Ideals of a matrix ring
Let D be a division ring and R = M_n(D) the ring of n×n matrices over D. The headline: R is a simple ring — its only two-sided ideals are 0 and R itself. This is dramatically false for commutative quotient rings, where ideals are everywhere. Matrix rings have almost none, and that scarcity is the whole point.
The proof is hands-on. Let E_ij be the matrix unit with a 1 in position (i,j). If a two-sided ideal I contains any nonzero matrix A, it contains some entry a = A_kl ≠ 0. Then E_ik A E_lj sits in I and equals a·E_ij, and since a is invertible in D we can multiply to clean it up to E_ij. Once one matrix unit is in I, all of them are, and they sum to the identity — so I = R.
Work in M_3(D). Suppose A in I with A_21 = a != 0.
E_12 * A * E_13 picks out entry A_21 = a and reseats it:
= a * E_13 in I.
Left-multiply by (a^-1) * E_11 (allowed, scalar a^-1 in D):
(a^-1) E_11 * (a E_13) = E_13 in I.
Conjugating E_13 by permutation matrix units gives E_11, E_22, E_33;
their sum is the identity matrix => 1 in I => I = M_3(D).
Moral: one nonzero element of an ideal already forces the whole ring.Left ideals are not two-sided
These column modules are all isomorphic to one another, and any simple module over M_n(D) is isomorphic to this one column module D^n. So a matrix ring has, up to isomorphism, exactly one simple module. That single fact is the seed of the whole classification to come.
The opposite ring
When order matters, left and right are genuinely different, and the bookkeeping device for switching sides is the opposite ring R^op: same underlying set and addition, but new product a ∗ b := ba. A left R-module is the same data as a right R^op-module. Transpose gives a slick fact: M_n(D)^op ≅ M_n(D^op), because (AB)^T = B^T A^T flips the order for you.