1

Let $L/Q$ be a finite Galois extension, the fundamental theorem shows that there is a one-to-one correspondence between $H\subset Gal(L/Q)$ and $L^H \subset L $. My question is, suppose $S \cong H $, is it necessarily true that $L^H \cong L^S$? This seems to be a correct statement to me, but I don't see any immediate proof. I was trying to write $L = Q(\alpha_i)$ and fix $\sigma \in H$ with $f$ denoting the isomorphism between $H$ and $S$. Then if $\sigma = (\alpha_1\alpha_2\alpha_3...\alpha_n)$ and $f(\sigma) = (\beta_1\beta_2\beta_3...\beta_n)$, we can construct the map $\phi$ that maps $\alpha_i$ to $\beta_i$. But I'm not sure how to make this part rigorous. Any helps?

The One
  • 834

1 Answers1

4

No.

In $\Bbb Q[\sqrt 2,\sqrt 3]$, we have the non-isomorphic subfields $\Bbb Q[\sqrt 2]$ and $\Bbb Q[\sqrt 3]$ (In one of them, the polynomial $X^2-(1+1)$ has a root, in the other it doesn't), but the corresponding subgroups are isomorphic (cyclic of order $2$).

Hagen von Eitzen
  • 374,180
  • just curious; what if $L$ is a splitting field of $Q$ for some irreducible and separable polynomial $p(x) \in Q[x]$? Then all the roots come from the same polynomial, and the example you described is avoided. Are the subfields isomorphic in that case? – The One May 09 '21 at 20:21