Let be a ring and an integer. Prove that every two-sided ideal of is equal to for some two-sided ideal .
First, let be a two-sided ideal. Let consist of all elements such that for some and . We will first show that is an ideal of , and then show that .
Let and be in , with . Recall from this previous exercise that is the matrix whose entry is the entry of and all other entries are zero.
Since is a two sided ideal, we have that and are in . Moreover, , and the entry of this matrix is . Thus is closed under subtraction.
Since is an ideal, we have . Thus , and is an additive subgroup of .
Now it suffices to show that for all and , . To that end, note that , and that the entry of is ; thus . Similarly, since , . Hence is an ideal.
It is clear that ; To show the other direction, let . Since each , we have for some matrix , for each and , and for some and . In particular, for each and . Note then that , since is an ideal. Thus , and we have .
Now we need to show that if is a two-sided ideal, then is a two-sided ideal.
To that end, let and be in . Then , so that . Hence . Thus is closed under subtraction.
Since , we have . Thus is an additive subgroup of .
Finally, let ; it suffices to show that . Recall that , where . Since is a two-sided ideal, . Thus . Similarly, . Thus is a two-sided ideal.