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In the part Proof of III, Pg. No. 9, under proof of Proposition 3, Chapter I, it says the following : Let $x$ be a nonzero element of $m$, and form the ring $A_x$ of fractions of the form $\frac{y}{x^n}$, with $y \in A$, and $ n \geq 0$ arbitrary. Then, Condition (ii) implies that $A_x = K$ ; indeed if not, $ A_x$ would not be a field,... Here, Condition (ii) is the following: $A$ has a unique non-zero prime ideal, where $A$ is a Noetherian integral domain.

I understand till before the part in bold, can someone please help me to understand the part in bold? I attach the images of the pages of relevant section below (pardon me for the bad image quality):

Local Feilds, Section 2

Section 2, Chapter I, Local Fields

P-addict
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  • is it beacuse elements in $m$, the maximal ideal of $A$ are not invertible in $A$, and so they won't be invertible in $A_x$ ? – P-addict May 06 '19 at 06:53

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$A$ is a domain with unique nonzero prime ideal. Hence $A_x$ is a domain too, and it's prime ideals are in bijective correspondence with those of $A$ not containing $x$. But this means that the only prime ideal in $A_x$ is the zero ideal, hence it is a field.

asdq
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