Exercise 8.2.3

Proof.

(a)

Let $A$ be the set of the subfields of $L$ containing $K_1$ and $K_2$. Then $A\ne \varnothing$, since $L\in A$.

The intersection of the subfields of $L$ containing $K_1$ and $K_2$ is a subfield of $L$ containing $K_1$ and $K_2$, thus $\underset{X\in A}{\bigcap <!--\; FUNCTION\; APPLICATION\; -->}X$ is an element of $A$, and this is the smallest element of $A$ for inclusion.

So there exists a smallest subfield of $L$ containing $K_1$ and $K_2$, that is $K_1{K}_2$.

(b)

Suppose that $K_1=F({\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_n),K_2=F({\beta }_1,{\beta }_2,\cdots ,{\beta }_m)$.

$K=K_1{K}_2$ contains $K_1$ and $K_2$, so contains $F$ , and also ${\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_n,{\beta }_1,{\beta }_2,\cdots ,{\beta }_m$, therefore $K\supset F({\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_n,{\beta }_1,{\beta }_2,\cdots ,{\beta }_m)$.

Conversaly, as $K=F({\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_n,{\beta }_1,{\beta }_2,\cdots ,{\beta }_m)$ is a subfield of $L$ containing $K_1=F({\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_n)$ and $K_2=F({\beta }_1,{\beta }_2,\cdots ,{\beta }_m)$, $K\in A$, so $K$ contains the smallest element of $A$, which is $K_1{K}_2$.

Conclusion: $K_1{K}_2=F({\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_n,{\beta }_1,{\beta }_2,\cdots ,{\beta }_m)$.