- #1

mathbalarka

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It is pretty well-known that there are distinct fields over $\Bbb Q$ with isomorphic Galois groups. Take, for example, $K_1 = \Bbb Q(\sqrt{2})$ and $K_2 = \Bbb Q(\sqrt{5})$. Both have Galois groups of order $2$ as easily can be seen and there is only one group of order $2$ which is $\Bbb Z_2$.

Now take some algebraically closed field $k$. I suspect that if two field extensions $K/k(t)$ and $K'/k(t)$ are Galois over $k(t)$ with isomorphic galois groups, it's necessary that $K = K'$. Is this true?

In particular, take up $\Bbb C(z)$ and let $$\mathcal{G}(K/\Bbb C(z)) \cong \mathcal{G}(K'/\Bbb C(z))$$ Is it possible to prove/disprove that $K = K'$?

Here is a possible way to do it using non-classical galois theory : $\Bbb C(z)$ forms the function field of meromorphic functions of single variable over $\Bbb P^1$, thus any Galois group of some Galois extension is realized as monodromy of coverings $X \to \Bbb P^1$ and $X_0 \to \Bbb P^1$. These are both galois coverings, so $\Bbb P^1 \cong X/G$ and $\Bbb P^1 \cong X_0/G$ G being the monodromy of both (the galois groups are isomorphic). So $X/G \cong X_0/G$. Now I just have to show that $X \cong X_0$, but I am not sure whether that holds. So I am rather partial to a standard classical approach to this problem.

Any help is appreciated

Balarka

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