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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

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2 Answers 2

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For finite $F$, you have the following result:

If $F$ is a finite field, and $K$ is a finite extension of $F$, then there exists $a \in K$ such that $K=F(a)$.

This means that there is no intermediate field containing $a$. Thus, any union of intermediate fields does not contain the element $a$ (hence it does not contain $K$).

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  • $\begingroup$ So essentially there is no difference between the case $|F|=\infty$ and $|F|<\infty$? $\endgroup$
    – pie
    Mar 30, 2018 at 10:59
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    $\begingroup$ My argument works for all simple extensions. However, there exist non-simple extensions, such as $\Bbb F_p(x,y) / \Bbb F_p(x^p,y^p)$, for $p$ a prime number. In such extensions, you need a different argument. In particular, the proof you posted in your question does not work. $\endgroup$
    – Crostul
    Mar 30, 2018 at 11:04
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$\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.

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  • $\begingroup$ But $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})$. $\endgroup$
    – pie
    Mar 30, 2018 at 10:53
  • $\begingroup$ @pie, sorry, very bad example, I have changed it. $\endgroup$ Mar 30, 2018 at 10:58
  • $\begingroup$ But why write $K=\mathbb{Q}(\sqrt[4]{2}, i \sqrt[4]{2})$ as $\Q(\sqrt[4]{2}, i \sqrt[4]{2}, -\sqrt[4]{2}, -i\sqrt[4]{2})$ since $\{\sqrt[4]{2}, i \sqrt[4]{2}, -\sqrt[4]{2}, -i\sqrt[4]{2}\}$ has to be a basis for $K$, in order to write like that, but its not. $\endgroup$
    – pie
    Mar 30, 2018 at 12:08
  • $\begingroup$ @pie what do you mean by a basis? And in any case the symbol $F(a, b, c, \dots)$ is commonly used to denote the smallest extension of $F$ that contains $a, b, c, \dots$, with no further assumption on $a, b, c, \dots$. $\endgroup$ Mar 30, 2018 at 12:13
  • $\begingroup$ Ok, but there is a theorem in Patrick Morandi's book that says that if $K$ is a finite extension of $F$, then $K$ is finitely generated over $F$. That means that $K$ has a finite $F$-basis. Will my proof be valid then? $\endgroup$
    – pie
    Mar 30, 2018 at 13:13

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