Loading web-font TeX/Math/Italic

The Stacks project

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


Comments (2)

Comment #8987 by on

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

There are also:

  • 2 comment(s) on Section 10.115: Noether normalization

Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.