Lemma 37.31.2. Let $A$ be a domain of finite type over a field $k$. Let $n \geq 2$. Let $g_1, \ldots , g_ n \in A$ be elements such that $V(g_1, g_2)$ has an irreducible component of dimension $\dim (A) - 2$. Then there exist $c_1, \ldots , c_ n \in k$ such that the elements

\[ x_1, \ldots , x_ n, \sum g_ ix_ i, \sum c_ ig_ i \in \text{Frac}(A)(x_1, \ldots , x_ n) \]

are algebraically independent over $k$.

**Proof.**
The algebraic independence over $k$ means that the morphism

\[ T = \mathop{\mathrm{Spec}}(A[x_1, \ldots , x_ n]) \longrightarrow \mathop{\mathrm{Spec}}(k[x_1, \ldots , x_ n, y, z]) = S \]

given by $y = \sum g_ ix_ i$ and $z = \sum c_ ig_ i$ is dominant. Set $d = \dim (A)$. If $T \to S$ is not dominant, then the image has dimension $< n + 2$ and hence every irreducible component of every fibre has dimension $> d + n - (n + 2) = d - 2$, see Varieties, Lemma 33.20.4. Choose a closed point $u \in V(g_1, g_2)$ contained in an irreducible component of dimension $d - 2$ and in no other component of $V(g_1, g_2)$. Consider the closed point $t = (u, 1, 0, \ldots 0)$ of $T$ lying over $u$. Set $(c_1, \ldots , c_ n) = (0, 1, 0, \ldots , 0)$. Then $t$ maps to the point $s = (1, 0, \ldots , 0)$ of $S$. The fibre of $T \to S$ over $s$ is cut out by

\[ x_1 - 1, x_2, \ldots , x_ n, \sum x_ ig_ i, g_2 \]

and hence equivalently is cut out by

\[ x_1 - 1, x_2, \ldots , x_ n, g_1, g_2 \]

By our condition on $g_1, g_2$ this subscheme has an irreducible component of dimension $d - 2$.
$\square$

## Comments (0)