Let \(F\) be a field and \(n\ge1\). Does there exist a surjective ring homomorphism \[ M_{n+1}(F)\longrightarrow M_n(F)? \] Give a proof of your answer.
Solution
For matrix units, write \(E_{ij}\) for the matrix having a single \(1\) in position \((i,j)\) and zeros elsewhere.
1. For every \(m\ge1\), the ring \(M_m(F)\) is simple.
Let \(I\) be a nonzero two-sided ideal of \(M_m(F)\). Choose a nonzero matrix \(A=(a_{ij})\in I\), and choose indices \(i,j\) with \(a_{ij}\neq0\).
For arbitrary \(r,s\), \[ E_{ri}AE_{js}=a_{ij}E_{rs}\in I. \] Since \(a_{ij}\) is invertible in the field \(F\), multiplying by the scalar matrix \(a_{ij}^{-1}I_m\) shows that \[ E_{rs}\in I. \] Thus every matrix unit lies in \(I\), and therefore \(I=M_m(F)\).
2. If a surjective ring homomorphism\[ \varphi:M_{n+1}(F)\longrightarrow M_n(F) \]existed, then \(\varphi\) would be injective.
The kernel of \(\varphi\) is a two-sided ideal of \(M_{n+1}(F)\). Because the target ring is nonzero and \(\varphi\) is surjective, \(\ker\varphi\) is a proper ideal. By step 1, the only proper ideal is \(0\). Hence \(\ker\varphi=0\).
3. Under the assumption of step 2, the matrices\[ P_i=\varphi(E_{ii}), \qquad 1\le i\le n+1, \]are nonzero pairwise orthogonal idempotents in \(M_n(F)\).
Since \[ E_{ii}^2=E_{ii}, \] each \(P_i\) is idempotent. Step 2 shows that \(\varphi\) is injective, so \(P_i\neq0\). For \(i\neq j\), \[ E_{ii}E_{jj}=0=E_{jj}E_{ii}, \] and therefore \[ P_iP_j=0=P_jP_i. \]
4. The vector space \(F^n\) cannot admit \(n+1\) nonzero pairwiseorthogonal idempotent endomorphisms.
Suppose \(P_1,\ldots,P_{n+1}\) were such idempotents. Each \(\operatorname{im}P_i\) is nonzero because \(P_i\neq0\).
These images form a direct sum. Indeed, if \[ v_1+\cdots+v_{n+1}=0, \qquad v_i\in\operatorname{im}P_i, \] then applying \(P_j\) gives \[ v_j=0, \] because \(P_jv_j=v_j\) and \(P_jv_i=0\) for \(i\neq j\).
Hence \[ \operatorname{im}P_1 \oplus\cdots\oplus \operatorname{im}P_{n+1} \] is a direct sum of \(n+1\) nonzero subspaces of the \(n\)-dimensional space \(F^n\), which is impossible.
5. Therefore the answer is\[ \boxed{\text{No}.} \]
4. Thus no such homomorphismexists.
6. Q.E.D.
Step 5 gives the required answer and proof.