Claim: $A$ is a ring such that every primary ideal is prime if and
only if $A$ is absolutely flat.
Let us say a ring is a PP(primary is prime) ring if every primary ideal is
prime.
Suppose $A$ is absolutely flat, then $A$ is PP. (exercise 3 in page 55 of "Introduction to Commutative Algebra" by Atiyah & Macdonald)
We only need to show that if $A$ is PP then $A$ is absolutely flat.
Notice that every primary ideal of $S^{-1}A$ is of the form $S^{-1}\mathfrak{q}$
where $\mathfrak{q}$ is a $\mathfrak{p}$-primary ideal such that
$\mathfrak{p}\cap S=\emptyset$. And if $J/I$ (here $J\supset I$) is a
primary ideal of $A/I$ then every zerodivisor in $A/J=(A/I)/(J/I)$ is
nilpotent, so $J$ is is primary ideal of $A$.
Now it is easy to show that, if $A$ is PP, then $S^{-1}A$ and
$A/I$ is PP for a multiplicatively closed subset $S$ and an ideal
$I$ and $\mathfrak{m}^2=\mathfrak{m}$ for every maximal ideal
$\mathfrak{m}$.
Based on these properties we are able to prove that PP rings are absolutely
flat.
We first show every prime ideal in $A$ is maximal. Suppose not, there are
two distinct prime ideals $\mathfrak{p}\subset \mathfrak{m}$ where
$\mathfrak{m}$ is maximal. Now $B=(A/\mathfrak{p})_{\mathfrak{m}}$ is
a PP, local domain(but not a field) we also use $\mathfrak{m}$ to
denote the maximal ideal of $B$. Let $0\neq b\in \mathfrak{m}$,
suppose $\mathfrak{q}$ is a minimal prime containing $b$, then
$C=B_{\mathfrak{q}}$ is a PP, local domain(not a field) and the only maximal ideal of
$C$ is minimal over the ideal $(b)=bC$ so $(b)$ is primary and hence
maximal, $(bC)^2=bC$, thus $bC=0$. It is impossible. We have proved
that every prime ideal in $A$ is maximal.
Let $\mathfrak{m}$ be any prime ideal in $A$, then $A_{\mathfrak{m}}$
is PP. If $A_{\mathfrak{m}}$ is not a field, pick any nonezero $x\in
\mathfrak{m}A_{\mathfrak{m}}$, then $(x)$ is primary hence equals
$\mathfrak{m}A_{\mathfrak{m}}$, so $\mathfrak{m}A_m=0$,
contradiction.
Hence every prime ideal $\mathfrak{m}$ in $A$ is maximal and
$A_{\mathfrak{m}}$ is a field, so $A$ is absolutely flat. We are done.
If $A \subset B$ and $B$ is a field, then $A$ must at least be an integral domain (subrings of fields are integral domains). So (b) is impossible.
(a) is possible, though. For example, if $A$ is a field then $B = A[x]/(x^2)$ is a ring which is a finitely generated $A$-module and therefore integral over $A$. But $B$ is not a field.
Best Answer
This is a good observation. The key here is that we want multiplication and our homomorphisms to be compatible with our addition operation. When we study groups, we have a set $G$ equipped with a single binary operation $G\times G\to G$ satisfying elementary properties: associativity, existence of inverses, and existence of an identity element. When we study rings $A$, we have two operations $+$ and $\times$, namely addition and multiplication. Usually $(A,+,\times)$ is required to be an Abelian group under $+$ and a (often commutative) monoid under $\times$ (inverses might not exist under multiplication, but the other properties of a group are respected).
The natural question remains: how should the operations interact with each other? If there is no interaction between the multiplicative and additive structures of the ring, we might as well separately study $(A,+)$ and $(A,\times)$ as an Abelian group and a monoid, respectively. So, we require that $\times$ "preserve" the group structure of $(A,+)$. A sensible interpretation of this is the following: the map $\phi: A\times A\to A$ given by $(x,y)\mapsto x\times y$ should be "like a homomorphism of $(A,+)$ to itself." With suitable modification, it is. If we fix $x$, then the map $$ \phi_x:(A,+)\to (A,+)$$ has $$ \phi_x(y+z)=\phi(x,y+z)=x(y+z)=xy+xz=\phi_x(y)+\phi_x(z)$$ and if we fix $y$, the map $$ \phi^y:(A,+)\to (A,+)$$ has $$ \phi^y(x+z)=\phi(x+z,y)=(x+z)y=xy+zy=\phi^y(x)+\phi^y(z).$$ So we have a pair of group homomorphisms associated with right and left multiplication.