Structure of a group of units of a ring composed of the direct product of two subrings

Question

If the direct product of two subrings is a ring, is the group of units of that ring a direct product of the group of units of each subring? How can you see this?

Answer 1

If $R$ and $R'$ are two unital rings, then $R\times R'$ is a unital ring under the following component-wise operations: $$+: (R\times R') \times (R \times R') \to R \times R'$$ given by $(a,b) + (c,d) = (a+c,b+d)$, and

$$\cdot: (R\times R') \times (R \times R') \to R \times R'$$ given by $(a,b)\cdot (c,d) = (ac,bd)$.

Its additive identity is $(0_R,0_{R'})$ and its multiplicative identity is $(1_R,1_{R'})$ and we have:

$$(a,b)\cdot(c,d)=(c,d)\cdot(a,b)=(1_R,1_{R'}) \implies (ac,bd)=(ca,db)=(1_R,1_{R'})$$

Hence, $ac=ca=1_R$ and $bd=db=1_{R'}$ which means that if $(a,b) \in \mathrm{U}(R \times R')$ then $a \in \mathrm{U}(R)$ and $b \in \mathrm{U}(R')$. In other words, $\mathrm{U}(R\times R') \subseteq \mathrm{U}(R)\times\mathrm{U}(R')$

Also, on the other hand, if $a \in \mathrm{U}(R)$ and $b \in \mathrm{U}(R')$ is a unit, we'll see that

$$(a^{-1},b^{-1})\cdot (a,b)=(a,b)\cdot (a^{-1},b^{-1})=(1_R,1_{R'})$$

Therefore, $$\mathrm{U}(R\times R') = \mathrm{U}(R)\times\mathrm{U}(R')$$

Answer 2

Let $R$ and $S$ be two subrings of a ring $T$ with identity $1$. Let $R^*$, $S^*$ and $(R \times S)^*$ be the group of units of $R$, $S$ and $R \times S$ respectively.

Let $(r,s) \in (R \times S)^*$. Then, there exists $(r',s') \in R \times S$ such that $$1_{R \times S} = (1,1) = (r,s) * (r',s') = (r * r', s * s').$$ So, $r * r' = s * s' = 1 \Rightarrow r \in R^*$ and $s \in S^*$. So, $(R \times S)^* \subseteq R^* \times S^*$.

Conversely, let $r \in R^*$ and $s \in S^*$. Then, there exist $r' \in R$ and $s' \in S$ such that $r * r' = s * s' = 1$. So, $$(r,s) * (r',s') = (r * r', s * s') = (1,1) = 1_{R \times S}.$$ Thus, $(r,s) \in (R \times S)^*$. So, $R^* \times S^* \subseteq (R \times S)^*$.

Hence, $R^* \times S^* = (R \times S)^*$.