Aluffi's proof that $\det(AB)=\det(A)\det(B)$ for commutative rings

Question

In Aluffi he proves that $\det(AB)=\det(A)\det(B)$ for commutative rings. I am not following the proof and hoping that someone will clarify it for me.

Earlier this was proved for fields. So to generalize this to commutative rings he gives the following argument:

First express this as a function of matrices in indeterminate entries.

$$ \det \pmatrix{ x_{11} & ... & x_{1n} \\ \vdots & \ddots & \vdots \\ x_{n1} & \dots &x_{nn}} \det \pmatrix{ y_{11} & ... & y_{1n} \\ \vdots & \ddots & \vdots \\ y_{n1} & ... &y_{nn}}= \det\pmatrix{ x_{11} y_{11} + \dots + x_{1n}y_{n1} & ... & x_{11} y_{1n} + \dots + x_{1n}y_{nn}\\ \vdots & \ddots & \vdots \\ x_{n1} y_{11} + \dots + x_{nn}y_{n1} & ... & x_{n1} y_{1n} + \dots + x_{nn}y_{nn}} $$

This can be translated into a polynomial identity and if we can show that this is true for $\mathbb{Z}[x_{11},...,x_{nn},y_{11},...,y_{nn}]$ then it must hold for any commutative ring since $\mathbb{Z}$ is initial in $\bf{Ring}$.

He then states that this holds for the field of fractions of $\mathbb{Z}[x_{11},...,x_{nn},y_{11},...,y_{nn}]$, so it must also hold for $\mathbb{Z}[x_{11},...,x_{nn},y_{11},...,y_{nn}]$.

My questions about this are:

1. How does $\mathbb{Z}$ being initial in $\bf{Ring}$, help us when we are proving an identity that holds for $\mathbb{Z}[x_{11},...,x_{nn},y_{11},...,y_{nn}]$?
2. My second question is about the field of fractions argument. Doesn't that immediately give a proof for all integral domains without reference $\mathbb{Z}[x_{11},...,x_{nn},y_{11},...,y_{nn}]$?

Answer 1

If $A=(a_{ij})$ and $B=(b_{ij})$ are in $M_n(R)$, where $R$ is a commutative unitary ring, then there is a homomorphism $\phi:\mathbb{Z}[x_{11},...,x_{nn},y_{11},...,y_{nn}]\to R$ such that $\phi(n)=n\cdot 1_R$ for $n\in\mathbb Z$, $\phi(x_{ij})=a_{ij}$ and $\phi(y_{ij})=b_{ij}$. If one proves that $\det(XY)=\det X\det Y$ for the matrices $X=(x_{ij})$, respectively $Y=(y_{ij})$, then apply $\phi$ to this relation and get $\det(AB)=\det A\det B$. But you know that $\det(XY)=\det X\det Y$ holds for matrices $X,Y$ whose entries are in a field. This is why the book suggests to look at these matrices in the field of fractions of the ring $\mathbb{Z}[x_{11},...,x_{nn},y_{11},...,y_{nn}]$.

Answer 2

$\newcommand{\x}{\mathbf{x}}\newcommand{\y}{\mathbf{y}}$Second question: Yes, by appealing to the field of fractions, you can prove that $\det(AB) = \det(A) \det(B)$ for square matrices $A, B$ over any integral domain.

First question:
Here is the main point: Consider the ring $A = \Bbb Z[x_1, \ldots, x_n] = \Bbb Z[\x]$. Suppose you have two polynomials $f(\x), g(\x) \in A$ such that $f(\x) = g(\x)$. Then, for any ring $R$, we can interpret $f(\x)$ and $g(\x)$ as polynomials in $R[\x]$. Moreover, $f(\x)$ and $g(\x)$ continue to be equal as polynomials in $R[\x]$. Thus, by considering the evaluation map, you can substitute $x_1, \ldots, x_n$ as elements from the ring $R$ and you have an identity.
For example, we have the polynomial equality $$(x_1 - x_2)(x_1 + x_2) = x_1^2 - x_2^2$$ in $\Bbb Z[x_1, x_2]$. This leads to the fact that $$(a - b)(a + b) = a^2 - b^2$$ in any (commutative and unital) ring.

The formal proof of this is obtained by using the fact that given elements $a_1, \ldots, a_n \in R$, there is a well-defined (unique) homomorphism $\Bbb Z[\x] \to R$ such that $x_i \mapsto a_i$. (The initial-ity of $\Bbb Z$ in $\mathbf{Ring}$ determines the map on the level of the constants uniquenely.)

To apply it to your case, note that the $\det(AB)$ and $\det(A)\det(B)$ can be viewed as polynomials in $\Bbb Z[\x, \y]$ (when you set all the entries of $A$ and $B$ to be interdeterminates) and you have shown that they are equal there. Thus, they are equal over any ring.

Answer 3

Answer: A "Bourbaki-proof": You may use the exterior product to prove the formula for any two matrices $A,B$ (of the same rank) over any commutative unital ring:

Let $E:=R\{e_1,..,e_n\}$ be a free $R$-module of rank $n$ and let $\def\End{\operatorname{End}} A,B\in \End_R(E)$ be $n \times n$-matrices with coefficients in $R$ ($R$ any commutative unital ring). It follows $$\wedge^n A, \wedge^n B \in \End_R(\wedge^n E)$$ are $R$-linear endomorphisms of the rank one free $R$-module $\wedge^n E \cong R e_1\wedge \cdots \wedge e_n:=R e$. It has the property that

$$\wedge^n A(ue) = \det(A) ue$$

is "multiplication with the element $\det(A)\in R$".

From this it follows since $\wedge^n$ is a functor (see the below link) that

$$ \det(AB) := \wedge^n (A \circ B) = \wedge^n A \circ \wedge^n B := \det(A) \det(B). $$

Prove that $\det(T) = \det(T_{w_{1}})\det(T_{w_{1}})\cdots \det(T_{w_{k}})$ where $W_{i}$ are $T$-invariant subspaces of $V$ - Trouble at last step.