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Suppose we are given two graded (commutative) rings $A$ and $B$ and a graded homomorphism $\psi:A\rightarrow{B}$ between them. Suppose moreover that $\psi$ is surjective in each degree i.e. that $\psi(A_n)=B_n$ for all $n$. If $\psi$ is injective in degrees $0$ and $1$, must it also be injective everywhere else? (all relations in $A$ come from these two degrees). If not in general, does it make a difference if we assume $B$ is an integral domain? (or even more specifically if it is a polynomial ring with coefficients in an integral domain).

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2 Answers 2

The answer is indeed no, for both cases : take $A=R[X,X^2Y]$, graded by degree in $X$ and $B=R[X]$, also graded by degree in $X$, and $\psi:A\to B$ sending $Y$ to $0$. This is a graded morphism between two integral domain (as long as $R$ is one), that is surjective in each degree, injective in degrees 0 and 1 (since every polynomial containing $Y$s must contain twince as much $X$s) but not an isomoprphism.

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Ok thanks for the answer! –  Axiom May 9 '12 at 15:15

The answer is no. Consider $A = \mathbb{R}[t]$, and $B = \mathbb{R}[t]/(t^k)$, both graded by degree. The homorphism $\psi : A \to B$ is the natural projection, which is surjective in each degree. For any choice of $k \geq 1$, we have $\psi$ injective up to degree $k-1$, but not injective for degree $k$ and higher.

About the follow-up question, if $B$ is an integral domain, let me think about that a little more.

Hope this helps!

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This doesn't work : the natural projection homomorphism is not a graded morphism (e.g. $t^{k+1} \mapsto t$ lowers the degree by $k$. –  Samuel T May 3 '12 at 12:04

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