Show that the ring \(M_n(F)\) of all \(n\times n\) matrices over a field \(F\) has no proper nonzero two-sided ideals.
Solution
Let \(E_{pq}\) denote the matrix unit whose \((p,q)\) entry is \(1\) and whose other entries are \(0\).
1. Let \(I\subseteq M_n(F)\) be a nonzero two-sided ideal. There are\(A=(a_{ij})\in I\) and indices \(i,j\) such that\[ a_{ij}\neq0. \]
Because \(I\) is nonzero, choose a nonzero matrix \(A\in I\). At least one entry of a nonzero matrix is nonzero; call such an entry \(a_{ij}\).
2. For every \(p,q\),\[ a_{ij}E_{pq}\in I. \]
Since \(I\) is a two-sided ideal, \[ E_{pi}AE_{jq}\in I. \] Multiplication by \(E_{pi}\) keeps only row \(i\) of \(A\) and moves it to row \(p\), while multiplication by \(E_{jq}\) keeps only column \(j\) and moves it to column \(q\). Hence \[ E_{pi}AE_{jq}=a_{ij}E_{pq}. \]
3. Every matrix unit \(E_{pq}\) belongs to \(I\).
The scalar \(a_{ij}\) is nonzero in the field \(F\), so it is invertible. By step 2, \[ a_{ij}E_{pq}\in I. \] Because \(I\) is a left ideal and \[ a_{ij}^{-1}I_n\in M_n(F), \] we obtain \[ E_{pq} = (a_{ij}^{-1}I_n)(a_{ij}E_{pq}) \in I. \]
4. The ideal \(I\) is all of \(M_n(F)\).
By step 3, each diagonal matrix unit \(E_{pp}\) lies in \(I\), so \[ I_n=\sum_{p=1}^n E_{pp}\in I. \] For every \(B\in M_n(F)\), the two-sided ideal property then gives \[ B=BI_n\in I. \] Thus \(I=M_n(F)\).
5. The only two-sided ideals of \(M_n(F)\) are\[ \boxed{0\text{ and }M_n(F)}. \]
The zero ideal is a two-sided ideal. Step 4 shows that every nonzero two-sided ideal is the whole ring.
6. Q.E.D.
Step 5 is the required conclusion.