Characterize the two-sided ideals of a matrix ring

Let R be a ring and n \geq 2 an integer. Prove that every two-sided ideal of M_n(R) is equal to M_n(J) for some two-sided ideal J \subseteq R.


First, let I \subseteq M_n(R) be a two-sided ideal. Let J \subseteq R consist of all elements r \in R such that r = (A)_{i,j} for some A \in I and 1 \leq i,j \leq n. We will first show that J is an ideal of R, and then show that I = M_n(J).

Let x = (A)_{p,q} and y = (B)_{s,t} be in J, with A,B \in I. Recall from this previous exercise that E_{p,q}AE_{s,t} is the matrix whose (p,t) entry is the (q,s) entry of A and all other entries are zero.

Since I is a two sided ideal, we have that E_{1,p}AE_{q,2} and E_{1,s}BE_{t,2} are in I. Moreover, E_{1,p}AE_{q,2} - E_{1,s}BE_{t,2} \in I, and the (1,2) entry of this matrix is x-y. Thus J is closed under subtraction.

Since I is an ideal, we have 0 \in I. Thus 0 \in J, and J is an additive subgroup of R.

Now it suffices to show that for all r \in R and x \in J, xr, rx \in J. To that end, note that (rI)A = rA \in I, and that the (p,q) entry of rA is rx; thus rx \in J. Similarly, since Ar \in I, xr \in J. Hence J \subseteq R is an ideal.

It is clear that I \subseteq M_n(J); To show the other direction, let A = [x_{i,j}] \in M_n(J). Since each x_{i,j} \in J, we have x_{i,j} = (B_{i,j})_{a_{i,j},b_{i,j}} for some matrix B_{i,j} \in I, for each i and j, and for some a_{i,j} and b_{i,j}. In particular, x_{i,j}E_{i,j} = E_{i,a_{i,j}}A_{i,j}E_{b_{i,j},j} \in I for each i and j. Note then that A = \sum_{1 \leq i,j \leq n} x_{i,j}E_{i,j} = \sum_{0 \leq i,j \leq n} E_{i,a_{i,j}}A_{i,j}E_{b_{i,j},j} \in I, since I is an ideal. Thus M_n(J) \subseteq I, and we have I = M_n(J).

Now we need to show that if J \subseteq R is a two-sided ideal, then M_n(J) \subseteq M_n(R) is a two-sided ideal.

To that end, let A = [x_{i,j}] and B = [y_{i,j}] be in M_n(J). Then x_{i,j}, y_{i,j} \in J, so that x_{i,j} - y_{i,j} \in J. Hence A+B = [x_{i,j} - y_{i,j}] \in M_n(J). Thus M_n(J) is closed under subtraction.

Since 0 \in J, we have 0 \in M_n(J). Thus M_n(J) is an additive subgroup of M_n(R).

Finally, let T = [t_{i,j}] \in M_n(R); it suffices to show that AT, TA \in M_n(J). Recall that AT = [c_{i,j}], where c_{i,j} = \sum_{k=1}^n a_{i,k}b_{k,j}. Since J \subseteq R is a two-sided ideal, c_{i,k} \in J. Thus AT \in M_n(J). Similarly, TA \in M_n(J). Thus M_n(J) \subseteq M_n(R) is a two-sided ideal.

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