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Let $E/F$ be a finite Galois extension. Let $K$ be a function field of transcendence degree one over $F$. Let $KE$ be the compositum of $K$ and $E$. Why is $KE/K$ also finite and Galois?

Also, why is $[KE:K]\leq [E:F]$?

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Also, why is $[KE:K]\leq[E:F]$? –  Alan Lee Mar 2 '12 at 16:16
Why put the question in a comment, instead of in the body? –  Arturo Magidin Mar 2 '12 at 17:14
P.S. What you are considering is better described as a "lift" than a compositum. –  Arturo Magidin Mar 2 '12 at 17:41
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1 Answer

Since $E$ is finite Galois over $F$, it is the splitting field of some separable polynomial $f(x)\in F[x]$. Let $\alpha_1,\ldots,\alpha_n$ be the roots of $f(x)$; then $E=F[\alpha_1,\ldots,\alpha_n]$.

Now, $f(x)$ is also a separable polynomial over $K[x]$ (since $\gcd(f,f') = 1$ over $F[x]$, hence over $K[x]$). And we have $$KE = K(F[\alpha_1,\ldots,\alpha_n]) \subseteq (KF)[\alpha_1,\ldots,\alpha_n] = K[\alpha_1,\ldots,\alpha_n]\subseteq KE,$$ hence $KE$ is the splitting field of $f(x)$ over $K$ (it is generated by the roots of $f(x)$ over $K$). Thus, $KE$ is Galois over $K$.

By the same argument, if $a_1,\ldots,a_n$ is an $F$-basis for $E$, then $E\subseteq K[a_1,\ldots,a_n]$, hence $KE\subseteq K[a_1,\ldots,a_n]$. Thus, $\dim_K(KE)\leq \dim_K(K[a_1,\ldots,a_n])\leq n$, so $$[KE:K] = \dim_K(KE) \leq n = [E:F].$$

Note. The fact that $K$ is of transcendence degree $1$ over $F$ is irrelevant. If $E/F$ and $K/F$ are field extensions contained in a common field $L$, then $[KE:K]\leq [E:F]$ always holds; and if $E/F$ is algebraic and Galois, then so is $[KE:K]$. You don't need to assume finiteness of $E/F$, as you can replace the single polynomial $f(x)$ with a family of polynomials.

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Under what circumstances is $[KE:K]=[E:F]$? –  Alan Lee Mar 2 '12 at 18:45
@Alan Lee: This will happen if $K$ and $E$ are linearly disjoint; I think that if you let $L$ be the algebraic closure of $F$ in $K$, then you need $[LE:L]=[E:F]$. –  Arturo Magidin Mar 2 '12 at 19:05
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