TITLE: Module homomorphism surjective but not injective
QUESTION [7 upvotes]: It's a well-known exercise in commutative algebra to show that if an A-module endomorphism of a noetherian module M is surjective, it's also injective. 
What are examples that the statement is wrong if we drop the Noetherian hypothesis?

REPLY [8 votes]: The simplest example is probably that of a countable dimensional $\Bbb F$ vector space $V$. Fix a basis $\{b_i\mid i\in \Bbb N\}$
There are two linear transformations which come in handy:

the one given by $A(b_i)=b_{i+1}$
the one given by $B(b_i)=b_{i-1}$ for $i>0$ and $B(b_0)=0$.

If you check these out, you should see that $BA=Id_V$, so $B$ is surjective, but since $B(b_0)=0$ it's clearly not injective.
While we're at it, you might as well notice that $A$ is necessarily injective, but it's clearly not surjective since no multiple of $b_0$ is in its image.

I'd also like to toss out that the exercise is fairly standard for noncommutative rings too :) Modules with this property are called hopfian modules, with the dual notion being "cohopfian" of course. The above example gives us both a nonHopfian an a noncoHopfian $\Bbb F$ module.