Why does localization look so close to quotienting? Consider the ismorphism theorem:
For a ring map $\phi: R_1 \rightarrow R_2$, let $\ker\phi \equiv \{ r_1\in R_1 : \phi(r_1) = 0 \}$. Then, $\ker \phi$ is an ideal of $R_1$, and there exists an epi-mono factorization of $\phi$ into $R_1 \twoheadrightarrow R_1 /\ker\phi \hookrightarrow R_2$ where $\hookrightarrow$ is injective and $\twoheadrightarrow$ is surjective.
Now consider the similar theorem for localization, where I denote by $R \hat \times S \equiv S^{-1}R$ for notational suggestivity. Now I write down:
For a ring map $\psi : R_1 \rightarrow R_2$, Let $\operatorname{rek}\psi \equiv \{ r_1 \in R_1 : \psi(r_1) = 1 \}$. Then $\operatorname{rek}\psi$ is a multiplicative subset of $R_1$ and there exists a mono-epi factorization of $\psi$ into $R_1 \hookrightarrow R_1 \hat \times \psi \twoheadrightarrow R_2$
I can build a table:
- $\phi \leftrightarrow \psi$.
- $\ker \leftrightarrow\operatorname{rek}$.
- injection $\leftrightarrow$ surjection.
- $/ \leftrightarrow \hat\times$
- ideal $\leftrightarrow$ multiplicative subset.
- $0 \leftrightarrow 1$.
to convert from the quotienting into localization. Is there some "deep" going on here for this duality? This $0 \leftrightarrow 1$ business makes me hopeful that there might be something deeper / categorical lurking in the background.
EDIT: I had only commutative rings in mind when I wrote this. Please feel free to take assumptions on $R$ as required (Commutative, Noetherian, for example), if that allows us to explain this "duality".
Best Answer
Fabio gives a very nice answer to your question, but doesn't directly address an important point of confusion in your original post/comments, so I'm adding this answer for posterity. In general, a map $\psi:R_1\rightarrow R_2$ will absolutely not induce an epimorphism $\text{rek}(\psi)^{-1}R_1\twoheadrightarrow R_2$, even if we take the stronger definition $\text{rek}(\psi)=\psi^{-1}(R_2^\times)$ given by Fabio. For instance, if every element of $\text{rek}(\psi)$ is already a unit in $R_1$, then we will just have $\text{rek}(\psi)^{-1}R_1=R_1$, and so using this it is easy to come up with examples where the induced map is not epi.
For instance, take $R_1=\mathbb{Q}$, and $R_2$ any field extension of $\mathbb{Q}$ with a non-trivial automorphism $\alpha$ that fixes $\mathbb{Q}$ pointwise, with $\psi:R_1\hookrightarrow R_2$ the inclusion map. Then $\text{rek}(\psi)=\mathbb{Q}^\times$ , so $\text{rek}(\psi)^{-1}\mathbb{Q}=\mathbb{Q}$ and the induced map to $R_2$ is just $\psi$, which is certainly not an epimorphism. (E.g. $\alpha\circ\psi=\text{id}_{R_2}\circ\psi$ but $\alpha\neq\text{id}_{R_2}$).
Indeed, the polynomial ring example you give in the comments of your post does not hold in general either. If we let $R_1=\mathbb{R}[x]$ and $R_2=\mathbb{R}[x,y]$, with $\psi:R_1\hookrightarrow R_2$ again the inclusion map, then once again $\text{rek}(\psi)=R_1^\times$ but $\psi$ is certainly not epi.
The problem in all of these examples is that $R_2$ can be very big compared to the image of $R_1$; hopefully the above examples clarify that point. (Note however, that – provided $R_2\neq\{0\}$ – the map $R_1\hookrightarrow \text{rek}(\psi)^{-1}R_1$ will still be injective, even if we use Fabio's stronger definition of $\text{rek}(\psi)$, because no element of $\text{rek}(\psi)$ can be a zero-divisor in $R_1$.)