An application of universal property of localization

Question

I am reading Bosch's AG. This is the reasoning:

Proposition 8. The canonical homomorphism $\tau: R \longrightarrow R_S$ from a ring $R$ to its localization by a multiplicative system $S \subset R$ satisfies $\tau(S) \subset\left(R_S\right)^*$ and is universal in the following sense: Given any ring homomorphism $\varphi: R \longrightarrow R^{\prime}$ such that $\varphi(S) \subset\left(R^{\prime}\right)^*$, there is a unique ring homomorphism $\varphi^{\prime}: R_S \longrightarrow R^{\prime}$ such that the diagram is commutative.

Furthermore, if $\varphi: R \longrightarrow R^{\prime}$ satisfies the same universal property as $\tau$ does, then $\varphi^{\prime}: R_S \longrightarrow R^{\prime}$ is an isomorphism.

Lemma 9. Let $R$ be a ring, $F=\left(f_i\right)_{i \in I}$ a family of elements in $R$, and $S \subset R$ the multiplicative subset generated by $F$, i.e. $S$ consists of all (finite) products of members in $F$. Then, fixing a system of variables $T=\left(t_i\right)_{i \in I}$, there is a canonical isomorphism

$$ R_S \xrightarrow{\sim} R[T] /\left(1-f_i t_i ; i \in I\right) . $$

In particular, for a single element $f \in R$, there is a canonical isomorphism $R_f \simeq R[t] /(1-f t)$.

Proof. The canonical ring homomorphism $\varphi: R \longrightarrow R[T] /\left(1-f_i t_i ; i \in I\right)$ sends all elements of $S$ to units. Thus, it factorizes over a well-defined ring homomorphism $\varphi^{\prime}: R_S \longrightarrow R[T] /\left(1-f_i t_i ; i \in I\right)$. On the other hand, it is easily checked that $\varphi$ satisfies the universal property of a localization of $R$ by $S$. Therefore $\varphi^{\prime}$ must be an isomorphism.

My question:

What does "On the other hand,..." mean? Does it mean I should check the "Furthermore..."? If so, I think it is hard to check "Given any ring homomorphism $\psi: R\longrightarrow R'$ such that $\psi (S)\subset\left(R^{\prime}\right)^*$, there is a unique ring homomorphism $\varphi':R[T] /\left(1-f_i t_i ; i \in I\right)\longrightarrow R'$ such that the diagram is commutative" since it is "any". On the contrary, I can construct the inverse map and check that it is indeed the inverse of $\varphi$ without too much difficult. Why should we use the universal property of localization?

Answer 1

If $\psi \colon R \to R'$ is any ring homomorphism such that $\psi(S) \subset (R')^*$, by the universal property of the polynomial ring $R[T]$ there exists a unique ring homomorphism $\psi' \colon R[T] \to R'$ such that $$ \psi' \circ \iota = \psi, $$ where $\iota \colon R \to R[T]$ is the canonical inclusion, and $\psi'(t_i) = \psi(f_i)^{-1}$ for all $i \in I$.

Note that, for each $i \in I$, $\psi'(1-f_it_i) = 1-\psi(f_i)\psi(f_i)^{-1} = 0$; i.e. $(1-f_it_i : i \in I) \subset \ker \psi'$.

Thus, by the universal property of the quotient ring $R[T]/(1-f_it_i : i \in I)$ there exists a unique ring homomorphism $\psi'' \colon R[T]/(1-f_it_i : i \in I) \to R'$ such that $$ \psi'' \circ \pi = \psi', $$ where $\pi \colon R[T] \to R[T]/(1-f_it_i : i \in I)$ is the canonical quotient map.

Finally, note that $\varphi = \pi \circ \iota$ (by definition) and $$ \psi'' \circ \varphi = \psi'' \circ \pi \circ \iota = \psi' \circ \iota = \psi. $$ In summary, for any ring homomorphism $\psi \colon R \to R'$ such that $\psi(S) \subset (R')^*$ there exists a unique ring homomorphism $\psi'' \colon R[T]/(1-f_it_i : i \in I) \to R'$ such that $\psi'' \circ \varphi = \psi$; that is, $\varphi$ satisfies the same universal property as $\tau \colon R \to R_S$ does. Thus, $\varphi'$ is an isomorphism.

In addition, to construct the inverse of $\varphi'$ you need to apply this same procedure to $\psi=\tau$, right? Hence, verifying the universal property is equivalent to finding the inverse of $\varphi'$.

Answer 2

Q: "How can I prove the uniqueness of the final one – Luca Hao Commented 3 hours ago"

A: Let $\phi: R \rightarrow R[t_i]/(\{1-f_it_i\}):=B$ be the canonical map and assume $f: R \rightarrow R_1$ is a map with $f(S) \subseteq R_1^*$. A map $f_1:B \rightarrow R_1$ commuting with $f$ must satisfy the following: For any $i$ we must have

$$f_1(1-f_it_i)=0$$

hence

$$f_1(1)=1=f_1(f_i)f_1(t_i)=f_if_1(t_i)$$

hence

$$f_1(t_i)=1/f_i$$

and it follows $f_1$ is uniquely determined by $f$.

C: can we deduce the final uniqueness by exactly the former two uniqueness – Luca Hao' Commented 1 hour ago

A: @Luca Hao - Since both rings $R_S$ and $B:=R[t_i]/(1−f_it_i)$ satisfy the universal property there is a canonical isomorphism $R_S≅B$ commuting with the maps from $R$ to $R_S$ and $B$. Note: You need to add the multiplicative unit $1$ to $S$ - this element is not included in $S$ with the above definition. – hm2020 Commented 1 min ago