explain why every subring of Zn is necessarily an ideal expl

Solution

Solution :

Let I be a subring of Z_n, n Z_n and x I. We consider three cases: n = 0, n > 0, and n < 0

Case 1: If n = 0, then nx = 0 I.

Case 2: If n > 0, then nx = x+x+...+x (n terms). Since I is closed under addition, this means nx I.

Case 3. If n < 0, then -n > 0, and nx = (-n)(-x) = -x-x-...-x (-n terms).

Since I is closed under additive inverses, -x I. Also, since I is closed under addition, (-n)(-x) = -x-x-...-x I.

Thus, in all cases, I is an ideal.

OR

Let S be a subring of Z_n. Since S is a ring, (S,+) is a group. If m Z, s S , then adding s by m times gives s++s=sm. So for all mZ_n, sS , we have sm S. Hence, S is an ideal.