# Understanding a proof of determinant of products for commutative rings

I am working in Aluffi's Algebra: Chapter 0 text. The following is a part of the proof that $$\det(AB) = \det(A)\cdot\det(B)$$ for any commutative ring $$R$$.

In order to verify this over every ring it suffices to verify the "universal" identity obtained by writing out the claimed equality for matrices with indeterminate matrices. For example, for $$n = 2$$, the statement

$$\det\left(\begin{matrix}x_1 & x_2\\ x_3 & x_4\end{matrix}\right)\det\left(\begin{matrix}y_1 & y_2\\ y_3 & y_4\end{matrix}\right) = \det\left(\begin{matrix}x_1y_1+x_2y_3 & x_1y_2 + x_2y_4\\ x_3y_1 + x_4y_3 & x_3y_2+x_4y_4\end{matrix}\right)$$

translates into the identity

$$(x_1x_4 - x_2x_3)(y_1y_4-y_2y_3) = (x_1y_1+x_2y_3)(x_3y_2+x_4y_4) - (x_1y_2 + x_2y_4)(x_3y_1 + x_4y_3)$$

Since this identity holds in $$\mathbb{Z}[x_1,\ldots,y_4]$$, it must hold in any commutative ring for any choice of $$x_1,\ldots,y_4$$: indeed $$\mathbb{Z}$$ is initial in $$\mathsf{Ring}$$.

I worked the details out on my own but I didn't use the fact that $$\mathbb{Z}$$ is initial.

If we let $$A = (x_{ij}),B=(y_{ij})$$ be $$n\times n$$ matrices of indeterminates, then define a polynomial in $$2n^2$$ variables given by

$$d(A,B) = \det A\det B-\det AB$$

Given two matrices $$M=(a_{ij}),N=(b_{ij})\in\text{Mat}(n;R)$$, there is an evaluation function

$$e:\mathbb{Z}[x_{11},\ldots,x_{nn},y_{11},\ldots,y_{nn}]\rightarrow R$$

sending $$f\mapsto f(a_{11},\ldots,a_{nn},b_{11},\ldots,b_{nn})$$. Since $$e$$ is a ring homomorphism, $$e(0) = 0$$ and $$d = 0$$--this I worked out too--$$e(d) = 0$$ which shows the determinant formula in any commutative ring.

My questions are:

1. How is the fact that $$\mathbb{Z}$$ is initial in $$\mathsf{Ring}$$ used in the original proof?

2. When does it come into play in my argument?

Let $r_1, ... , r_t$ be elements of a unital commutative ring $R$. Let $A$ be a ring, and let $\phi: A \rightarrow R$ be a ring homomorphism.
Then $\phi$ always extends to a ring homomorphism $A[X_1, ... , X_n] \rightarrow R$ given on monomials by
$$a X_1^{m_1} \cdots X_n^{m_n} \mapsto \phi(a)r_1^{m_1} \cdots r_n^{m_n}$$
This is where $\mathbb{Z}$ being initial comes in: in order to define $e$, you need a ring homomorphism $\mathbb{Z} \rightarrow R$ to begin with.