Mathematics

@arctictern So, suppose we consider $x = (1 2 3 4)$ then we need to construct a y such that $yxy{-1} = x^{-1}$ so we constructed and got $y = (1 4)(2 3)$ we consider $H = <x>$ and $K = <y>$. Then $HK = H \rtimes K = Z_4 \rtimes Z_2 = D_8$ so we are done.

Ted Shifrin

I am not remotely surprised that tern got an answer correct :D

arctic tern

@Steve if that plane were literally the xy-plane and those two vectors were (1,0) and (0,1), you'd be able to tell just by looking right?

Adeek

By construction the semi-direct product isn't trivial

arctic tern

5:21 AM

@Adeek looks fine. (you mean only one 2-sylow up to conjugacy I presume)

Adeek

yeah @arctictern

Steve

@arctictern oh lol

Mike Miller

@PVAL I see. That field is indeed somewhat troubled. I'm happy to be somewhere that feels safer.

Ted Shifrin

Sleep well, @MikeM. Bye for now, all.

Adeek

cya @TedShifrin

usukidoll

5:22 AM

that's enough abstract algebra for me today

I'll do the rest tomorrrow evening x.x!

Adeek

@arctictern if we have $f(x) = x^4 + ax^2 + b \in Q[x]$ where $b \neq 0$ then if $\alpha$ is a root then $\sqrt(b) / \alpha$ is a root ?

arctic tern

yes

Adeek

why ? is it just algebra ?

regular algebra I mean ?

arctic tern

If $u$ is a root of $u^2+au+b$ then $b/u$ is another root. apply with $u=\alpha^2$

vieta's

Adeek

you mean sqrt(b) / u ?

arctic tern

5:33 AM

no

$\alpha$ is a root of $x^4+ax^2+b$
$\alpha^2$ is a root of $u^2+au+b$
$u^2+au+b$ factors as $(u-\alpha^2)(u-b/\alpha^2)$
$x^4+ax^2+n$ factors as $(x^2-\alpha^2)(x^2-b/\alpha^2)$
so $\sqrt{b/\alpha^2}=\sqrt{b}/\alpha$ is another root of $x^4+ax^2+b$

Adeek

I see

thanks

Adeek

5:46 AM

@arctictern if the polynomial f(x) is given as before I want to prove the galois group is either $\{e\},Z_2,Z_4 ,Z_2 x Z_2, or D_8$

is there a fast way to do it ?

arctic tern

cases: (a) u^2+au+b=(u-p)(u-q) so splitting field is sqrts of p and q adjoined to Q so extension has degree 1,2 or 4 so galois group has size 4. (b) u^2+au+b is irreducible, roots are p+/-sqrt(q) (go from there)

Adeek

oke thanks

@arctictern maybe I have an idea. Does the field extension $Q(\alpha,\sqrt(b))$ contain all the roots of f(x) ?

arctic tern

it is the splitting field yes

Adeek

why ?

I mean it contain the roots $\alpha$ and $-\alpha$ and $\sqrt(b) / \alpha$ but why does it contain all roots ?

arctic tern

6:02 AM

If $\alpha$ is a root then the polynomial factors as $(x+\alpha)(x-\alpha)(x+\sqrt{b}/\alpha)(x-b/\sqrt{\alpha})$. all roots are expressible in terms of $\alpha$ and $\sqrt{b}$ and conversely $\alpha$ and $\sqrt{b}$ are expressible in terms of roots.

Adeek

how did you factor this so quickly ?

arctic tern

did you read the multiline derivation I gave above?

combine that with difference of squares

Adeek

I see

I see very nice

Adeek

6:17 AM

@arctictern I guess $[Q(\alpha) : Q] = 1,2,3 \ or\ 4$ can we eliminate 3 ?

arctic tern

yes

Adeek

I mean if we can eliminate 3 then I think it follows smoothly

why does it follow smoothly ?

arctic tern

the splitting field is a tower of two quadratic extensions

first one adjoins a root of u^2+au+b, then one adjoins the square root of the first thing we adjoined

Adeek

oh

oh I see we could also do it using that thing we proved before that irreducible poly of degree 3 is either all real or 1 real and 2 complex.

no that doesn't work

your method is better

Adeek

6:45 AM

@arctictern can you give me an example of polynomial of degree 4 with trivial galois extension ?

x^4

with $b \neq 0$

(x^2-1)^2

this has trivial galois group ?

arctic tern

it factors over Z so...

Adeek

6:47 AM

I see

Adeek

7:20 AM

@arctictern I was trying to find polynomials of degree 4 which gives galois group $\{e\},Z_2,Z_2xZ_2,Z_4$ I found all but I can't seem to find one for $Z_4$

nvm I found it

$f(x) = x^4 + 4x^4 + 2$