Module homomorphism surjective but not injective 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?
 A: 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.
A: Consider the Prüfer $p$-group $\mathbb{Z}(p^\infty)$, which is artinian. If you quotient by the minimal nonzero subgroup, you get a group isomorphic to $\mathbb{Z}(p^\infty)$. So you can define an endomorphism of $\mathbb{Z}(p^\infty)$ which is surjective but not injective.
You could also use a free module of infinite rank, say $M=R^{\mathbb{N}}$, with base $(e_n)_{n\in\mathbb{N}}$ and define $f(e_n)=e_{n+1}$, then use again an isomorphism of the image with $M$.
Another, perhaps surprising, example, is that of a finitely generated module with the property. Consider a ring $R$ such that the free right $R$-module $R_R$ is isomorphic to $R\oplus R$. So, if $f\colon R\to R\oplus R$ is this isomorphism, $p\colon R\oplus R\to R$ is the projection on the first component and $i\colon R\to R\oplus R$ is the injection into the first component, we see that
$$
p\circ f\colon R\to R
$$
is surjective but not injective, while
$$
f^{-1}\circ i\colon R\to R
$$
is injective but not surjective.
The standard example of such a ring is $R=\operatorname{End}(V)$, where $V$ is an infinite dimensional vector space over a field $F$. Rings where $R^m\cong R^n$ implies $m=n$ (for $m,n\in\mathbb{N}$) are said to have invariant basis number. Every commutative ring has IBN, but noncommutative ones may not have this property. There are, for any $k\in\mathbb{N}$, examples of rings such that $R^m\cong R^n$ if and only if $m\equiv n\pmod{k}$.
