Equalizers in CRing This is a follow-up to this question of Martin Brandenburg.
Let $B$ be a ring (in this post "ring" means "commutative ring with $1$") and $A$ a subring. 
We say that $A\subset B$ is an equalizer if there are two morphisms of rings from $B$ to $C$ which coincide precisely on $A$. 
(One also says that $A\subset B$ is a kernel, or that $A\to B$ is regular, but I find the phrase "is an equalizer" more descriptive.)
As pointed out by Martin, this is equivalent to the condition that $b\in B$ and $b\otimes 1=1\otimes b$ in $B\otimes_AB$ imply $b\in A$.
As also pointed out by Martin, $A\subset B$ is an equalizer if $B$ is faithfully flat over $A$. (This follows from Proposition 13, Chapter 1, Section 2, Subsection 11, and from Proposition 13, Chapter 1, Section 3, Subsection 6 in Nicolas Bourbaki, Algèbre commutative: Chapitres 1 à 4, Masson, Paris 1985. Subquestion: are there better proofs?)
The following question seems unavoidable:

Is $A\subset B$ an equalizer if and only if the functor $B\otimes_A-\ $ reflects exactness?

The phrase "$B\otimes_A-\ $ reflects exactness" means: If $M\to N\to P$ is a complex of $A$-modules and if $B\otimes_AM\to B\otimes_AN\to B\otimes_AP$ is exact, then $M\to N\to P$ is exact.
Here is a nano-answer: Assume that all maximal ideals of $A$ are principal, that $B$ is flat over $A$ and that $A\subset B$ is an equalizer. Then $B$ is faithfully flat over $A$. 
Proof. If $B$ was not faithfully flat, by Tag 00HP in the Stacks Project, there would be a maximal ideal $(a)$ of $A$ such that $aB=B$, that is, $a$ is a unit of $B$. In $B\otimes_AB$ we would get 
$$
1\otimes a^{-1}=a^{-1}a\otimes a^{-1}=a^{-1}\otimes aa^{-1}=a^{-1}\otimes1.
$$
(Subquestion: If all maximal ideals of $A$ are principal, does it follow that $A$ is a principal ideal ring?)
To get an example of an equalizer $A\subset B$ where $B$ is not flat over $A$, consider the inclusion $\mathbb Z\subset\mathbb Z\times\mathbb Z/(2)$. It is clear that $(\mathbb Z\times\mathbb Z/(2))\otimes_{\mathbb Z}-\ $ reflects exactness.
The co-equalizers in CRing are precisely the surjective epimorphisms. For instance $\mathbb Z\to\mathbb Q$ is a monomorphism and an epimorphism, but is neither an isomorphism, nor an equalizer, nor a surjection, nor a co-equalizer. Moreover $\mathbb Q$ is flat, but not faithfully flat, over $\mathbb Z$, and $\mathbb Q\otimes_{\mathbb Z}-\ $ does not reflect exactness.
EDIT. Another question:

Is the composition of two regular monomorphisms a regular monomorphism?

 A: Incomplete answer. Too long for a comment.
Here is a candidate for an equalizer $A\subset B$ which would not reflect exactness. 
More precisely I'll define an inclusion $A\subset B$ and prove that $B\otimes_A-\ $ does not reflect exactness. I believe that $A\subset B$ is an equalizer, but I'm unable to prove it. All I can do is to rewrite the condition that $A\subset B$ is an equalizer in a more elementary (but more technical) form.
Let $K$ be a field; let $X_1,X_2,Y_1,Y_2$ be indeterminates; set 
$$
A:=K[X_1,X_2],
$$
$$
B':=K[X_1,X_2,Y_1,Y_2],
$$
$$
g:=X_1Y_1+X_2Y_2-1\in B',
$$
$$
B:=B'/(g).
$$ 
The natural morphism from $A$ to $B$ being clearly injective, we can, and will, view $A$ as a subring of $B$. 
The functor $B\otimes_A-\ $ does not reflects exactness because we have 
$$
B\ \underset{A}{\otimes}\ \frac{A}{(X_1,X_2)}\simeq\frac{B}{X_1B+X_2B}=0,
$$ 
the last equality being justified by the fact that 
$$
1=X_1y_1+X_2y_2\in X_1B+X_2B,
$$ 
where $y_i$ is the image of $Y_i$ in $B$. 
As observed in the question, the conditions
$(1)\ A\subset B$ is an equalizer
$(2)\ b\in B$ and $b\otimes 1=1\otimes b$ in $B\otimes_AB$ imply $b\in A$
are equivalent.
We shall rewrite $(2)$ in a more explicit form. 
First let's write $B\otimes_AB$ as a quotient of a ring of multivariate polynomials with coefficients in $K$. Write $y_i$ for the image of $Y_i$ in $B$. Note that $B\otimes_AB$ is generated, as a $K$-algebra, by $X_1,X_2$ and the four elements 
$$
y_{i1}:=y_i\otimes 1,\quad y_{i2}:=1\otimes y_i
$$ 
for $i=1,2$. Thus $B\otimes_AB$ as a quotient of 
$$
C':=K[X_1,X_2,Y_{11},Y_{21},Y_{12},Y_{22}],
$$ 
where $Y_{11},Y_{21},Y_{12},Y_{22}$ are indeterminates. In other words, we have a natural surjective $K$-algebra morphism from $C'$ to $B\otimes_AB$. The kernel of this morphism is generated by $X_1Y_{11}+X_2Y_{21}-1$ and $X_1Y_{12}+X_2Y_{22}-1$. Write $C$ for the quotient of $C'$ by the ideal generated by the two above polynomials.
Second let's describe the $A$-algebra morphisms from $B$ to $C$ corresponding to the morphisms $b\mapsto b\otimes1$ and $b\mapsto 1\otimes b$ from $B$ to $B\otimes_AB$. These morphisms $B\to C$ come from $A$-algebra morphisms $B'\to C'$, which we denote by $f\mapsto f_{(1)}$ and $f\mapsto f_{(2)}$, and which are characterized by $Y_{i(j)}:=Y_{ij}$, that is $f\mapsto f_{(1)}$ maps $Y_1$ to $Y_{11}$ and $Y_2$ to $Y_{21}$, and similarly for $f\mapsto f_{(2)}$. In this notation the polynomials $X_1Y_{11}+X_2Y_{21}-1$ and $X_1Y_{12}+X_2Y_{22}-1$ above are respectively equal to $g_{(1)}$ and $g_{(2)}$. 
Note that the monomials in $X_1,X_2,Y_1,Y_2$ which are not divisible by $X_2Y_2$ give rise to a $K$-basis of $B$.
Then it's not hard to see that the condition $(3)$ below is equivalent to $(1)$ and $(2)$ above:

$(3)$ Let $f\in B'$ be a polynomial containing no monomial of the form $X_1^iX_2^j$, and no monomial of the form $X_2Y_2\mu$ where $\mu\in B'$ is a monomial. Assume that $$f_{(1)}-f_{(2)}=h_1g_{(1)}+h_2g_{(2)}$$ for some $h_1,h_2$ in $C'$. Then $f=0$.

Again, I'm unable to prove or disprove $(3)$.
A: Here is a partial answer. 
Let's keep the notation in the question, and add the convention that all tensor products are taken over $A$.
I claim:

If $B\otimes-\ $ reflects exactness, then $A\subset B$ is an equalizer.

Let $A\subset B$ be an arbitrary inclusion of rings. It suffices to prove
$(1)\ A\subset B$ is an equalizer if and only the complex $A\xrightarrow\iota B\xrightarrow fB\otimes B$, where $\iota$ is the inclusion and $f$ the morphism $b\mapsto b\otimes1-1\otimes b$, is exact.
$(2)$ The complex 
$$
B\xrightarrow{B\otimes\iota}B\otimes B\xrightarrow{B\otimes f}B\otimes B\otimes B
$$ 
is exact.
As observed in the question, $(1)$ holds.
To prove $(2)$, let $\sum b_i\otimes b'_i$ be in $\ker(B\otimes f)$. We have
$$
\sum\ b_i\otimes\Big(b'_i\otimes1-1\otimes b'_i\Big)=0.\label3\tag3
$$ 
It suffices to show 
$$
\sum b_i\otimes b'_i=(B\otimes\iota)\left(\sum b_ib'_i\right).
$$ 
This equality reduces first to
$$
\sum b_i\otimes b'_i-\sum b_ib'_i\otimes1=0,
$$ 
and then to 
$$
\sum\ b_i\ \Big(1\otimes b'_i-b'_i\otimes1\Big)=0.
$$ 
But this follows from $(\ref3)$.
