groups-rings-fields

[barren sierra]

which also has Z/3 as a subgroup

[south patrol]

I think the thing is that complex conjugation commutes with those automorphisms of L fixing Q

since L will be formed by appending real elements to Q (since otherwise complex conjugation would be an automorphism and it wouldn't be of degree 3)

So we have two generators of order 2 and 3 which commute, hence they generate Z/6

[barren sierra]

ah

and then is what I said before enough justification to state Z/3 and Z/2 exist as subgroups in the first place?

(I already showed the extension is degree 6)

[south patrol]

Tbf any group order 6 will have those subgroups lol

[barren sierra]

fair enough

but does my explanation make sense?

then we have Q(omega) has two automorphisms: the identity and the map omega -> omega^-1 which forms a cyclic group of order 2
this is the part I am most curious about being right

[south patrol]

But ye it does to me, was gonna say yeah yours is more descriptive than just saying 'groups order 6'

[barren sierra]

ok cool

[south patrol]

I guess also you can point out that w's minimal polynomial over L and Q coincide too as x^2 + x + 1 and so yeah the only automorphism is complex conjugation

[barren sierra]

wait what

w's minimal polynomial over L and Q coincide too as x^2 + x + 1
agreed

and so yeah the only automorphism is complex conjugation
how

[south patrol]

Ok ngl I'm sleepy that was pointless to say, I mean automorphisms of L(w)/L but then that's clear ig

[pastel cliff]

$R = \mathbb{Q}[i] = {a+bi:a,b \in \mathbb{Q}}$

what do multiplicative inverses look like in this

[cloud walrus]

μ₂ (pomodoro role)

[south patrol]

Well you can compute them like you would normally over C right

[pastel cliff]

like evalute the system of eqns you get from trying to solve xy = 1 for x,y \in R

[south patrol]

Oh i meant like \frac{1}{a+ib} = \frac{a-ib}{a^2 + b^2} right

[pastel cliff]

O

ok i knew that but i dont remember how to arrive at that form

actually nvm

[south patrol]

Multiply numerator and denom through a-ib ye

[pastel cliff]

yeyeye

[barren sierra]

conjugate go brr yea

[pastel cliff]

very dumb question but why do we need to get the i out of the denom in the first place

oh wait

cuz if we dont it's not contained in the set

and wouldn't be an inv.

[south patrol]

ye we wanna get coefficients of 1,i in Q if ygm

Or rather like

[pastel cliff]

im picking up what you're putting down

[south patrol]

I assumed that's why you meant by "what do inverses look like" anyway ig

[pastel cliff]

not written by me but is this valid?

[next obsidian]

Any char 0 field contains Q

Look up what a prime sub field is

[sullen island]

why did we need this to show that there exists a maximal nilpotent ideal? doesnt zorns lemma assert that?

[next obsidian]

No

Or uhhh

Oh I see, Zorn’s gives you a maximal one

But you want a nilpotent ideal such that all other nilpotent ideals is contained in it

[sullen island]

ye im not understanding the rationale behind this line of the proof

[next obsidian]

Zorn’s doesn’t immediately give you that

It just gives you one with no larger nilpotent ideal

[sullen island]

uhhhmmm

[next obsidian]

There’s a difference between
J such that if I is nilpotent, I < J

[sullen island]

we just need a largest dont we?

[next obsidian]

And a maximal nilpotent ideal which Zorn’s gives

Largest = contains all others

[sullen island]

ohhhhh

[next obsidian]

Not maximal

[sullen island]

i was taking largest = m,aximal

i see, thank you

[next obsidian]

Nah

[sullen island]

where is this from btw

ive never heard this

[next obsidian]

I mean I think this isn’t really precise, but this is what it means more often than not

Really you should say “maximum” or “largest, which means…”

But just in my memory people use largest to mean a maximum element, not just a maximal one

[sullen island]

ah okay

[pastel cliff]

elements in Z[i] dont have multiplicative inverses right (besides obvious choices)

and only some in Q[i] do...?

im working on this

equivalent claim is me thinking abt it

[thorn delta]

is F = Q[i]?

and only some in Q[i] do...?
all nonzero elements of Q[i] would have multiplicative inverses by virtue of being a field

[pastel cliff]

should've specified but yes

but ok that first claim

is it like subtly saying that beta is invertible in Z[i]

meaning that it must be +/-1 or +/-i

[thorn delta]

ah okay, no. they mean multiplicative inverse of beta in Q[i]

[pastel cliff]

so the inverse of a gaussian integer living in Q[i] ok

[thorn delta]

well ye, its like if u is a rational then there are integers, p, q such that u = p/q = pq^{-1}.

[pastel cliff]

this is still weird tho

hol up

methinks contradiction

i try

or contrapositive...?

ok nvm, any hints might be appreciated

[thorn delta]

are you trying to prove the "equivalent claim"?

[pastel cliff]

sorta?

i mean that was just me rephrasing it to try and make sense

[thorn delta]

To get the equivalent claim from the first one, they are just multiplying both sides of u = alpha beta^-1 by beta

[pastel cliff]

i mean yeah lol that;s how i wrote it

but idk if it;s any easier to prove

u got any ideas on how to approach :/

[thorn delta]

like easier to prove without assuming the first claim?

[pastel cliff]

https://tenor.com/view/cat-kitten-cute-kitty-pussy-cat-gif-16510039

you can ignore the equiv claim, that's just me talking to myself, the first one is thee one im being asked to prove

[thorn delta]

ah okay, its a clearing denominators thing

you have u = (p/q, a/b), and you want to multiply u by something that gets rid of the denominators

[pastel cliff]

qb is obvious choice but doesnt i get in the way of simplifying that out

maybe i just messed up before, i'll try and work it out again

ok the i does get in the way

[thorn delta]

wdym?

[pastel cliff]

like

it gets in the way of simplifying or trying to solvee

an element of Q[i] is a/b + x/y i

[thorn delta]

yes, i agree

so far

and a/b + x/y i = (1/(by))(ay + bxi)

[pastel cliff]

:O

im being silly

i think

,ti

[cloud walrus]

The current time for nitezba is 01:41 AM (EDT) on Thu, 28/04/2022.

[pastel cliff]

a;sdfa;ks

1/by is the invertible beta

ay + bxi is the gaussian integer

that's it isn't it...

[thorn delta]

ye

[pastel cliff]

if im asked to prove $\mathbb{Q}[x]$ isn't a field, what am i to assume x is...

[cloud walrus]

μ₂ (pomodoro role)

[pastel cliff]

it can't be just an arbitrary Q right?

[thorn delta]

probably an indeterminate. i.e. its the ring of polynomials with coefficients in Q

[pastel cliff]

🤦‍♂️

https://tenor.com/view/cat-hammer-annoyed-giant-gif-24029759

what the fuck is this proof tho

oh wait nvm

🔨

but wouldn't it be better to say that it's not necessarily in Q[x], since there might be a chance it is

[thorn delta]

there is a difference between Q[u] where u is an element of some field extension and Q[x] where x is indeterminate. x^-1 is never in Q[x]

[pastel cliff]

[thorn delta]

this is something that can be proven too. If p(x) \in Q[x] is the inverse of x, then xp(x) = 1 (can you get a contradiction from here?). solution: || xp(x) = 1 while degree(p(x)) + 1 = degree(p(x)) + deg(x) = degree(xp(x)) = deg(1) = 0 so degree(p(x)) < 0 which is impossible. ||

[pastel cliff]

last one i'll ask help on but If k is a field, then k[x] = k(x).

just a nudgee might help

[thorn delta]

are you being asked to prove this?

because ||it isn't true||

[pastel cliff]

prove or disprove

a brief nap later i realize it's false

[thorn delta]

i suppose how obvious that is depends on how these are defined. also a nap from 2am to 3am

[pastel cliff]

it wasn't a willing nap

i'll try and prove it and be back soon

i need to come back and actually understand this better later but i think this is right

Q(pi) is necessarily a field ofc, but Q[pi] doesnt have multiplicative inverses for all elts

[next obsidian]

Hurb

Why pick an example that’s harder on yourself?

[thorn delta]

again the uninstantiated use of x should prob be treated as indeterminate

[next obsidian]

Let x just be an indeterminate

Polynomials vs rational functions, obviously the former isn’t a field

[pastel cliff]

it's my proof

[thorn delta]

ahem ||you've already done the work||

[pastel cliff]

i forgor 💀

didnt carry through the nap

but yeah i realized lol

[next obsidian]

So I mean secretly

Q(pi) is the same as Q(x)

And similar for [] but that’s because pi is transcendental over Q

This kinda just makes your life harder tho to prove it

[pastel cliff]

masochism

thank you tho <3

[pastel cliff]

isn't this just first iso?

im being asked to prove it but like

@thorn delta im only pinging bc of the hour but can you verify this proof please 👉 👈

[proud bear]

I don't think Z/4Z and (Z/2Z)^2 are isomorphic as rings because they aren't even isomorphic as groups

[chilly ocean]

The claim isn't true and not sure what your proof is supposed to show (assuming the usual definitions)

[proud bear]

I think it is. You can take || R=S and let f be some endomorphism that isn't an iso for example ||

[chilly ocean]

|| trivial ring ||

[proud bear]

|| ah hmm ||

[pastel cliff]

it's a negation of For isomorphic commutative rings $R$ and $S$, then any ring homomorphism $f:R \to S$ is an isomorphism.

[cloud walrus]

μ₂ (pomodoro role)

[pastel cliff]

bc i suspected it to be false...?

[chilly ocean]

It's true in one special case

[pastel cliff]

when

[chilly ocean]

What do you think

[pastel cliff]

i mean my gut tells is me false so im not really sure

[chilly ocean]

Say R = S for simplicity

[pastel cliff]

wait do you mean my negated claim is true in one case or the original claim is true?

[pastel cliff]

actually back of the book says that this claim is false

[chilly ocean]

How do you define com ring homomorphisms

[pastel cliff]

how you'd expect with the condition that f(1) = 1

[chilly ocean]

Then that claim is true even for non-trivial cases

[chilly ocean]

This one

Because take S = R = Z

Then f(1) = 1

And f(n) = nf(1) = n

So any homomorphism of Z must be an isomorphism

[pastel cliff]

ok i can see that

i will have to think about that in the morning bc i think i agree

in which case the book might be wrong but i dont have any errata to check

[chilly ocean]

Originally I thought about R = S = trivial ring

Because if you don't demand that f(1) = 1 then you can just map eveeything to 0

And only trivial ring remains

[silent osprey]

f(1_S)=1_R is true when a ring homomorphism is onto - you can prove this

for all rings

[chilly ocean]

@silent osprey

[pastel cliff]

ye

[pastel cliff]

and this is one to one

[chilly ocean]

Actually, what?

[pastel cliff]

if i may bother with one last thing, this claim is true right...?

[chilly ocean]

Are you talking about general homomorphisms or the isomorphic ones

[pastel cliff]

like you can't necessarily generalize a group homomorphism to a ring homomorphism right? i cant find an example but i cant see that being true

general

[silent osprey]

general homomorphism

[chilly ocean]

Embed Z in Q

It's not onto

Am I missing something

[pastel cliff]

wait that's not a ring homomorphism

hol up

how is it not

am i dense

[chilly ocean]

There exist functions f:R to R such that f(x+y) = f(x)+f(y) but f isn't measurable

So in particular is not a homomorphism of rings

[chilly ocean]

To answer this

[pastel cliff]

ok wait but how is this not a ring homomorphism

[chilly ocean]

It is a ring homomorphism

[pastel cliff]

[chilly ocean]

It's actually a monomorphism

[chilly ocean]

Sorry, I should have said, not continuous, since you might not know what measurable means.

[pastel cliff]

i do not unfortunateley

[chilly ocean]

You can construct solutions like that using axiom of choice

Consider R as a vector space over Q

Then any such f is a linear map on this space

You can use axiom of choice to pick some basis of it

And then map basis vectors to basis vectors in some non-trivial way

That's how you construct solutions to this

Which are not continuous

You can require f(1) = 1 too if you want

[proud bear]

f: Z[i] -> Z where f(a+bi)=a+b should also be a counterexample to this right?

[chilly ocean]

Yeah

[pastel cliff]

sorry but im still thinking about this, you can take out the n only bc theyre isomorphic to begin with right

[chilly ocean]

No

You can take out n because it's a homomorphism

f(nx) = nf(x) in any ring

Using the property f(x+y) = f(x)+f(y) you can prove that

[chilly ocean]

Why the last sentence true. ?

[chilly ocean]

how can i see that C[x,y]/(xy) and C[x]\times C[y] are isomorphic after inverting some elments

I think if I invert x

[tall jay]

What about f being an isomorphism guarantees that o(G) is coprime to m?

I'm doing the forward direction of this proof, and I'm not really sure what I can use

[quaint ivy]

perhaps it's best to prove the contrapositive

what happens if m and |G| aren't coprime

[tall jay]

Assume m and |G| to be not coprime. Then, there exists some integer that divides both of them?

[quaint ivy]

yes, that's the definition of two numbers not being coprime

using that you should be able to show that f isn't injective

by taking elements of a certain order (I'm not sure how much you can assume here, but you can always do this since finite abelian groups are products of cyclic groups)

[tall jay]

So, there exists some integer that divides m and |G|, so we have some x in G such that px=1?

[quaint ivy]

sort of

you probably want to find some x in G such that f(x)=mx=0 (where 0 is the identity in the abelian group G)

try showing this for the cyclic case (Z/nZ)

and then see how the argument carries to the general case (like I said any finite abelian group is isomorphic to some product Z/nZ x ... x Z/mZ)

[frail zealot]

ah i was about to ask if they were all isomorphic to some product

[tall jay]

It should just be some x of order o(G)

right?

[quaint ivy]

yes https://en.wikipedia.org/wiki/Abelian_group#Classification

[quaint ivy]

something smaller than the order of G

like, if m=rk and o(G)=rt, then any x of order t should work

[tall jay]

So basically, if $gcd(m,n) \neq 1,$ there's some prime $p$ such that $p|m$ and $p|o(G)$. Since G is a finite abelian group, there's an $x* \neq 0$ such that $mx* = 0$.

[cloud walrus]

beeswax

[tall jay]

Is this what you're saying

[quaint ivy]

yes

and that shows f isn't injective since f(0)=0

[tall jay]

GoT it

[quaint ivy]

so by contrapositive you have the converse of this

[tall jay]

Wait, where does the p dividing m and o(G) come into play

This is my proof: if $gcd(m,n) \neq 1,$ there's some prime $p$ such that $p|m$ and $p|o(G)$. Since G is a finite abelian group, there's an $x* \neq 0$ such that $mx* = 0$. But $f(0)=0$ for any $m$, so injectivity fails. Since the contrapositive is true, then the forward direction holds

[cloud walrus]

beeswax

[tall jay]

n is the order of G I guess

And I don't see how we use p divides m and n

[frail zealot]

it doesn't have to be prime |G| could also divide m and it would still not be an isomorphism

[tall jay]

So it can be any integer k that divides both m and o(G). But how does it play into the mx*=0 and f(0)=0

[frail zealot]

Z/2Z for example the only numbers that would not be coprime are 2^n

Z/3Z 3^n

and so on

the reason being take an element from Z/3Z and mutiply it by 3^n

it would be zero

cause it's divisible by 3

after you show this base case

you move on to the products

which is slightly more complicated but not by much

Z/nZ x Z/mZ

the order is nm

the numbers not coprime are n^k *m^j

[split mantle]

Hi, does anyone know how to prove that (x^2 - yz, xy - zt) is not a prime ideal in k[x, y, z, t]. Should I just keep playing with the combinations until I could come up with an counterexample? Is there a more systematic way to do it?

[next obsidian]

It suffices to show that the image of (xy - zt) isn’t prime in k[x,y,z,t]/(x^2 - yz)

The main benefit to this IMO is that you can replace x with sqrt(yz) because that’s what you’re saying x is, something squaring to yz

This might make finding some factorization of xy - zt more clear

To be clear I don’t have a proof, but that’s my first instinct

[split mantle]

Thanks, I'll think about it

[tall jay]

Essentially the same question from earlier, but I want to prove that
f:G->G defined by f(x) = mx (G is a group, m is an integer)
is an isomorphism, with the assumption that the order of G is d and
gcd(m,d)=1. Also, G is finite and abelian

Trying to use the gcd assumption, I have this so far

since gcd(m,d)=1, we have integers u,v: um + vd =1. So:
x = (um+vd)x = umx +**vdx** =  umx + 0 = umx.```

I don't know how this contributes to proving isomorphic, because since G is finite, f is surjective iff bijective. So I don't get why I can't just

f(x)=f(y)
mx=my
x=y

[terse crystal]

It’s an isomorphism because it has an inverse

um+vd=1 for some integers u and v, then g—>ug is the inverse of your homomorphism

[tall jay]

Does g->ug fean f(g)=ug

[tall jay]

I don’t follow this

[tall jay]

Im just not sure how to use the gcd assumption in proving the bijection part

[robust pollen]

If $R$ is a ring, then $\operatorname{Hom}_R(-, R)$ is a (contravariant) functor from right to left $R$-modules. Does it reflect isomorphisms, i.e. can we conclude from $\operatorname{Hom}_R(M, R) \cong \operatorname{Hom}_R(N, R)$ that $M \cong N$?

[cloud walrus]

expectTheUnexpected

[gritty sparrow]

Hom(Q,Z) and Hom(R,Z) should both be trivial

[robust pollen]

Gotcha. What if M and N are fgp though?

[gritty sparrow]

What does fgp stand for?

[robust pollen]

finitely generated and projective. It should be true there, I guess

[gritty sparrow]

Yeah

[wooden ember]

Idk why but this exercise made me feel smart

Prove that if K c L is an algebraic extension of odd degree, then K(a) = K(a^2) for all a in L\K

The proof is just so simple but magical I love it

[next obsidian]

Is it just that if it weren’t then K(a) is degree 2 over K(a^2)?

[wooden ember]

Yeah K(a) is at most degree 2 over K(a^2) but it has to be of odd degree so it’s degree 1

I know it’s easy but I struggled on this for 1h and love the simplicity of it in the end

[next obsidian]

Sometimes the solution is the “dumbest” thing

[wooden ember]

Yeah

[next obsidian]

Just like, the simplest thing you decide is too dumb to work

[wooden ember]

I was trying to write a explicitly in terms of a^2 by trying some expressions with the minimal poly of a and it wasn’t working out

[next obsidian]

I had a proof once where I like took a set theoretic cover of a field

And the proof involved intersecting with that

It felt so stupid because that isn’t even a sub field anymore

[wooden ember]

Monke brain

[next obsidian]

I thought my proof was surely wrong and I checked Bourbaki (in French lol) and their proof was literally the same

[wooden ember]

Bourbaki must have been pretty monke brain too then

[next obsidian]

🦧

[chilly ocean]

Let $f=X^8-3$ and let $\alpha$ be its real root. I need to prove that $\mathbb{Q}(\alpha^4)$ is the only quadratic subextension of $\mathbb{Q}(\alpha)$.

So I set $K=\mathbb{Q}(\sqrt{D})$ with D a square free positive integer and I think I proved that 3 always ramifies in $\mathcal{O}_K$, so $D=3D'$ with $D'$ and 3 being coprime, but I don't know how to conclude, can someone tell me if I'm on the right track at least?

[cloud walrus]

Gio

[sharp turret]

can anyone tell me what F_H means in the 2nd question there?

[next obsidian]

The fixed field of H I assume

[chilly ocean]

Fixed field i assume

[next obsidian]

Jinx

[sharp turret]

thanks 👍

that makes sense

[strong yacht]

I have an idea for this problem, but I don't know much advanced theory, so what I'm going to propose might not work

[chilly ocean]

the factorization I found is $3\mathcal{O}_k = (\sqrt{D},3)^2$, I don't know if that helps

[cloud walrus]

Gio

[strong yacht]

This is what I came up with @chilly ocean , I've skipped many steps and not everythign is justified

I think something like this could work

[next obsidian]

Mans is out here checking #groups-rings-fields solving literally every Galois theory problem asked

After you take your Galois theory exam or whatever you’re gonna be the most powerful Galois theorist on earth

[strong yacht]

Ahahah, to be honest I've grown quite fond of it, it has become my favourite course in undergrad maths

[strong yacht]

You might know some more theory about this that I don't so feel free to disregard what i sent, if that is the case

[fiery berry]

In my lecture notes we defined the Vector Space generated by some set M via the following: It is the vector space of all maps $f : M \rightarrow \mathbb{K} , f(m) = 0$ for almost all m. As we know every element can be written as linear combination $f = \sum_{m \in M} f(m) \delta_m$. Now it says that we can write this aswell as $f = \sum_{m \in M} \lambda_m m $ where $\lambda_m \in \mathbb{K}$ and $\lambda_m = 0$ for almost all m. and says that we can think of that space as the finite linear combinations of elements of M with coefficients in K. I can't make the connection between these two ways of expressing that space. The first combination is a function from M to K, the second is just some formal sum. How should i think of this and why is that natural?

In case it's relevant. The way where this becomes relevant for me is in the definition of regular representations

[cloud walrus]

chrisply

[chilly ocean]

lambda_m is f(m) and delta_m is m

[fiery berry]

ok thats what i thought

but i mean its not like there is an isomorphism between some spaces right?

or is the space of these formal sums a vector space aswell?

well

oh lol i htink it is

or is it?

yeah nvm this was kind of a dumb question. I just wanted to make extra sure that i have the right picture going on. Especially since i have been a bit rusty in algebra

[chilly ocean]

I'm too stupid to find things out on my own so I need any help I can get, thank you very much

[strong yacht]

I doubt that very much, maths is just hard

I've added some more justification to what I said before

I don't quite have that the traces of sqrt(d) \alpha^i are always zero, but I think it's true from playing around on wolfram alpha

[barren sierra]

how do I leverage 3 to do 4?

[lethal dune]

what is the galois group of that extension?

[barren sierra]

$\text{Gal}(\mathbb{Q} (\sqrt[8]{2}, i) / \mathbb{Q}) = \left\langle \sigma, \tau \middle| \sigma^8 = \tau^2 = \text{id},~ \sigma \tau = \tau \sigma^3 \right\rangle$

so like

not quite dihedral

[lethal dune]

τ=id?

[barren sierra]

whoops

fixed

[cloud walrus]

Spamakin🎷

[barren sierra]

ok yea now it's 100% correct

so I know I'm looking for normal subgroups of this Galois group (call it G)

because I know all the extentions are already seperable

[lethal dune]

can you show the previous groups are subgroup of G?

also normal

[barren sierra]

but how do I know those are the only ones?

[lethal dune]

about that

[lethal dune]

are you sure about $\sigma \tau = \tau \sigma^3$

[barren sierra]

yea

$\sigma$ maps $\sqrt[8]{2} \to \zeta_8 \sqrt[8]{2}$ while fixing $i$ and $\tau$ maps $i \to -i$ and fixes $\sqrt[8]{2}$.

[cloud walrus]

Spamakin🎷

[barren sierra]

but you must get that $\sigma$ maps $\zeta_8 \to -\zeta_8$ and $\tau$ maps $\zeta_8$ to $\zeta_8^7$

[cloud walrus]

Spamakin🎷

[barren sierra]

and it all checks out

[lethal dune]

$\sigma \tau( i) = \sigma( -i) = \zeta_8^7$

$\tau \sigma^3(i) = \tau (\zeta_8^5) = \zeta_8^3$

[lethal dune]

is what I am getting

[barren sierra]

sigma fixes i

$\tau \sigma^3(i) = \tau (-i) = -i$ and $\sigma \tau( i) = \sigma( -i) = -i$

[cloud walrus]

Spamakin🎷

[barren sierra]

also this is literally straight from Dummit and Foote

so I shall take them as correct

[lethal dune]

ah okay

[barren sierra]

so suppose I figure out if $\mathbb{Z}/8\mathbb{Z}, D_8, Q_8$ are normal or not in this not-quite-dihedral group

[cloud walrus]

Spamakin🎷

[barren sierra]

how do I enumerate through other groups

is there a quick way to do this using Galois theory?

[strong yacht]

This might help

https://groupprops.subwiki.org/wiki/Subgroup_structure_of_semidihedral_group:SD16

https://math.stackexchange.com/questions/85172/fastest-way-to-compute-subfields-of-mathbbq-sqrt82-i-which-are-galois?rq=1

This as well

[barren sierra]

So you determined the splitting field is $\mathbb{Q}(i, \sqrt[4]{2})$ right?

[cloud walrus]

Spamakin🎷

[barren sierra]

that's good

[sharp turret]

yeah thats what i got for splitting field

[barren sierra]

typically when people ask these questions

they want the Galois group identified as some more "basic" or common group

so your Galois Group really looks like what common group?

[sharp turret]

so like a dihedral group or something

id have to think about it

[barren sierra]

how many groups of 4 elements are there?

[strong yacht]

The Galois group is not correct

Complex conjugation should be an automorphism too

It should be D_8

[barren sierra]

it's over Q(i)

so i must be fixed

so conjugation would not work

[strong yacht]

Ahh yeah good point 👍

[barren sierra]

hint there are 2

[sharp turret]

uhh z4 is order 4

[barren sierra]

cool

and then the Klein 4 group https://groupprops.subwiki.org/wiki/Klein_four-group

so which one of these does your Galois group look like?

[sharp turret]

unfamiliar with klein group 1sec

[barren sierra]

Klein 4-group is Z/2Z x Z/2Z

[sharp turret]

uhh seems to look like z4 to me

is that right

[barren sierra]

bingo

[sharp turret]

cool thanks yeah that makes sense

[barren sierra]

so you would write $\text{Gal}_{\mathbb{Q}(i)}(F) \simeq \mathbb{Z}/4\mathbb{Z}$

[cloud walrus]

Spamakin🎷

[south patrol]

ig this is also a Kummer extension more generally right which is cute

[barren sierra]

a wut

I think I'm learning about those next week

[south patrol]

Oh so like if you have a field K of characteristic p say containing an nth root of unity (with n coprime to p, or any n if char 0) then adjoining a root of x^n - a gives you an extension whose Galois group is cyclic

In this case the important thing is Q(i) has a primitive fourth root of unity and we're appending a root of x^4 = 2

[barren sierra]

ooooo

[south patrol]

So the general theory is like oh this is cyclic

[barren sierra]

ok I've seen those before just didn't know those have a name

[south patrol]

Oh fair lol

oh yeah okay so yes the main ting on this is that the galois group of Q(i,2^1/4)/Q(i) embeds in like uh

oh the group of 4th roots in Q(i) so Z/4Z

[tardy yacht]

I know that in the finite dimensional case, V* tensor W is isomorphic to Hom(V,W), given by f tensor w to (v to f(v)w). Also, in the infinite dimensional case I know this can be false. But what about when W is finite dimensional and V is infinite dimensional? Does the map still work? Thanks

[upper cape]

Anyone have tips for computing the (left) maximal ideals of a group algebra?

I've tried computing every ideal manually, but realized that will take many hours. Is there a better way of doing this?

[tardy yacht]

<@&286206848099549185>

[alpine island]

Have you tried #❓how-to-get-help

[frail zealot]

no one in this channel has helpers as ping

[chilly ocean]

helpers still see it, you know

[oblique leaf]

so was this a rulebreak or not?

[frail zealot]

not that i can see but it might as well be equivalent to shouting into the void

[frail zealot]

it's kinda hard to suggest anything when people don't know exactly what you're doing like what have you tried

[upper cape]

Fair enough, currently I've just been approaching this by a pretty much bruteforce strategy

Wondering if there is some theory which is built to deal with these sorts of questions

[frail zealot]

depending on your ring you should be able to compute classes of ideals at once

my strong suit is def not rings... but idk if you're looking for a quick solution and you don't actually need to do it by hand you could use sage or something

i recently started using sage and it's amazing

[upper cape]

how do you use it?

[frail zealot]

install, type sage into terminal and it opens a python environment like if you were to type python into your terminal but i has a collection of tools for dealing with algebra objects. i've only used it for group theory but you can create your objects and then like find ideals for example very quickly

https://doc.sagemath.org/pdf/en/reference/rings/rings.pdf

i'm not sure which doc group algebra will be in

[upper cape]

ahh that sounds great! thanks!

[frail zealot]

https://doc.sagemath.org/pdf/en/reference/algebras/algebras.pdf

actually they're in the algebras doc

section 5.9

yw ! i hope it helps @upper cape

[full panther]

it is true that if f(x) in F[x] has a splitting field E, then g(x) = a*f(x) for nonzero a in F has the same splitting field E, right?

it doesnt change the set of roots

[chilly ocean]

I need to find all abelian subextensions of the splitting field of $f=X^8-3$ over $\mathbb{Q}$, so I computed its maximal abelian subextension, which I will call $E$, which should be $\mathbb{Q}(\zeta_8,\sqrt{3})$.

I think I showed that $E$ can't be the entire Galois group since $G_{\mathbb{Q}f/\mathbb{Q}} = G{\mathbb{Q}f/\mathbb{Q}(\zeta_8)} \rtimes G{\mathbb{Q}_f/\mathbb{Q}(\sqrt[8]{3})} $ and that $\mathbb{Q}(\zeta_8,\sqrt{3})$ is a subextension of $E$, and that $E$ contains $\mathbb{Q}(\zeta_8,\sqrt{3})$.

I used the fact that $\mathbb{Q}_f/\mathbb{Q}(\zeta_8)$ is cyclic of degree 8 to narrow E down to either $\mathbb{Q}(\zeta_8,\sqrt{3})$ or $\mathbb{Q}(\zeta_8,\sqrt[4]{3})$ but in order to show that the second one is impossible I had to prove that it is normal "by hand" and I had to find two elements which don't commute by using a presentation of the Galois group I had found before.

I don't think that this is the quickest way of doing this and it's probably not even correct, does anyone have any better ideas?

[cloud walrus]

Gio

[chilly ocean]

I can show more of my work if necessary

[stoic rose]

This map is an isomorphism whenever W is finite dimensional, and V can be whatever

Note that the map is always defined, it's just not an isomoprhism in general

[terse crystal]

Yes

I mean, not f, you can use a different letter, like p. p(g)=ug. Then p is the inverse homomorphism of f

[sullen island]

the radical of A is defined as the set of all elements in A that annihilate every simple (and hence semisimple) A-module, but why is every A-module annihilated by rad(A) necessarily a semisimple A-module?

[sullen island]

nvm got it

[oblique leaf]

companion matrices are already in rational canonical form right?

[terse crystal]

Yeah

It is a single block itself

[oblique leaf]

ok thank you! just making sure

[chilly ocean]

How would one go about finding all of the irreducible monic polynomials contained in F_3[x] of degree less than 4?

It's anything which can't be factored into roots with coefficients 0, 1 or 2, correct?

[terse crystal]

Just one root

f is irreducible of order 3 iff it doesn’t have a root

[chilly ocean]

Wouldn't that be nearly all permutations of such polynomials? My highschool algebra is lacking...

[terse crystal]

?

Just find all $a_{0},a_{1},a_{2}$ such that $a_{0}+a_{1}x+a_{2}x^{2}+x^{3} \neq 0$

[cloud walrus]

Cogwheels of the mind

[chilly ocean]

Yes, my root finding skills are just nonexistent

Is there a general algorithm/method to doing so?

[terse crystal]

$\begin{pmatrix}1&0&0\1&1&1\1&-1&1\end{pmatrix}\begin{pmatrix}x\y\z\end{pmatrix}= \begin{pmatrix}0\-1\1\end{pmatrix}$

[cloud walrus]

Cogwheels of the mind

[terse crystal]

$F_{3}^{3}-$ set of solutions

[cloud walrus]

Cogwheels of the mind

[chilly ocean]

I see. Thank you.

[wooden ember]

any hint on how i might go about showing that x^(p-1)+x^(p-2)+...+x+1 is irreducible over R for p prime?

i know how to show it over Q but doing anything over R always seems troublesome to me

[jagged meadow]

well, it's false isn't it?

[wooden ember]

that's completely possible

this is in the context of finding $[\mathbb{R}(e^{\frac{2\pi i}{p}}):\mathbb{R}]$

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

and i just sort of assumed it would be $p-1$ like it is for $\mathbb{Q}$ but it could be lower sure

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

just not sure how to approach it

[jagged meadow]

irreducible polynomials in R have degree 1 or 2 though

[wooden ember]

am i being stupid

could you quickly explain why, itll come back to me if ive seen it before

this is sounding more and more familiar by the second 😂

[jagged meadow]

pick a complex root of your polynomial

say z

[wooden ember]

ahhhhhh

conjugate pairs

of course

[jagged meadow]

yeah

[wooden ember]

alright thanks lol

im a fool

[jagged meadow]

I was almost questioning myself haha

[wooden ember]

this is even something i proved in a linear algebra course im pretty sure lol

small brain moment

so was the condition of taking prime primitive roots of unity superfluous then?

probably to trick us like i got tricked

[jagged meadow]

most likely

[strong yacht]

I don't think you need the full strength of the theorem that every real polynomial has a complex root, and its complex conjugate is also a root, to do this question @wooden ember

$(X-\zeta)(X-\zeta^{-1})=X^2-(\zeta+\zeta^{-1})X+1$ and $\zeta+\zeta^{-1}$ is real (e.g. geometric reason). So your extension is degree 2.

[cloud walrus]

Greenman

[wooden ember]

yeah, but it always serves as motivation to think of the polynomial you put out

thanks though

[strong yacht]

Ah right, I can see that and the discussion you had was good. My motivation was more the picture of the complex roots of unity

[wooden ember]

yeah i see what you mean. I didnt really to try and build up the polynomial by forcing X-zeta to be a factor, ill keep that in mind in the future

[lapis temple]

Hi all! Has anyone read J.A. Greens paper on Characters of general linear groups?

[wooden ember]

is it generally a hard problem to show that two simple extensions are field-isomorphic (or not)? Even for algebraic extensions, i just completed an exercise to show that $\mathbb{Q}(i)\not\cong\mathbb{Q}(\sqrt{-5})$ and it involved introducing what i assume is what ive heard people refer to as the integral ring of $\mathbb{Q}[i]$ and $\mathbb{Q}[-\sqrt{-1}]$. It seems like problems of this type can easily require your to delve further into algebraic number theory, am i wrong?

[cloud walrus]

limbostar

[rigid depot]

Can someone help me with this? (Didn't know whether I should put this in #point-set-topology or here but I'll go with here)

Let $G$ be a profinite and procyclic group (i.e. $G$ is profinite and there is some $g \in G$ such that $\overline{\langle g \rangle} = G$).
I wanna prove that the open subgroups in $G$ are precisely the ones of the form $G^n$ for some $n \in \mathbb{N}$.

To see that $G^n$ is an open subgroup, note that $G/G^n$ is a finite set, hence $G^n$ has finite index in $G$. Since $G$ is profinite (more generally, compact) the closure of $G^n$ is the intersection of all open subset $U \subset G$ that contain $G^n$. Hence if $x \in \overline{G^n}$, then every open neighborhood $U$ of $x$ has non-empty intersection with $G^n$. So there is some $v = xu \in G^n$ where $u \in U$. Hence $x = vu^{-1} \in G^n$, so $\overline{G^n} \subset G^n$. Hence $G^n$ is closed and has finite index and thus is open.

I'm having trouble showing that every open subgroup is of this form though. Some help (and possibly corrections on my first part of the proof) is greatly appreciated!

[cloud walrus]

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

[ionic mural]

hey guys! quick question if you don't mind

i am reading the proof of the following theorem

and don't understand something

why is the minimal polynomial of this form, and why is it separable?

any help would be appreciated :)

[gritty sparrow]

The coefficients of this polynomial are symmetric polynomials in the images of alpha under the automorphisms so it is easy to verify that each coefficient is invariant under each automorphism, hence by the galois correspondence the coefficient belongs to L. Minimality is easy to prove. It is separable since L/F is galois hence in particular it is separable.

[ionic mural]

why is minimality easy? i don't immediately see it

[gritty sparrow]

Let p(x) be the minimal polynomial. Alpha is a root, and applying some automorphism to both sides of p(alpha)=0 we see that the image of alpha under any automorphism is a root of the minimal polynomial

[ionic mural]

awesome thanks chmister. that wasn't bad at all

[gritty sparrow]

Np

[wooden ember]

so how would you go about showing it in this case?

Similar shenanigans to what i was describing above?

[cloud walrus]

Greenman

[strong yacht]

In this particular instance, you can do a brute force calculation if you hate yourself.

[cloud walrus]

Greenman

[wooden ember]

yeah i see

[strong yacht]

I meant cube root 3 there lol on the left

[wooden ember]

i mean it's gross but it's not so bad

[strong yacht]

Right, but the problem does not scale well when you ask similar questions for higher degree roots.

[wooden ember]

i was thinking more about applying deeper theory to problems that seem innocent like this

[strong yacht]

Perfect, because there IS wonderful theory to cover exactly this stuff

[wooden ember]

i dont know how i feel about problems that seem easy needing complicated results

i swear my opinion on the matter switches on a weekly basis

[strong yacht]

I know exactly what you mean

It seems obviously true that something like the cube root of 3 should not be expressible as a linear combination of powers of cube root of 2, but it's not obvious

https://kconrad.math.uconn.edu/blurbs/galoistheory/tracenorm.pdf

Check out Keith Conrad's expository paper on traces and norms. You can view field extensions as vector spaces, where the elements of the field extensions are linear maps from the vector space to itself (by multiplication).

[wooden ember]

ah that looks interesting

will definitely give it a read

[strong yacht]

The really cool part is that the trace of an element (defined in the paper) is essentially equivalent to the coefficient of the second highest power of an element's minimal polynomial

And this trace is k-linear (where k is the field we extended from), so it makes investigation equations like cuberoot(3) = a+bcuberoot(2)+c*cuberoot(4) MUCH easier

[wooden ember]

characteristic polynomial wants to know your address

[strong yacht]

There's also so much more than this too. For instance, somebody asked in this chat (and the algebraic number theory chat) about whether sqrt(d) is in Q(eighthroot(3)) if and only if d=3, in the algebraic number theory chat they talked about ramification of primes and stuff to rule out a lot of d

But yes to answer your original question, absolutely. A lot of this stuff is non-trivial

[pastel cliff]

is there an elegant way to do this

actually is there a general tactic for disproving isomorphisms

my first idea is to just make a multiplication table and show the obvious but i dont wanna do that

I_4 is Z/4Z which is ugly but not my choice

[proud bear]

F has no zero divisors because it's a field. what about Z/4Z?

[pastel cliff]

2

[proud bear]

well

[pastel cliff]

[proud bear]

i guess that's similar to what you were going to do

[pastel cliff]

yeah but ig that's word-able

there's no general strategy?

[rigid depot]

I guess the same argument works for showing that F_q and Z/qZ are not isomorphic as commutative rings unless q is a prime number, since Z/qZ always has zero divisors for non-prime q

[proud bear]

i guess a general strategy is to look at properties that are preserved under isomorphism. like zero divisors, units, order, etc. another way is to try to directly construct an isomorphism and show that leads to a contradiction. (for example, if f: Z->Z[x] is an isomorphism, then we must have f(1)=1, f(2)=f(1+1)=2, etc., and by induction f(n)=n for any integer n. but then nothing is mapped to x, so Z is not isomorphic to Z[x])

[pastel cliff]

thank you

can someone give me a nudge for how to prove that four-elt. fields must be isomorphic

it cant be hard but im missing sumn

[fiery berry]

I am confused by one of my exercises once again: They ask me to prove that a map $\rho : G \rightarrow End(V)$ that satisfies $\rho(gh) = \rho(g)\rho(h)$ is a representation of G iff there exists a $g_0 \in G$ such that $\rho(g_0) \in GL(V)$. So if $\rho$ is already a representation this is obvious. I don't see where i need the existence of such a $g_0$ to show the other direction. Can't i just say that $id_V = \rho(gg^{-1}) = \rho(g)\rho(g^-1)$ and immediately conclude that $\rho(g)$ is invertible and thus that $\rho$ is a group homomorphism from $G$ to $GL(V)$ I just don't get what this g0 is supposed to be doing here.

[cloud walrus]

chrisply

[fiery berry]

ah lol i need it to be able to say that rho(1) is indeed the identity. wow i write it out here and all of a sudden i see what i am missing

[pastel cliff]

writing things out has a way of doing that

[fiery berry]

i'll just blame my ongoing corona infection that im so pepega

[pastel cliff]

any hints tho

[gritty sparrow]

The easy way is to just try and put a field structure by hand. The hard way is to say that any field of four elements is the splitting field of x^4-x over F_2 hence they are isomorphic

[pastel cliff]

idk what a splitting field is

by hand it is

ive been trying contradiction to no avail

[gritty sparrow]

Ah, unfortunately that is the only “conceptual” way I know how to prove it. By hand isn’t so bad in this case though since it is only 4 elements

[pastel cliff]

what if i show that in two arbitrary fields of order 4, the same equations must hold

like 1+1 = 0 and so on

is that enough?

[barren sierra]

How do I answer this last part?

is it enough to say no because every group of order 4 has a subgroup of order 2

which must correspond with a proper subfield? (By the Galois correspondence)

[pastel cliff]

i guess i mean like, in order for a four element set to be a field it must be constructed a certain way and only that way, does that guarantee iso

[barren sierra]

yea

Basically build it up in such a way

that there is only one possibility

[pastel cliff]

[barren sierra]

so what are some things you know about the order of 1?

etc etc

[pastel cliff]

is the uniqueness of said constructioon implied?

[barren sierra]

yea because you show basically every choice is forced by the field axioms and the fact you have 4 elements

[wooden ember]

huh so that's how you systematically define a field norm

interesting

[long obsidian]

Is there a part of the Galois correspondence that allows one to count the number of distinct intermediate fields of a specific size?

[chilly radish]

By size you mean the degree of the extension?

If so then yea you can just count the subgroups with that index in the galois group

The correspondence gives you
[E^H:E^G]=[G:H] where E is the galois extension. Take G to be the galois group of E/F and you get that E^G=F by definition

Idk if there's a better way to do this

[tardy yacht]

is there a way to construct an inverse when V is infinite dimensional? I am aware of how to do it when the dimension of V is finite, but if its not it would involve an infinite sum

[chilly radish]

Infinite dimension != Infinite sum. Infinite sums don't make sense unless you have a norm anyways

[tardy yacht]

Yeah I know, i was referencing a possible inverse where you take a basis of V and its dual, and define the inverse to be f(phi)=Sum_i(ei* tensor phi(ei) ), which would not make sense with infinite dimensional V

[chilly radish]

Well, given a basis each element can be uniquely written as a finite linear combination of some basis elements, so you could still define a map in a similar way

If i'm understanding you correctly

I think I get you know

[tardy yacht]

Yeah so the issue is that the definition on the inverse relies on taking the entire basis of V and summing it, but this doesn't make sense if V is inifinite dimensional

[chilly radish]

Only finitely many of the e_i will be mapped to a nonzero element by phi

Since its codomain is finite dimensional

So the sum you mentioned will always be finite

Wait

No i'm saying nonsense sorry

I have to think about this for a bit

[tardy yacht]

phi could simply send things to others that are linearly dependent but not necessarily zero right?

[chilly radish]

Exactly

Phi is supported on some finite dimensional subspace, so you could choose a basis for that subspace and send phi to the corresponding tensor sum, but this will most likely not be well-defined because there's no unique completement to the kernel

You could define the inverse on the basis elements of Hom(V,W). Fix a basis e_i of V and f_j of W, then the basis elements are some phi_ij which send e_i to f_j and otherwise send all other basis elements to 0. These get mapped to e_i* \otimes f_j, then every linear map is a finite linear combination of these. I think this would work but I fail to see where finite-dimensionality of W comes into play

like if anything i'd think V would need to be finite dimensional

[tardy yacht]

Yeah i dont see it either...

[idle frigate]

I've got a question,
if ф on V over K is an endomorphism and ф² = id why must char(K) ≠ 2 so that ф is definitely diagonalizable.
Like what is the intutive reasoning for this.

[chilly radish]

Actually, I think it is correct. Say WLOG your space is some F^n then V^* \otimes F^n\cong (V^\ast)^n, and Hom(V,F^n) is isomorphic to (V^\ast)^n as well via just composing with the projection onto the i-th coordinate. I'm still not sure if what I described would give an iso in this case tho

[stoic rose]

You only need a basis for W to construct the inverse. Specifically, if $(b_i)$ is a basis for $W$ and $(b_i^)$ is the dual basis, the inverse is given by $f \mapsto \sum_i b_i^ \otimes f(b_i)$.

[cloud walrus]

radiateur-man

[chilly radish]

The original question was talking about Hom(V,W) and V*\otimes W when W is finite dimensional and V maybe isn't. What you described is the case where V is finite dimensional and W maybe isn't

[next obsidian]

This is just not true afaict

You can take F_2 x F_2 and just do the map (x,y) -> (y,x)

[idle frigate]

the characteristic of F_2 is 2 tho

the pre-requisite is ≠ 2

[next obsidian]

That wasn’t really communicated

[idle frigate]

then I am sorry

[next obsidian]

It looked you’re asking why char(K) ≠ 2

Anyway, then idk

[idle frigate]

Yeah, that's the question, why must it be ≠ 2

[next obsidian]

But that’s false

[chilly radish]

It mustn't as chmonkey just proved

[next obsidian]

I literally produced a counterexample

[chilly ocean]

minimal polynomial is (t - 1)(t + 1) and diagonalizability is equivalent to the minimal splitting into distinct linear factors

[idle frigate]

I think I might be phrasing something wrong

[chilly ocean]

you are

[idle frigate]

So what I mean is why does exactly 1 + 1 ≠ 0 result in that ф is diagonalizable. I see that there would be a counter example if 1 + 1 = 0, but I don't see why exactly 1 + 1, and not for example 1 + 1 + 1 = 0

[chilly ocean]

char not 2 implies these are distinct factors

[cloud walrus]

dumbo

[next obsidian]

No, you need them to be distinct irreducible

In that case this is the Chinese remainder theorem

For a counterexample, take f = g = x

[idle frigate]

and char = 2 would imply they are the same?

[next obsidian]

Yes it is sufficient by the Chinese remainder theorem

[idle frigate]

alright I am gonna think about it, thanks!

[chilly ocean]

yes

there is probably a more elementary way to do it, but that's the first thing that came to mind

ah, i remember

[chilly radish]

You can just directly prove that 1,-1 are eigenvalues and their eigenspaces split the space

[chilly ocean]

try writing every vector as a sum of a 1-eigenvector and a -1-eigenvector

sniped me

[chilly radish]

You need 2 to be invertible for this proof to divide by 2 for the decomposition

Lol

[idle frigate]

I already did that one

I think I might be missing something

[stoic rose]

Oh yeah I misread, just take my answer with V and W swapped then @tardy yacht

When V is infinite dimensional it's not an isomorphism

[tardy yacht]

That's what I was thinking, thanks

[marsh goblet]

hi all
i'm trying to do this exercise in lang but struggling a bit with trying to prove that the kernel of the map from A/B to A^f/B^f is contained within A_f/B_f

It seems like the kernel will be a subset of A_f/B, which is not really what I want...any hints would be much appreciated :))

[rapid slate]

Hello
I have been given an assignment to classify all groups of order 36. We have been taught the concept of a semi-direct product but i can't seem to use it properly for the assignment.
Can someone outline a procedure? I don't have much ideas (except for the number of sylow subgroups it can have and what are the abelian ones)?

[prisma ibex]

this outlines it nicely: https://math.berkeley.edu/~wodzicki/257/G36.pdf

[rapid slate]

Thanks for the PDF
I tried reading it but i got a bit lost in the first paragraph itself
Why are they considering the action of G on the left cosets of G/P?

[cloud walrus]

limbostar

[tough raven]

In general, that map from $V* \otimes W \to Hom(V, W)$ is injective and its range is all finite-rank maps. Thus, if either $V$ or $W$ is finite-dimensional, it is an isomorphism.

(Indeed, if $\phi_i \in V*, w_i \in W$ for $i$ from $1$ to $n$, then $\sum_i (\phi_i \otimes w_i)(v)$ is in the span of $w_1, \dotsc, w_n$ so range of $\sum_i \phi_i \otimes w_i$ is finite-dimensional. If $T \colon V \to W$ with range $W_0$ having $w_1, \dotsc, w_n$ as a basis, there are maps $\nu_1, \dotsc, \nu_n \colon W_0 \to k$ taking ``coordinates in the $w_i$ basis'' (i.e. the dual basis to $w_i$). If $\phi_i = \nu_i \circ T$, then $\sum_i \phi_i \otimes w_i \in V* \otimes W$ maps to $T$.)

[cloud walrus]

Raghuram

[terse crystal]

We consider the exact sequence 0->Z->Z->Z/pZ->0 and apply Ext(-,Z/pZ) on it perhaps?

[maiden breach]

I'm trying to do a problem but I'm not really sure what it's actually asking:
Given $\beta = \sqrt{a} + \sqrt{b}$ for some $a \neq b$ positive square free integers, find the polynomials in $\beta, \beta^{-1}$ with rational coefficients such that $\sqrt{a} = f(\beta)$ and $\sqrt{b} = g(\beta)$.
I'm confused by what 'polynomials in $\beta, \beta^{-1}$ with rational coefficients' means. Since the polynomials have rational coefficients, where does `in $\beta, \beta^{-1}$ come in? The section is on field extensions.

[cloud walrus]

Grenadilla

[long obsidian]

If I have a group G of order |G|=2^4 *17 can sylow theory tell me anything about the number of subgroups of a given size?

For instance, I feel that there is a unique group of over 2^4. But I'm not sure if I can also say something about the number of groups of order 17 by similar reasoning.

[terse crystal]

There is a unique sylow 2-group and a unique sylow 17-group. Because number of sylow 2-groups are 1+2k |17 for some k, you can’t have 17 sylow 2-groups so it’s unique. Similarly, number of sylow 17-groups is 1+17u |16 for some u, so u has to be 0

[woven delta]

Why can't you have 17 Sylow 2 groups?

[terse crystal]

Maybe we can…?

[woven delta]

I agree re Sylow 17 groups

Okay so this is going to be a semi direct product then

Okay so are there nontrivial semidirect products of a group G of order 16 with a group H of order 17? This is an action of G on H, or a map from G to Aut(H). Aut(H) is cyclic of order 16

It seems like there are nontrivial actions

[long obsidian]

Thank you I think there are 17 slow 2 groups.

I wanna compute the Galois group of f(x)=x^3-3x+1 over Q. I.e. take a splitting field and compute Gal(F/Q)

I figure that as the degree is 3 and it's potential only rational roots are 1 and -1, but neither are roots, so by gauss's lemma the polynomial is irreducible in Q. Then [F\Q]=degree(f)=3. Then by the correspondence |Gal(F\Q)|=[F\Q]=3. But Z/3Z is the unique group of order 3 up to isomorphism. So Gal(F/Q) is Z/3Z up to isomorphism

I saw a solution where someone said it was A_3 but I'm pretty sure they were just being fancy and thinking about how the roots are permuted, but in the end A_3 isomorphic to Z/3Z

I'm not sure if my reasoning is valid. If this is all correct then someone veryifying it for me would be appreciated.

[strong yacht]

An irreducible polynomial of degree n must have Galois group that is a transitive subgroup of S_n, and that's all you can say just from the degree. It is not true that its size must be exactly n: take x^3-2 whose Galois group is S_3, a group with six elements.

In other words you have more work to do to conclude that your Galois group is Z/3Z

[pastel cliff]

is there anything useful that comes out of quotienting K by the img(f) in a homomorphism f: G -> K

[lethal cipher]

I could really use help understanding this part of the proof. I don't understand why we can find subgroups of gal(L/Q) where G_i is normal in G_i-1 and they have index 2. This is not clear to me

Actually, the normal aspect just comes from solvability. The index 2 part is the one I'd like clarification on

[spiral wolf]

http://www.math.huji.ac.il/~arielweiss/wp-content/uploads/2019/03/Galois-theory-chapter-11.pdf

Consider giving this page a read -- in particular, it shows that an irreducible cubic has Galois group A3 when the discriminant is a square number, and S3 otherwise.

[dull root]

If I have an nxn matrix with all 1's in the diagonal, and all the other entries were 1/2, is there an easy way to calculate this determinate?

[spiral wolf]

Based off a few mathematica tests, it looks like the answer is (n+1)/2^n

But I couldn't tell you why... this post is the closest I could find
https://mathoverflow.net/questions/271379/on-the-determinant-of-a-class-symmetric-matrices

[lethal cipher]

<@&286206848099549185>

[dull root]

Double checking: If I have a matrix, I know i can add or subtract rows, but I can also add or subtract columns right??

and this is due to the fact that the transpose of a matrix has same determinant?

[spiral wolf]

You mean when doing row-reduction?

[dull root]

yes

[spiral wolf]

If you're solving a system of linear equations, then no

Column reduction will get you a different answer

If you just want the rank of the matrix or to find out if it's non-singular, then you're fine

Let $f(x) = x^5-x-1$. Show this polynomial is irreducible mod 5 and conclude it is irreducible over $\mathbb{Q}$.

[cloud walrus]

Espio

[spiral wolf]

I've done the first part, show it's irreducible mod 5, but I'm not sure how to conclude it's irreducible over Q. What theorem is relevant here?

[dull root]

I'm having trouble seeing that if I have a invertible Gram matrix, then the vectors or linearly independent.

[cursive temple]

cant you use gauss lemma

so it suffices to show irreducibility over Z

[spiral wolf]

For finite-dimensional real vectors in {\displaystyle \mathbb {R} ^{n}}\mathbb {R} ^{n} with the usual Euclidean dot product, the Gram matrix is {\displaystyle G=V^{\top }V}{\displaystyle G=V^{\top }V}, where {\displaystyle V}V is a matrix whose columns are the vectors {\displaystyle v_{k}}v_{k} and {\displaystyle V^{\top }}{\displaystyle V^{\top }} is its transpose whose rows are the vectors {\displaystyle v_{k}^{\top }}{\displaystyle v_{k}^{\top }}.

If the Gram matrix is invertible, then V is invertible also, and thus the vectors are linearly independent.

[dull root]

${\displaystyle \mathbb {R} ^{n}}\mathbb {R} ^{n} with the usual Euclidean dot product, the Gram matrix is {\displaystyle G=V^{\top }V}{\displaystyle G=V^{\top }V}, where {\displaystyle V}V is a matrix whose columns are the vectors {\displaystyle v{k}}v{k} and {\displaystyle V^{\top }}{\displaystyle V^{\top }} is its transpose whose rows are the vectors {\displaystyle v{k}^{\top }}{\displaystyle v{k}^{\top }}$

[cloud walrus]

MasakaBakana

[spiral wolf]

Hold up lol

I copied from Wikipedia

Basically G = V transpose * V, where V is the matrix whose columns are the vectors generating G

If the Gram matrix is invertible, then V is invertible also, and thus the vectors are linearly independent.

[dull root]

ahhh i seeee

yes, i didn't realize the gram matrix can be written like that

[spiral wolf]

So if my polynomial is irreducible mod 5, it is also irreducible over Z? Where does that come from?

[pastel cliff]

chinese remainder theorem should apply generally right? like for two ideals I,J of a ring R, it's true that $R/(I \cap J) \cong R/I \times R/J$?

[cloud walrus]

μ₂

[spiral wolf]

The post you replied to was weeks ago, just so you know

[pastel cliff]

i know

i was rereading a discussion i had with john a while ago

moreso just linking myself to it but asking something somewhat independent of it

[pastel cliff]

actually I and J would need some requirement right

like what's the equivalent of coprimeness for ideals

disjointness?

[cursive temple]

i just mean that it suffices to show irreducibility over Z

[spiral wolf]

Definitely, I’m just not sure how to show irreducibility over Z. The question asks to show irreducibility over Z/5Z first but I don’t know how to use that.

[cursive temple]

showing that reducibility over Z implies reducibility over Z/5Z seems nicer

[proud bear]

if G and K are just groups or rings, im(f) might not be a normal subgroups/an ideal. however, if G and K are abelian groups, vector spaces (or R-modules), K/im(f) is called the cokernel of f. it kind of "measures" the surjectivity of f in the same way the kernel measures injectivity of f

[pastel cliff]

wait when is im(f) not a normal subgroup

[proud bear]

f: Z/2Z->S_3 where 0->id, 1->(1 2) for example. i'm pretty sure {id, (1 2)} is not normal is S_3

[devout crow]

comaximality (ideals I and J of R are comaximal if I + J = R)

[strong yacht]

@spiral wolf yes it is true that irreducible mod p implies irreducible over Q (if the lead coefficient is not divisible by p), this is what your question means by "conclude"

[strong yacht]

Very nice in fact, I'm pretty sure just the same factorisation would work

[spiral wolf]

Gotcha. Does this theorem have a name or is it just what it is?

[ember field]

are F-algebras also F-modules??

[chilly ocean]

you tell me

[strong yacht]

To be honest with you, I'm not sure if it has a specific name. One thing to note: if irreducible over some mod p, then irreducible over Q, but the converse is NOT true. To memory: i think x^4+1 is reducible for every prime but irreducible over Q

[ember field]

I think it is true

[chilly ocean]

Hello, is it possible to find the solution of this equation analytically?

I'm assuming that it's possible with lambert w function but problem is that septic function

This is the primary form of the equation

I already found what I wanted by just plugging into wolfram alpha, but just curious

[strong yacht]

Sweet Jesus

[pastel cliff]

is that fucking johnny sins chilling in #groups-rings-fields

your last one was amazing bro

[strong yacht]

Thanks! I'm just trying to get in character for my next performance

[lethal cipher]

Disregard, I figured it out. It just had to do with the definition of solvabiliy, which was way easier than I was making it out to be

[toxic zephyr]

anyone have any examples of a ring homomorphism with domain Z[i] with the principal ideal of some gaussian integer as the kernel? say, ker(φ)=(1+i) for example?

[chilly ocean]

sure, take any principal ideal in Z[i] and consider the quotient map

[toxic zephyr]

true... i was looking to see if there were any less trivial ones because i wanted to do an example of utilizing the fundamental ring homomorphism theorem to find a ring isomorphic to Z[i]/(1+i)

at least im pretty sure thats using it correctly

[tribal moss]

You can explicitly define e.g. the map into Z/(2) that maps a+bi to 0 if a and b have the same parity, and 1 if the parity is different, and show from first principles that it's a homomorphism.

[next obsidian]

Shouldn’t Z[i]/(i + 1) = Z?

Like you’re just claiming that i = -1 now, but maybe it’s more complicated

I would do this, write this as

Z[x]/(x^2 + 1, x + 1)

It’s Z/2Z

Nvm

:)

Oh sure duh

i = -1

So i^2 = (-1)^2

So -1 = 1

[tribal moss]

And indeed (x²+1,x+1) contains (x²+1) - (x-1)(x+1) = 2.

[toxic zephyr]

can you explain what you mean by this? im a little lost

[next obsidian]

I mean when you mod out by (i + 1) you’re in effect setting i + 1 = 0

So in the quotient i = -1

This is just a way to get an idea as to what ring a quotient should be, this is why Z[x]/(x^2 + 1) ≈ Z[i], you’re saying x^2 = -1 aka x = i

Then you construct a map Z[x] -> Z[i] sending x to i and then apply 1st iso

In this case though because i satisfies the relation i^2 = -1 the map Z[i] -> Z sending i to -1 isn’t a ring homomorphism mainly because you’d need for 1 = -1

So instead you could think “oh it should go to Z/2Z” and the map is sending i to -1 = 1

And then show that this works or whatever

[next obsidian]

Or you do this computation

First you quotient by x+1, which says x = -1

So then you’re looking at Z/((-1)^2 + 1) = Z/(2)

[toxic zephyr]

im not sure where Z[x]/(x^2 + 1, x + 1) is coming from

if anything i would think its Z[x]/(x^2-2x+2)

[next obsidian]

Because Z[x]/(x^2 + 1) ≈ Z[i]

The specific isomorphism sends x to i

So now you apply the 3rd iso

You know that under this identification you’re looking at

(Z[x]/(x^2 + 1))/(x^2 + 1, x + 1)/(x^2 + 1)

The reason is that the ideal (i + 1) pulls back to (x^2 + 1, x + 1) under the map Z[x] -> Z[i] sending x to i

[tribal moss]

There's one of the isomorphism theorems that says (R/I)/f(J) = R/(I+J) where f is the quotient map R -> R/I.

This can be used to swap around quotienting by x²+1 and x+1.

[toxic zephyr]

is this material usually taught somewhat early on in an abstract algebra course?

bc none of this sounds familiar and im wondering if its because my class is being taught out of order

[next obsidian]

Not really

It’s not usually taught like this, you figure it out or someone explains it to you

At least that’s how I learned this

And I bet lots of others

[tribal moss]

Yeah, sounds about that.

I'm sure textbook authors try to make it such applications somewhat clear when they present quotients and the isomorphism theorems, but it's one of those things that just need a fair amount of examples to gel -- and the isomorphism theorems tend to come so early that you haven't seen many relevant example yet at that point.

[ember field]

If there is a field extension of degree m*n, when does there exist an extension of degree m in between?

[barren sierra]

Ok so I feel with the theorems in the second picture the problem in the first picture should be easy but idk I don't know where to start

I don't really get the relevance of the fact that L/F is a splitting field

I feel (b) => (a) is just immediate from (3)?

[lethal cipher]

Okay, so I have a question. Say there is a chain of field extensions $F\subset K_1 \subset K_2 \subset L$, and let $M$ be the Galois closure. If $\sigma \in Gal(M/F)$, is it true that $[K_2:K_1]=[\sigma(K_2):\sigma(K_1)]$? If so, why?

[cloud walrus]

dackid

[lethal cipher]

Okay, it should be good now.

[lethal cipher]

It's true. I was able to prove it to myself

[barren sierra]

anyone have any ideas?

[grand sigil]

Probably has something to do with every element of f being written as a linear combination

I no sure

[barren sierra]

hm

[chilly ocean]

Is every irreducible f(x) separable?

[lethal cipher]

It depends on the characteristic of F

The only time you'll run into issues is when char(F)=p where p is prime

[cloud walrus]

dackid

[lethal cipher]

Hope that helps

[next obsidian]

x^p - t is irreducible but not separable over F_p(t)

In fact the roots are all the same

[muted breach]

hey instead of blocking text out, you could just use parentheses

[chilly ocean]

Wdym

[muted breach]

oh just that a few days ago i also saw you posting a screenshot of a book and blocking irrelevant text out and i thought

you could just use parentheses to indicate the relevant part of the text

then less effort i guess

well i'm assuming that's what you're trying to do lol

[devout crow]

how should I think of the quotient ring $$\frac{C(\bR)}{\langle \mathrm{id} \rangle}$$? I know it's equivalence classes of $f \sim g \iff f-g$ is linear, but what does this ring look like?

[cloud walrus]

xdres

[chilly ocean]

are you sure that's the equivalence relation here?

[devout crow]

oh wait no it'll just be "has a root at 0"

so the elements of the quotient are just determined by their value at 0? so it's iso to R?

[chilly ocean]

i'm not sure about that. you're saying that if f has a root at 0, then f(x) = xh(x) for some continuous function h. but what if f is not differentiable?

if you were instead working with C^\infty(R) or even the ring of analytic functions, i'd see it, but i don't believe it here

[devout crow]

hmm yeah true, e.g. cbrt(x) I guess?

[chilly ocean]

that works

[tribal moss]

The ideal is exactly the functions where f(0)=0 and f'(0) exists.

[ember field]

if a field has no finite extensions does that make it alg. closedd ??

[tribal moss]

Yes; a polynomial without a root always gives rise to a finite extension.

[deep sky]

I’m still getting used to finding and adequately labeling quotient rings. Is it appropriate to describe Z[x]/(2x^2 -4, 4x-5) as F_2 for terms of degree 2 or higher, F_4 for the linear term, and just regular Z for the constant term?

[barren sierra]

Anyone have any ideas? I took another stab at it and made no progress

[tribal moss]

No. The ideal contains 8(2x²-4) = 16x²-32 and also contains (4x-5)(4x+5) = 16x²-25. Subtracting these, we see that the ideal contains 7, so unless something collapses it further down to (1), everything will end up modulo 7.

(I found this by attempting polynomial division of 2x²-4 by 4x+5, but multiplying the numerator by whatever is necessary to keep the coefficients of the quotient polynomial integers).

Then, since we're working modulo 7, we can multiply 4x+5 by 2 to get 8x+10 which the same as x-4 modulo 7, so x=4 in the quotient ring.

And therefore the quotient ring is just Z/7Z.

(We can check that 2x²-4 indeed maps to zero in Z/7Z when we send x to 4).

[deep sky]

This maybe just indicates I don’t fully understand what’s allowed and not allowed, but does the quotienting by the 4x-5 alone at least bring us to F_4 for all terms besides the constant?

My understanding is based entirely on converting any instance of a coefficient 4 or higher to a lower degree coefficient of 5, and then similar for any 2 coeff to a 4 two degrees lower

by setting one of the monomials in each divisor equal to the other one

[tribal moss]

Not really. That would be the case if we were quotienting by 4x instead of 4x-5.

[deep sky]

But isn’t the difference just that the modulus causes a roll over in the constant term’s value

[tribal moss]

Well yes, but because that difference is there you can't just say that the x^1 coefficient works modulo 4, because that would mean, for example, that 1x and 5x are the same thing.

[deep sky]

Yeah, I guess I was glossing over the carry-over arithmetic

Would it make sense to describe it as F_4[x] with the property of any multiple of 4 adding 5 to the term one degree under, or at that point is the arithmetic different that it’s silly to phrase it with reference to F_4, and you should just describe the elements from scratch?

[tribal moss]

It's very much different. As soon as we have a polynomial with nonzero constant term in the ideal, all of the nice levels in the original polynomial ring begin to blend together, and we can no longer speak meaningfully about the degree of a term or coefficient.

Also beware that F_4 does not mean what you think it means.

You mean Z/4Z, the integers modulo 4.

F_4 means the field with 4 elements, which is a different ring.

[deep sky]

Even when you adjoin x to it?

Ohhh they’re only equal when n from F_n is a prime right?

[tribal moss]

Yes, F_p is the same a Z/pZ when p is prime.

[deep sky]

Got it, thank you!

[barren sierra]

dumb question

part c should say sqrt(D) right?

not sqrt(2)?

[sharp sonnet]

yeah

[barren sierra]

so these are the automorphisms I've found

I just don't know how to get the corresponding fixed fields

[barren sierra]

nvm I used this

so I conjectured $\mathbb{Q}(\zeta_8) = \mathbb{Q}(i, \sqrt{2})$

[cloud walrus]

Spamakin🎷

[lethal cipher]

That's a pretty good conjecture

[barren sierra]

ya

now I just gotta find minimal polynomials => calculate traces

[barren sierra]

Actually wait

[barren sierra]

What does this formula have to do with trace from linear algebra

[lethal cipher]

It looks like it's because it's the sum of the independent basis vectors @barren sierra , which seems to be similar to taking the sum of the diagonals in the matrix

[barren sierra]

Hm I see

[lethal cipher]

That's what I see at face value at least.

[barren sierra]

ok so intermediate fields are $\mathbb{Q}(i), \mathbb{Q}(\sqrt{2}), \mathbb{Q}(\sqrt{-2})$

[cloud walrus]

Spamakin🎷

[barren sierra]

The respective minimal polynomials are $x^2 +1, x^2 - 2$ and $x^2 + 2$. But then by part (d) all of these have trace 0 since $a_{d - 1} = 0$ for all of these.

[cloud walrus]

Spamakin🎷

[barren sierra]

is that right? that feels wrong.

I guess by part (a) we can't have that $\mathbb{Q}(\alpha) = K$ since $\text{Tr}_{K/\mathbb{Q}}(\alpha) \in \mathbb{Q}$

[cloud walrus]

Spamakin🎷

[lethal cipher]

Question, what is Emb_F(K,L) in your definition?

[tough raven]

It turns out to be equal to the trace of “multiplication by α” viewed as a F-linear map on the F-vector space K.

I think

[barren sierra]

the set of field homomorphisms from K into L that fix F

[lethal cipher]

Hmm, so why is that not the same as the Galois Group of L over F?

[barren sierra]

cause not every homomorphism is an automorphism

also you can't compose them without restriction

[chilly ocean]

Do kernels in the category of unital rings exist?

[hidden haven]

There is no 0 object, so you can't define kernels

[chilly ocean]

But you have initial objects

Which is Z

[hidden haven]

How do you use that to define kernel?

[chilly ocean]

Just replace the 0 object by initial object in the definition

[hidden haven]

Then there may not be a 0 map between A and B for arbitrary rings

Are you not defining kernel of f as equalizer of f and 0?

[chilly ocean]

Yes, but kernel could still exist, for example any morphism has a cokernel which is always trivial

[hidden haven]

No, you can't say equalizer of f and 0 exists if there is no 0 map

Z/2Z → Z/2Z there cannot exist a 0 map here

[chilly ocean]

I'm defining kernel using the definition on nlab, not kernel pair

So it's a certain pullback

[hidden haven]

oh

Ye then kernels exist because the category of unital rings has all pullbacks

[chilly ocean]

Oh. How do they look like

[hidden haven]

Look at the construction of pullbacks as subobjects of products

Like for the map R → zero ring, the kernel would be R x Z

Which looks weird lmao

[chilly ocean]

Oh. So R x Z but quotient of it

No wait

[hidden haven]

Quotients appear in pushouts

[chilly ocean]

{(r, n) : f(r) = n•1_S} right

[hidden haven]

yes

[chilly ocean]

Thank you for your help

[hidden haven]

[runic hemlock]

I'm trying to prove that $|\text{Aut}(L/K)|$ divides $[L:K]$ for any finite extension $L/K$, but have no progress. Any ideas?

[cloud walrus]

Porphyrion

[runic hemlock]

I'm avoiding using Gal because L/K isn't Galois.

[thorn delta]

i think you can measure the failure of |Aut(L/K)| = [L : K] using the fixed field of Aut(L/K)

L will be galois over the fixed field of Aut(L/K), so things are nicer

actually hmm

[runic hemlock]

oh yeah that solves

Thanks!

[thorn delta]

ah okay nice, npnp

[runic hemlock]

This follows from Grothendieck's Generic Freeness Lemma. Assume that $m$ doesn't intersect $\bZ-{0}$. Then by the lemma there is a nonzero integer $x$ so that $K_x=K$ is a free $\bZ_x$-module (this is localization). But it's also a vector space over $\bQ$, and it's not hard to show that a nonzero $\bQ$-vector space isn't free over $\bZ_x$ (for example, in a free $\bZ_x$ module you can't divide certain elements by a prime $q\nmid x$)

So $m\cap \bZ\neq {0}$, which implies $K$ has positive characteristic, so by regular Nullstellensatz it is a finite extension of $\bZ/p$ hence finite .

[cloud walrus]

Porphyrion

Porphyrion

[modern hawk]

hi, quick question, I got told that this wasn't a proper definition, could someone help me make it a proper one as I'm a little unsure what is wrong with it.

The \textit{Number Theoretic Transform} is a generalized form of the \textit{Discrete Fourier Transform} over the ring $R_q = \bZ_q[X] / \langle X^n + 1 \rangle$. NTT is defined as the following: let $\omega$ be the $n$th root of unity for the polynomial

[Xi = \sum^{n-1}{j=0} x_j \cdot \omega^{ij}. ]

[cloud walrus]

fisu

[dull root]

I have this following question: So if I have a field F_2, it is clear to me that the polynomial x^4 + x + 1 is irreducible over F_2. However, how can I see that it is reducible over F_4?

So, I know that if F_4 is a simple extension of F_2 by a root of a irreducible deg 2 polynomial over F_2.

The only irreducible deg 2 polynomial over F_2 is x^2 + x + 1, so if I take a root, ie I take alpha such that alpha^2 = alpha + 1, then F_2(alpha) = F_4. However, how do I find the irreducible polynomials of small degrees (deg 1,2,3) in F_4? So in this notation above, F_4 = {0,1,\alpha, \alpha^2 = \alpha + 1}

[stoic rose]

What does the last sentence mean? "let omega be the nth root of unity for the polynomial ...", that sentence doesn't make sense gramatically.

[stoic rose]

One way to see that it cannot be irreducible in F_4 is to note that the polynomial splits in F_16, which has only degree 3 over F_4

[dull root]

That works, but I also want to get practice trying to find irreducible polynomials of small degree too. It was easy to do in F_2, but I don't see how to do it in other finite fields

[stoic rose]

For degree 2 and 3 irreducible is equivalent to having no roots, so you just have to check for all elements of F_4 that they are not a root.

[dull root]

Any deg 1 poly is irreducible trivially. Over F_4, there are 3^2 = 9 of them right?

sicne the coefficeints can be 1, alpha, or alpha^2

[stoic rose]

The constant coefficient could be 0 so 3*4

[dull root]

nvm i see what you are saying

but sometihng ike alpha x + alpha is not irreducible

[stoic rose]

To be honest though, it's pretty rare to work with explicit elements of finite fields other than Z/pZ, it's usually much more convenient to use Galois theory arguments.

[stoic rose]

It is

Since alpha is a unit

[dull root]

But x+1 divides alpha x + alpha

[stoic rose]

Yes but they differ by a unit. Irreducible means that whenever f=ab either a or b is unit.

Just like 2 is divisible by -2 in Z but it's still a prime number

[dull root]

i see

[stoic rose]

Or X+1 is divisible by (X+1)/2 in Q[X] but it's still an irreducible polynomial

But we often restrict to monic polynomials though to get a single representative for each equivalence class of (multiplying by units)

[dull root]

So if we only look at irredicble monic polynomaisl over F_4 of degree 1, There is

x, x+1, x + \alpha, x+ alpha^2, only four right?

[stoic rose]

Exactly

[dull root]

The product of any two of these gives a reducible poly of degree 2, and any other monic polys of degree two I find are automatically irreducible?

[stoic rose]

Yes

[dull root]

Eariler you said deg 2 irreducible poly over F_4 is equivalent to having no roots. I don't see why that is the case

Over F_2, I see that

[stoic rose]

That's true over any field

[dull root]

oh yea thats a brain fart

[stoic rose]

Because if it's reducible it has a degree 1 factor and thus a root

[dull root]

because a deg 2 can only split as 2 deg 1

yes

[stoic rose]

Same for degree 3

[dull root]

not for 4, since 4 can split as 2 deg 2

So over F_2 we could list out those deg 2 polynomails that dont have a root by making sure they have an odd number of terms. This only gives x^2 + x + 1

Since the only two coeffiiencts are 1 and 0

But F_4 doesnt have a similar criterion

[stoic rose]

Not really, you have to check for each of the four elements of F4 if it's a root

[dull root]

So for example, if I considered this reducible deg 2 polynomial (x+1)(x + alpha^2) = x^2 + (alpha + 2) x + alpha + 1, I don't see what this coefficient alpha + 2 in F_4 is. F_4 = {0, 1, alpha, alpha^2 = alpha +1} so all the other coefficients are in there

[stoic rose]

Remember that 2=0

[dull root]

Since the char(F_4) = 2, we have 2 = 0 right?

[stoic rose]

Yes

[dull root]

I see that x^4 + x + 1 must factor into 2 degree 2 irreducubile monic polynomials of F_4. However, there are 6 such irreducible monics polynomials over F_4, so 36 total combinations... I know what all 6 irreduicble polynomials look like, but how do I see which ones factor x^4 + x + 1 without manually checking all 36 combinations?

[stoic rose]

For each of the 6 irreducible monic polynomials you can check if it divides x^4 + x + 1 using polynomial long division

And then if it does you automatically get the other factor

[dull root]

Is -1 =1 in a field of power 2?

[stoic rose]

You mean characteristic 2?

[dull root]

yea

[stoic rose]

Yes

[dull root]

ty, it was still painful to go through them, but it worked