Lemma 9.26.10. Let $k'/k$ be a finite extension of fields. Let $k'(x_1, \ldots , x_ r)/k(x_1, \ldots , x_ r)$ be the induced extension of purely transcendental extensions. Then $[k'(x_1, \ldots , x_ r) : k(x_1, \ldots , x_ r)] = [k' : k] < \infty$.

Proof. By multiplicativity of degrees of extensions (Lemma 9.7.7) it suffices to prove this when $k'$ is generated by a single element $\alpha \in k'$ over $k$. Let $f \in k[T]$ be the minimal polynomial of $\alpha$ over $k$. Then $k'(x_1, \ldots , x_ r)$ is generated by $\alpha , x_1, \ldots , x_ r$ over $k$ and hence $k'(x_1, \ldots , x_ r)$ is generated by $\alpha$ over $k(x_1, \ldots , x_ r)$. Thus it suffices to show that $f$ is still irreducible as an element of $k(x_1, \ldots , x_ r)[T]$. We only sketch the proof. It is clear that $f$ is irreducible as an element of $k[x_1, \ldots , x_ r, T]$ for example because $f$ is monic as a polynomial in $T$ and any putative factorization in $k[x_1, \ldots , x_ r, T]$ would lead to a factorization in $k[T]$ by setting $x_ i$ equal to $0$. By Gauss' lemma we conclude. $\square$

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