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


[transcendence degree of subring of polynomial ring](https://mathoverflow.net/questions/110060/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 $R$ ? 



Is a polynomial ring integral over this subring ? 
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 $A= \sum\_{j=0}^mRx\_j$. 


