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
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