Definition of exact sequence

Question

Take a sequence $A \to B \to C \to 0$ of modules over a commutative ring. How would one show that it is exact? I understand the necessary surjectivity of $B \to C$, but what about the first map?

Answer 1

You also need to show that $\textrm{im} (A \to B) = \textrm{ker} (B \to C)$. It would be easier if you just remembered that condition because then the surjectivity follows from that, namely

$$\textrm{im} (B \to C) = \textrm{ker} (C \to 0) = C$$

Answer 2

Remember that "exact" means "At each point, the kernel of the next map equals the image of the previous map." So saying $A\rightarrow B\rightarrow C\rightarrow 0$ is exact is saying:

• $im(B\rightarrow C)=ker(C\rightarrow 0)$, and

• $im(A\rightarrow B)=ker(B\rightarrow C)$.

Now the former is equivalent to saying "$im(B\rightarrow C)=C$", since the kernel of any map from $C$ to $0$ is all of $C$.

The latter, meanwhile, isn't simplifiable: you just have to check that indeed $im(A\rightarrow B)=ker(B\rightarrow C)$.

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Note that exactness is a property, not of the sequence of modules, but of the sequence of maps between modules. For instance, suppose $A=B=C=\mathbb{Z}$ (as $\mathbb{Z}$-modules). Then if we let $id$ be the identity map and $z: x\mapsto 0$, we have:

• $A\xrightarrow{id}B\xrightarrow{id}C\rightarrow 0$ is not exact, but

• $A\xrightarrow{z}B\xrightarrow{id}C\rightarrow 0$ is exact.