# Graded rings and their localizations

Let $A$ be a $\mathbb{Z}_{\geq 0}$-graded ring, $f \in A$ - homogenious, and $I \subset A$ - homogenious ideal. Let $A_f$ be its localization, and $A_{(f)}$ - subring of elements of degree 0. How to show, that

$$(A/I)_{(f)} = A_{(f)}/(I A_f \cap A_{(f)})?$$

It is used in Hartshorne, Algebraic geometry, section 1.3, proposition 3.4

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Yes, you are right. – user46336 Nov 11 '12 at 19:16

## 1 Answer

From the canonical surjection $A\to A/I$ you get a canonical surjection $A_f\to (A/I)_f$. In degree $0$ this is a surjection from $A_{(f)}$ to $(A/I)_{(f)}$. Now find its kernel.

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Is it possible to prove it using some universal property? – user46336 Nov 11 '12 at 19:18
$(A/I)_f = A_f/IA_f$ is trivial, since it's universal map, with $f$ is invertible and vanishing $I$. How to prove for this subring of degree 0? – user46336 Nov 11 '12 at 19:44
Then all the problem is trivial, if you want to say so. If you know that two graded rings are (graded) isomorphic, then their degree zero parts are also isomorphic. – user26857 Nov 11 '12 at 19:52