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Let $K$ be a field of characteristic zero and $K(x)$ the field of rational functions with coefficients in $K$. Let $K(u)$ denote the subfield of $K(x)$ generated by $u \in K(x)$ over $K$.

My questions are following:

  1. Suppose that $u=x^{2}$. Prove that $K(x)/K(x^{2})$ is Galois and find all elements of the Galois group Gal$(K(x)/K(x^{2}))$.

My attempt: Let $f(X)=X^2-u \in (K(u))[X]$. $f(X)$ is linear, and thus irreducible in $(K[X])[u]=(K[u])[X]$. Gauss’ Lemma tells us that the polynomial $f(X)$ is also irreducible over $(K(u))[X]$. Moreover, $f(X)=(X-x)(X+x)$ then $f(X)$ is separable over $k(u)$ and $\pm x$ are roots of $f(X)$. This means that, $K(x)$ is the splitting field for separable polynomial $f(X)$ over $K(u)$. Therefore $K(x)/K(x^{2})$ is Galois. In addition, $[K(x):K(x^{2})]=2$. Can we conclude that Gal$(K(x)/K(x^{2}))\simeq \Bbb Z/2\Bbb Z$? I am aware that Gal$(K(x)/K) \simeq PGL_2(K)$ but I couldn't figure out what are exactly elements of Gal$(K(x)/K(x^{2}))$?

  1. Suppose that $u=x^{2}-x$. Prove that $K(x)/K(x^{2}-x)$ is Galois and find all elements of the Galois group Gal$(K(x)/K(x^{2}-x))$.

My attempt: Similar to that in above discussion, I found that $K(x)$ is the splitting field for separable polynomial $g(X)=X^2-X-x^2+x$ over $K(x^{2}-x)$. Therefore $K(x)/K(x^{2}-x)$ is Galois and we can write Gal$(K(x)/K(x^{2}-x))\simeq \Bbb Z/2\Bbb Z$? The similar question arises, for $\sigma\in $ Gal$(K(x)/K(x^{2}-x))$, can we determine $\sigma$?

  1. Prove that $K(x^{2}) \cap K(x^{2}-x)=K$.

It is clear to me that $K \subseteq K(x^{2}) \cap K(x^{2}-x)$. But I can't prove the reverse inclusion.

Any help would be much appreciated.

user26857
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user
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    I am sure that you find the discussion in this related thread helpful. – Jyrki Lahtonen Apr 14 '16 at 10:08
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    Considering that the arguments for parts 1 and 2 are correct, I'd close this as a duplicate of the question linked above (which has $x^2+x$, but the change is irrelevant). What do you think? – egreg Apr 14 '16 at 10:53
  • Yes, you can close this if necessary. Actually I was searching for answers on our site but I couldn't find any. Jyrki's suggestion is really helpful. – user Apr 14 '16 at 11:14

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