What does it mean the $k : 1$ ?

[cloud walrus]

mns

[simple mulch]

ping please

[thorn delta]

just a guess, but maybe it means it means the fibers of phi, phi^-1(g) for g in G2, each have k elements.
or in other words, |ker phi| = k

here, k : 1 is read as "k-to-1"

just my interpretation tho. thats not standard terminology/notation or anything

[simple mulch]

you guessed correctly, I must've said he has (| ker phi| = k) in parenthesis

but can you elaborate on it?

For instance, the fibers of phi

[thorn delta]

the fibers of a homomorphism are cosets of the kernel. So in particular, the cardinality of the fibers is the same as the cardinality of the kernel
since |gH| = |H| for any subgroup H of a group G

here we have k = |phi^-1(g)| = |g (ker phi)| = |ker phi|

[zenith trellis]

Ok I think I have a good idea now. Since $\phi$ is an automorphism, $\phi(xy) = \phi(x)\phi(y)$ for any $(x,y)\in G^2$. Then, $(yx)^3 = y^3 x^3$. If I use that with $y = a$, I get $\phi(ax) = (ax)^3 = axaxax = ax^3$ and then $axaxa = ax^2$, $xaxa = x^2$, $axa = x = x^3$, $x^2 = e$. We are done because this property brings directly the fact that $G$ is abelian.

[cloud walrus]

Silfer

[zenith trellis]

If you're not convinced, for any $(x,y)\in G^2$, $(xy)^{-1} = xy = y^{-1} x^{-1}$ but since $x^{-1} = x$ and $y^{-1} = y$ we have $xy = yx$.

[cloud walrus]

Silfer

[thorn delta]

the idea here is that if c = phi(g), you can get all of the other elements of h in G1 with phi(h) = c, by multiplying elements of the kernel to g. For example, if k is in the kernel of phi, then
phi(gk) = phi(g)phi(k) = c1 = c

[simple mulch]

Thank you very much!

[thorn delta]

npnp

[simple mulch]

interesting

[thorn delta]

if you know first isomorphism theorem, this is the principle that the construction of the map G1/ker -> G2 relies on

if H = ker phi, an element of G1/H is a coset gH for g in G1. phi(gh) = phi(g) for any h in H = ker(phi), so it follows that phi'(gH) = phi(g) is a well defined map G1/H -> G2

[simple mulch]

hum ok

I am stuck in the exercise, you might be able to give a hint

Let $G_1$ be a finite group, $p$ prime such that $p \mid |G_1|$, $H$ a subgroup of $G_1$; $\phi : G_1 \rightarrow G_2$ an epimorphism $k:1$. Show that $|H| = k'|\phi(H)|$ for some $k' \mid k$.

[cloud walrus]

mns

[simple mulch]

I thought about using the restriction of $\phi$ to $H$ and use the following lemma: If $f : G \rightarrow L$ is an homomorphism and $K < \phi(G) = \Im(\phi)$, then $|G:\phi^{-1}(K)| = |\phi(G) : K|$

[cloud walrus]

mns

[simple mulch]

but then we get $|H| = \frac{|\phi^{-1}(H)}{|H|}|\phi(H)|$

[cloud walrus]

mns

[thorn delta]

group homomorphism question

[zenith trellis]

yep because phi(x) = axa with a² = e again

[hollow mica]

I know that for a finite dimensional vector space V, it is naturally isomorphic to its double dual V**

this isomorphism being: v -> [f -> f(v)], where v ∈ V and f ∈ V*

Since it is an isomorphism, I was wondering what the map in the other direction is?

[chilly ocean]

It sends a functional on V* to the vector at which it is the evaluation map. I think you're going to have to choose bases to write it down explicitly.

[hollow mica]

wait actually, showing that the above map is an isomorphism implies that all elements of V** are evaluation maps, so I can just say

given f ∈ V**, let v ∈ V be such that f(g) = g(v) for all g ∈ V*. Then the map sending f to v is an isomorphism

this feels circular but I don't see anything wrong with it

[chilly ocean]

That's exactly what I just wrote down.

[hollow mica]

oh lol

How can I use this to show that a linear map (on finite-dim spaces) f: V -> W is the same as its double transpose f**: V** -> W**

nowhere can I find a proof of this fact because it is "obvious"

[chilly ocean]

It's obvious once you figure out the precise statement.

[cloud walrus]

rakko

[chilly ocean]

This is straightforward.

[cloud walrus]

rakko

[pastel cliff]

do i smell correspondence thm?

[wooden ember]

yes i smell it too

[pastel cliff]

so is it reasonable to try and find something that R is isomorphic too and whose subgroups are more easily known/found?

asking bc i cant find such a ring just yet but it's probably some silly product

[wooden ember]

nah i wouldnt think of it that way

once you show that I is an ideal it's not too hard to see what R/I looks like

try to think about what it means for two matrices to be equivalent mod I here

also you should have a zero bottom left

not a 1

or you wouldnt even have a ring

[pastel cliff]

oop

[long nebula]

ty!

[hollow mica]

don't you want to show that f**(ev_v) = ev_(f(v)) ? I don't see why the φ ∈ W* is relevant

oh wait nvm

I see why

wait nvm

whyd you do that

[chilly ocean]

They're elements of W^**, so they're equal if they agree everywhere on W^*.

That's what I'm checking.

[hollow mica]

oh I see

is the definition of T* the map that makes the above diagram commute?

if so don't you need to show that it is unique

[hollow mica]

and how do we know such a map T* exists and is well defined

[chilly ocean]

Try proving it!

You can do so using just the condition <v, Tw> = <T^*v, w>.

It's also immediate from the definition here.

[cloud walrus]

rakko

[chilly ocean]

The key word is "isomorphism" here.

That's what ensures there is one and only one such map, and that it is given by what I wrote.

[chilly ocean]

(Although, in proving this you're probably going to be using the properties of an inner product that make the map \varphi above an isomorphism, so they're really the same proof.)

[formal ermine]

is the stabilizer of this group operation $t = 1$

[cloud walrus]

illuminator3

illuminator3

[hollow mica]

wait is it unique because, if it were anything else than the image would change so the diagram wouldn't commute

[chilly ocean]

I don't understand your argument.

[hollow mica]

we found a map (call it f) that makes the diagram commute

[chilly ocean]

It's unique because, if it exists, it must be of the form I wrote.

[hollow mica]

now if a different map g also made the diagram commute, then f(x) \neq g(x) for some x... oh I was gonna conclude the proof here but f(x) and g(x) can be mapped to the same thing under phi so this doesnt work

[chilly ocean]

The stabilizer of the group action? The term "stabilizer" typically only applies to elements of the set the group is acting on. Are you looking for the intersection of all such stabilizers (i.e. the kernel of the action)?

The stabilizer of the origin is the entire group R^\times, not just {t = 1}.

[formal ermine]

ah wait nvm

yeah

it's asking me to find all stabilizer of the group action

bit confused how exactly I'm supposed to specify them

like isn't it { R^\times, { 1 } }

set of all stabilizers

[chilly ocean]

"Group of all stabilizers"?

[formal ermine]

set

typo

[chilly ocean]

If it's asking you to find all the stabilizers, it's probably asking you to write down the stabilizer for each point in R^2.

[chilly ocean]

It's unique because if it exists it's of the form I wrote.

[formal ermine]

yeah but for any point that's not the origin it's just 1, no?

[chilly ocean]

Yes.

[formal ermine]

and for the origin it's R^\times

[chilly ocean]

Yes.

[formal ermine]

do I just write down what I just said in the last two messages

as the answer

[chilly ocean]

Yes.

[formal ermine]

thanks

[chilly ocean]

Provide proof if you need to, but that is the answer.

[formal ermine]

if $|G| = p^mk$, $p$ prime and $p \not|: k$ then $G$ is not simple because by sylow 1 it has a p-sylow subgroup and by sylow 2 that subgroup is normal?

[cloud walrus]

illuminator3

[formal ermine]

oh wait

I have to show that the sylow subgroup is unique, right?

[celest cairn]

Idk what I did wrong :/

[delicate bloom]

not sure what you're trying to do

the nth cyclotomic polynomial is the minimal polynomial that has all the roots of unity that are order exactly n, for instance phi_4(x)=x^2+1 = (x-i)(x+i)

seems like you're confusing stuff like phi_4(x)=x^2+1 with x^4-1=(x+i)(x-i)(x+1)(x-1)

$$x^n-1=\prod_{d|n} \Phi_d(x)$$

[cloud walrus]

Merosity

[celest cairn]

Yea, I was using a different formula. Here’s something I did a while ago.

[delicate bloom]

your formula is not well defined

what's $\zeta_k$?

[cloud walrus]

Merosity

[delicate bloom]

you can do something like $$\Phi_n(x)=\prod_{\substack{1 \le k \le n \ \gcd(k,n)=1}} (x-e^{i \frac{2\pi}{n}k})$$ which has $\varphi(n)$ terms in it

[cloud walrus]

Merosity

[celest cairn]

Ok, I guess I thought this formula might be a little too advanced for me to understand. (I’m in 11th grade.)

[delicate bloom]

instead of just summing from k=1 to n, you need to check that gcd(k,n)=1 too, or you don't put that term in

so for instance for Phi_8(x) you only include the terms with k as 1, 3, 5, or 7

[celest cairn]

Ok

[delicate bloom]

I think I see your problem now, assumingyou got it from here: https://mathworld.wolfram.com/CyclotomicPolynomial.html there's a little prime on the product symbol

that means the product only goes over the integers relatively prime to n

[obsidian sleet]

Cyclotomic polynomial is product of the smaller cyclotomic ones

so for divisors of n

is another way to do this

[chilly radish]

i've never seen this notation wow

[delicate bloom]

I've seen it several times, unfortunate they put it in the wolfram mathworld article so subtly like this

[obsidian sleet]

Seek dummit and foote the holy bible

[delicate bloom]

I'm assuming this is where he got the formula from and is confused lol, but can't know for sure until he confirms

[obsidian sleet]

oh mero already said what i said above

in pain

[celest cairn]

Yes, I actually got it from Wikipedia

https://en.m.wikipedia.org/wiki/Cyclotomic_polynomial

[delicate bloom]

mobius inversion to get a formula is pretty cool, but requires more work unfortunately

[celest cairn]

Thanks for the help guys! 🙂

[delicate bloom]

you're welcome, I strongly suggest not learning math off wikipedia and finding a book and/or course to go through

[celest cairn]

Ok 👍

[celest cairn]

First time applying the formula so idk if I did it right. I did get the correct answer though.

[south patrol]

looks good

[celest cairn]

Sweet, thx.

[south patrol]

Do you get where the formula comes from?

[celest cairn]

No, not really.

[south patrol]

The point is that you take a product over the primitive (fourth) roots of unity - these are those roots of unity which generate the set of fourth roots of unity

So in this case, the fourth roots are 1,-1,i,-i. Now if you just take powers of 1 or -1, you only get at most two of the fourth roots

Whereas with i and -i, you get all four as you can check

In fact, it turns out that the primitive roots are all obtained by taking some primitive root (so i here) and taking it to the power k where k is coprime to 4, so in this case you wind up with i and -i as expected

[south patrol]

Proving this is kinda group theoretical and idk how much group theory you've done

[celest cairn]

Not much lol

[south patrol]

Sure i mean nws but this is kinda what it is doing

the gcd(k,n) =1 thing is just doing this automatically for you

[celest cairn]

Ah alright, thanks. I’ll be back in a few, I’m going to try and find
$\Phi_8(x)$

[cloud walrus]

Sapphire Gaming

[tribal drum]

can someone help me with this? i dont understand how to actually write the subgroup, i can just see that g^7 = 1

[ocean jolt]

I think you would just list the different g^x, x=0..6 (or 1..7)

[tribal drum]

oh thats it?

[ocean jolt]

that's what I would do

[tribal drum]

how about this?

[hazy portal]

Let θ : G → H be a group isomorphism. Then ord(θ(g)) = ord(g) for all g ∈ G.
how to show ord(θ(g)) = ord(g)

[chilly ocean]

Did you try anything at all?

I feel like this one is just screaming at you if you write out what "order of theta(g)" means.

[hazy portal]

yea

idk how to start

[chilly ocean]

I just told you.

[hazy portal]

order of theta(g) is first element of theta(g) is the identity ?

[chilly ocean]

No.

You need to go back to the definition of order before you do this exercise.

The order of an element g of a group is the smallest positive integer n with g^n = e.

[tall mountain]

I saw an example that used the fact that you can construct a field automorphism for Q(√2) that sends a + b√2 to a - b√2, and as a consequence of this √2 is irrational, since the automorphism fixes elements of Q but it doesn't fix √2. is it just me or is this circular reasoning, since I feel like we have already used the fact that √2 is irrational at some point (like convincing ourselves that Q(√2) and Q are not the same field)

[hazy portal]

i know how to show ord(g)=e

idk how to show ord(theta(g))=e

[chilly ocean]

"ord(g) = e" and "ord(theta(g)) = e" make no sense.

The order of an element is a positive integer.

[chilly ocean]

That is to say, ord(g) is the positive integer such that g^(ord(g)) = e and there is no positive integer n < ord(g) with g^n = e.

[hazy portal]

yea

but how to ord(θ(g)) = ord(g)?

[chilly ocean]

Using the definition I just wrote.

Try computing (theta(g))^(ord(g)), for example.

Try anything.

[thorn delta]

Another way to think about it is ord(g) is the order of the cyclic subgroup generated by g. You can use this and think lagranges theorem to make some progress

[chilly ocean]

I get the feeling the direct proof is much, much more preferable in this case.

Just for the sake of simplicity.

[hazy portal]

i still dont understand how to start this

[chilly ocean]

Really?

[hazy portal]

i mean

what should i do first to get there

[chilly ocean]

Using this definition, check that ord(g) is the integer such that (theta(g))^(ord(g)) = e and such that (theta(g))^n = e implies n >= ord(g).

You're obviously going to have to use the fact that theta is an isomorphism at some point.

(Although you really only need that it's an injective homomorphism for this problem. Take that as a hint if you need.)

[celest cairn]

Yea idk what I did wrong cause this is obv incorrect.

[dim plinth]

Alright I'm struggling with ideals again. What exactly does it mean for an ideal to be generated by more than one element? We're working only with ideals of polynomials in my class, but I'm wondering if the concept exists for ideals in general.

The book says an element p in the ideal generated by f1, f2, ..., fn can be written as p = p1f1 + p2f2 + ... + pnfn, but I don't understand how that makes sense as far as ideals go.

I tried thinking about it in terms of integers, and thinking about the "ideal generated by 2 and 5", then the elements are anything of the form 2x + 5y, but then there would be members in that ideal that are not divisible by 2 nor 5 (e.g. 7).

Could someone set me straight on what it means for an ideal to be generated by multiple elements?

[main needle]

been trying to do this combinatorically but the equation is very uncrackable, any better ways to do it ?

[delicate bloom]

you're right

[plain solstice]

an ideal is generated by certain elements just means everything in that ideal can be written as a linear combination of those two members? the elements of the ideal are divisible by the smallest positive element of the ideal

oh you're talking about polynomials

i thought this was ideals of integers

well ideals contained in Z

[fading holly]

ok so im gonna fail abstract algebra, but im trying my best.. for this question, it would not be a group because i * i = i^2 which fails closure, right?

[sonic parcel]

i² = -1 is in the group

[fading holly]

wait does i always mean imaginary number?

[lunar ledge]

yeah generally

[fading holly]

hmm ok i will assume she means imaginary then lol im not exactly sure

[sonic parcel]

multiplying by i can be seen as a rotation 90 degrees anticlockwise

that should help you decide whether you think it is a group or not

[fading holly]

i think that means its a group right?

[dim plinth]

does what you said hold for ideals of polynomials too?

i figure a ring is a ring

[fading holly]

how do i show associativity though, or do I really need to try every single combination to prove it

[sonic parcel]

you can do that, or show it's a subgroup of a known group

[fading holly]

umm is the second one easier cus i really dont wanna write out like a million equations

but can i really show its a subgroup if i havent even shown its a group

[plain solstice]

use the subgroup criterion: a subset of a group is a group iff for any x,y belonging to the subster, xy and x^-1 belong to the set as well

[next obsidian]

I mean you’re defining multiplication using the multiplication in C

That is associative

So the operation is associative

[ripe basalt]

does anyone have a copy of dummit and foote who could possibly check a typo for me?

actually ill just type it out

actually i kind of dont want to type it out :x

[plain solstice]

a polynomial ideal is generated by polynomials f1, ...f,n if every member of the ideal can be written as a linear combination of those polynomials, in analogy with integers. sorry for the late response @dim plinth

Similarly every element in the polynomial ideal is divisible by the polynomial with the smallest degree

[dim plinth]

ok so ive got that definition down now. but now what is the monic generator of that ideal?

[plain solstice]

By Euclidean divison for polynomials

Basically it is the smallest polynomial in that ideal

Then you use Euclidean division to show that it’s monic

[dim plinth]

smallest meaning lowest degree right

[plain solstice]

Yes

Honestly the precise details of the proof are escaping me rn im opening my book

For integers it’s very simple

Take any element of an integer ideal

[dim plinth]

so if i just have a set of polynomials f1,..,fn, can i immediately say anything about the monic generator of the ideal generated by them?

[plain solstice]

Well

It’s going to be the GCD of those polynomials

I think

[dim plinth]

right ok

yeah no that's right that's actually the solution i wrote out to one of my problems lol

ok

i think it's sinking in now. ty

[plain solstice]

Just think about the integer case first

[dim plinth]

wait actually lets go back to integers

[plain solstice]

Take any element k of an integer ideal

[dim plinth]

right so lets say we have the integer ideal generated by 2 and 5

[plain solstice]

Use the Euclidean algorithm with P the smallest element of that ideal that’s positive

K = qP + r

r is less than p

So it has to be zero

Meaning every element of the ideal can be written as a linear combination of the smallest element

So it generates the entire ideal

In this case the linear combination is just one term

[dim plinth]

i see

so does the example im thinking of for integers work? can we say there's an integer ideal generated by 2 and 5?

[plain solstice]

So in the case you mentioned previously that ideal is actually the entire ring of integers

Because the GCD of 2 and 5 is one

[dim plinth]

thats what i thought might be the case

yep

ok

that makes a lot more sense now

[plain solstice]

Awesome

[dim plinth]

and for instance the ideal generated by 2 and 4 is just 2Z because gcd(2,4) = 2

[plain solstice]

Yes

[dim plinth]

ok i think im slowly getting ideals

a bit late because my homework's already done but hey

understanding is better than not understanding

[plain solstice]

You can repeatedly use the Euclidean algorithm to compute the minimal generator

I think that’s what it’s called

[dim plinth]

yeah i need to review the euclidean algorithm too

[plain solstice]

Because there’s more than one generator usually

[dim plinth]

but tomorrow is discussion/review day and tonight i must grind

ty for the help!

[plain solstice]

Yeah I’ve been there

Np

[wild solar]

can anyone help me and tell me where to find the proof of this statement(that this homomorphism is an isomorphism)? I'm having severe trouble understanding my professor....but I can't seem to find anything online. I'm taking an introduction to rings and fields class.

[delicate bloom]

kind of hard to read but looks like the chinese remainder theorem

[wild solar]

oh, thank you! I'll search it online

[delicate bloom]

you're welcome, good luck

[median pawn]

Hey

[cloud walrus]

Hausdorff

[median pawn]

where p is prime

every element of GL_2(Z/pZ) corresponds to a basis of (Z/pZ) x (Z/pZ)

for the first vector, there are p^2 - 1 non-zero choices

the second vector cannot be zero, and cannot be a scalar multiple of the first vector

why are there p^2 - p options for the second vector?

[cloud walrus]

Hausdorff

[chilly ocean]

The second column just needs to not be a scalar multiple of the first.

So for the second column you count how many scalar multiples of the first column there are and subtract.

[chilly ocean]

No. This would imply that v = 0.

If av = bv and a is not equal to b, then (a - b)v = 0, which implies that v = 0.

[median pawn]

Right, thanks!

[ripe basalt]

Anyone know how to show that if $H$ is normal in $G$ and $G=HK$ with both $H$ and $K$ solvable then $G$ is also solvable?

[cloud walrus]

*-algebra

[ripe basalt]

i know i can take a subnormal series like $1 \lhd H \lhd HK$

[cloud walrus]

*-algebra

[ripe basalt]

and $H$ is solvable so there's actually a series like $1 \lhd H_0 \lhd \cdots \lhd H_r = H$ where the quotients are abelian

[cloud walrus]

*-algebra

[ripe basalt]

i was thinking of modding out H in the series $1 \lhd H \lhd HK$

[cloud walrus]

*-algebra

[ripe basalt]

so you get like $1 \lhd HK/H$

[cloud walrus]

*-algebra

[ripe basalt]

maybe this is iso to K and gives a the solvable series to $HK/H$? if that's true i think it solves the problem

[cloud walrus]

*-algebra

[median pawn]

if I have a group action of G on X, and G/H is a quotient of G, is there a natural way of making G/H act on X? I suppose (g + H).x := g.x should work.

[chilly ocean]

Try checking that that is well-defined.

You'll see what you end up needing.

A fancy way of looking at it is that you're trying to pass a homomorphism G -> Sym(X) to one on the quotient G/H -> Sym(X).

Sym(X) being the set of bijections X -> X.

[chilly ocean]

H and G/H (= HK/H = K / (H \cap K) by the second isomorphism theorem) are both solvable, the first by assumption and the second since K is (hence any of its quotients are). It's not hard to prove that, if a group has a normal subgroup which is solvable and whose quotient is solvable, then the group itself must be solvable.

[ripe basalt]

ah yes i was working on it in the silence and i got up to the "quotients of K are also solvable"

okay the quotients of K being solvable is pretty much the correspondence theorem, or 4th iso

whatever u want to call it

[chilly ocean]

I like to call it the lattice isomorphism theorem.

I can never remember which one is which, except for the first one.

[ripe basalt]

i have trouble remembering the entirety of some of the longer ones

like the one we used here actually

which u remembered the name of

2nd iso theorem

[chilly ocean]

I googled it.

[ripe basalt]

ah

[chilly ocean]

All I remembered was that there was some theorem that had to do with quotienting a group product by one of the factors.

[ripe basalt]

im studying for a midterm so i wont have that luxury 😦

that is exactly how i remember it too though

[chilly ocean]

Good luck.

[ripe basalt]

i am having a really hard time solving literally any problems

so i guess i will need the luck

[chilly ocean]

It was the same for me in group theory.

Never really got the hang of it after a certain point. Felt way too unintuitive.

[ripe basalt]

yes i mean i see a problem and i just stare at it

ill either have no idea, or ill have about 4 different ones

and im not sure which one will be the best

all of that sucks when ur taking a test with time pressure

[chilly ocean]

One more reason time-pressure tests are awful.

Add it to the list.

[ripe basalt]

yes i was told the midterm was up until pretty much the D&F section on field theory. So the beginning of the book until field theory. the midterm is also only 50 minutes long

[chilly ocean]

What the hell?

[ripe basalt]

and the problem i just gave you was a "practice" problem the professor used for this midterm a couple years ago

the one about the product being solvable

yes im just extremely worried

but oh well, i think studying is helping by an epsilon

[gusty thistle]

What 😮😮

Including all the module stuff ?

[ripe basalt]

oh my bad, chapters 10 through 12 not in there

so it's 1 through 13 minus those 3

[celest cairn]

Idk what I did wrong…

[proud fox]

nothing is wrong other the missing 8 instead of n on top of pi

[formal ermine]

if $R$ is a ring and $I \subseteq R$ an ideal, then $R/I = \Set{x + I | x \in R}$?

[cloud walrus]

illuminator3

[elder wave]

yes

well that is the underlying set

[rigid cave]

if the top row is exact here, then does this imply that the bottom "row" is too?

[cloud walrus]

Chmoki ✓

[rigid cave]

where k and e are mono

like is ker f in im g?

[rigid cave]

by bottom "row" I mean A --> D --> E

[south patrol]

If we take A = B = C = 0 then like

The top row is exact and the bottom row is exact iff f is monic

[rigid cave]

right okay hmm

bruh my screenshot thingy never works

My original question was why the top row being exact is equivalent to the bottom row being it too (maybe I should move to the topology channel tho?)

[south patrol]

Lol K-theory nice

[glossy crag]

I'm reading about Wedderburn theory in Knapp's Advanced Algebra and I think I've processed the main theorem successfully (R semisimple=>direct product of matrix rings), however I have three questions.

1. Why are products of matrix rings over division rings right semisimple? Is it the same proof as for showing M_n(D) is left semisimple (one considers the left ideals of matrices with <=1 non-zero column), just now with rows instead?
2. Why in the decomposition of R is there only 1 factor? The number of factors is equal to the number of isomorphism classes of simple R-modules, so equivalently, why must there be only 1 isomorphism class of R-modules in this situation?
3. Why must each D_i be finite-dimensional, explicitly? It makes intuitive sense, I just want an explicit derivation of this fact. Could we say "R contains a copy of each D_i (multiples of the identity matrix) and if one were infinite-dimensional, so would be R"?

[formal ermine]

what do they mean when they say "observe the group operation of G acting on X". like what do they want me to do?

[chilly ocean]

Look at it.

Ponder it.

Feel it. Become one with it.

[formal ermine]

wut

[chilly ocean]

Context?

[next obsidian]

Absorb it

[formal ermine]

"Let G be a group of order 36. Consider the group operation of G acting on its 3-Sylow-Groups"

[next obsidian]

They’re just saying that’s the group action

Presumably the rest of the problem asks you to prove things about this group action

So that’s just to define which one you’re gonna be proving stuff about

[formal ermine]

yea, they want me to define a homo G -> S_4 and conclude that G isn't simple

yeah alright

was a bit confused about the wording

thanks

[glossy crag]

Why is it whenever I post a question in this channel I get no replies? Is everyone here only adept at answering the most basic of algebra questions?

[simple mulch]

If $f : G_1 \rightarrow G_2$ is an epimorphism. Can we say that $Im f = G_2$ right?

More generally, given $f : X \rightarrow Y$ surjective. Then the image of f is Y

[gusty thistle]

still insane

[south patrol]

depends on your category

lol

epimorphisms needn't be surjective in general e.g. for rings this fails

but if you mean surjective then yes, essentially by definition

[hidden haven]

1. yes
2. R is simple, but a product A × B of 2 non trivial rings cannot be simple (A × 0 and 0 × A are ideals
3. Yes, but also if D_i is inf dim, then so is M_n(D_i), and that injects (as a vector space) into R

[glossy crag]

2. Shit, that's right.
   Thanks.

[glossy crag]

1. Quick follow-up on the theorem: the proof only shows that left ssrings are right ss. For the converse, I assume one can prove the exact same decomposition theorem replacing "left" with "right" everywhere, such a decomposition then would also be left ss, making all right ss rings also left ss. Right?

[hidden haven]

Yes, or you could also use apply the argument to the opposite ring

[glossy crag]

Even better.

[hidden haven]

If R is right ss, then R^op is left ss so this decomposition holds, then apply op again and products remain products, matrix rings remain matrix rings

[glossy crag]

Yep.

Thanks

[cloud walrus]

illuminator3

[formal ermine]

nvm. figured it out

[zenith trellis]

We have a finite group $(G,\cdot)$ that is non abelian. Show that if I pick randomly (uniform probability of picking each one) 2 elements $x,y\in G$, the probability to have $xy = yx$ is under $\frac{5}{8}$.

[cloud walrus]

Silfer

[next obsidian]

Oh man, commuting probability stuff is neat

[zenith trellis]

yeah

[next obsidian]

My friend wrote a paper about it and proved some neat result

Anyway IIRC you basically like

You count stuff

Lol

So like any element commutes with itself

[zenith trellis]

IIRC ? what's that ?

[next obsidian]

If I recall correctly

[zenith trellis]

oh ok

[next obsidian]

The center will contribute like

If the center is like say 25% of G

Then in the pairs you get like

Uhhhh

I mean in a pair (g,h) if g or h is in the center

[zenith trellis]

using the center is indeed something we'll do

[next obsidian]

They commute

And you can write down a bunch of numbers and it just like pops out IIRC

This isn’t that helpful 😞

[zenith trellis]

the first thing is to write down that probability

[next obsidian]

My point is it isn’t like some secret theorem or anything

You just kinda look at G x G and write down what you know always commutes

[zenith trellis]

like a sum indexed by the elements of G

[next obsidian]

And add it up or whatever

Like you need to show the center is > 1/2 of the group

And then you’re done

[zenith trellis]

Actually you'll need the following :

• Z(G)
• The group of elements that commute with x for x in G
• Lagrange's theorem
• If k|n and k is not n then k<=n/2

[zenith trellis]

but that's impossible

[next obsidian]

I guess so, but that isn’t that much I think

Right if the probability is > 5/8

Or you could do it directly

I’m just gonna stop talking because I’m not being helpful

[zenith trellis]

|G| is a multiple of |Z(G)|

so you can't have |G| < 2|Z(G)| since G is not abelian

[zenith trellis]

you have ideas but you just have to pick the right ones lol

[lethal dune]

If I remember correctly this problem has been worked out in Gallian

[zenith trellis]

tbh i was not able to solve this by myself

[next obsidian]

I proved it with a friend like 4 years ago, but I forget all the details

[south patrol]

classic

[delicate bloom]

ooh cool didn't know about this

[next obsidian]

You can say even more

5/8 probability iff the (inner? Outer?) automorphism group is the Klein four

[south patrol]

probs inner innit

that's cool tho

cause ye ig if inner is Z/4 then it is cycic so group is abelian right lol

[formal ermine]

how do I show that $S_3$ and $\bZ_2 \times \bZ_3$ are the only groups with order 6?
we had this one lemma in the lecture that if $|G| = pq$, p and q prime with p doesn't divide q - 1 and $p < q$ then $G \cong \bZ_p \times \bZ_q$ but I can't use that here because 2 doesn't not divide 2

[cloud walrus]

illuminator3

[agile burrow]

if |G| = 6, then by Cauchy it contains an element x of order 3 and an element y of order 2

The subgroup <x> is index 2, hence normal. Then G = <x> <y>, hence G is a semidirect product of <x> = Z/3Z and <y> = Z/2Z

You can classify the possible semidirect product structures by hand

[south patrol]

I think I remember a more elementary way, i'll try to recall it

[formal ermine]

we haven't done cauchy yet

[south patrol]

You can just do an elementary counting argument

[formal ermine]

is there no shorter way

[south patrol]

I'm trying to find my way lol like

[formal ermine]

the Z6 one is easy

because if G is cyclic

then Z6

[south patrol]

I think one way is (again by counting) take an element order 2, say, <g>, and then consider action of G on the cosets of <g> (by left multiplication say)

Then that gives a homomorphism into S3 and I think you can show that the kernel is trivial if G isn't abelian

But probably the most elementary way I know (which is basically Walter's argument too) is to assume G isn't order 6, then take an element g and element h of order 2, 3 respectively (by cuonting) and then G = <g,h> by considering order. So the structure is determined by seeing how these relate to one another and a few computations give you the result

[agile burrow]

don't you need Cauchy to prove Sylow theorems?

[south patrol]

didn't they delete that thing they said about sylow

[formal ermine]

idk what I was doing with sylow there

I thought I had something but that was utter nonsense

[formal ermine]

no?

[south patrol]

feels like overkill for smth here

you don't need Cauchy but Cauchy is way easier to prove lol and is a corollary anyway

[agile burrow]

how do you get elements of order 2 and 3 by counting

[south patrol]

count

if there's an element order 6 then we are done; suppose not the case. Then there are either 0,2 or 4 elements of order 3 so we are done unless all elements are of order 2, in which case we have a contradiction on other grounds

[agile burrow]

yeah ok, I guess that works

[south patrol]

I can't think of what the most elementary argument is that not all elements are order 2 lol but an easy way would be that it forms a vector space over Z/2 and hence would have to be of order 2^n

lol

[south patrol]

But yes afaik you can make this work

[ripe basalt]

the semi direct product of Z_3 with Z_2 is iso to S_3

[agile burrow]

ok

i still like my way better

[ripe basalt]

well its just S_3 and C_6

[agile burrow]

yes, I think that's what we were showing

[south patrol]

lol

[ripe basalt]

ah. the semi direct product thing you can show that the only semidirect product of Z_2 with Z_3 must be through the trivial homomorphisms

so its iso to Z_2 x Z_3 and thus C_6

yeah classifying groups like this blows

how do you ever know when youre done i have no idea

[next obsidian]

It isn’t that hard sometimes like

With semi direct products for small groups you can usually show it’s just a semi direct product of G and H

And then you can compute the automorphism group and see all the maps

This doesn’t like get you far for bigger groups but like groups of order 8 is pretty manageable

[ripe basalt]

ya

[main needle]

For $n\neq 6$ if $\sigma \in S_n$ is of order 2, then the size of the conj. class of $\sigma$ is equal to the size of conj. class of $(1 2)$ if and only if $\sigma$ is a 2-cycle

I tried to do this combinatorically but the equation is uncrackable, any easier ways to do it ?

[cloud walrus]

ru0xffian

[pastel cliff]

what are the prime and maximal ideals of a product ring

an easy guess is just the product of prime/maximal ideals in R and S for R x S, but i dont think those would necessarily be prime

[void cosmos]

i think this just follows from two cycles having the same strucutre <--> they are conjugates

[main needle]

doesn't really help much here.

I tried to do the combinatoric argument, which I am sure there is a way do it but it's kinda annoying

[void cosmos]

yes i agree they are

so

[next obsidian]

It’s product of a prime / maximal with the entire ring

In the other factors

So it ends up being the disjoint union of the set of primes / maximal

Cuz they’re like p x S or R x p

These are in bijection with primes in R and primes of S

Split up based on what side p is

[pastel cliff]

why does one have to be the entire ring

[next obsidian]

Prove it

Haha

[void cosmos]

why doesnt this follow from that they both have the same cycle type?

or we can just use some orbit-stab so like

the number of k-cycles in S_n is n!/(n-k)!*k

[elder wave]

they wouldn't, take some r not in R, some s not in S then (r,0)*(0,s) = 0 is in R x S

[next obsidian]

Okay here’s why in a different sense

Suppose you have I x J

Don’t quote me on this, so work it out

But I’m pretty sure the quotient would be iso to R/I x R/J

Unless one of these is 0, it can’t be an integral domain

This is probably what Timo said but phrased in terms of quotients

[elder wave]

I think from that you also get the result how they look

[pastel cliff]

proved this in a diff part yeah

[next obsidian]

Oh okay

Then yeah if I or J isn’t all of R

The quotient isn’t an ID

[elder wave]

since then you either have a domain or a field

[next obsidian]

Because products of nontrivial rings can’t be

[elder wave]

yeah

nice

[pastel cliff]

https://tenor.com/view/mindblown-excuse-me-what-me-dr-phil-gif-14053511

[void cosmos]

lmao i dont get why phi and (12) having the same structure just not implies they are conjugate together and conjugate the same elements

[elder wave]

We get that (R_1 x R_2) / I x J is a domain if I x J is prime

try combining that with the isomorphism chmonkey mentioned and work it out

[pastel cliff]

actually isnt this enough to show that one of them must be the full ring

[elder wave]

Not sure what you mean by that

(You can ping me btw)

But that alone is not enough to show that

[upper cape]

Does anyone know if it is possible to characterize the field of fractions of an integral domain as a categorical limit?

I know that if R is an integral domain, then this universal property defines the field of fractions F

But I have been unable to translate this into a limit of somekind

I was hoping that this would turn out to correspond to a limit in the category of rings, but I'm not sure that it works

[next obsidian]

So for one it’s the limit of R_f

Where R_f is the localization wrt {1,f,…,}

Err colimit

When you take this over all f non-zero

I think you can write it as a colimit more directly tho

[sweet sundial]

can somebody explain to me why the answer is 10

and if the invariant tables are non-isomorphic

please

<@&286206848099549185>

please help

[timid yarrow]

not much of an algebraist... looking at PSL(2, R) = SL(2, R) - {pm I}... what is the identity element of this, then?

wait... gapped by \ vs. /

is = SL(2,R) / {pm I} not \setminus...

this makes much more sense

[next obsidian]

Indeed

Alternatively you can say it’s like 2x2 matrices with positive determinant

[timid yarrow]

that's a little more intuitive to work with for me lmfao

[ripe basalt]

Hey if G = AB where A, B are abelian subgroups of G, how would i show that $[G,G] \subset [A,B]$?

[cloud walrus]

*-algebra

[timid yarrow]

[ripe basalt]

im actually trying to show that they're equal

but one way is obvious

oh wait there's a theorem like

if G/N is abelian then [G,G] <= N

yeah...shit

so then i just need to show G/[A,B] is abelian

can't i just say if g,h are elements in the quotient group then $gh[A,B] = [A,B] = hg[A,B]$

[cloud walrus]

*-algebra

[ripe basalt]

oh yeah cuz $gh = hg[g,h]$. nice

[cloud walrus]

*-algebra

[barren sierra]

yup

[fading holly]

does this C denote any specific group?

like S is symmetric and U is something else

[next obsidian]

It's a subgroup of G

it's defined right there, it's the set of x in G such that xg = gx

this is called the commutator of g, because it's the set of elements of G which commute with G, e.g. xg = gx

[fading holly]

is little g any element of G? it just pops out of nowhere

[next obsidian]

OH!

g should be a

whoopsie

[fading holly]

oh u think so?

[next obsidian]

definitely

[fading holly]

ok thanks lol ig my prof made some mistakes

[lunar ledge]

C is the centralizer of a, no?

Oh chmonkey said that already I'm blind

[past path]

This problem seems reasonable enough to solve. Could someone tell me if there's a typo in the definition of I?

Are the sets A meant to be indexed up to m or up to n?

I get that there could be products involving the trivial ring, but the way this problem is phrased it could technically be that the ones that are left out are on the wrong end, since it specifically says A_i is an ideal of R_i and not of some R.

[chilly ocean]

You are really worrying way too much. Yes, m should be n here, it's a typo.

[wooden ember]

specifically for point 2 here, am i correct in arguing simply that Ann(Ext^i(M,N)) contains Ann(M) without passing through x-a and y-b? I get that thinking in terms of (x-a,y-b) will probably be useful for the rest of the exercise but for point 2 here i feel like you can just argue for each r in Ann(M)

cause if i take the multiplication by r map phi: M->M, then Hom(phi,N) is the 0 map and this lifts to 0 maps on Hom(P., N) for a projective resolution P. of M, and thus induces a 0 map Ext^i(phi, N) and cohomology, but this map is also multiplication by r

[rustic minnow]

Just so I am understanding the definition of irreducible elements correctly I was wondering if I take the polynomial ring Z[x] for example, then I would have that 2 would be an irreducible element as it has no inverse, but 4 would not be an irreducible element as 4 can be obtained by multiplying two non-inverse elements, namely 2*2.

Somewhat related I was reading about irreducible polynomials (for ex p(x) ) and there was a claim about p(x) not having the constant term being equal to zero. Then this is clear I presume if p(x) is part of a field since then any non-zero element will be a unit in the polynomial and if equal to zero it cannot be irreducible. Is my understanding correct?

[south patrol]

The fact 2 is irreducible isn't just because it's not invertible so I'm not sure what you meant by that

And I'm not too sure what you meant by "any non-zero element will be a unit in the polynomial"

[rustic minnow]

Well 2 is also irreducible following the definition of irreducibility by not being able to factor it into two non constant polynomials for instance if thats what you mean. For the other part I meant any non-zero constant polynomial is a unit in F[x] when working with a field.

[formal ermine]

@south patrol I thought about the groups of order 6 thing a bit more and came up with this: \
if $G$ has an element of order $6$ then it must be isomorphic to $\bZ_6$. \
so assume that $G$ has no element of order $6$. it must then have an element $a$ of order $3$. \
note that if $c \in G$ and $c \not \in \langle a \rangle$, then $c$ has order $2$. \
choose $b \in G$ with $b \not \in \langle a \rangle$. since $G = \langle a \rangle \cup b\langle a \rangle$, the elements of $G$ are $\Set{e, a, a^2, b, ba, ba^2}$. then $ab = (ab)\inv = b\inv a\inv = ba^2$. \
this is enough information to completely fill out the group table of $G$: (group table picture) therefore it must be $S_3$.

[cloud walrus]

illuminator3

[formal ermine]

is that rigorous

[south patrol]

Yes

That works

If I've read it correctly lol

[formal ermine]

epic

[south patrol]

But yes it is more or less the same as semidirect product way

Just more elementary which is nice

[formal ermine]

yeah the semidirect product was introduced in the homework exercise before this one

like this is 3 and it was introduced in 2

[south patrol]

Ah

[formal ermine]

so I don't fully understand it yet lol

gonna check it out after I've submitted this homework

[pastel cliff]

kinda random but for a field $F$, is $F \cong F[x]/\ang{x}$

[cloud walrus]

stμ₂dying

[pastel cliff]

i think yes right bc in that quotient all nonconstant polynomials get sent to 0

so you're just left with elements of the field

[chilly ocean]

Yes.

[pastel cliff]

rakko not very ppl it seems

[chilly ocean]

The map F[x] -> F sending x to zero has kernel <x> and is surjective.

That's a quick proof.

[chilly ocean]

The dark blue was hard to read.

And I don't want to use light mode.

[pastel cliff]

can i get a hint for showing that $\Q[x] / \ang{x(x-1)} \cong \Q \times \Q$

[cloud walrus]

stμ₂dying

[pastel cliff]

smells of first iso of course

but not sure what the map would be

[warm urchin]

Can i get a hint on this question ? not sure where to start , tried to find some property not perserved by isomophism but i feel like thats the wrong approach.\

Let $F_{m}$ and $F_{n}$ be free groups on m and n generators, respectively. \
Show $F_{m} \cong F_{n}$ if and only if $m = n$ \

[cloud walrus]

Susilian

[chilly ocean]

Map p(x) to (p(0), p(1)).

[pastel cliff]

roots of x(x-1)...

bleh

surjectivity should be obvious right

[chilly ocean]

It should be the easier part.

Given r and s in Q, you want to find a polynomial p(x) in Q[x] with p(0) = r and p(1) = s. For example, ||p(x) = r(1 - x) + sx|| works.

[south patrol]

One way to do this is to think about the universal property of the free group on a set

[pastel cliff]

goober question but this is true right

like you can just do that with cartesian products and no funny business happens

[chilly ocean]

Yes, this is fine.

[elder wave]

Only unfunny business happened

Btw nitezba did you figure out the rest of the solution from yesterdays question

[chilly ocean]

I personally wouldn't even prove that the map I wrote down is a homomorphism.

[elder wave]

It was a neat problem

[chilly ocean]

It's obvious.

[pastel cliff]

going back to it soon jumping around questions

i need to change that username damn

[pastel cliff]

crank prof, bro took points off cuz once cuz i didnt write a claim as a full sentence

[chilly ocean]

My condolences.

I've had similarly terrible professors in the past. I eventually started writing my problem sets like an asshole.

[pastel cliff]

precisely what im doing

[chilly ocean]

I'm thinking of the kind of professor who would dock marks if you didn't prove every little thing. Even easy set theory equalities, sometimes.

A homework assignment should not start with 7-8 lemmas.

[chilly ocean]

This was an upper year course, by the way. Not a first year baby course.

[pastel cliff]

"We begin by restating Euclid's proof of infinite primes..."

[elder wave]

I got taken off a mark for using Z_p as notation for Z/pZ (this was an intro course and no one knew p adics were a thing)

[chilly ocean]

This should at most get a comment to use the other notation. Your grader was just being a dick.

[pastel cliff]

going back to that question from yesterday timo, im not sure i see how we get from I x J not being an ID to saying that one of the ideals has to be R or S

[elder wave]

If I x J is prime then the quotient by it is a domain

now remember the isomorphism and that every product of rings without a trivial component has zero divisors

[pastel cliff]