Zariski Dense Argument in Linear Algebra

1.

Let $k$$k$ be a fixed algebraic closed field and consider the ring $M_{n,n}(k)$$M_{n,n}(k)$​.

Remark. For any integral domain $R$$R$ you could consider the algebraic closure of $Q(R)$$Q(R)$ and the inclusion.

The notation $M^{\vee }$$M^\vee$ refer to the adjugate matrix.

The aim of this blog is to prove that for any matrix $M,N\in M_{n,n}(k)$$M,N\in M_{n,n}(k)$, $(MN{)}^{\vee }=N^{\vee }{M}^{\vee }$$(MN)^\vee=N^\vee M^{\vee}$​ via classic algebraic geometry.

This proof is well-known to those familiar with basic algebraic geometry, it is beautiful and general.

We know that ${\mathbb{A}}_k^{n^2}\cong M_{n,n}(k)$$\mathbb A_k^{n^2}\cong M_{n,n}(k)$ and define a function $f:{\mathbb{A}}_k^{n^2}\times {\mathbb{A}}_k^{n^2}\to {\mathbb{A}}_k^{n^2}$$f:\mathbb A_k^{n^2}\times \mathbb A_k^{n^2}\to \mathbb A_k^{n^2}$

$$\begin{array}{}\text{(1)}& (M,N)⟼(MN{)}^{\vee }-N^{\vee }{M}^{\vee }\end{array}$$

which is continuous since it is a polynomial function as well.

Notice that for invertible matrices, we have $\det M=M{M}^{\vee }⟹\det M{M}^{-1}=M^{\vee }$$\det M=MM^\vee\implies\det M M^{-1}=M^\vee$.

Hence

$$\begin{array}{}\text{(2)}& (MN{)}^{\vee }=\det MN(MN{)}^{-1}=\det N{N}^{-1}\det M{M}^{-1}=N^{\vee }{M}^{\vee }.\end{array}$$

Thus $f{|}_X=0$$f|_X=0$, where $X:=\{(M,N)\in {\mathbb{A}}_k^{n^2}\times {\mathbb{A}}_k^{n^2},\det MN\ne 0\}$$X:=\{(M,N)\in \mathbb A_k^{n^2}\times \mathbb A_k^{n^2},\det MN\ne 0\}$ is an open set with respect to the Zariski topology since $\det MN=\det M\det N$$\det MN=\det M\det N$ is a polynomial function. Notice that ${\mathbb{A}}_k^{n^2}\times {\mathbb{A}}_k^{n^2}\cong {\mathbb{A}}_k^{2n^2}=V(0)$$\mathbb A_k^{n^2}\times \mathbb A_k^{n^2}\cong\mathbb A_k^{2n^2}=V(0)$ is an irreducible space, hence $X$$X$ is dense. Now we know that $0$$0$ is closed and $X\subseteq f^{-1}(0)$$X\subseteq f^{-1}(0)$ is a closed set as well, $X$$X$ is dense implies that

$$\begin{array}{}\text{(3)}& f^{-1}(0)={\mathbb{A}}_k^{n^2}\times {\mathbb{A}}_k^{n^2}\end{array}$$

2.

We could use Zariski Dense argument to prove Cayley–Hamilton theorem for integral domain as well.

First, let's consider the affine space ${\mathbb{A}}_k^{n^2}$$\mathbb{A}_k^{n^2}$ corresponding to $M_{n,n}(k)$$M_{n,n}(k)$, where $k$$k$ is an algebraically closed field.

Note that this space is irreducible because ${\mathbb{A}}_k^{n^2}=V(0)$$\mathbb{A}_k^{n^2} = V(0)$ and $(0)$$(0)$ is a prime ideal, thus all non-empty open sets are dense (here we are considering the Zariski Topology).

Consider $f:{\mathbb{A}}_k^{n^2}\to {\mathbb{A}}_k^{n^2}$$f: \mathbb{A}_k^{n^2} \rightarrow \mathbb{A}_k^{n^2}$, $A\mapsto p_A(A)$$A \mapsto p_A(A)$, note that diagonalizable matrices are contained in $f^{-1}(0)$$f^{-1}(0)$.

And diagonalizable matrices with $n$$n$ distinct eigenvalues is equivalent to requiring the characteristic polynomial to be separable, that is, the discriminant is not $0$$0$. This means matrices with $n$$n$ distinct eigenvalues form an open set $U$$U$, thus dense.

Therefore $U\subseteq f^{-1}(0)$$U \subseteq f^{-1}(0)$, and since $0$$0$ is closed, this implies $f^{-1}(0)$$f^{-1}(0)$ is closed, and since $U$$U$ is dense, we have $f^{-1}(0)={\mathbb{A}}_k^{n^2}$$f^{-1}(0) = \mathbb{A}_k^{n^2}$.

However, from here we can generalize this to work for any ring because you can first consider the fraction field of the ring and then embed $R^{n^2}$$R^{n^2}$ into ${\mathbb{A}}_k^{n^2}$$\mathbb{A}_k^{n^2}$.