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 $F_1$, $F_2$ and $K$ be fields of characteristic $0$ such that $F_1 \cap K = F_2 \cap K = M$, the extensions $F_i / (F_i \cap K)$ are Galois, and $[F_1 \cap F_2 : M ]$ is finite. Then is $[F_1 F_2 \cap K : M]$ finite?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

No. First, the extension $\mathbb{C}(z)/\mathbb{C}$ is Galois for any $z$ that is transcendental over $\mathbb{C}$: if $p(z)$ is a nonconstant polynomial, and $\alpha$ is a root and $\beta$ is a nonroot, then the automorphism $\sigma\colon z\mapsto z+\alpha-\beta$ maps $z-\alpha$ to $z-\beta$, so that $p(z)$ has a factor of $z-\alpha$ and no factor of $z-\beta$ but $\sigma p$ has a factor of $z-\beta$, so $p\neq\sigma p$. Thus, no polynomial is fixed by all automorphisms. If $\frac{p(z)}{q(z)}$ is a rational function with $q(z)$ nonconstant, then a similar argument shows that we can find a $\sigma$ that "moves the zeros" of $q$, so that $p(z)/q(z)$ will have a pole at $\alpha$ but no pole at $\beta$, while $\sigma p/\sigma q$ has a pole at $\beta$. Thus, the fixed field of $\mathrm{Aut}(\mathbb{C}(z)/\mathbb{C})$ is $\mathbb{C}$, hence the extension is Galois.

Now, let $x$ and $y$ be transcendental over $\mathbb{C}$. Take $F_1=\mathbb{C}(x)$, $F_2=\mathbb{C}(y)$, $K=\mathbb{C}(xy)$, all subfields of $\mathbb{C}(x,y)$. Then $M=\mathbb{C}$, so $F_i/M$ is Galois; $F_1\cap F_2=\mathbb{C}$, so $[F_1\cap F_2\colon M]=1$. But $F_1F_2\cap K = \mathbb{C}(x,y)\cap \mathbb{C}(xy) = \mathbb{C}(xy)$, and $\mathbb{C}(xy)$ is of infinite degree over $M=\mathbb{C}$

share|improve this answer

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.