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?


1 Answer 1


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.


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