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Problem: Let $R$ be an integral domain with fraction field $K$. Assume that $x \in R$ or $x^{-1} \in R$ for all $x \in K^{\times}$. Prove that $R$ is a local ring (the ring $R$ is said to be a valuation ring of $K$).

Attempt: I don't really know how to begin. There a different possibilities to show that $R$ is a local ring. Either showing it has a unique maximal ideal, or showing that $R \setminus R^{\times}$ is an ideal, or that the sum of two non-units is a non-unit.

I take $r \in R$ and $x \in R \setminus R^{\times}$. I want to show that $rx \in R \setminus R^{\times}$. I can consider $rx$ as an element in the fraction field $K$. Then either $(rx)^{-1} \in R$ or $rx \in R$. But the second is trivial. So I'm not really sure how to use my assumption.

Any help is appreciated.

Kamil
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1 Answers1

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Highlights:

Prove that the set of all non-units in $\;R\;$, call it $\;M\;$, is an ideal: that $\;M\;$ is closed under multiplication with $\;R\;$ is pretty easy. Now, if $\;a,b\in M\;$ , then you must prove $\;a+b\in M\;$. But assuming $\;a\neq0\;$ (otherwise this is trivial), we can write in $\;K\;$ :

$$a+b=a(1+a^{-1}b)\implies \text{ if there exists}\;r\in R\;\;s.t.\;\;a(1+a^{-1}b)r=1$$

then $\;a\;$ is a unit in $\;R\;$ if $\;1+a^{-1}b\in R\;$ , so it must be $\;1+a^{-1}b\notin R\;$, from which we get that $\;a^{-1}b\notin R\;$, and thus $\;ab^{-1}\in R\;$ , and finally

$$a+b=b(1+ab^{-1})\in R\implies a+b\in M$$

otherwise $\;b\;$ is a unit and contradiction.

Fill in details above now.

DonAntonio
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  • I do not understand why you make this implication: 'if there exists $r \in R$... How does this follow from what you said before? – Kamil Jul 18 '19 at 07:07
  • Are u assuming that $a + b $ is a unit, and then you want to derive a contradiction? Shouldn't the $r$ be in $K$ then? Because you are working in $K$. – Kamil Jul 18 '19 at 07:09