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Let $M$ be a finitely generated graded module over a graded ring $R$. Let $\mathcal{F}$ be the corresponding coherent sheaf on $\operatorname{Proj} R$. There is a natural map of graded $R$-modules $$\phi \colon M \to \Gamma^*(\mathcal{F}) := \bigoplus_{n} \Gamma(\operatorname{Proj} R, \mathcal{F}(n)).$$ If I recall Ravi Vakil's notes correctly, $M$ is called saturated if $\phi$ is an isomorphism.

Is there a term (perhaps semi-saturated, or some such) for modules $M$ such that $\phi$ is injective?

This concept is appealing for several reasons. For one thing, it is easier to test "semi-saturatedness" than saturatedness; e.g., unless I am mistaken, $\phi$ is automatically injective if $M$ admits any positive-degree homogeneous nonzerodivisor. For another, at least if $R$ is a polynomial ring, $M = R/I$ is "semi-saturated" iff $I$ is a saturated ideal of $R$. (Note that the definition of "saturated ideal" is different from the definition given above for "saturated module", and I do not think the two are equivalent for ideals.)

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I don't know a name for this, but I have a comment. This is quite close to being torsion free (or more precisely, $R_+$-torsion free) if $\mathcal{F}$ is already S2. Ok, for example, lets suppose that $R$ is normal and $\mathcal{F}$ is reflexive on $X$ (this is equivalent to S2, depending on your choice of terms). Then $\Gamma^*(\mathcal{F})$ is the S2-ification = reflexification of $M$. There's a paper of Hartshorne called something like "Generalized divisors on Gorenstein schemes" where he points out that in the context I described, the injectivity is equivalent to being torsion free. – Karl Schwede Jul 18 '11 at 20:11
You can make sense of this even without the reflexification statements too. – Karl Schwede Jul 18 '11 at 20:12
up vote 6 down vote accepted

To elaborate on Karl's comment:

Let $m$ be the irrelevant ideal of $R$, then there is a short exact sequence:

$$0 \to H_m^0(M) \to M \to \Gamma^*(\mathcal{F}) \to H_m^1(M) \to 0$$

(see Eisenbud's book, Theorem A4.1, p. 693). Here $H_m^i(M)$ denote the local cohomology modules. So the map is injective precisely when $H_m^0(M):= \cup_n (0:_M m^n) = 0$. I believe such module is called $m$-torsion-free (don't know a reference off hand, may be Brodmann-Sharp's book on local cohomology?).

Also, it is equivalent to $m$ contains a non-zerodivisor on $M$ (may be that's what you meant in the last paragraph?)

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