环论中局部化环$R_M$的概念解析与相关习题解答请求

Hey there! Let's start by breaking down exactly what $R_M$ is, since that's your core question, then we can unpack the problem you're working through.

First, let's recall the general idea of ring localization: when we take the quotient ring (localization) of a commutative ring $R$ with respect to a multiplicative set $S$ (a subset closed under multiplication that includes the ring's identity), we're essentially building a new ring where every element of $S$ becomes invertible.

In your specific case, the multiplicative set is $S = R - M$—that's all elements of $R$ that aren't in the maximal ideal $M$. Let's make this concrete with what $R_M$ looks like:

• Elements of $R_M$ are fractions of the form $\frac{a}{s}$, where $a \in R$ and $s \in S$ (so $s \notin M$).
• We consider two fractions equivalent if $\frac{a}{s} = \frac{b}{t}$ when there exists some $u \in S$ such that $u(at - bs) = 0$ in $R$—this is just the standard way to avoid "dividing by zero" inconsistencies.
• Ring operations work exactly like regular fraction arithmetic:
  • Addition: $\frac{a}{s} + \frac{b}{t} = \frac{at + bs}{st}$
  • Multiplication: $\frac{a}{s} \cdot \frac{b}{t} = \frac{ab}{st}$

This ring $R_M$ is called the localization of $R$ at the maximal ideal $M$, and it's a classic example of a local ring (a ring with exactly one maximal ideal)—which is exactly what part (a) asks you to prove!

Here's the full problem you're tackling, formatted for clarity:

Let $R$ be a commutative ring with identity and $M$ a maximal ideal of $R$. Let $R_M$ be the ring of quotients of $R$ with respect to the multiplicative set $R-M= {s\in R \mid s\notin M}$. Show the following:

a) $M_M={\dfrac{a}{s} \mid a\in M, s\notin M}$ is a unique maximal ideal of $R_M$.

b) The fields $R/M$ and $R_M/M_M$ are isomorphic.

To give you a quick intuition for these parts:

• For (a): Any element of $R_M$ not in $M_M$ looks like $\frac{s}{t}$ where $s,t \notin M$. Since $M$ is maximal, $s$ has an inverse modulo $M$, and since $st \notin M$, $\frac{t}{s}$ is the inverse of $\frac{s}{t}$ in $R_M$. So every element outside $M_M$ is invertible, which means $M_M$ is the only maximal ideal.
• For (b): The natural isomorphism maps $a + M \in R/M$ to $\frac{a}{1} + M_M \in R_M/M_M$. You just need to verify this map is well-defined, a ring homomorphism, and bijective (since both are fields, showing it's non-trivial is enough to confirm it's an isomorphism).

If you want a step-by-step walkthrough of either proof, just let me know!

备注：内容来源于stack exchange，提问作者user1242284