Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $[KE:K]\leq [E:F]$?

share|improve this question
    
Also, why is $[KE:K]\leq[E:F]$? –  Alan Lee Mar 2 '12 at 16:16
1  
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
add comment

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.