The Cartesian product of two rings is a ring

Let R and S be rings. Prove that the direct product R \times S is a ring under componentwise addition and multiplication. Prove that R \times S is commutative if and only if R and S are commutative. Prove that R \times S has an identity if and only if R and S have identities.


We already know that R \times S is an abelian group under componentwise addition. Now if r_i \in R and s_i \in S, we have (r_1,s_1)((r_2,s_2)(r_3,s_3)) = (r_1,s_1)(r_2r_3,s_2s_3) = (r_1(r_2r_3),s_1(s_2s_3)) = ((r_1r_2)r_3,(s_1s_2)s_3) = (r_1r_2,s_1s_2)(r_3,s_3) = ((r_1,s_1)(r_2,s_2))(r_3,s_3). So componentwise multiplication is associative. Moreover, (r_1,s_1)((r_2,s_2) + (r_3,s_3)) = (r_1,s_1)(r_2+r_3,s_2+s_3) = (r_1(r_2+r_3), s_1(s_2+s_3)) = (r_1r_2 + r_1r_3, s_1s_2 + s_1s_3) = (r_1r_2,s_1s_2) + (r_1r_3,s_1s_3) = (r_1,s_1)(r_2,s_2) + (r_1,s_1)(r_3,s_3). Thus componentwise multiplication distributes over componentwise addition on the left; similarly, it distributes on the right. Thus R \times S is a ring.

If R and S are commutative, then (r_1,s_1)(r_2,s_2) = (r_1r_2,s_1s_2) = (r_2r_1,s_2s_1) = (r_2,s_2)(r_1,s_1), so that R \times S is commutative. If R \times S is commutative, then (r_1r_2,s_1s_2) = (r_1,s_1)(r_2,s_2) = (r_2,s_2)(r_1,s_1) = (r_2r_1,s_2s_1). Comparing entries, we have r_1r_2 = r_2r_1 and s_1s_2 = s_2s_1, so that R and S are commutative.

If R and S have 1s, then (1,1)(r,s) = (1r,1s) = (r,s) and (r,s)(1,1) = (r1,s1) = (r,s), so that (1,1) \in R \times S is an identity. Suppose (u,v) \in R \times S is an identity. Then for all r \in R and s \in S, we have (ur,vs) = (u,v)(r,s) = (r,s) = (r,s)(u,v) = (ru,sv). Then ur = ru = r and vs = sv = s; by the uniqueness of identities, both R and S have a 1.

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