Consider the ring $R=M_2(\Bbb Q)$ of $2\times 2$ matrices with entries in $\Bbb Q$. Suppose $S$ is a subring of $R$ containing the center of $R$, and suppose $S$ is an integral domain. I am trying to show that $S$ must be a field.
It is well-known that the center $Z$ of $R$ consists of scalar matrices, so $Z \cong \Bbb Q$. Thus if I can show that I must have $S=Z$, then I will be done, but I can't see whether this is true. Any hints? Thanks in advance.
Best Answer
Basically, your assumption means that $S$ is a $\mathbb{Q}$-subalgebra of $R$ which is an integral domain. Since $R$ is finite dimensional over $\mathbb{Q}$, so is $S$.
But we have the following theorem:
Thm. Let $K$ be a field, and let $A$ be a finite dimensional commutative (unital associative and commutative) $K$-algebra. If $A$ is an integral domain, then $A$ is a field.
Proof. Let $a\in A\setminus\{0\}$, and consider $\ell_a: x\in A\mapsto as\in A$. Since $A$ is a $K$-algebra, this map is $K$-linear endomorphism of the $K$-vector space $A$ . Since $a\neq 0$ and $A$ is an integral domain, $\ell_a$ is injective. Since $A$ is finite dimensional, $\ell_a$ is also surjective. IN particular, there eixsts $a'\in A$ such that $a'a=1$, hence $a$ is invertible and $A$ is a field.