# transcendence degree of subring of polynomial ring

Hi all

Quick question that I think is true but haven't been able to prove.

Suppose there is a subring $R \subset k[x_1, \ldots, x_r]$ containing field $k$ and generated by homogenous polynomials $p_1, \ldots, p_t$. Now assume the radical of the homogenous ideal of $k[x_1, \ldots, x_k]$ generated by the $p_i$ is the maximal ideal $(x_1, \ldots, x_r)$. Then the transcendence degree of $R$ over $k$ is $r$.

• Let $X$ be affine $r$-space over $k$ and let $Y$ be affine $t$-space over $k$, and let $f:X \rightarrow Y$ be the $k$-map defined by $x \mapsto (p_1(x),\dots,p_t(x))$. The hypothesis is that $f^{-1}(0) = 0$ at the level of geometric points, so by openness on the source for the locus of points isolated in their fibers (semi-continuity of fiber dimension on the source) there is a dense open $U$ in $X$ so that $f|_U$ is quasi-finite. Its schematic image is Spec($R$), so $U \rightarrow {\rm{Spec}}(R)$ is quasi-finite and dominant. Thus, trdeg$_k$(Frac($R$)) is ${\rm{dim}}(R) = {\rm{dim}}(U) = r$. – user27056 Oct 19 '12 at 2:50
• Doesn't your hypothesis imply at once that the fraction field of the normalization of your ring is $k(x_1,\dots,x_r)$? – Mariano Suárez-Álvarez Oct 19 '12 at 6:10
• @Mariano: What's your argument? I think you're mixing up "ideal generated" in a bigger ring with "algebra generated" over $k$. The assertion that you claim follows "at once" seems to be false. Try $p_i = x_i^2$ with $1 \le i \le t = r$. – user27056 Oct 19 '12 at 14:08
• @xbnv: You can find arguments there: mathoverflow.net/questions/110250/…. So, Mariano's comment gives an alternative (affirmative) answer to the question. – Jeff Oct 22 '12 at 21:45

Set $R=k[p_1,\dots,p_t]$, $S=k[x_1,\dots,x_r]$ and $\mathfrak{m}=(x_1,\dots,x_r)$ (the maximal irrelevant ideal of $S$). We know that the ideal $(p_1,\dots,p_t)S$ is $\mathfrak{m}$-primary. A graded version of the dimension inequality formula for local rings gives $\dim S\le \dim R+\dim S/(p_1,\dots,p_t)S$ and thus we obtain that $r\le \dim R$. On the other hand, by using Noether normalization we get that $\dim R\le r$.
• Maybe it's worthy to mention that any subring of $k[x_1,\dots,x_r]$ which contains $k$ has the Krull dimension at most $r$. – user26857 Oct 19 '12 at 21:36
• @navigetor23: Why is that true for $k$-subalgebras that aren't finitely generated (as arise in Nagata's counterexamples to Hilbert's $N$th problem for suitable $N$)? – user27056 Oct 20 '12 at 6:52