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

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.

• In standard notation you want to prove that $R_M$ is local and $MR_M$ is its maximal ideal. – user26857 Feb 10 at 18:42

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.
• 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) – davidh Feb 10 at 8:48
• @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 $s' \in M$.) – Alex Wertheim Feb 10 at 8:53