Let $S = k[x_1,\ldots,x_n]$ be a polynomial ring over an infinite field $k$, let $S_{+}$ denote the irrelevant ideal of $S$ and let $I$ be a homogeneous ideal of $S$. I want to show that if $\dim(S/I) = 0$, then $I^\text{sat} = S$.
Here it says that
Since $\dim(/)=0$, $I$ contains a power of the irrelevant ideal
This would imply that $S_{+} \subseteq \sqrt{I}$ and therefore $1 \in I^\text{sat}$. (Would these steps be correct so far?)
However, I don't see why $I$ must contain a power of the irrelevant ideal if $\dim(S/I)= 0$ ?
Let $\newcommand{\m}{\mathfrak m}\m$ denote the irrelevant ideal of $S$ (note that $\m$ is maximal). Since $\dim(S/I)=0$, prime ideals containing $I$ are maximal. Recall that for a graded ring the nilradical is equal to the intersection of all homogeneous prime ideals (see [1]). Combining the above facts, we see that the nilradical of $S/I$ is equal to $\m/I$, i.e., the radical ideal of $I$ is $\m$. In a Noetherian ring, this is equivalent to $\m^r\subseteq I\subseteq \m$ for some $r$.