Lemma 9.6.9. Let $k$ be a field and let $E/k$ and $F/k$ be field extensions. Then there exists a common field extension $M/k$, i.e., an extension field such that there exist maps $E \to M$ and $F \to M$ of extensions of $k$.
Proof. We only prove this when $E$ is a finitely generated field extension of $k$; the general case follows from this by a Zorn's lemma type argument (details omitted).
First, suppose that $E$ is a simple extension of $k$. By Lemma 9.6.8 this means either $E = k(t)$ is the rational function field or $E = k[t]/(P)$ for some irreducible polynomial $P \in k[t]$. In the first case, we take $M = F(t)$ the rational function field with obvious maps $E \to M$ and $F \to M$. In the second case, we choose an irreducible factor $Q$ of the image of $P$ in $F[t]$ and we take $M = F[t]/(Q)$ with obvious maps $E \to M$ and $F \to M$.
If $E = k(\alpha _1, \ldots , \alpha _ n)$, then by induction on $n$ we can find an extension $M/k$ and maps $F \to M$ and $k(\alpha _1, \ldots , \alpha _{n - 1}) \to M$. By the simple case discussed in the previous paragraph, we can find an extension $M'/k(\alpha _1, \ldots , \alpha _{n - 1})$ and maps $M \to M'$ and $k(\alpha _1, \ldots , \alpha _ n) \to M'$. Then $M'$ viewed as an extension of $k$ works. $\square$
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