Let $A$ be a $k$-algebra of Krull dimension $1$, where $k$ is a field. Let $n\geq 2$ a natural number and $\frak{p}$ a prime ideal of $A^{\otimes n}$ which is not maximal. Consider the ring homomorphisms \begin{align*} f_i: A&\to A\otimes_k\cdots\otimes_k A\\ a &\mapsto 1\otimes\cdots\otimes a \otimes\cdots \otimes1 \end{align*}
for $i=1,...,n$, where the $a$ appears at the $i$-th entry.
Is it true that at least one of the $f_{i}^{-1}(\frak{p})$ is zero? What if $A$ is a PID and $k$ is algebraically closed?
Note that $\frak{p}\subset\frak{m}$ for some maximal ideal, so dim$A=1$ implies it is enough to show that one of the inclusions $f_{i}^{-1}(\frak{p})\subseteq$ $f_{i}^{-1}(\frak{m})$ is strict.
Remark: my question comes from geometry, since it is equivalent to asking if at least one of the projections Spec$(A)\times\cdots\times $Spec$(A)\to $Spec$(A)$ is dominant when restricted to an irreducible closed subscheme of dimension $>0$. I would really like to have a direct algebraic proof, but if you can easily answer invoking a known geometric result (with reference) that's also fine.
If $\operatorname{Spec} A$ is irreducible, this result follows. Assume there exists a positive-dimensional irreducible closed subscheme $X\subset \prod_i \operatorname{Spec} A$ with no projection dominant. Then $f_i(X)$ is irreducible as it's the continuous image of an irreducible set, and since it cannot contain any dense set, it must be a point. As $X\subset \bigcap_i f_i^{-1}(f_i(X))$, this implies that $X$ is a point, contradiction.
If $\operatorname{Spec} A$ is not irreducible, then the image of $X$ cannot be dominant, as $X$ is irreducible and the continuous image of an irreducible set is again irreducible.