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Let $R$ be a domain and let $M \subset R$ be a maximal ideal. Let $K$ be the quotient field of $R$.

Let $T = \{\frac{r}{s} , s\notin M\} \subseteq K$, and $M_1 = \{\frac{m}{s},s \notin M, m \in M\} \subseteq T$.

I want to show $M_1$ is a maximal ideal in T, and moreover it is the only unique ideal in T. I'm lost as to where to begin on this problem. Thanks.

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  • $\begingroup$ In standard notation you want to prove that $R_M$ is local and $MR_M$ is its maximal ideal. $\endgroup$ – user26857 Feb 10 at 18:42
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To show that $M_{1}$ is a maximal ideal of $T$, we need to show that any ideal $I$ of $T$ which properly contains $M_{1}$ is all of $T$. By the definition of $I$, any such ideal $I$ necessarily contains an element of the form $r/s$ with $r, s \notin M$. But by the definition of $T$, $s/r \in T$, whence $r/s$ is a unit in $T$ and so $I = T$, as desired.

Moreover, the above argument also shows that any ideal of $T$ which is not contained in $M_{1}$ is all of $T$. Thus, any proper ideal of $T$ is contained in $M_{1}$. Thus, if $Q$ is a maximal ideal of $T$, then $Q$ is contained in $M_{1}$, hence equal to $M_{1}$. This proves that $M_{1}$ is unique.

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  • $\begingroup$ This might be a stupid question, but why $s\notin M$ and $s' \notin M$ implies $ss'\notin M$ (to show $M_1$ is ideal) $\endgroup$ – davidh Feb 10 at 8:48
  • $\begingroup$ @davidh: this is a consequence of the fact that $M$ is a maximal ideal of $R$, hence prime. (The condition that $s, s' \notin M$ implies $ss' \notin M$ is the contrapositive of the condition that $ss' \in M$ implies $s \in M$ or $s' \in M$.) $\endgroup$ – Alex Wertheim Feb 10 at 8:53

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