I am still trying to understand the concept of a natural or canonical isomorphism and I actually think I understand the idea but maybe somebody could help me in the following:

Let $R$, $A$ be rings (commutative with unit) and $S\subset R$ a multiplicatively closed subset, $\tau_A:R\rightarrow A$ a ring homomorphism satisfying $\tau_A(S)\subset A^*$ and $f:A\rightarrow R_S$ an isomorphism. Let $\tau_S:R\rightarrow R_S$ be the ring homomorphism associated to $R_S$.

Is it true in this situation that the natural morphism $\phi:R_S\rightarrow A$ (satisfying $\phi\circ\tau_S=\tau_A$) actually is an isomorphism? As $f$ can be an arbitrary, unnatural isomorphism (meaning $\tau_A\neq f^{-1}\circ \tau_S$) it feels like it's not true but then again I can't find a counterexample.

So maybe someone can give a hint, thanks already!

  • $\begingroup$ It's rather because $\tau_A$ can be arbitrary that $\phi$ need not be an isomorphism. For instance let $S=\{1\}$, $A=R$ and $\tau_A$ a not-iso endomorphism $\endgroup$ – Max Apr 30 '18 at 17:50
  • $\begingroup$ just as a small note, you may not want to call the morphism $\phi$ "natural" in this context. It is more that $\phi$ is induced by a universal property. Calling things "natural isomorphism" (like the isomorphism between a finite dimensional vector space and its double dual) often hints that there is some natural transformation of functors going on (even if that isn't explicit what the functors are) $\endgroup$ – user171177 Apr 30 '18 at 21:48

Consider for example $$R=k[x_1, x_2, x_3, \dots], A=k(x_1, x_2, x_3, \dots, y)$$ where $k$ is a field and $x_1, x_2, x_3, \dots, y $ are indeterminates, $S=R\setminus\{0\}$ and $\tau_A$ is the obvious inclusion (every $x_i$ is sent to $x_i$). Then $\tau_S$ is the inclusion of $R$ into its fraction field, which is $R_S=k(x_1, x_2, x_3, \dots)$, and $\phi$ is the obvious inclusion of this field to $A$ (again, $x_i$ goes to $x_i$). In particular, $\phi$ is not an isomorphism, because $y$ is not in the image.

However, there is an isomorphism $f:R_S \rightarrow A$: Just send $x_1$ to $y$ and each $x_i$ to $x_{i-1}, \;\;i\geq 2$.

Remark: In case you don't believe in non-Noetherian rings, you can obtain a Noetherian example of the same type from this by replacing $R$ by $R$ localized at, say, $S'=k[x_2, x_3, x_4, \dots]\setminus \{0\}$, so that $R_{S'}=K[x_1]$ for a field $K=k(x_2, x_3, x_4, \dots)$.


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