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I am trying to solve exercise 5 of Chapter Integral dependence and valuations'' from the book of Atiyah and MacdonaldIntroduction to Commutative algebra''.

Let $A$ be a subring of $B$ such that $B \setminus A$ is closed under multiplication. Show that $A$ is integrally closed in $B$.

My attempt to solve the problem:

Let $z \in B \setminus A$. We need to show that $z$ can not be a root of a polynomial $ f \in A[x]$. I can proof it if the degree of $f$ equals to $2$. Indeed, let $z^2 + az + b = 0$, where $a, b \in A$. Then $-a-z \in B$ is also a root of this polynomial and hence $a(-a-z)=b \in B$, however $B \setminus A$ is closed under multiplication that is the contradiction.

José Carlos Santos
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Alexey Milovanov
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1 Answers1

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Let $b\in B$ be integral over $A$. So, it satisfies a polynomial with coefficients from A, say $$b^{n}+a_{n-1}b^{n-1}+\cdots+a_{1}b+a_{0}=0.$$ So from that you have $$-a_{0}=b(b^{n-1}+a_{n-1}b^{n-2}+\cdots+a_{1}) \in A.$$ But $B-A$ is multiplicatively closed so either $b\in A$ (we are done then) or $b^{n-1}+a_{n-1}b^{n-2}+\cdots+a_{1} \in A$. Then iterate the process. Eventually you will get $b\in A$.

user26857
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Riju
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    The proof depends on the polynomial being monic - but that hypothesis is never explicitly used above. This dependency should be made explicit (e.g. by doing the base case of the induction). – Bill Dubuque Jul 15 '17 at 16:10