Proving Things About Rings Using Things About Vector Spaces All rings below are assumed to be commutative and having an identity.
$\newcommand{\bw}{\bigwedge}\newcommand{\R}{\mathbf R}\newcommand{\mc}{\mathcal}$
Consider the following problem:

Problem 1. Let $M$ and $N$ be $n\times n$ matrices with entries from a field $F$ of characteristic $0$. Then
  $$\det(MN)=\det M\det N\tag{1}$$

The above can be neatly proved using exterior algebras (I have hidden the details of this since this is not important here. Doing this seems to reduce the length of the post. Long posts tend to decrease the number of readers.):

We know that given given a linear operator $T$ on an $n$-dimensional vector space based on the field $F$, we have a induced map $\bw^k T:\bw^kV\to \bw^k V$. If $k=n$, then since $\dim(\bw^n V)=1$, we kwow that there is a unique constant $c$ such that $\bw^n T=c I$, where $I$ is the identity map on $\bw^n V$.
It is easily shown that $c$ is same as the determinant of the matrix of $T$ which respect to any basis of $V$. Now if $T, S\in \mathcal L(V)$, then
$$
\det(ST)(v_1\wedge \cdots \wedge v_n)=\bigwedge^n(TS)(v_1\wedge \cdots \wedge v_n)
\\=TS(v_1)\wedge\cdots TS(v_n)=\bw^n T(Sv_1\wedge \cdots \wedge Sv_n)
=\det S\ \bw^n T(v_1\wedge\cdots\wedge v_n)
\\=\det S\det T(v_1\wedge\cdots\wedge v_n)
$$

Of course, equation (1) remains true even if we have the entries of $M$ and $N$ coming from a ring. I am not aware of any analog of exterior algebras for arbitrary rings. So I have a workaround.
Consider the polynomial ring $A=\R[x_{ij},y_{ij}:\ 1\leq i,j\leq n]$ in $2n^2$ indeteminates $x_{ij}$ and $y_{ij}$. Let $\mc M$ be the $n\times n$ matrix whose $i,j$-th entry is $x_{ij}$ and $\mc N$ be the matrix whose $i, j$-th entry is $y_{ij}$.
Consider the polynomials $P, Q$ and $R$ in $A$ defined as $P=\det(\mc M\mc N)$, $Q=\det \mc M$, and $R=\det \mc N$. Now let $S\in A$ be defined as $P-QR$.
We know from (1) that $S$ evaluates to $0$ identically and thus $S$ must be the zero polynomial in $A$.
Note that $S$ can be regarded as a polynomial in $Z:=\mathbb Z[x_{ij}, y_{ij}:\ 1\leq i, j\leq n]$.
Now suppose we have matrices $M$ and $N$ with entries from a ring $B$.
By the universal property of polynomial rings, we have a unique ring homomorphism $f:Z\to B$ which carries $x_{ij}$ to the $i,j$-th entry of $M$ and $y_{ij}$ to the $i,j$-th entry of $N$.
Now since $S$ is the zero polynomial, we have $f(S)=0$. But $f(S)=\det(MN)-\det M\det N$. And thus we have proved (1) in the general setting of rings without doing much symbolic computation.
Does the above make sense? It seems a bit artificial. Is there a more direct way (of course, apart from playing with symbols).
This approach (provided all is well) can be used to prove other similar results.
Thanks.
 A: There does exist a notion of exterior powers of modules over arbitrary (commutative) rings, and they can be defined almost exactly the same way as for vector spaces.  Here is a reference that goes through the details.  Using this definition, the proof that determinants are multiplicative over a field generalizes straightforwardly to arbitrary rings.
However, the argument you make using $\mathbb{Z}[x_{ij},y_{ij}]$ is quite clever, and is a handy proof method to keep in mind.  As a minor note, instead of using $\mathbb{R}[x_{ij},y_{ij}]$, you can also just use the fact that $\mathbb{Z}[x_{ij},y_{ij}]$ embeds in its field of fractions $\mathbb{Q}(x_{ij},y_{ij})$ and apply (1) to the matrices over that field with entries $x_{ij}$ and $y_{ij}$.  Alternatively, if you choose $2n^2$ algebraically independent complex numbers $a_{ij}$, $b_{ij}$, then the map $\mathbb{Z}[x_{ij},y_{ij}]\to\mathbb{C}$ sending $x_{ij}$ to $a_{ij}$ and $y_{ij}$ to $b_{ij}$ is injective.  This means that it actually suffices to just know (1) for matrices over $\mathbb{C}$.
As you mention, this trick can be used to prove many other facts about matrices over arbitrary rings.  The best-known application is the following very neat proof of the Cayley-Hamilton theorem.  As in the previous paragraph, the Cayley-Hamilton theorem for matrices with entries in an arbitrary ring can be reduced to the case of matrices with entries in $\mathbb{C}$.  The Cayley-Hamilton theorem is easy to prove for diagonalizable matrices, and it is not too difficult to prove that diagonalizable matrices are dense in all matrices over $\mathbb{C}$ (considering matrices as elements of $\mathbb{C}^{n^2}$ equipped with the usual topology).  Since polynomials are continuous, it follows that the Cayley-Hamilton theorem holds for all complex matrices, and hence for all matrices over any ring.
