Surjective homomorphisms $\mathbb{Z} \oplus \mathbb{Z} \rightarrow \mathbb{Z} \oplus \mathbb{Z}$ are isomorphisms. Is there an easy way to prove that any surjective homomorphism $\mathbb{Z} \oplus \mathbb{Z} \rightarrow \mathbb{Z} \oplus \mathbb{Z}$ is an isomorphism? 
I was told that this holds more generally for any direct sum of $\mathbb{Z}$. This is easy in the case of $\mathbb{Z} \rightarrow \mathbb{Z}$ since any homomorphism has $1 \mapsto n$ for some $n$. 
 A: Given any Noetherian module $M$, a surjection $f:M\to M$ is always an isomorphism. To see why, consider 
$$
ker(f)\subset ker(f^2)\dots ker(f^n)\subset\dots
$$
As $M$ is Noetherian, this chain stabilizes at, say $ker(f^k)$. Let us examine $ker(f^k)\cap im(f^k)$. Assume $x$ is in this intersection; then $f^k(x)=0$, and we can write $f^k(y)=x$ for some $y\in M$. Then $f^k(f^k(y))=f^k(x)=0$, so $y\in ker(f^{2k})=ker(f^k)$. Thus $x$ is in fact zero. 
But now $f$ is surjective, so $f^k$ is also surjective; hence $ker(f^k)\cap im(f^k)=ker(f^k)=\{0\}$, showing that $f^k$ (and hence $f$ as well) is injective. 
Now you just have to show that $\mathbb{Z}\oplus\mathbb{Z}$ is Noetherian!
A: I'm not sure what your background is so I'll offer a most elementary way to reason it specific to $\mathbb{Z}$, although as ThomasGrubb explained there are far more general results (see Nakayama's Lemma).
If $1 \oplus 0 \rightarrow a \oplus b$ and $0 \oplus 1 \rightarrow c \oplus d$ then the homomorphism is completely defined by $x \oplus y \rightarrow (ax + cy) \oplus (bx + dy)$
In matrix notation, 
$\begin{bmatrix}
x \\
y
\end{bmatrix} \rightarrow
\begin{bmatrix}
a & c \\
b & d
\end{bmatrix} \begin{bmatrix}
x \\
y
\end{bmatrix}
$
Clearly the matrix is invertible iff our original homomorphism is.  
The problem thus boils down to showing that a ($2 \times 2$) integer matrix has full column rank iff it is invertible.  This is easy, because if it has full column rank  then we can find $x_1,y_1$ such that
$$\begin{bmatrix}
1 \\
0
\end{bmatrix} =
\begin{bmatrix}
a & c \\
b & d
\end{bmatrix} \begin{bmatrix}
x_1 \\
y_1
\end{bmatrix}
$$ and similarly $x_2, y_2$ such that 
$$\begin{bmatrix}
0 \\
1
\end{bmatrix} =
\begin{bmatrix}
a & c \\
b & d
\end{bmatrix} \begin{bmatrix}
x_2 \\
y_2
\end{bmatrix}
$$
Thus 
$$\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} =
\begin{bmatrix}
a & c \\
b & d
\end{bmatrix} \begin{bmatrix}
x_1 & x_2 \\
x_2 & y_2
\end{bmatrix}
$$
The above argument immediately generalizes to $\bigoplus_n \mathbb{Z}$.
