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 $R$.

finiteover $R$. – Mahdi Majidi-Zolbanin Oct 21 '12 at 23:22