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?
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5Yes that's correct. – Noah Olander Jun 05 '16 at 19:16
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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.
Arthur
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Yes you are right. I think the easiest way to formalize this is showing that A satisfies the universal property of the localization.
svelaz
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5This is so much better (by lightyears) than the other answer, because it is trivial to show that the identity map $A \to A$ satisfies the universal property. Namely it clearly maps $S = A - \mathfrak m$ to units and any map $A \to B$ trivially factors uniquely over the identity map. – MooS Jun 07 '16 at 05:15
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2But to see that $S=A-\mathcal{m}$ is mapped to units by the identity, you would still have to use the first part of the other answer, right? Namely, you would have to proof that $S=A-\mathcal{m}$ consists of units only. – Guenterino Feb 02 '22 at 09:58