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Assume $A$ is a discrete valuation ring with quotient field $K$ and maximal ideal $\mathfrak{m}$. If $S$ is a local ring containing $A$ and contained in $K$ with maximal ideal containing $\mathfrak{m}$, how do I show that $S = A$?

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Let $x\in S$. Then there are two possibilities: $x\in A$, or $x\notin A$. If the latter happens then $x^{-1}\in A$ and then $x^{-1}\in\mathfrak m$, so $x^{-1}\in\mathfrak m_S$. But $x\in S$, so $1=xx^{-1}\in\mathfrak m_S$, a contradiction.

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