If $D$ is an integral domain, show $M_n(D)$ is a prime ring

Question

So for $M_n(D)$ to be a prime ring it means that if $A$,$B$ are ideals of $M_n(D)$ and $AB=0$ then either $A=0$ or $B=0$.

The fact that $D$ is an integral domain seems quite crucial, because this means that if $a,b \in D$ and $ab=0$ then either $a=0$ or $b=0$.

Anyway, lets let $A,B$ be ideals of $D$ s.t. $AB=0$.

$AB = \{ab : a \in A, b \in B\}$

I have also deduced that if $A$ is an ideal of $M_n(D)$, then $A$ is of the form $M_n(I)$ where $I$ is an ideal of $D$.

I'm gonna need some advice from this point.. Thanks everyone!

Answer 1

Let's prove the contrapositive: if $A \ne 0$ and $B \ne 0$, then $AB \ne 0$.

Using your fact, let $A = M_n(I)$ and $B = M_n(J)$ for some ideals $I,J$ of $D$. If $A \ne 0$ and $B \ne 0$, then there are nonzero elements $x \in I$ and $y \in J$, so that $x\mathbf{1} \in A$ and $y\mathbf{1} \in B$, where $\mathbf{1}$ is the identity matrix (so $x\mathbf{1}$ is the matrix with $x$ on the diagonal and $0$ elsewhere). Then $xy\mathbf{1} = (x\mathbf{1})(y\mathbf{1}) \in AB$ is nonzero since $D$ is an integral domain.

Answer 2

Assume that $A,B \ne 0$.

Let $M = (a_{ij})\in A$ with $M \ne 0$. There exists an element $a_{rs} \ne 0$.

Denote $E_{ij}$ the matrix with $1$ as the position $(i,j)$ and $0$ elsewhere. We have

$$a_{rs}E_{11} = E_{1r}ME_{s1} \in A$$

Similarly, take $N = (b_{ij})\in B$ with $b_{kl} \ne 0$ so $b_{kl}E_{11} \in B$.

Therefore $$0 \ne (a_{rs}b_{kl})E_{11} = (a_{rs}E_{11})(b_{kl}E_{11}) \in AB$$

Note that $a_{rs}b_{kl} \ne 0$ because $D$ is an integral domain.

Answer 3

Another way: with not much effort you can show that $M_n(I)M_n(J)=M_n(IJ)$.

Then $M_n(I)M_n(J)=M_n(IJ)=M_n(0)$ implies $I=0$ or $J=0$.

Superficially it looks like we’re not using elements of $R$, but you would probably prove my initial claim with elements.