This "answer" is meant to supplement Tom Leinster's answer, and is really in response to Martin Brandenburg's comment below Tom's answer, where he asks for an example of a continuous (i.e., limit-preserving) functor $\textbf{CRing} \to \textbf{Set}$ that is not a right adjoint. Adam Epstein's idea suggested the following possibility.

Choose, for each infinite cardinal $\alpha$, a field $F_\alpha$ of that cardinality (say, for definiteness, the characteristic zero algebraically closed field of transcendence degree $\alpha$ over $\mathbb{Q}$), and put $A_\alpha = \mathbb{Z} \times F_\alpha$. The non-trivial quotient rings (corresponding to regular epis) of $A_\alpha$ are precisely $F_\alpha$ and quotient rings of $\mathbb{Z}$.

This has the following consequence: for any (commutative) ring $R$ of cardinality less than $\alpha$, there is exactly one map $f: A_\alpha \to R$. For if we have a (regular epi)-mono factorization $A_\alpha \to Q \to R$ where $Q \to R$ is monic, then the possibilities $Q = A_\alpha$ and $Q = F_\alpha$ are ruled out, hence the factorization must take the form

$$A_\alpha \stackrel{\text{epi}}{\to} \mathbb{Z}/(n) \stackrel{\text{mono}}{\to} R$$

where $(n)$ is uniquely determined as the annihilator of the identity in $R$.

Now form the functor

$$G = \prod_{\alpha \in \text{Card}} \hom(A_\alpha, -): \textbf{CRing} \to \textbf{Set}$$

As soon as $\alpha \gt \text{Card}(R)$, we have that $\hom(A_\alpha, R)$ is a one-element set. Thus for each $R$, $G(R)$ is a set even though $G$ itself is a class-sized product. Being a product of continuous functors, $G$ is continuous. But $G$ cannot be representable (just by simple cardinality considerations; e.g., $G(A_\alpha)$ has size greater than $\alpha$, for any $\alpha$, since algebraically closed fields have lots of automorphisms).

This "answer" is meant to supplement Tom Leinster's answer, and is really in response to Martin Brandenburg's comment below Tom's answer, where he asks for an example of a continuous (i.e., limit-preserving) functor $\textbf{CRing} \to \textbf{Set}$ that is not a right adjoint. Adam Epstein's idea suggested the following possibility.

Choose, for each infinite cardinal $\alpha$, a field $F_\alpha$ of that cardinality (say, for definiteness, the characteristic zero algebraically closed field of transcendence degree $\alpha$ over $\mathbb{Q}$), and put $A_\alpha = \mathbb{Z} \times F_\alpha$. Each non-trivial quotient ring (corresponding to a regular epi) of $A_\alpha$ either contains a copy of $F_\alpha$, or is a quotient ring of $\mathbb{Z}$ (possibly $\mathbb{Z}$ itself).

This has the following consequence: for any (commutative) ring $R$ of cardinality less than $\alpha$, there is exactly one map $f: A_\alpha \to R$. For if we have a (regular epi)-mono factorization $A_\alpha \to Q \to R$ where $Q \to R$ is monic, then the possibility where $Q$ contains a copy of $F_\alpha$ is ruled out, hence the factorization must take the form

$$A_\alpha \stackrel{\text{epi}}{\to} \mathbb{Z}/(n) \stackrel{\text{mono}}{\to} R$$

where $(n)$ is uniquely determined as the annihilator of the identity in $R$.

Now form the functor

$$G = \prod_{\alpha \in \text{Card}} \hom(A_\alpha, -): \textbf{CRing} \to \textbf{Set}$$

As soon as $\alpha \gt \text{Card}(R)$, we have that $\hom(A_\alpha, R)$ is a one-element set. Thus for each $R$, $G(R)$ is a set even though $G$ itself is a class-sized product. Being a product of continuous functors, $G$ is continuous. But $G$ cannot be representable (just by simple cardinality considerations; e.g., $G(A_\alpha)$ has size greater than $\alpha$, for any $\alpha$, since algebraically closed fields have lots of automorphisms).

This "answer" is meant to supplement Tom Leinster's answer, and is really in response to Martin Brandenburg's comment below Tom's answer, where he asks for an example of a continuous (i.e., limit-preserving) functor $\textbf{CRing} \to \textbf{Set}$ that is not a right adjoint. Adam Epstein's idea suggested the following possibility.

For any set $S$, let $\mathbb{Z}[S]$ be the free commutative ring on $S$, aka the polynomial algebra with coefficients in $\mathbb{Z}$ and indeterminates drawn from $S$. Let $A_S$ denote the localization of $\mathbb{Z}[S]$ where all elements except those belonging to $\mathbb{Z}$ are inverted. It follows that the only quotient rings (corresponding to regular epis) of $A_S$ are $\mathbb{Z}$ and its quotient rings.

This has the following consequence: for given $S$ and any (commutative) ring $R$, there is exactly one non-injective map $f: A_S \to R$. For, any such $f$ factors uniquely as

$$A_S \stackrel{\text{epi}}{\to} \mathbb{Z}/(n) \stackrel{\text{mono}}{\to} R$$

where $(n)$ is uniquely determined as the annihilator of the identity in $R$.

Now, for each cardinal $\alpha$, let $S_\alpha$ be a set of that cardinality (for example, we could take $S_\alpha$ to be the von Neumann cardinal $\alpha$ itself, considered as a set), and form the functor

$$F = \prod_{\alpha \in \text{Card}} \hom(A_{S_\alpha}, -): \textbf{CRing} \to \textbf{Set}$$

The main thing to notice is that for any ring $R$, as soon as $\alpha \gt \text{Card}(R)$, we cannot have an injection $A_{S_\alpha} \to R$, in which case $\hom(A_{S_\alpha}, R)$ is a one-element set. Thus for each $R$, $F(R)$ is a set even though $F$ itself is a class-sized product. Being a product of continuous functors, $F$ is continuous. But $F$ cannot be representable (just by simple cardinality considerations; e.g., $F(A_{S_\alpha})$ has size at least $\alpha^\alpha$, for any $\alpha$).

This "answer" is meant to supplement Tom Leinster's answer, and is really in response to Martin Brandenburg's comment below Tom's answer, where he asks for an example of a continuous (i.e., limit-preserving) functor $\textbf{CRing} \to \textbf{Set}$ that is not a right adjoint. Adam Epstein's idea suggested the following possibility.

Choose, for each infinite cardinal $\alpha$, a field $F_\alpha$ of that cardinality (say, for definiteness, the characteristic zero algebraically closed field of transcendence degree $\alpha$ over $\mathbb{Q}$), and put $A_\alpha = \mathbb{Z} \times F_\alpha$. The non-trivial quotient rings (corresponding to regular epis) of $A_\alpha$ are precisely $F_\alpha$ and quotient rings of $\mathbb{Z}$.

This has the following consequence: for any (commutative) ring $R$ of cardinality less than $\alpha$, there is exactly one map $f: A_\alpha \to R$. For if we have a (regular epi)-mono factorization $A_\alpha \to Q \to R$ where $Q \to R$ is monic, then the possibilities $Q = A_\alpha$ and $Q = F_\alpha$ are ruled out, hence the factorization must take the form

$$A_\alpha \stackrel{\text{epi}}{\to} \mathbb{Z}/(n) \stackrel{\text{mono}}{\to} R$$

where $(n)$ is uniquely determined as the annihilator of the identity in $R$.

Now form the functor

$$G = \prod_{\alpha \in \text{Card}} \hom(A_\alpha, -): \textbf{CRing} \to \textbf{Set}$$

As soon as $\alpha \gt \text{Card}(R)$, we have that $\hom(A_\alpha, R)$ is a one-element set. Thus for each $R$, $G(R)$ is a set even though $G$ itself is a class-sized product. Being a product of continuous functors, $G$ is continuous. But $G$ cannot be representable (just by simple cardinality considerations; e.g., $G(A_\alpha)$ has size greater than $\alpha$, for any $\alpha$, since algebraically closed fields have lots of automorphisms).