Proof.
Assumption (a) implies there is a unit u of A whose image in K lies in the maximal ideal of B. Then u is a nonzerodivisor of R and for every a \in A there exists an n such that u^ n a \in R. It follows that A = R_ u.
Let \mathfrak m_ A be the Jacobson radical of A. Let x \in \mathfrak m_ A be a nonzero element. Since \dim (A) = 1 we see that K = A_ x. After replacing x by x^ n u^ m for some n \geq 1 and m \in \mathbf{Z} we may assume x maps to a unit of B. We see that for every b \in B we have that x^ nb in the image of R for some n. Thus B = R_ x.
Let z \in R. If z \not\in \mathfrak m_ A and z does not map to an element of \mathfrak m_ B, then z is invertible. Thus x + u is invertible in R. Hence \mathop{\mathrm{Spec}}(R) = D(x) \cup D(u). We have seen above that D(u) = \mathop{\mathrm{Spec}}(A) and D(x) = \mathop{\mathrm{Spec}}(B).
Case (b). If x \in \mathfrak m_ A, then 1 + x is a unit and hence 1 + x \in R, i.e, x \in R. Thus we see that \mathfrak m_ A \subset R \subset A. In fact, in this case A is integral over R. Namely, write A/\mathfrak m_ A = \kappa _1 \times \ldots \times \kappa _ n as a product of fields. Say x = (c_1, \ldots , c_ r, 0, \ldots , 0) is an element with c_ i \not= 0. Then
x^2 - x(c_1, \ldots , c_ r, 1, \ldots , 1) = 0
Since R contains all units we see that A/\mathfrak m_ A is integral over the image of R in it, and hence A is integral over R. It follows that R \subset A \subset B as B is integrally closed. Moreover, if x \in \mathfrak m_ A is nonzero, then K = A_ x = \bigcup x^{-n}A = \bigcup x^{-n}R. Hence x^{-1} \not\in B, i.e., x \in \mathfrak m_ B. We conclude \mathfrak m_ A \subset \mathfrak m_ B. Thus A \cap \mathfrak m_ B is a maximal ideal of A thereby finishing the proof.
\square
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