4
$\begingroup$

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.

Thank you in advance.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – RCT Mar 17 '17 at 21:24
  • $\begingroup$ @RCT yeah, I'm not even sure what a "right adjoint" is, so doesn't really help me much. $\endgroup$ – ALannister Mar 17 '17 at 21:25
  • 1
    $\begingroup$ 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. $\endgroup$ – RCT Mar 17 '17 at 21:28
4
$\begingroup$

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 $R_2$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ what does this have to do with isomorphism, though? $\endgroup$ – ALannister Mar 17 '17 at 21:43
  • $\begingroup$ @ALannister: What the argument above shows is not merely an isomorphism, but an equality: $U(R_1 \times R_2) = U(R_1) \times U(R_2)$. $\endgroup$ – Alex M. Mar 17 '17 at 21:45
  • 2
    $\begingroup$ 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 $U(R_{1} \oplus R_{2}) \subseteq U(R_{1})\oplus U(R_{2})$? $\endgroup$ – ALannister Mar 17 '17 at 21:56
  • 1
    $\begingroup$ okay. Got it. Then, of course, it's induction time. $\endgroup$ – ALannister Mar 17 '17 at 21:58
  • 1
    $\begingroup$ @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 $i=k$. $\endgroup$ – Alex M. Mar 18 '17 at 17:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.