Let $A\subset B$ be rings, and suppose that $B\setminus A$ is closed under multiplication. Show that $A$ is integrally closed in $B$. (Atiyah and MacDonald, Introduction to Commutative Algebra, Chapter 5, Exercise 7)
I tried localizing at $B\setminus A$, but this did not seem to work. Neither did a direct application of the definition of "integral dependence". Any suggestions?
Take $x\in B$ which is integral over $A$ and write $x^n+a_1x^{n-1}+\cdots +a_n=0$ with $a_i\in A$ and $n$ the least possible. Then $x(x^{n-1}+a_1x^{n-1}+\cdots+a_{n-1})=-a_n\in A$. If $x$ is not in $A$, then $ x^{n-1}+a_1x^{n-1}+\cdots+a_{n-1}\in A$, contradiction.