For a finite extension $K/F$ of degree $>1$, prove that $K$ is not the union of finitely many proper intermediate fields.

Let $K/F$ be a finite extension of degree $>1$. Prove that $K$ is not the union of finitely many proper intermediate fields.

Since $K/F$ is a finite extension, we can write $K=F(a_1,a_2,\ldots,a_n)$ for some $a_i\in K$. First consider the case $|F|=\infty$. Then $a_1+a_2+\cdots +a_n \in K$ which does not belong to any of the proper intermediate fields since otherwise $K=F(a_1+a_2+\cdots +a_n)\subset M$, where $M$ is an intermediate field. A contradiction. So $K$ is not the union of finitely many proper intermediate fields.

Can anyone show me how to proceed in the case of $|F|<\infty$.

Thanks

Answer

$\newcommand{\Q}{\mathbb{Q}}$$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$I do not think your argument is quite correct. For instance for $F = \Q$ and $K = \Q(\sqrt[4]{2}, i \sqrt[4]{2}, -\sqrt[4]{2}, -i\sqrt[4]{2})$, but $\sqrt[4]{2}+ i \sqrt[4]{2} -\sqrt[4]{2} -i\sqrt[4]{2} = 0$.

Or to take Crostul's example $K = \Bbb F_p(x,y) / \Bbb F_p(x^p,y^p)$, with $F = \Bbb F_p(x^p,y^p)$, we have $K = F(x, y)$, but $K/F$ is not a simple extension, so that $F(x + y) \subsetneq K$.

Rather, in the case when $F$ is infinite, proceed by induction on $\Size{K:F}$.

If there are no intermediate $L$ such that $F \subsetneq L \subsetneq K$, we are done. This covers the induction basis when $\Size{K:F}$ is $1$, or a prime number.

Otherwise, let $L$ be a maximal (with respect to inclusion) such intermediate extension. By induction, $L$ is not a union of proper subfields. Let $a \in L$ be an element which is not in any proper subfield of $L$.

It follows that $L$ is the only proper subfield of $K$ in which $a$ is contained. For, if $a \in N$, with $N \ne L$ a proper subfield, then $N$ does not contain $L$, as $L$ is maximal, so $N \cap L$ is a proper subfield of $L$, and it contains $a$.

Let $b \in K \setminus L$.

Now $K$ is a union of finitely many proper intermediate fields $M_{1}, M_{2}, \dots, M_{n}$, and each of the infinite elements $a + \lambda b$, for $\lambda \in F$, lie in one of them. Then there must be a proper intermediate fields $M = M_{i}$, for some $i$, which contains infinite elements of the form $a + \lambda b$, for $\lambda \in F$, It follows that $a, b \in M$. Thus $a \in M \ne L$, a contradiction.