Answer to Question #203328 in Abstract Algebra for Raghav

Question

If U(R) denotes the group of units of a ring R, show that

U(R₁ × R₂) = U(R₁) × U(R₂) for rings R₁ and R₂.

Answer 1

Let 1 denote the unity element in each of the rings R_i. So, (1, 1) is the unity element in R₁$\times$R₂, since if (a₁, a₂) ∈ R1$\times$R2, then

(a₁, a₂)(1, 1) = (a₁ · 1, a₂ · 1) = (a₁, a₂),

and

(1, 1)(a₁, a₂) = (1 · a₁, 1 · a₂) = (a₁, a₂). 

Now, suppose that (u₁, u₂) ∈ U(R₁$\times$R₂). Then, since this element is a unit in the ring R₁$\times$R₂, then by definition, this means there exists an element (x₁, x₂) ∈ R₁$\times$R₂ such that 

(u₁, u₂)(x₁, x₂) = (1, 1). 

But we have, by definition of multiplication in the direct sum of rings,

(u₁, u₂)(x₁, x₂) = (u₁x₁, u₂x₂)

 That is, we have

(u₁, u₂)(x₁, x₂) = (u₁x₁, u₂x₂) = (1, 1)

So, for each i, 1 ≤ i ≤ n, we have u_ix_i = 1 = x_iu_i . That is, x_i ∈ R_i and x_i is a multiplicative inverse to u_i . Thus u_i ∈ U(R_i). Therefore, (u₁, u₂) ∈ U(R₁)$\times$U(R₂). Therefore, we have

U(R₁$\times$R₂) ⊆ U(R₁)$\times$U(R₂).

To prove the other direction of containment, suppose that (y₁, y₂) ∈ U(R₁)$\times$U(R₂). This means

y_i ∈ U(R_i). So, there exists a w_i ∈ R_i such that y_iw_i = 1 = w_iy_i .

Consider the element (w₁, w₂) ∈ R₁$\times$R₂. Then we have

(y₁, y₂)(w₁, w₂) = (y₁w₁, y₂w₂) = (1, 1),

and

(w₁, w₂)(y₁, y₂) = (w₁y₁, w₂y₂) = (1, 1).

Thus, (w₁, w₂) is a multiplicative inverse for the element (y₁, y₂) in R₁$\times$R₂. In particular, we have (y₁, y₂) ∈ U(R₁)$\times$U(R₂). Therefore,

U(R₁)$\times$U(R₂) ⊆ U(R₁$\times$R₂).

Thus, we have now shown U(R₁$\times$R₂) = U(R₁)$\times$U(R₂).