Let \(R=M_n(F)\) be the ring of \(n\times n\) matrices over a field \(F\). If \(S\) is a ring and \[ h:R\to S \] is a ring homomorphism, show that \(h\) is either injective or zero.
Solution
Let \(E_{ij}\) denote the standard matrix unit with a \(1\) in position \((i,j)\) and zeros elsewhere.
1. Every nonzero two-sided ideal of \(R=M_n(F)\) is all of \(R\).
Let \(I\subseteq R\) be a nonzero two-sided ideal and choose \[ 0\neq A=(a_{rs})\in I. \] There are indices \(p,q\) with \(a_{pq}\neq0\). For arbitrary \(i,j\), \[ E_{ip}AE_{qj}=a_{pq}E_{ij}\in I. \] Because \(a_{pq}\) is invertible in \(F\), multiplication by the scalar matrix \(a_{pq}^{-1}I_n\) gives \(E_{ij}\in I\). Hence \[ I_n=E_{11}+\cdots+E_{nn}\in I. \] Since an ideal containing the identity is the whole ring, \(I=R\).
2. The kernel of \(h\) is a two-sided ideal of \(R\).
The kernel is an additive subgroup. If \(A\in\ker h\) and \(B\in R\), then \[ h(BA)=h(B)h(A)=0, \qquad h(AB)=h(A)h(B)=0. \] Thus \(BA,AB\in\ker h\).
3. Therefore\[ \boxed{h=0\quad\text{or}\quad h\text{ is injective}}. \]
By step 2, \(\ker h\) is a two-sided ideal of \(R\), so step 1 gives \[ \ker h=R \qquad\text{or}\qquad \ker h=0. \] In the first case \(h\) is the zero homomorphism. In the second case \(h\) is injective.
4. Q.E.D.
Step 3 is the required dichotomy.