Lemma 10.57.7. 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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