Saturation of homogeneous ideal

Question

Let $I \subset S=k[x_0,...,x_n]$ be a homogenous ideal. The saturation of $I$, $\bar{I}$ is defined to be $\{s \in S: \exists m \; s.t. \; \forall i \; x_i^m s \in I\}$

Is it true that $\bar{I}=(s \in S: s$ homogeneous and $\forall i \; s|_{x_i=1} \in I|_{x_i=1}$)? (where $s|_{x_i=1}$ means the polynomial obtained by setting $x_i=1$ in $s$, and $I|_{x_i=1}$ defined to be $\{s|_{x_i=1}: s\in I\}$).

The reason I am asking this is that in a set of notes on algebraic geometry I am reading, the author seems to have used this implicitly, yet I cannot show this. Any help is appreciated!

Answer 1

Your question is equivalent to the following statement:

Suppose deg $f$ $\geq$ deg $g$, $f$, $g$ are homogeneous then $f|_{x_i=1}=g|_{x_i=1}$ if and only if $f=x_i^kg$ for some $k\in \mathbb{N}$

Since we can divide $f$ and $g$ by $x_i$ if necessary, we can assume both $f$ and $g$ are not multiples of $x_i$

Then the statement can be proved by noticing $f=x_i^lf|_{x_i=1}(x_0/x_i,..., x_n/x_i)$ where $l$ is the degree of $f|_{x_i=1}$