# How to prove that $\mathrm{Proj}\left(B/J\right)$ is isomorphic to $\mathrm{Proj}\left(A/J\right)$ if $I\subset J$?

Let $B$ be a graded ring with positive degrees, and let $I$ and $J$ be homogeneous ideals of $B$. We suppose that there exists $N$ such that $I\cap B_{n}=J\cap B_{n}$ for all $n\ge N$.

How to show that if $I\subset J$, the natural arrow $\mathrm{Proj}\left(B/J\right)\hookrightarrow\mathrm{Proj}\left(B/I\right)$ is an isomorphism?

• You mean to write $I \cap B_n = J \cap B_n$. Check equality on the distinguished affine open sets $D(f) = \operatorname{Spec} B_{(f)}$.
– Hoot
Apr 26, 2015 at 18:28
• See Algebraic Geometry by R. Hartshorne, Chapter II, Ex. 2.14 (c). May 6, 2015 at 19:22
• @Hoot Would you please write about this in more detail in an answer? The textbook also says about checking equality on the distinguished affine open sets, but I don't think I've seen an example of that being done.
– Jake
May 16, 2015 at 0:48
• @Jake Sure. Give me a moment.
– Hoot
May 17, 2015 at 19:24

## 1 Answer

The morphism in question comes from the surjection $B/I \to B/J$. If $f \in B$ is homogeneous of positive degree then we get $(B/I)_f \to (B/J)_f$ and then $((B/I)_f)_0 \to ((B/J)_f)_0$, both surjective. So the only question is injectivity. Say $b/f^n$ maps to $0$ where $b \in B_{n \deg f}$ (everything should be mod $I$, really). Then for $m \gg 0$ we have $f^mb \equiv 0 \bmod{J}$, so $f^mb \in J_{(n+m)\deg f}$ and for $m$ large enough the hypothesis tells you that this is the same as $I_{(n+m)\deg f}$ and it follows that $b/f^n$ is already zero in $((B/I)_f)_0$.