Lemma 37.31.1. Let $K/k$ be a geometrically irreducible and finitely generated field extension. Let $n \geq 1$. Let $g_1, \ldots , g_ n \in K$ be elements such that there exist $c_1, \ldots , c_ n \in k$ such that the elements

\[ x_1, \ldots , x_ n, \sum g_ ix_ i, \sum c_ ig_ i \in K(x_1, \ldots , x_ n) \]

are algebraically independent over $k$. Then $K(x_1, \ldots , x_ n)$ is geometrically irreducible over $k(x_1, \ldots , x_ n, \sum g_ ix_ i)$.

**Proof.**
Let $c_1, \ldots , c_ n \in k$ be as in the statement of the lemma. Write $\xi = \sum g_ ix_ i$ and $\delta = \sum c_ ig_ i$. For $a \in k$ consider the automorphism $\sigma _ a$ of $K(x_1, \ldots , x_ n)$ given by the identity on $K$ and the rules

\[ \sigma _ a(x_ i) = x_ i + a c_ i \]

Observe that $\sigma _ a(\xi ) = \xi + a \delta $ and $\sigma _ a(\delta ) = \delta $. Consider the tower of fields

\[ K_0 = k(x_1, \ldots , x_ n) \subset K_1 = K_0(\xi ) \subset K_2 = K_0(\xi , \delta ) \subset K(x_1, \ldots , x_ n) = \Omega \]

Observe that $\sigma _ a(K_0) = K_0$ and $\sigma _ a(K_2) = K_2$. Let $\theta \in \Omega $ be separable algebraic over $K_1$. We have to show $\theta \in K_1$, see Algebra, Lemma 10.47.12.

Denote $K'_2$ the separable algebraic closure of $K_2$ in $\Omega $. Since $K'_2/K_2$ is finite (Algebra, Lemma 10.47.13) and separable there are only a finite number of fields in between $K'_2$ and $K_2$ (Fields, Lemma 9.19.1). If $k$ is infinite^{1}, then we can find distinct elements $a_1, a_2$ of $k$ such that

\[ K_2(\sigma _{a_1}(\theta )) = K_2(\sigma _{a_2}(\theta )) \]

as subfields of $\Omega $. Write $\theta _ i = \sigma _{a_ i}(\theta )$ and $\xi _ i = \sigma _{a_ i}(\xi ) = \xi + a_ i \delta $. Observe that

\[ K_2 = K_0(\xi _1, \xi _2) \]

as we have $\xi _ i = \xi + a_ i \delta $, $\xi = (a_2 \xi _1 - a_1 \xi _2)/(a_2 - a_1)$, and $\delta = (\xi _1 - \xi _2)/(a_1 - a_2)$. Since $K_2/K_0$ is purely transcendental of degree $2$ we conclude that $\xi _1$ and $\xi _2$ are algebraically indepedent over $K_0$. Since $\theta _1$ is algebraic over $K_0(\xi _1)$ we conclude that $\xi _2$ is transcendental over $K_0(\xi _1, \theta _1)$.

By assumption $K/k$ is geometrically irreducible. This implies that $K(x_1, \ldots , x_ n)/K_0$ is geometrically irreducible (Algebra, Lemma 10.47.10). This in turn implies that $K_0(\xi _1, \theta _1)/K_0$ is geometrically irreducible as a subextension (Algebra, Lemma 10.47.6). Since $\xi _2$ is transcendental over $K_0(\xi _1, \theta _1)$ we conclude that $K_0(\xi _1, \xi _2, \theta _1)/K_0(\xi _2)$ is geometrically irreducible (Algebra, Lemma 10.47.11). By our choice of $a_1, a_2$ above we have

\[ K_0(\xi _1, \xi _2, \theta _1) = K_2(\sigma _{a_1}(\theta )) = K_2(\sigma _{a_2}(\theta )) = K_0(\xi _1, \xi _2, \theta _2) \]

Since $\theta _2$ is separably algebraic over $K_0(\xi _2)$ we conclude by Algebra, Lemma 10.47.12 again that $\theta _2 \in K_0(\xi _2)$. Taking $\sigma _{a_2}^{-1}$ of this relation givens $\theta \in K_0(\xi ) = K_1$ as desired.

This finishes the proof in case $k$ is infinite. If $k$ is finite, then we can choose a variable $t$ and consider the extension $K(t)/k(t)$ which is geometrically irreducible by Algebra, Lemma 10.47.10. Since it is still be true that $x_1, \ldots , x_ n, \sum g_ ix_ i, \sum c_ ig_ i$ in $K(t, x_1, \ldots , x_ n)$ are algebraically independent over $k(t)$ we conclude that $K(t, x_1, \ldots , x_ n)$ is geometrically irreducible over $k(t, x_1, \ldots , x_ n, \sum g_ ix_ i)$ by the argument already given. Then using Algebra, Lemma 10.47.10 once more finishes the job.
$\square$

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