Lemma 10.57.6. Suppose $S$ is a graded ring, $\mathfrak p_ i$, $i = 1, \ldots , r$ homogeneous prime ideals and $I \subset S_{+}$ a graded ideal. Assume $I \not\subset \mathfrak p_ i$ for all $i$. Then there exists a homogeneous element $x\in I$ of positive degree such that $x\not\in \mathfrak p_ i$ for all $i$.

**Proof.**
We may assume there are no inclusions among the $\mathfrak p_ i$. The result is true for $r = 1$. Suppose the result holds for $r - 1$. Pick $x \in I$ homogeneous of positive degree such that $x \not\in \mathfrak p_ i$ for all $i = 1, \ldots , r - 1$. If $x \not\in \mathfrak p_ r$ we are done. So assume $x \in \mathfrak p_ r$. If $I \mathfrak p_1 \ldots \mathfrak p_{r-1} \subset \mathfrak p_ r$ then $I \subset \mathfrak p_ r$ a contradiction. Pick $y \in I\mathfrak p_1 \ldots \mathfrak p_{r-1}$ homogeneous and $y \not\in \mathfrak p_ r$. Then $x^{\deg (y)} + y^{\deg (x)}$ works.
$\square$

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