Endomorphism Ring of $\mathbb Z\times\mathbb Z$

Detailed Solution

The abelian group $\mathbb Z\times\mathbb Z$ is a free abelian group of rank $2$. Every group endomorphism of $\mathbb Z\times\mathbb Z$ is therefore determined by the images of the standard generators $$ e_1=(1,0),\qquad e_2=(0,1). $$ Thus every endomorphism can be represented by a $2\times2$ integer matrix $$ A= \begin{pmatrix} a&b\\ c&d \end{pmatrix}, \qquad a,b,c,d\in\mathbb Z. $$ Under this correspondence, addition of endomorphisms corresponds to matrix addition, while composition corresponds to matrix multiplication. Hence $$ R\cong M_2(\mathbb Z), $$ the ring of $2\times2$ matrices with integer entries. Matrix multiplication is not commutative in general. For example, consider $$ A= \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}, \qquad B= \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}. $$ Then $$ AB= \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}, $$ whereas $$ BA= \begin{pmatrix} 0&0\\ 0&0 \end{pmatrix}. $$ Therefore, $$ AB\neq BA. $$ Thus $M_2(\mathbb Z)$, and hence $R$, is noncommutative. Option C is false because $R$ does have nonzero zero divisors. For the matrices above, $$ A\neq0,\qquad B\neq0,\qquad AB\neq0, $$ so these particular matrices do not give zero divisors in that order. However, taking $$ C= \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}, \qquad D= \begin{pmatrix} 0&0\\ 1&0 \end{pmatrix}, $$ gives $$ CD= \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}, $$ so this pair also does not give zero. Instead, take $$ E= \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}, \qquad F= \begin{pmatrix} 0&0\\ 0&1 \end{pmatrix}. $$ These are both nonzero, but $$ EF= \begin{pmatrix} 0&0\\ 0&0 \end{pmatrix}. $$ Hence $R$ has nonzero zero divisors. Option D is false because not every nonzero matrix is invertible. For example, $$ \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix} \neq0 $$ has determinant $0$, so it is not a unit in $M_2(\mathbb Z)$. Therefore, the correct statement is that $R$ is a noncommutative ring. Thus the correct option is $\boxed{\mathrm{B}}$.