In the following find appropriate necessary and sufficient c

In the following, find appropriate necessary and sufficient conditions (with argumentation): Under which conditions on R, the polynomial ring R[x_1, x_2, ..., x_n] has the identity? Under which conditions on R, the ring R[x_1, x_2, ..., x_n] is commutative? Let direct sum of rings Ri. Under which conditions on the direct summands Ri, the ring R has the identity? Under which conditions on the itVs, the ring R is commutative? Let R = R_1 circle plus R_2 circle plus circle plus R_n be a direct sum of rings R_i. Under which conditions on the direct summands R_i, the rings R_i has the identity? Under which conditions on the R_i\'s, the ring R is commutative?

Solution

9.3 (a) If the ring R has the identity element 1 (multiplicative identity), then the polynomial ring R[x₁,x₂,…,x_n] has the identity and it is 1.

If the ring R is commutative, then the polynomial ring R[x₁,x₂,…,x_n] is also commutative.

(b) R is a direct sum of rings R_i for i = 1,2,3,…,n

Therefore, R is a ring.

If r R, then r = (r₁, r₂, …, r_n) where r_i R_i for all i = 1,2,3,…,n

If all the rings R_i has multiplicative identity, then the ring R has the identity. Let ei denote the identity element of ring R_i for all i = 1,2,3,…,n; then R has the identity element e = (e₁, e₂, …, e_n)

Proof: Let r R, then r = (r₁, r₂, …, r_n) where r_i R_i for all i = 1,2,3,…,n

re = (r₁, r₂, …, r_n)(e₁, e₂, …, e_n) = (r₁e₁, r₂e₂, …, r_ne_n) = (r₁, r₂, …, r_n) = r

er = (e₁, e₂, …, e_n)(r₁, r₂, …, r_n) = (e₁r₁, e₂r₂, …, e_nr_n) = (r₁, r₂, …, r_n) = r

Therefore, e is the identity element of R. (Proved)

If all the rings R_i are commutative, then the ring R is also commutative.

Proof: Let x, y R; then x = (x₁, x₂, …, x_n), y = (y₁, y₂, …, y_n) where x_i, y_i R_i for all i = 1,2,3,…,n

xy = (x₁, x₂, …, x_n)(y₁, y₂, …, y_n)

= (x₁y₁, x₂y₂, …, x_ny_n)

= (y₁x₁, y₂x₂, …, y_nx_n)    Because rings R_i are commutative, hence, x_iy_i = y_ix_i for all i = 1,2,3,…,n

= (y₁, y₂, …, y_n)(x₁, x₂, …, x_n) = yx

Therefore, for any x, y R; xy = yx

In other words, R is commutative. (Proved)