Lemma 10.115.3. Let $k$ be a field. Let $S = k[x_1, \ldots , x_ n]/I$ for some proper ideal $I$. If $I \not= 0$, then there exist $y_1, \ldots , y_{n-1} \in k[x_1, \ldots , x_ n]$ such that $S$ is finite over $k[y_1, \ldots , y_{n-1}]$. Moreover we may choose $y_ i$ to be in the $\mathbf{Z}$-subalgebra of $k[x_1, \ldots , x_ n]$ generated by $x_1, \ldots , x_ n$.

Proof. Pick $f \in I$, $f\not= 0$. It suffices to show the lemma for $k[x_1, \ldots , x_ n]/(f)$ since $S$ is a quotient of that ring. We will take $y_ i = x_ i - x_ n^{e_ i}$, $i = 1, \ldots , n-1$ for suitable integers $e_ i$. When does this work? It suffices to show that $\overline{x_ n} \in k[x_1, \ldots , x_ n]/(f)$ is integral over the ring $k[y_1, \ldots , y_{n-1}]$. The equation for $\overline{x_ n}$ over this ring is

$f(y_1 + x_ n^{e_1}, \ldots , y_{n-1} + x_ n^{e_{n-1}}, x_ n) = 0.$

Hence we are done if we can show there exists integers $e_ i$ such that the leading coefficient with respect to $x_ n$ of the equation above is a nonzero element of $k$. This can be achieved for example by choosing $e_1 \gg e_2 \gg \ldots \gg e_{n-1}$, see Lemma 10.115.2. $\square$

Comment #8987 by on

Maybe nitpicking, but shouldn't the case $n=1$ be handled separately? Just saying something like "for $n=1$, $S$ is a finitely-generated $k$-vector space; thus $S$ is finite over $k$."

Comment #9199 by on

I think the argument also works for $n = 1$.

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