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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 $R$ ?

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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 $R$. –  Mahdi Majidi-Zolbanin Oct 21 '12 at 23:22
    
@Ralph: I don't think that is correct. The correct statement is finite = integral + finite type. –  Mahdi Majidi-Zolbanin Oct 22 '12 at 0:13
    
Mahdi, you're right. I deleted my comment. –  Ralph Oct 22 '12 at 0:20
1  
@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. –  François Brunault Oct 22 '12 at 9:44

4 Answers 4

up vote 6 down vote accepted

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 $A= \sum_{j=0}^mRx_j$.

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That lemma is really nice! –  Jeff Oct 22 '12 at 21:37

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.

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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 $R$.

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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.

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