# galois theory

• Feb 21st 2008, 02:34 AM
edgar davids
galois theory
Can someone pls pls pls help me with this questions asap! Thanks so much

Edgar

Let L = C(s,t) where s and t are indeterminates (or more formally L is the field of fractions of C[s,t]). Let Φ: L → L be given by Φ(s) = t, Φ(t) = s, and let G be the group generated by Φ. Find the fixed field of G, justifying your answer.
• Feb 24th 2008, 09:24 PM
ThePerfectHacker
Quote:

Originally Posted by edgar davids
Let L = C(s,t) where s and t are indeterminates (or more formally L is the field of fractions of C[s,t]). Let Φ: L → L be given by Φ(s) = t, Φ(t) = s, and let G be the group generated by Φ. Find the fixed field of G, justifying your answer.

I never studied anything about these types of problems before so this is just a guess.

Say that,
$\alpha = a_0^0 + a_1s+...+a_n^0 s^n$
$a_0^1t+a_1^1 ts+...+a_n^1 ts^n$
....
$a_0^n t^n + a_1^n t^n s+..+a_n^n t^ns^n$.

It must have this form because, it cannot be that degree of t exceedes degree of s for that would make the polynomial of degree s exceedes degree of t which means it cannot be the same after applying the automorphism $\theta$ to it.

Now this value stays unchanged under $\theta$. The values across the diagnols can be anything while $a_k^j = a_j^k$.

For example,
$2+s+6s^2$
$t+5st+3s^2t$
$6t^2+3t^2s+9t^2s^2$

We can pair it as,
$2+(s+t)+6(s^2+t^2)+ 5st+3st(s+t)+9(st)^2$.

This seems to suggest that,
$L^G = C(st,s+t,s^2+t^2,s^3+t^3,...)$

But it turns out that $C(st,s+t.s^2+t^2,...) = C(st,s+t)$.

Because $s^n+t^n$ can be obtained from $st,s+t$ for $n\geq 2$. For example, $s^2+t^2 = (s+t)^2 - 2st$. And $s^3+t^3 = (s+t)^3 - 3st(s+t)$. And so on.

Thus. the fixed field under the group $G$ is $L^G = C(st,s+t)$.