See pages 71 and 72 of [Jou]
Lemma 37.32.1. 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 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
Observe that $\sigma _ a(\xi ) = \xi + a \delta $ and $\sigma _ a(\delta ) = \delta $. Consider the tower of fields
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 infinite1, then we can find distinct elements $a_1, a_2$ of $k$ such that
as subfields of $\Omega $. Write $\theta _ i = \sigma _{a_ i}(\theta )$ and $\xi _ i = \sigma _{a_ i}(\xi ) = \xi + a_ i \delta $. Observe that
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 independent 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
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.
$\square$
\[ x_1, \ldots , x_ n, \sum g_ ix_ i, \sum c_ ig_ i \in K(x_1, \ldots , x_ n) \]
\[ \sigma _ a(x_ i) = x_ i + a c_ i \]
\[ K_0 = k(x_1, \ldots , x_ n) \subset K_1 = K_0(\xi ) \subset K_2 = K_0(\xi , \delta ) \subset K(x_1, \ldots , x_ n) = \Omega \]
\[ K_2(\sigma _{a_1}(\theta )) = K_2(\sigma _{a_2}(\theta )) \]
\[ K_2 = K_0(\xi _1, \xi _2) \]
\[ K_0(\xi _1, \xi _2, \theta _1) = K_2(\sigma _{a_1}(\theta )) = K_2(\sigma _{a_2}(\theta )) = K_0(\xi _1, \xi _2, \theta _2) \]
[1] We will deal with the finite field case in the last paragraph of the proof.
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)