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Let $A \subset B$ be a finite extension of graded rings (so that $B$ is finite as a graded $A$-module). There is an induced finite morphism $\operatorname{Proj} B \to \operatorname{Proj} A$.

Is this morphism necessarily surjective?

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up vote 4 down vote accepted

If $\psi:A \to B$ is injective, then you can say that $\Psi : Proj(B) \to Proj(A)$ is dominant, that is has dense image.

Indeed, take an open set $D_+(f)\subset Proj(A)$ for some non-nilpotent graded element $f\in A$. We want to show that this open set intersects the image of $\Psi$.

To see this, we know that $\psi(f)$ is non nilpotent so that $D_+(\psi(f))$ is not empty. Therefore there exists a graded prime ideal $\mathfrak p \subset B$ such that $\mathfrak p \notin D_+(\psi(f))$.

Therefore $f \notin \psi^{-1}(\mathfrak p)$ and $\Psi(\mathfrak p)=\psi^{-1}(\mathfrak p) \in D_+(f)$, which concludes.

Remark. There is no need to assume finiteness for $\psi$.

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Interesting. Thanks. But does that mean that in general the answer to my question is 'no'? – Bonanza Aug 21 '11 at 22:54
The general answer for arbitrary extensions $A\subset B$ is no, as is shown immediately by the example $\mathbb Z \subset \mathbb Q$ (the gradation being trivial). If one adds the finiteness assumption, then it might become true, as a reasonable analogue of the Cohen-Seidenberg theorem for graded rings. Indeed, CS is a consequence of the going up, itself coming from Nakayama's lemma, which is known to admit a graded version. – Henri Aug 22 '11 at 8:34

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