Localizing at maximal ideals and the product

Question

Let $D$ be an integral domain, $M_{i}$, $i = 1,...,r$ be some of its mutually distinct maximal ideals, and $e_{i}$be positive integers for all $i$. Is it true in general that the extension of the product ideal $M_{1}^{e_{1}}\cdots M_{r}^{e_{r}}$ in the localized ring $D_{M_{1}}$ is the same as the extension of $M_{1}^{e_{1}}$ in $D_{M_{1}}$? (In short, coincide with $M_{1}^{e_{1}}D_{M_{1}}$ up to identification $D\leq D_{M_{1}}$?) If this type of statement is true with some additional conditions, would anyone give some precise statements and indications of proofs?

Answer 1

Yes, it's true. Here are the basic facts used:

1. For any morphism $f\colon A\to B$ the extension of ideals $I \mapsto I^e$ (from ideals of $A$ to ideals of $B$) is a multiplicative operation, that is: $(I\cdot J)^e = I^e \cdot J^e$.

2. An ideal $I$ extends to the total ideal $1$ for the map $A \mapsto S^{-1} A$ if and only if $I \cap S \ne \emptyset$.

3. For the map $A\mapsto A_{\mathfrak{m}}$ any maximal ideal $\mathfrak{m}' \ne \mathfrak{m}$ extends to $(1)$ ( since $\mathfrak{m}' \not \subset \mathfrak{m}$).