Let $R$ be a commutative ring. Then its ring of idempotents $I(R)$ consists of the idempotent elements of $R$, with the same multiplication as in $R$, but with the new addition $x \oplus y := x+y-2xy$. (This addition might look a bit mysterious, but when we identify idempotent elements with the clopen subsets of $\mathrm{Spec}(R)$ via $x \mapsto D(x)$, then $I(R)$ is nothing else than the boolean algebra of clopen subsets with multiplication $\cap$ and addition $\Delta$.)

We obtain a functor $I : \mathsf{CRing} \to \mathsf{Bool}$, where $\mathsf{CRing}$ denotes the category of commutative rings and $\mathsf{Bool}$ the category of boolean rings. My question is: Does this functor $I$ have a left or right adjoint? Or does $I(R)$ have any universal property?


In a boolean ring every element is idempotent and, since the characteristic is $2$, we have $x\oplus y=x+y$. So, for a boolean ring $B$, $I(B)=B$. If $i$ denotes the embedding functor $\mathsf{Bool}\to\mathsf{CRing}_2$, the category of rings with characteristic $2$. This suggest there is an adjunction between $i$ and $I$, because $I(i(B))=B$.

In the case of characteristic $2$, the $\oplus$ operation is the same as $+$, so the adjunction is almost obvious, because it's just composing with the inclusion $I(R)\hookrightarrow R$.

There's no general adjunction of the embedding $\mathsf{Bool}\to\mathsf{CRing}$, because there are no ring morphisms from a boolean ring to a ring with, say, characteristic $0$.

  • $\begingroup$ The function $\hat{\varphi}$ mentioned in your answer cannot be a ring homomorphism in general. Indeed, if $R$ has characteristic zero, then there cannot exist a (unital) ring homomorphism $i(B) \to R$ for any Boolean ring $B$. $\endgroup$ – Manny Reyes Jan 13 '15 at 15:58
  • $\begingroup$ @MannyReyes The map $\hat{\varphi}$ is built from a given ring homomorphism $\varphi\colon B\to I(R)$. If there's none, nothing bad happens. $\endgroup$ – egreg Jan 13 '15 at 16:01
  • $\begingroup$ There is an isomorphism of Boolean rings $\varphi \colon \mathbb{Z}/(2) \to I(\mathbb{Z})$, but no corresponding ring homomorphism $\hat{\varphi} \colon i(\mathbb{Z}/(2)) \to \mathbb{Z}$. $\endgroup$ – Manny Reyes Jan 13 '15 at 16:09
  • $\begingroup$ @MannyReyes Yes, you're completely right and I changed the answer accordingly. $\endgroup$ – egreg Jan 13 '15 at 16:17

This functor is represented by the free idempotent, or $\mathbb{Z}[x]/(x^2 - x) \cong \mathbb{Z} \times \mathbb{Z}$. The statement that the corresponding functor to sets in fact lifts to Boolean rings says that $\text{Spec } \mathbb{Z} \times \mathbb{Z}$ naturally has the structure of a Boolean ring object in $\text{Aff}$. This structure comes from the following general construction:

Let $C$ be a distributive category, by which I mean a category with finite coproducts and finite products such that the latter distribute over the former. The relevant example here is $\text{Aff}$. Let $1$ denote the terminal object, and let $2 = 1 + 1$. In $\text{Aff}$ the terminal object is $\text{Spec } \mathbb{Z}$ and hence $2$ is the spectrum of the free idempotent.

Now I claim that there is a natural functor (of distributive categories) from $\text{FinSet}$ to $C$ sending the one-element set to $1$. I further claim that $2$ is naturally a Boolean ring object in $\text{FinSet}$ (in fact the full subcategory of $\text{FinSet}$ on the powers of $2$ is precisely the Lawvere theory of Boolean rings, or equivalently of Boolean algebras), and this structure gets transported to a Boolean ring structure on $2$ in $C$.


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