Is $\operatorname{CRing}^\mathrm{op}$ a regular category, where $\operatorname{CRing}$ is the category of commutative rings (with unity, as usual)?
To be regular, a category must have finite limits, coequalizers of kernel pairs, and pullback-stable regular epimorphisms.
In the case of $\operatorname{CRing}^\mathrm{op}$, that would mean that if $A \to B$ is a regular monomorphism in $\operatorname{CRing}$, then so must $C \to B \otimes_{A} C$ for any commutative associative unital $A$-algebra $C$.
Of course, $A \to B$ must be a regular monomorphism. If we allow it to be any monomorphism, then $\mathbb{Z} \to \mathbb{Q}$ is monic (and epic, although not surjective), but $C \to \mathbb{Q} \otimes_{\mathbb{Z}} C$ is monic if and only if the additive group of $C$ is torsion-free.
If $A \to B$ is faithfully flat, then it is easily seen that $C \to B \otimes_{A} C$ must also be faithfully flat for any commutative associative unital $A$-algebra $C$. Also, all faithfully flat homomorphisms must be regular monomorphisms.
For the general case (where $A \to B$ is not necessarily faithfully flat, or even merely flat), I do not know the answer.
If $\mathbf{CRing}^\mathrm{op}$ were a regular category, then it would be possible to factorise every ring homomorphism as an epimorphism followed by a regular monomorphism; moreover that regular monomorphism must be the equaliser of the cokernel pair of the homomorphism we start with. This is not the case.
For simplicity, let $k$ be a field, let $A = k [t, x]$, let $B = k [t, y]$, and let $f : A \to B$ be the unique $k$-algebra homomorphism such that $f (t) = t$ and $f (x) = t y$. Then $f : A \to B$ makes $B$ an $A$-algebra: indeed, $B \cong A [y] / (t y - x)$. The cokernel pair of $f : A \to B$ is $B \otimes_A B \cong k [t, y_1, y_2] / (t y_1 - t y_2)$ with the two obvious injections $B \rightrightarrows B \otimes_A B$. A straightforward calculation shows that the equaliser of $B \rightrightarrows B \otimes_A B$ is the $k$-subalgebra $E \subseteq B$ generated by $\{ t y^n : n \ge 0 \}$. For general reasons, $E \hookrightarrow B$ is the smallest regular monomorphism through which $f : A \to B$ factors, so if $\mathbf{CRing}^\mathrm{op}$ were a regular category then $f : A \to E$ must be an epimorphism.
On the other hand, the image of $f : A \to B$ is the $k$-subalgebra $A' \subseteq B$ generated by $\{ t, t y \}$. Clearly, $A' \subsetneqq E$, but this is not quite the same as saying $f : A \to E$ is not an epimorphism. Regarding $E$ as an $A'$ algebra, we have (I think): $$\begin{align*} E & \cong A' [z_2, z_3, z_4, \ldots] / (t z_2 - t^2 y^2, t z_3 - t y z_2, t z_4 - t y z_3, \ldots) \\ & \cong A [z_2, z_3, z_4, \ldots] / (t z_2 - x^2, t z_3 - x z_2, t z_4 - x z_3, \ldots) \end{align*}$$ Consider $C = E / (t y)$. Besides the canonical quotient homomorphism $E \to C$, we have the $A'$-algebra homomorphism that sends $z_2, z_3, z_4, \ldots$ to $0$. Thus we have two distinct homomorphisms $E \rightrightarrows C$ that agree on $A'$ – so $f : A \to E$ is not an epimorphism, as claimed.