## Tag `00JS`

Chapter 10: Commutative Algebra > Section 10.56: Proj of a graded ring

Lemma 10.56.6. 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. $\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 12999–13006 (see updates for more information).

```
\begin{lemma}
\label{lemma-graded-silly}
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 $i$.
\end{lemma}
\begin{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.
\end{proof}
```

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