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

abstract-algebracommutative-algebralinear algebraring-theory

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 $\mathbb{Z}[x_{11},…,x_{nn},y_{11},…,y_{nn}]$?

Best Answer

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 $\mathbb{Z}[x_{11},...,x_{nn},y_{11},...,y_{nn}]$.

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