$V(X_1,\ldots,X_n)$ has empty interior.

Question

Let $A$ be a ring, $n\geq 1$ an integer and $Z=V(X_1,\ldots,X_n)\subset\mathbb{A}^n_A=\operatorname{Spec} A[X_1,\ldots,X_n]$. It is true that $\mathring Z=\emptyset$? Surely this is the case if $A$ is a field. (Reason: If $A$ is a field, then $(X_1,\ldots,X_n)$ is a closed point. But $\mathbb{A}^n_A$ is irreducible and therefore connected, so $Z=\{(X_1,\ldots,X_n)\}$ must have empty interior.)

Answer 1

It is certainly true when $A$ is integral, or even irreducible as one may base change with respect to $A\to A/\text{rad}(A)$ which is a homeomorphism on the underlying topological space.

If $A$ is noetherian write $Y=\text{Spec A}$ we have $\mathbb{A}^n_Y=\mathbb{A}^n_{Y_1}\cup...\cup \mathbb{A}^n_{Y_p}$ where the $Y_i$'s are the irreducible components of $Y$. In fact it is clear that those are the irreducible components of $\mathbb{A}^n_Y$.

Now $U$ is open in $\mathbb{A}^n_Y$ iff $U\cap \mathbb{A}^n_{Y_i}$ is open in each $\mathbb{A}^n_{Y_i}$ (because it suffices to check the corresponding fact for closed subsets and the union of finitely many closed subsets is closed). That means that $V(X_1,..., X_n)$ has empty interior whenever $A$ is noetherian.

In fact this gives you the general result as an open subset $U$ contained in $V(X_1,...,X_n)$ must be empty when restricted to $\mathbb{A}^n_Y$ where $Y$ is an irreducible component of $A$ and $U=\bigcup_i (U\cap \mathbb{A}^n_{Y_i})$, regardless of whether there are finitely many of them.