# Group of units of direct sum of rings is isomorphic to direct sum of the groups of units

Let $R_{1}$, $R_{2}$, $\cdots$, $R_{m}$ be rings with identity. I need to prove that the following group isomorphism holds:

$U(R_{1} \oplus R_{2} \oplus \cdots \oplus R_{n}) \simeq U(R_{1}) \oplus U(R_{2}) \oplus \cdots \oplus U(R_{n})$.

I surmise that induction is going to be necessary here, but I'm having trouble even just getting started to prove it for just the base case, where $n = 2$: $U(R_{1} \oplus R_{2}) \simeq U(R_{1}) \oplus U(R_{2})$.

I have absolutely no idea where to begin, so any kind of point in the right direction would be appreciated. Just be willing to answer lots of follow-up questions, please.

• This is probably not the kind of answer you're looking for (which is why I'm not posting it as an answer), but $U$ is a right adjoint (to the group ring functor), and so automatically preserves products. – RCT Mar 17 '17 at 21:24
• @RCT yeah, I'm not even sure what a "right adjoint" is, so doesn't really help me much. – ALannister Mar 17 '17 at 21:25
• No problem. Ask yourself: how is the multiplication defined on the direct sum? What is the identity element in the direct sum? What, then, does it mean to be a unit in the direct sum? I think when you write out all these definitions explicitly, you will see the claim quite clearly. – RCT Mar 17 '17 at 21:28

Keep in mind that for a finite number of rings, the finite direct sum is isomorphic to the finite direct product. It is easier, then, to think of $R_1 \times R_2$ rather than of $R_1 \oplus R_2$. In particular, the elements of $R_1 \times R_2$ are pairs of the form $(r_1, r_2)$, the multiplication is $(r_1, r_2) (s_1, s_2) = (r_1 s_1, r_2 s_2)$ and the multiplicative neutral element is $(1,1)$.
It then follows easily that $(r_1, r_2)$ is invertible in $R_1 \times R_2$ if and only if there exist $(s_1, s_2) \in R_1 \times R_2$ such that $(r_1, r_2) (s_1, s_2) = (s_1, s_2) (r_1, r_2) = (1,1)$, which is equivalent to saying that $(r_1 s_1, r_2 s_2) = (s_1 r_1, s_2 r_2) = (1,1)$, which in turn is equivalent to saying that $r_1 s_1 = s_1 r_1 = 1 \in R_1$ and $r_2 s_2 = s_2 r_2 = 1 \in R_2$, which finally is equivalent to saying that $r_1$ is invertible in $R_1$ and $r_2$ is invertible in $R_2$.
• @ALannister: What the argument above shows is not merely an isomorphism, but an equality: $U(R_1 \times R_2) = U(R_1) \times U(R_2)$. – Alex M. Mar 17 '17 at 21:45
• because what you're doing I guess is taking an element in $U(R_{1} \times R_{2})$, say $(r_{1},r_{2})$, and you're saying that by definition, then, for any $(s_{1},s_{2})$, we must have that $(r_{1},r_{2})(s_{1},s_{2}) = (s_{1},s_{2})(r_{1},r_{2}) = (1,1)$, which is equivalent to saying that $(r_{1}s_{1},r_{2}s_{2}) = (s_{1}r_{1},s_{2}r_{2})=(1,1)$. Just so I understand this, was this part establishing that $U(R_{1} \oplus R_{2}) \subseteq U(R_{1})\oplus U(R_{2})$? – ALannister Mar 17 '17 at 21:56
• @ALannister: Asume it for all $i \le k$ and then prove it for $k+1$ with: $$U(\color{blue} {R_1 \times \dots \times R_k} \times R_{k+1}) = \color{blue} {U(R_1 \times \dots \times R_k}) \times U(R_{k+1}) = \color{blue} {U(R_1) \times \dots \times U(R_k)} \times U(R_{k+1})$$ where we have used the induction hypothesis once for $i=2$ and once for $i=k$. – Alex M. Mar 18 '17 at 17:26