2

I am reading Huybrechts' Fourier-Mukai Transforms in Algebraic Geometry, which including the following sentence:

The category of coherent sheaves does not have enough injectives for the simple reason that a finitely generated module is usually too small to be injective.

We know that an $\mathcal{O}_X$-module $\mathcal{F}$ is injective if and only if $\mathcal{F}_x$ is injective $\mathcal{O}_{x,X}$-module for every $x\in X$ (the statement is true at least for nice schemes, e.g. Noetherian).

So, yes, we reduce to the question that a finitely generated module is injective or not. But why it is "too small to be injective"? We know zero module is injective and it is the smallest module... I am very confused...

quid
  • 42,835
sysMirror
  • 2,327

1 Answers1

5

The zero module is an exception and is not helpful for having enough injectives. Nonzero injective modules typically (though not always) have to be "large". This follows intuitively from the definition of an injective module: if $I$ is injective, then whenever $M$ is a module with a submodule $N$, every homomorphism $N\to I$ can be extended to $M$. This means $I$ must "have room" to accomodate any new elements that $M$ might have in a way compatible with the map from $N$, which typically makes it hard for $I$ to be finitely generated.

More concretely, a nonzero injective module $I$ over an integral domain which is not a field cannot be finitely generated. This follows from the fact that $I$ must be divisible (which is just the extension property in the case where $M$ is the ring and $N$ is a principal ideal); see Integral domain with a finitely generated non-zero injective module is a field for the full argument.

Eric Wofsey
  • 342,377