  Let $R$ be a commutative ring with $1$ and $n$ a natural number. We say that
a submodule $N$ of $R^n$ is semiprime if for every $f=(f_1,\ldots,f_n) \in R^n$
such that $f_i f \in N$ for $i=1,\ldots,n$ we have $f \in N$. Our main result
is that every semiprime submodule of $R^n$ is equal to the intersection of all
prime submodules containing it. It follows that every semiprime left ideal of
$M_n(R)$ is equal to the intersection of all prime left ideals that contain it.
For $R=k[x_1,\ldots,x_d]$ where $k$ is an algebraically closed field we can
rephrase this result as a Nullstellensatz for $M_n(R)$: For every
$G_1,\ldots,G_m,F \in M_n(R)$, $F$ belongs to the smallest semiprime left ideal
of $M_n(R)$ that contains $G_1,\ldots,G_m$ iff for every $a \in k^d$ and $v \in
k^n$ such that $G_1(a)v=\ldots=G_m(a)v=0$ we have $F(a)v=0$.
