Let M be the $n\times m$ matrix representing the injection $A^m \to A^n$. Define Di(M) to be the ideal generated by the determinants of all i-by-i minors of M. Let r be the largest possible integer such that Dr(M) has no annihilator (i.e. there is no nonzero element a∈A such that aDr(M)=0); I think r is usually called the McCoy rank of M.
Assume that $m>n$. We shall get a contradiction.
Choose a nonzero a∈A such that aDr+1(M)=0. By assumption, $a$ does not annihilate Dr(M), so there is some r-by-r minor that is not killed by $a$; we may assume it is the upper-left r-by-r minor. Thus, $r \leq n$, so that $r+1 \leq n+1\leq m$. Let M1, ..., Mr+1 be the cofactors of the upper-left (r+1)-by-(r+1) minor obtained by expanding along the bottom row. (This is well-defined even if the $r+1$-th row of $M$ does not exist, because these cofactors use only the first $r$ rows and the first $r+1$ columns of $A$, and $A$ has both since $r \leq n$ and $r+1 \leq m$.) Note, in particular, we know Mr+1 is the determinant of the upper-left r-by-r minor, so aMr+1≠0.
The claim is that the vector (aM1,...,aMr+1,0,0,...) (which we've already shown is non-zero) is in the kernel of M. To see that, note that when you dot this vector with the k-th row of M, you get $a$ times the determinant of the matrix obtained by replacing the bottom row of the upper-left (r+1)-by-(r+1) minor with the first r+1 entries in the k-th row. If k≤r, this determinant is zero because a row is repeated, and if k>r, this determinant is the determinant of some (r+1)-by-(r+1) minor, so it is annihilated by $a$. But since A is injective, the kernel of M is 0, and we have a contradiction.
Although in the body of your question you mention the action of $SU(2)$ on the Riemann sphere, the simplest (to my mind) answer to the question itself is to understand $SU(2)$ as the unit-norm quaternions acting by conjugation on the imaginary quaternions. It is easy to see that this defines a homomorphism $SU(2) \to SO(3)$, which is surjective (since both $SU(2)$ and $SO(3)$ are connected) and has kernel consisting of the quaternions $\pm 1$.
You might find this in one of Coxeter's books or in Elmer Rees's Notes on Geometry.
Best Answer
Dear Tim, on page 31 they consider a ring $A$ and two $A$- algebras defined by their structural ring morphisms $f:A\to B$ and $g:A\to C$. They then define the tensor product as a ring $D=B\otimes _A C$ and want to make it an $A$- algebra. For that they must define the structural morphism $A\to D$ and they claim that it is given by the formula $a \to f(a)\otimes g(a)$.This is false since that map is not a ring morphism. The correct structural map $A\to D$ is actually $a\mapsto 1_B\otimes g(a) =f(a)\otimes 1_C$.
PS: To prevent misunderstandings, let me add that Atiyah-MacDonald is, to my taste, the best mathematics book I have ever seen, all subjects considered.