This question is motivated by Mariano's comment on this question:

transcendence degree of subring of polynomial ring

Suppose $k$ is a field and the subring $R$ of the polynomial ring $k[x_1,...,x_r]$ is generated by homogeneous polynomials $p_1,...,p_t$ such that $\sqrt{(p_1,...,p_t)}=(x_1,...,x_r)$ in $k[x_1,...,x_r]$.

Question: Is it true that $k[x_1,...,x_r]$ is an integral extension of $R$ ?

  • $\begingroup$ Is the assumption that the $p_i$'s are homogeneous, necessary? Note that both Will's and Konstantin's answers prove more than asked. They prove that $k[x_1,\ldots,x_r]$ is finite over $R$. $\endgroup$ Oct 21, 2012 at 23:22
  • $\begingroup$ @Ralph: I don't think that is correct. The correct statement is finite = integral + finite type. $\endgroup$ Oct 22, 2012 at 0:13
  • $\begingroup$ Mahdi, you're right. I deleted my comment. $\endgroup$
    – Ralph
    Oct 22, 2012 at 0:20
  • 1
    $\begingroup$ @Mahdi : I think the assumption is necessary. Consider the blow up $f:\mathbf{A}^2 \to \mathbf{A}^2$ given by $f(x,y)=(xy,y)$. It satisfies $f^{−1}(1,1)=\{(1,1)\}$, but it is not integral. Now translate everything by $(1,1)$, and you get a map $f:\mathbf{A}^2 \to \mathbf{A}^2$ which satisfies $f^{−1}(0)=\{0\}$ but which is not integral. $\endgroup$ Oct 22, 2012 at 9:44

4 Answers 4


There is a nice lemma that relates ideals and subalgebras in graded rings that can be applied to the problem:

Lemma: Let $A=\bigoplus_{n \ge 0}A_n$ be a graded ring that is commutative (or graded commutative). Then for homogeneous elements $p_i \in A\; (i \in I)$ of positive degree are equivalent:

  1. $A/(p_i \mid i \in I)$ is a finitely generated $A_0$-module.
  2. $A$ is finite (and hence integral) over the subring $A_0[p_i \mid i \in I]$

Now the argument is similar to Konstantin's: There is $u>0$ such that $(x_1,...,x_r)^u \subseteq (p_1,...,p_t)$ [Atiyah-MacDonald, 7.16]. Hence there is an epimorphism $A/(x_1,...,x_r)^u \to A/(p_1,...,p_t)$ of $k$-algebras and since $A/(x_1,...,x_r)^u$ is finite-dimensional over $k$ the same holds for $A/(p_1,...,p_t)$. Thus $A$ is integral over $k[p_1,...,p_t]$ by the lemma.

Because I have no reference for the lemma at hand, I will show $1. \Rightarrow 2.$ what was used above. Let $x_1,...,x_m \in A$ be homogeneous representatives of positive degree of $A_0$-generators of $A/(p_i \mid i \in I)$ and let $a \in A$ be homogeneous of degree $n$. Then there are homogeneous $a_i \in A$ and $\alpha_j \in A_0$ s.t. $a = \sum_i a_ip_i + \sum_j \alpha_jx_j$. Because $x_i,p_j$ are of positive degree, $\deg(a_i) < n$ and by (a suitable) induction hypotheses we have $a_i \in \sum_{j=0}^mRx_j\;(x_0 := 1)$. Hence $a \in \sum_{j=0}^mRx_j$ and we conclude by induction $A= \sum_{j=0}^mRx_j$.


Consider the map from $\mathbb P^r$ with coordinates $(x_0:x_1:\dots:x_r)$ to a weighted $\mathbb P^t$ with coordinates $(y_0:y_1:....:y_r)$ with weights $w_0=1$ and $w_i$ is the degree of $p_i$ given by $y_0=x_0$, $y_i=p_i$. This map is well-defined, since if all $p_i$ vanish then all $x_i$ vanish. This map is proper, since it is projective.

The inverse image of the hyperplane $w_0=0$ is the hyperplane $x_0=0$. If you restrict to the complement of this hyperplane, you get a map from the affine scheme $\operatorname{Spec} k[x_1,...,x_r]$ to $\operatorname{Spec} k[y_1,...,y_t]$. This map is still proper, and now affine, thus finite, so $k[x_1,...,x_r]$ is finite over the Noetherian ring $k[y_1,...,y_t]$ so it is integral over that ring.


Let $A = k[x_1,\ldots,x_r]$, let $\mathfrak{m}$ be the ideal $(x_1,\ldots,x_r)$ in $A$ and let $I$ be the ideal $(p_1,\ldots,p_t)$ in $R$. The assumption $\sqrt{AI} = \mathfrak{m}$ implies that $\mathfrak{m}^u \subseteq AI$ for some integer $u \geq 1$. Hence $A / AI$ is a finite dimensional vector space over the ground field $k$.

Since the polynomials $p_i$ are homogeneous, $A / AI$ is actually a graded $k$-vector space. Choose homogeneous elements $v_1,\ldots, v_n \in A$ whose images span $A / AI$; then these images also generate $A / AI$ as a graded $R$-module, so we may write

$A = AI + \sum_{i=1}^n Rv_i$.

Let $Q = A / \sum_{i=1}^n Rv_i$. This is a positively graded $R$-module such that $QI = Q$, by construction. If $Q$ is non-zero, let $t \geq 0$ be least such that $Q_t \neq 0$; then for any $j \geq t$, $Q_jI$ consists of homogeneous elements of degree strictly greater than $t$ (since $\sqrt{AI} = \mathfrak{m}$ forces $\deg p_i \geq 1$ for all $i$). Then $QI \subseteq \oplus_{j > t} Q_j$, a contradiction. (This argument is called the graded Nakayama Lemma).

So in fact $Q = 0$ and $A = \sum_{i=1}^n Rv_i$ is a finitely generated $R$-module. Therefore $A$ is an integral extension of $R$.


I will try a quick answer based on the following well known

Theorem. (Graded Noether Normalization) Let $K$ be a field and $S$ a graded $K$-algebra. Set $r=\dim S$. For homogeneous elements $p_1,\dots,p_d\in S$ TFAE:

(a) $p_1,\dots,p_r$ is a homogeneous system of parameters;

(b) $S$ is an integral extension of $K[p_1,\dots,p_r]$;

(c) $S$ is a finite extension of $K[p_1,\dots,p_r]$.

(See Bruns and Herzog, Cohen-Macaulay Rings, Theorem 1.5.17.)

In our case $S=K[X_1,\dots,X_r]$ is a graded $K$-algebra of dimension $r$. Furthermore, $\sqrt{(p_1,\dots,p_r)S}=(X_1,\dots,X_r)$ means exactly that $p_1,\dots,p_r$ is a homogeneous system of parameters for $S$, and this is it.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.