use universal property of localization to show that $R_f\cong R[x]/(fx-1)$

Question

Let $R$ be a commutative ring with $1$ and let $D=\{\,f^n\,|\,n\geq0\,\}$ be the multiplicative set where $f\in R$ is not nilpotent. And the question is the topic. The following is my idea:

Let $\varphi:R[x]\to D^{-1}R$ defined by $r\mapsto r$ and $x\mapsto1/f$. Then clearly $\varphi$ is surjective so that $$ R[x]/\ker\pi\cong D^{-1}R $$ by the First Isomorphism Theorem. Via the isomorphism above, by the universal property of localization, the following diagram $$ \require{AMScd}\begin{CD} R @>>> R[x]/\ker\varphi\\ @. {_{\rlap{}}\style{display: inline-block; transform: rotate(30deg)}{{\xrightarrow[\rule{3.52em}{0em}]{}}}} @V\psi VV\\ @. R[x]/(xf-1) \end{CD} $$ commutes where the arrows are compositions of natural injections and natural projections: $$ \begin{cases} R\hookrightarrow R[x]\twoheadrightarrow R[x]/\ker\varphi \\R\hookrightarrow R[x]\twoheadrightarrow R[x]/(xf-1). \end{cases} $$ By the universal property of quotient rings, the following diagram $$ \require{AMScd}\begin{CD} R[x] @>>> R[x]/(xf-1)\\ @. {_{\rlap{}}\style{display: inline-block; transform: rotate(30deg)}{{\xrightarrow[\rule{3.52em}{0em}]{}}}} @VV\phi V\\ @. R[x]/\ker\varphi \end{CD} $$ commutes where the arrows are compositions of natural projections: $$ \begin{cases} R[x]\twoheadrightarrow R[x]/(xf-1). \\R[x]\twoheadrightarrow R[x]/\ker\varphi. \end{cases} $$ Now we get two unique homomorphisms $\psi$ and $\phi$ determined by the arrows from $R$ and $R[x]$ to $R[x]/(xf-1)$ and $R[x]/\ker\varphi$.

[1] Whether the isomorphism $(R[x]/(xf-1))/(\ker\varphi/(xf-1))\cong R[x]/\ker\varphi$ is useful or not to solve this problem.

[2] It remains to show that $\ker\pi\subseteq(xf-1)$, which is "not difficult to see that" stated by the authors, but I do not know how to move on. Is there a quick way to show this just by the universal properties of localizations and quotient rings? Since I have not learnt the universal property of polynomial rings as mentioned in other answers.

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Update: I have known the usage of the universal property of universal property of localizations, by @linkja and @Bill Dubuque 's help, as follows:

It is clear in the ring $R[x]/\langle xf-1\rangle$ we have that $\iota(1)=1+\langle xf-1\rangle$ and $\iota(f^n)^{-1}=x^n+\langle xf-1\rangle$ where $D=\{\,f^n\,|\,n\geq0\,\}$, so we can use the universal property of localizations to obtain the commutative diagram: $$ \require{AMScd}\begin{CD} R @>\sigma>> D^{-1}R\cong R[x]/\ker\varphi\\ @. {_{\rlap{}}\style{display: inline-block; transform: rotate(30deg)}{{\xrightarrow[\rule{3.52em}{0em}]{\iota}}}} @V\psi VV\\ @. R[x]/\langle xf-1\rangle \end{CD} $$ where $\iota=\psi\circ\sigma$. Since $\psi$ is surjective, it suffices to show that $\psi$ is also injective.

In fact, in this commutative diagram, $\psi$ is injective if and only if $\iota$ is injective.

Since $\iota$ is injective proved in this link, $\ker(\psi\circ\sigma)=\ker\iota=0$, i.e., $\forall\,r\in R/\{0\}$, $\psi(r)=\psi(\sigma(r))\ne0$. Since $0\not\in D$, $\forall\,d\in D\subseteq D^{-1}R$, $\psi(d)=\psi\circ\sigma(d)=\iota(d)\in(R[x]/\langle xf-1\rangle)^{\times}$, i.e., $\psi(d)$ is never a zero divisor.

Thus $\forall\,d^{-1}r\in D^{-1}R$, $$ \begin{aligned} \psi(d^{-1}r)=\psi(d)^{-1}\psi(r)&=0 \\(\text{since }\psi(d)\text{ is never a zero divisor})\ \ \text{iff}\ \ \psi(r)&=0 \\\text{iff}\ \ \iota(r)=\psi\circ\sigma(r)=\psi(r)&=0 \\(\text{since }\iota\text{ is injective})\ \ \text{iff}\ \ r&=0 \\\text{iff}\ \ d^{-1}r&=0, \end{aligned} $$ hence $\psi$ is injective. And since $\psi$ is also a surjective ring homomorphism, $\psi$ is an isomorphism, so finally we get $$ R_f=D^{-1}R\cong R[x]/\ker\varphi=R[x]/\langle xf-1\rangle, $$ as desired.

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Update: The above answer is not completed: it have not considered the case that $f$ is a zero divisor so $\iota$ is not necessarily injective. I have given a complete proof here: use universal property of localization to show that $R_f\cong R[x]/\langle xf-1\rangle$ where $f$ may be a zero-divisor but not nilpotent.

Answer 1

You want to show that $R[X]/(Xf-1)\cong D^{-1}R$, so you want to construct an isomorphism between them:

For this we work similar as you did in your question:

We prove that $R[X]/(Xf-1)$ is isomorphic to $D^{-1}R$ by the universal property of localization, for which it suffices to verify the properties i), ii), iii) below, by Corollary 3.2 in Atiyah and MacDonald's Introduction to commutative algabra.

We look at the obvious morphism $$\pi: R\to R[X] \to R[X]/(Xf-1), r\mapsto r +(Xf-1).$$

i) We see that for all $s\in D = \{f^n:n\geq 0\}$ that $s$ is invertible as its inverse is just $X^n+(Xf-1)$ by the given relation ($Xf = 1$).

ii) And we also see that if $\pi(r) = 0$ in $R[X]/(Xf-1)$ then $r \in (Xf-1)\subset R[X]$, so there is some $P(X) = a_n X^n+...+a_0\in R[X]$ such that $P(X)\cdot(Xf-1) = r$ so we have that $fa_nX^{n+1}+(fa_{n-1}-a_n)X^n+...+ a_0 = r$. comparing coefficients we get that $a_0 = r$ and all other coefficients are equal to zero. Inductively we see that $fa_0-a_1=0$ so $a_1=fa_0 = fr$ and therefore $a_n = f^nr$. At some point $a_n$ is zero and therefore $f^nr = 0$ for some $n$ and we gat that $sr = 0$ for some $s \in D$.

iii) Every element of $R[X]/(Xf-1)$ is of the form $\frac{\pi(r)}{\pi(f^n)}$ as $\frac{1}{\pi(f)} = X$.

Those are the same calculations you did (in part) in your question.

There, as you have written both rings as quotients of the polynomial ring in one variable it is enough to see that the ideals you divide out are the same. The calculation that they are the same ideals is precisely ii). Part iii) was your argument that $R[X]\to D^{-1}R$ is surjective and i) is just in the definition of your map.

Answer 2

It suffices to show that the map $\rho:R\to R[x]/\langle fx-1\rangle$ sending $r$ to the image of $r$ in $R[x]/\langle fx-1\rangle$ satisfies the universal property of localization of $R$ at the set $S=\{1,f,f^2,\dots\}$. Evidently, $\rho(f^k)$ is invertible for all $k\ge0$. Now suppose $A$ is a commutative ring and $\rho':R\to A$ is a map such that $\rho'(S)\subseteq A^\times$. We must show that there exists a map $g:R[x]/\langle fx-1\rangle\to A$ such that $g\circ \rho=\rho'$.

By the universal property of the polynomial algebra $R[x]$, there exists a map $\phi:R[x]\to A$ which sends $r\in R$ to $\rho'(r)$ and $x$ to $\rho'(f)^{-1}$. Then, $\phi(fx-1)=0$, hence $\phi$ factors as $$ R[x]\longrightarrow\frac{R[x]}{\langle fx-1\rangle}\stackrel{g}{\longrightarrow} A \, , $$ completing the proof.

Answer 3

Let $R$ be a commutative ring with $1$ and let $D=\{\,f^n\,|\,n\geq0\,\}$ be the multiplicative subset of $R$ where $f\in R$ is not nilpotent (but $f$ may be a zero divisor). Show that $R[x]/\langle xf-1\rangle\cong D^{-1}R$.

Let $\tau:R\to R[x]\to R[x]/\langle xf-1\rangle$ be defined by $\tau:r\mapsto r\mapsto r+\langle xf-1\rangle$, which is naturally be a ring homomorphism that sends $1$ to $\bar1$. Note that: $$ \begin{aligned} \bar x\bar f &=xf+\langle xf-1\rangle \\&=xf-(xf-1)+\langle xf-1\rangle \\&=1+\langle xf-1\rangle \\&=\bar1, \end{aligned} $$ hence $$ \bar1=\bar1^k=\big(\bar x\bar f\big)^k=\overline{x^k}\cdot\tau(f^k), $$ namely, $\forall\,f^k\in D=\{\,f^n\,|\,n\geq0\,\}$, $\tau(f^k)\in\big(R[x]/\langle xf-1\rangle\big)^{\times}$.

Now we have shown that $\tau$ is a ring homomorphism mapping $1$ to $\bar1$ and $d$ to some unit in the codomain for every $d\in D$, then by the universal property of localizations on commutative rings with identities, we have the following diagram of ring homomorphisms: $$ \require{AMScd}\begin{CD} R @>\sigma>> D^{-1}R\\ @. {_{\rlap{}}\style{display: inline-block; transform: rotate(30deg)}{{\xrightarrow[\rule{3.52em}{0em}]{\tau}}}} @V\psi VV\\ @. R[x]/\langle xf-1\rangle \end{CD} $$ where $\sigma$ is the localization map and that $\psi$ is uniquely determined by $\tau$ and $\sigma$. It remains to prove that $\psi$ is surjective and injective.

Let $g=\sum_jb_jx^j\in R[x]$. Then $$ \begin{aligned} \textstyle\overline{\sum_jb_jx^j} &=\textstyle\sum_j\overline{b_j}\ \bar x^j \\&=\textstyle\sum_j\tau(b_j)\psi(1/f)^j \\&=\textstyle\sum_j\psi(b_j/1)\psi(1/f)^j \\&=\textstyle\psi\left(\sum_jb_j/f^j\right) \\&\in\operatorname{im}\psi, \end{aligned} $$ so $\psi$ is surjective.

Note first that $$ \bar1=\psi(1/1)=\psi\left(\frac f1\cdot\frac1f\right)=\psi(f/1)\psi(1/f), $$ so $\psi(1/f)$ is a unit.

Let $r/f^n\in\ker\psi$ where $r\ne0$. Then sine $\psi(1/f)$ is a unit and hence never a zero-divisor, $\bar0=\psi(r/f^n)=\psi(r/1)\psi(1/f)^n$ implies $0=\psi(r/1)=\psi\circ\sigma(r)=\tau(r)$, i.e., $r\in\langle xf-1\rangle$, namely, $\exists\,g=\sum_{j=0}^mb_jx^j\in R[x]$ s.t. $r=g(x)\cdot(xf-1)$. Expanding $g$ in the RHS we obtain that $$ \begin{aligned} r &=\textstyle\left(\sum_{j=0}^mb_jx^j\right)(xf-1) \\&=b_mfx^{m+1}-b_0 \\&\phantom{=}+\left(\sum_{j=1}^m(b_{j-1}f-b_j)x^j\right). \end{aligned} $$ By comparing the coefficients on both sides we see that $r=-b_0$, $b_mf=0$ and $b_{j-1}f-b_j=0$ for any $j\in[1,m]$, so inductively we also have $b_j=-rf^j$ for $j\in[1,m]$ and $rf^{m+1}=b_mf=0$.

One should be careful, however, we can not directly conclude that $r=0\in R$ since $f$ may be a zero-divisor in $R$.

Recall that $\ker\sigma=\{\ x\in R\ |\ \exists\,d\in D\ \text{s.t.}\ xd=0\ \}$, hence $rf^{m+1}=0$ together with $f^{m+1}\in D$ implies $\sigma(r)=r/1=0/1$ in $D^{-1}R$, thus $r/f^n=(r/1)(1/f^n)=0/1\in D^{-1}R$, providing that $\psi$ is injective.

Thereby $\psi$ is a injective and surjective ring homomorphism, i.e., an isomorphism between $D^{-1}R$ and $R[x]/\langle xf-1\rangle$, so eventually we get $R_f:=D^{-1}R\cong R[x]/\langle xf-1\rangle$.