Can someone enlighten me on the following isomorphism described in "Exercises in Modern Algebra" by Professor A. Hattori (in Japanese)?

Let $\lbrace R^{\mu} \rbrace$ be an inverse system of rings (with units.) Then $(\varprojlim (R^{\mu}))^{\times} \cong \varprojlim (R^{\mu})^{\times}$ holds where $^{\times}$ denotes the group of units. The verification of this fact is easy so the proof is omitted.

I was not able to come up with a proof, so I googled and found one by Professor Igusa (Corollary 2.7. of http://people.brandeis.edu/~igusa/Math101aF07/Math101a_notesBall.pdf).

I know I have to be satisfied with this, but I don't think the proof is "easy": the techniques used hardly fit into first-year algebra courses.

Is there an easy trick that Professor Hattori had in mind which I am overlooking?

  • $\begingroup$ I think the isomorphism is pretty clear if we view the inverse limit as the subring of the direct product satisfying the appropriate transition condition on the coordinates. $\endgroup$
    – anon
    Nov 25, 2012 at 6:49
  • $\begingroup$ @anon: Maybe I am dumb: I don't recognise it as pretty clear. Can you be more specific on this? $\endgroup$
    – eltonjohn
    Nov 25, 2012 at 6:59
  • $\begingroup$ You can use the > character to create a block quote, like in email. P.S. Would the ring-theory tag be appropriate here? $\endgroup$
    – user856
    Nov 25, 2012 at 9:24
  • $\begingroup$ Hello eltonjohn. Have you considered registering your account? I believe this is also you. $\endgroup$ Nov 25, 2012 at 9:44
  • 1
    $\begingroup$ @eltonjohn "it's me" is correct. Well, perhaps the moderators can help, I notified them. It might be more convenient for you to have your questions all associated to one user. (if you want to look at them again in the future) $\endgroup$ Nov 25, 2012 at 11:08

3 Answers 3


Whether this counts as an "easy trick" is up to you, but the group of units is a functor $\text{Ring} \to \text{Grp}$ which has a left adjoint, namely the group ring construction. Any functor with a left adjoint preserves limits.

  • $\begingroup$ Thank you very much indeed. To me your explanation looks a "slick manoeuvre". Probably not extremely slick if one knows that the group ring construction is left adjoint to the group of units construction, but I was not aware of that anyway. $\endgroup$
    – eltonjohn
    Nov 25, 2012 at 9:46
  • 1
    $\begingroup$ Another way to verify this is to use the fact that arbitrary limits can be defined in terms of products and equalizers, so to prove that the group of units functor preserves limits (which is stronger than the desired statement) it suffices to verify that it preserves products and equalizers, both of which are relatively straightforward. $\endgroup$ Nov 25, 2012 at 9:54
  • $\begingroup$ Thank you again for additional enlightenment. Thanks to this forum, I have extensively broadened my horizon. $\endgroup$
    – eltonjohn
    Nov 25, 2012 at 10:06
  • $\begingroup$ I just checked anon(ymous?)'s answer to close this, as I can do check for only one. At the same time I am grateful to Qiaochu for reminding me of adjoint functors. $\endgroup$
    – eltonjohn
    Feb 13, 2015 at 13:05

It is standard in the proof of existence of inverse limits to provide a concrete construction of the inverse limit as a subring of the direct product of the system's rings satisfying the appropriate transition conditions. That is, if $\varphi_{\mu\lambda}:R^\mu\to R^\lambda$ are the transition morphisms of the system,

$$\varprojlim R^\mu\cong\{(r^\mu):\underbrace{\forall\mu,\lambda,~\varphi_{\mu\lambda}(r^\mu)=r^\lambda}_{\text{transition conditions}}\}\subseteq \prod_\mu R^\mu. \tag{$\square$}$$

We use $\subseteq$ to denote subring above. It is straightforward to check that the middle set is a subring; for convenience we will refer to it by $L$ (for "limit").

For good measure, you may want to convince yourself that (a) the assignment of groups of units to rings is functorial in the latter, and (b) this functor distributes through arbitrary direct products.

If $(r^\mu)\in L$ is a unit, then there is a $(s^\mu)\in L$ for which $1_{L}=(1_{R^\mu})=(r^\mu)(s^\mu)=(r^\mu s^\mu)$ and hence the coordinate $r^\mu$ is invertible for each $\mu$. Thus if $(r^\mu)\in L^\times$ then $(r^\mu)\in\prod_\mu (R^\mu)^\times$; but since $(r^\mu)$ is in $L$ it satisfies the transition conditions, and hence it is contained in the subgroup of $\prod_\mu(R^\mu)^\times$ which is isomorphic to $\varprojlim(R^\mu)^\times$ as per $(\square)$ (although working in a different category, the idea works just the same). Conversely, if $(r^\mu)$ is in the designated subgroup of $\prod_\mu(R^\mu)^\times$ (itself a subset of $\prod_\mu R^\mu$) which is isomorphic to $\varprojlim (R^\mu)^\times$, then each coordinate $r^\mu$ is a unit in $R^\mu$ with inverse say $s^\mu$, and so $(r^\mu)(s^\mu)=1_L$ shows $(r^\mu)$ is invertible and in $L$ (again since it satisfies the TCs).

Thus the copy of $\varprojlim (R^\mu)^\times$ sitting inside $\prod_\mu(R^\mu)^\times$ sitting inside $\prod_\mu R^\mu$ is equal to the units of the copy of $\varprojlim R^\mu$ sitting inside $\prod_\mu R^\mu$.

  • $\begingroup$ Thank you very much indeed. Yes, I was trying to come up with a proof that deploys direct construction of limits. Your explanation is wonderful: I didn't have to read between the lines! $\endgroup$
    – eltonjohn
    Nov 25, 2012 at 9:30

Let $R$ be a diagram of rings.

If $u\in\varprojlim R$ is a unit then clearly each $\pi_\bullet(u)\in R_\bullet$ is a unit, since the $\pi$ are ring homomorphisms. Moreover, if the two units $u,u'$ have equal projections $\pi_\bullet(u)=\pi_\bullet(u')$ then their multiplication homomorphisms $m_u,m_{u'}:\varprojlim R\to\varprojlim R$ would have the same components on each $R$ and thus would be equal by the universal property of a limit; then $u=m_u(1)=m_{u'}(1)=u'$ follows.

Now say $u$ satisfies $\pi_\bullet(u)$ is a unit for each $u$. The $m_{\pi_\bullet(u)}:R_\bullet\cong R_\bullet$ assemble to a natural automorphism of the diagram $R$; it follows from the universal property of limits (for an alternative view, it follows from the highly similar fact that "limit" is a functor) that the limit $\varprojlim m_{\pi_\bullet(u)}$ is an isomorphism. However, this is nothing but $m_u$. Therefore $u$ is a unit.

Thus, the canonical map $(\varprojlim R)^\times\to\varprojlim(R^\times)$ is an isomorphism.

To elaborate on a different perspective offered by Qiaochu; really, we want to show $(-)^\times:\mathsf{Ring}\to\mathsf{Grp}$ is a continuous functor. This would follow if we procured a left adjoint... and indeed, the group ring functor $\Bbb Z[-]:\mathsf{Grp}\to\mathsf{Ring}$ is left adjoint to $(-)^\times$. For there is an obvious group embedding $\eta:G\to(\Bbb Z[G])^\times$ and if $\phi:G\to R^\times$ is a group homomorphism then for $f:\Bbb Z[G]\to R$, $f^\times\eta=\phi$ iff. $f([g])=\phi(g)$ for all $g$ iff. $f(\sum_i n_i[g_i])=\sum_i n_i\phi(g_i)$ for all $\sum_i n_i[g_i]\in\Bbb Z[G]$; it is then not too hard to check a unique (unital) ring homomorphism $f$ exists making the right diagram commute.

For yet another perspective I learned from Martin Brandenburg: the composite $U(-)^\times:\mathsf{Ring}\to\mathsf{Grp}\to\mathsf{Set}$ is representable. It is isomorphic to $\mathsf{Ring}(\Bbb Z[x,x^{-1}],-)$ since a ring homomorphism $f:\Bbb Z[x,x^{-1}]\to R$ is determined entirely by a choice of $f(x)$ and $f(x^{-1})$ such that $f(x)\cdot f(x^{-1})=1$, so they are determined entirely by a choice of unit $f(x)\in R$ (since inverses are unique). Clearly $\mathsf{Ring}(\Bbb Z[x,x^{-1}],R)\ni f\sim f(x)\in R^\times$ defines a natural isomorphism of functors. Therefore, as all representables are continuous, $U(-)^\times$ is continuous. But $U$ in fact reflects limits, so it follows $(-)^\times$ is continuous.


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