Localization of a local ring

Question

If $A$ is a local ring, $\mathfrak{m}$ is its maximal ideal, then is $A_{\mathfrak{m}}\cong A$ ? That's, in my opinion, because every denominator is invertible. Am I right?

Answer 1

Yes, you are right. More details below.

If you have an element $a \in A$ that is not invertible, then $1 \notin (a)$, so $(a) \neq A$. Therefore $(a)$ is clearly contained in some maximal ideal, and since there is only one in this case, $(a)\subseteq \mathfrak m$. This yields $a \in \mathfrak m$. Thus every element in $A\setminus \mathfrak m$ is invertible.

Now, for the localisation, there is a canonical map $A \to A_{\mathfrak m}$ given by $a \mapsto \frac{a}{1}$. We show that it is an isomorphism by constructing an inverse. Take $\frac{a}{b} \in A_{\mathfrak m}$ (technically this element should be written $(a, b)$). Since $b \notin \mathfrak m$, the inverse $b^{-1}$ exists, and we have by the definition of localisation that $$\frac{a}{b} = \frac{ab^{-1}}{bb^{-1}} = \frac{ab^{-1}}{1}$$ so we may construct an inverse map $A_{\mathfrak m}\to A$ that takes $\frac ab$ to $ab^{-1}$. By the above equalities, this is clearly an inverse to the canonical localisation map.

Answer 2

Yes you are right. I think the easiest way to formalize this is showing that A satisfies the universal property of the localization.