Are locally integral extensions integral?

Question

Let $R$ be a commutative ring with unit and let $A \subseteq B$ be commutative $R$-algebras. Suppose $B_P$ is integral over $A_P$ for all primes $P$ of $R$. Is $B$ then necessarily integral over $A$ ?

Answer 1

Let $x\in B$. Then $x$ is integral over $A_P$, so there exist $a_i\in A$ and $s_i\in R-P$ such that $x^n+(a_{n-1}/s_{n-1})x^{n-1}+\cdots+a_0/s_0=0$. Clearing the denominators we find an element $s_P\in R-P$ such that $s_Px$ is integral over $A$. Since the ideal of $R$ generated by all elements $s_P$ equals $R$ there is a (finite) linear combination $\sum_Pr_Ps_P=1$ with $r_P\in R$. Multiplying this by $x$ we get $x=\sum_Pr_P(s_Px)$, a sum of integral elements over $A$, and thus $x$ itself is integral over $A$.