Radical of a homogeneous ideal If $S=\bigoplus_{d\ge 0} S_d$ is a graded ring and $\mathfrak a\subset S$ a homogenous ideal, I'm trying to prove this implication:

$\sqrt {\mathfrak a}=S_+=\bigoplus_{d\gt 0}S_d\implies S_d \subset \mathfrak
 a$ for some $d\gt 0$.

My attempt of solution
I know that $x_i^d\in \sqrt {\mathfrak a}$ for $i=0,\ldots,n$ and $d\gt 0$, but I couldn't go further.
I don't have much experience solving graded ring questions.
I really need help.
Thanks a lot.

I think you are supposed to be assuming Noetherian here, which is what the other users are pointing out. 
Let me assume that $\mathfrak{a}\subseteq S:=k[x_0,\ldots,x_n]$ and let you generalize. If $\sqrt{\mathfrak{a}}=S_+$, then for all $i$ there exists $m_i$ such that $x_i^{m_i}\in\mathfrak{a}$. Let $M=(n+1)\max\{m_i\}$. We claim that $S_M\subseteq\mathfrak{a}$. Indeed, it suffices to show that each monomial $x_0^{e_0}\cdots x_n^{e_n}\in \mathfrak{a}$ for $e_0+\cdots+e_n=M$, since these span $S_M$ as a $k$-space. But, note that since $e_0+\cdots+e_n=M$, at least one of the $e_i$'s must be at least as big as $\max\{m_i\}$, and thus at least one of $x_i^{e_i}\in\mathfrak{a}$ from where the conclusion follows.
