Show that the ring of all \(n\times n\) matrices over a field has no two-sided ideals other than \(0\) and the whole ring.
Solution
Let \(k\) be the field and write \[ R\coloneqq M_n(k). \] For \(1\leq r,s\leq n\), let \(E_{rs}\) denote the standard matrix unit.
1. Let \(I\subseteq R\) be a nonzero two-sided ideal. Then \(I\) containsat least one matrix unit \(E_{rs}\).
Choose a nonzero matrix \[ A=(a_{ij})\in I. \] There are indices \(p,q\) with \[ a_{pq}\neq0. \] For arbitrary indices \(r,s\), two-sided ideal closure gives \[ E_{rp}AE_{qs}\in I. \] A direct multiplication of matrix units gives \[ E_{rp}AE_{qs} = a_{pq}E_{rs}. \] In particular, taking any fixed \(r,s\) and multiplying by the scalar \(a_{pq}^{-1}\in k\) shows \[ E_{rs}\in I. \]
2. Every matrix unit \(E_{rs}\) belongs to \(I\).
The computation in step 1 holds for arbitrary \(r\) and \(s\), while the same nonzero entry \(a_{pq}\) of \(A\) is used throughout. Therefore \[ E_{rs}\in I \] for every pair of indices.
3. Every nonzero two-sided ideal \(I\) equals \(R\).
The matrix units \[ \{E_{rs}:1\leq r,s\leq n\} \] form a \(k\)-basis of \(M_n(k)\). By step 2, they all lie in \(I\). Since an ideal is in particular an additive subgroup closed under multiplication by scalar matrices, it contains every \(k\)-linear combination of the matrix units. Hence \[ I=R. \]
4. The only two-sided ideals of \(M_n(k)\) are\[ \boxed{0\text{ and }M_n(k)}. \]
The zero ideal is a two-sided ideal, and step 3 shows that every nonzero two-sided ideal is the whole ring.
5. Q.E.D.
Step 4 is the required classification.