No monomorphism $\mathbb{Z}^3\rightarrow \mathbb{Z}^2$ Should be simple enough, yet I can't show that there are no monomorphisms $\mathbb{Z}^3\rightarrow \mathbb{Z}^2$. (It is true, right?)
 A: Hint If $f(x): \mathbb Z^3 \to  \mathbb Z^2$ is any morphism then 
$$f(1,0,0), f(0,1,0), f(0,0,1) \in \mathbb Z^2 \subset \mathbb Q^2$$
Then, they must be linearly dependent over $\mathbb Q$ since $ \mathbb Q^2$ is a 2 dimensional $\mathbb Q$-vector space. It is trivial to prove that linear dependence over $\mathbb Q$ implies linear dependence over $\mathbb Z$.
Alternately $f$ is given by a $2 \times 3$ matrix with rational entries. Then the reduced row echelon form of $A$ has all rational entries so $\ker(A)$ contains a vector with all entries rational. Prove now that you can find another vector with all entries integer.
A: Hint: Show that given any $x, y, z \in \mathbb Z^2$ there exist non-zero integers $a, b, c \in \mathbb Z$ such that
$$ax + by + cz = (0, 0)$$
(You should be able to give an explicit formula for $a, b, c$ in terms of the entries of $x, y, z$)
Next show that if you had a monomorphism $\phi\colon\mathbb Z^3 \to \mathbb Z^2$ then there would be no such $a, b, c$ for the vectors $\phi(0, 0, 1), \phi(0, 1, 0), \phi(0, 0, 1)$.
A: If we have such that monomorphism $$\phi:\mathbb Z^3\to\mathbb Z^2$$ then according to first theorem of homomorphism we get $\mathbb Z^3\leq\mathbb Z^2$.
