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Tag 00JT

Lemma 10.56.7. Let $S$ be a graded ring. Let $\mathfrak p \subset S$ be a prime. Let $\mathfrak q$ be the homogeneous ideal of $S$ generated by the homogeneous elements of $\mathfrak p$. Then $\mathfrak q$ is a prime ideal of $S$.

Proof. Suppose $f, g \in S$ are such that $fg \in \mathfrak q$. Let $f_d$ (resp. $g_e$) be the homogeneous part of $f$ (resp. $g$) of degree $d$ (resp. $e$). Assume $d, e$ are maxima such that $f_d \not = 0$ and $g_e \not = 0$. By assumption we can write $fg = \sum a_i f_i$ with $f_i \in \mathfrak p$ homogeneous. Say $\deg(f_i) = d_i$. Then $f_d g_e = \sum a_i' f_i$ with $a_i'$ to homogeneous par of degree $d + e - d_i$ of $a_i$ (or $0$ if $d + e -d_i < 0$). Hence $f_d \in \mathfrak p$ or $g_e \in \mathfrak p$. Hence $f_d \in \mathfrak q$ or $g_e \in \mathfrak q$. In the first case replace $f$ by $f - f_d$, in the second case replace $g$ by $g - g_e$. Then still $fg \in \mathfrak q$ but the discrete invariant $d + e$ has been decreased. Thus we may continue in this fashion until either $f$ or $g$ is zero. This clearly shows that $fg \in \mathfrak q$ implies either $f \in \mathfrak q$ or $g \in \mathfrak q$ as desired. $\square$

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\begin{lemma}
\label{lemma-smear-out}
Let $S$ be a graded ring.
Let $\mathfrak p \subset S$ be a prime.
Let $\mathfrak q$ be the homogeneous ideal of $S$ generated by the
homogeneous elements of $\mathfrak p$. Then $\mathfrak q$ is a
prime ideal of $S$.
\end{lemma}

\begin{proof}
Suppose $f, g \in S$ are such that $fg \in \mathfrak q$.
Let $f_d$ (resp.\ $g_e$) be the homogeneous part of
$f$ (resp.\ $g$) of degree $d$ (resp.\ $e$). Assume $d, e$ are
maxima such that $f_d \not = 0$ and $g_e \not = 0$.
By assumption we can write $fg = \sum a_i f_i$ with
$f_i \in \mathfrak p$ homogeneous. Say $\deg(f_i) = d_i$.
Then $f_d g_e = \sum a_i' f_i$ with $a_i'$ to homogeneous
par of degree $d + e - d_i$ of $a_i$ (or $0$ if $d + e -d_i < 0$).
Hence $f_d \in \mathfrak p$ or $g_e \in \mathfrak p$. Hence
$f_d \in \mathfrak q$ or $g_e \in \mathfrak q$. In the first
case replace $f$ by $f - f_d$, in the second case replace
$g$ by $g - g_e$. Then still $fg \in \mathfrak q$ but the discrete
invariant $d + e$ has been decreased. Thus we may continue in this
fashion until either $f$ or $g$ is zero. This clearly shows that
$fg \in \mathfrak q$ implies either $f \in \mathfrak q$ or $g \in \mathfrak q$
as desired.
\end{proof}

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