= Solution
Each entry of $\operatorname{char}_A(A)$ is a polynomial in the entries of $A$, so their simultaneous zero set $Z$ is <Zariski closed>. A matrix in $D(\Delta)$ has distinct eigenvalues and is diagonalizable. If $A=S\operatorname{diag}(\lambda_1,\ldots,\lambda_n)S^{-1}$, then
$$
\operatorname{char}_A(A)
=S\operatorname{diag}(\operatorname{char}_A(\lambda_1),\ldots,
\operatorname{char}_A(\lambda_n))S^{-1}=0.
$$
Thus $D(\Delta)\subseteq Z$. Affine space is irreducible because its coordinate ring is an integral domain; by part (i), its nonempty open subset $D(\Delta)$ is dense. The closed set $Z$ containing it must be all of affine space. This proves the <Cayley-Hamilton theorem>:
$$
\boxed{\operatorname{char}_A(A)=0\quad\text{for every }A.}
$$
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