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I've been fairly confused about this question (which isn't homework or anything, just something I was wondering about for myself), and coming up with different answers:

Let $U : \mathbf{Rings} \to \mathbf{Groups}$ be the group of units functor. Note that $U$ has left adjoint $L$ given by the functor $G \mapsto \mathbb{Z}[G]$. Is $U$ monadic?

I think I have the answer now, but given the confusion I'd like to get confirmation.

Answer: No. Suppose that $U$ were monadic. Then that would imply that for every ring $R$, the map $\varepsilon_R : \mathbb{Z}[R^*] \to R, e_u \mapsto u$ would be the coequalizer of a parallel pair and therefore surjective. However, consider the case $R = \mathbb{C}[t]$; then $R^* = \mathbb{C}^*$, and so the map $\varepsilon_R : \mathbb{Z}[\mathbb{C}^*] \to \mathbb{C}[t], e_\lambda \mapsto \lambda$ is certainly not surjective, giving a contradiction. $\quad\square$

So, I think the reason the functor doesn't satisfy the monadicity theorem is: even though $\mathbf{Rings}$ has coequalizers of all parallel pairs, $U$ must not respect coequalizers of all $U$-contractible pairs.


Come to think of it, I wonder if it would also be possible to do a more direct proof by showing that $\mathbb{C}[t]^* \simeq \mathbb{C}^*$ as $UL$-algebras, yet $\mathbb{C}[t] \not\simeq \mathbb{C}$, so $U$ cannot induce an equivalence of categories. I seem to keep getting bogged down in the details of checking what the $UL$-actions are on both sides.

(I'm also curious about whether the induced functor $\mathbf{Rings} \to \mathbf{Groups}_{UL}$ is essentially surjective. Looking at it, I can't see any particular reason to expect any group $G$ with an action $\mathbb{Z}[G]^* \to G$ to be a group of units of a ring, yet I also can't immediately come up with a $UL$-algebra not in the essential image. But this part of the question isn't that important to me.)

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  • $\begingroup$ Oh, I just realized - another proof would be: monadic functors must reflect isomorphisms, yet $U$ applied to the non-isomorphism $\mathbb{C} \hookrightarrow \mathbb{C}[t]$ is an isomorphism. $\endgroup$ – Daniel Schepler Aug 16 '18 at 19:27
  • $\begingroup$ Which I guess also answers my question about $\mathbb{C}[t]^* \simeq \mathbb{C}^*$ as $UL$-algebras... $\endgroup$ – Daniel Schepler Aug 16 '18 at 19:45

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