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One of the hardest modules I took at Imperial was MATH70061 Commutative Algebra taught by Yankı Lekili. What made the module so challenging was the fact that we had to submit solutions to 6-8 tedious homework problems every 2 weeks. In addition, the course covered quite a broad selection of material, which we had to stay on top of. The lecture notes and homework sheets are publicly available on Yankı's homepage. To my surprise, Problem 5 of sheet IV was a glaring outlier in the set of problems we had to solve, as the solution was frankly so easy it could've been explained to a kindergartener. The problem asks for a proof of the simple identity $$A_f\cong A[x]/(fx-1).$$ Here $A$ is any commutative ring (with identity, of course), $f$ is any element of $A$, and $A_f$ denotes the localisation with respect to the multiplicative set $$\{1,f,f^2,f^3,\ldots\}.$$ Hence, the elements of $A_f$ are quotients $a/f^k$, where $a\in A$ and $k\in\mathbb{N}$. If $f$ is nilpotent, we encounter division by zero in this definition, so $A_f$ collapses to $0$. On the other hand, if $f$ is not nilpotent, $A_f$ behaves pretty much as one would expect. Note that the problem is intuitively obvious: on the right hand side we are adjoining a new variable to $A$, which we force to be a multiplicative inverse of $f$ by making $fx-1$ vanish (i.e. $fx=1$). I encourage the reader to try proving the statement now, because this is one of those problems, where formulating the proof yourself is way more fruitful than trying to decode a proof someone else wrote.

The easy proof

Here is the obvious proof of the identity. Recall that it suffices to prove that $A[x]/(fx-1)$ satisfies the universal property of localisation. More specifically, we want to find a homomorphism $\varphi\colon A\to A[x]/(fx-1)$ such that for every homomorphism $\psi\colon A\to B$ mapping $f$ to a unit in $B$, there is a unique morphism $\theta\colon A[x]/(fx-1)\to B$ making the below triangle commute. The first thing we can do, is use the universal property of polynomial algebras to obtain a homomorphism $\varphi_0\colon A\to A[x]$ and a unique homomorphism $\theta_0\colon A[x]\to B$ such that $\theta_0(x)=\psi(f)^{-1}$ and which makes the below triangle commute. Then, use the 1st isomorphism theorem to obtain a homomorphism $\varphi'_1\colon A[x]\to A[x]/\ker(\theta_0)$ and a unique homomorphism $\theta_1\colon A[x]/\ker(\theta_0)\to B$ making the below triangle commute. Finally, note that $\theta_0(fx-1)=\psi(f)\psi(f)^{-1}-1=0$, so $\varphi_1$ splits as Setting $\varphi=\varphi_1\circ\varphi_0$ and $\theta=\theta_1\circ\rho$ gives what we want, completing the proof.

The abstract proof

After writing the first proof down, I thought to myself that there should be an abstract proof that works by somehow "composing" the different universal properties. Then I remembered the fact that given two pairs of adjoint functors $$\begin{gather} L_1\colon\mathcal{C}\to\mathcal{D},\quad R_1\colon\mathcal{D}\to\mathcal{C}\\ L_2\colon\mathcal{D}\to\mathcal{E},\quad R_2\colon\mathcal{E}\to\mathcal{D} \end{gather}$$ with $L_1\dashv R_1$ and $L_2\dashv R_2$, the compositions are adjoint: $$(L_2\circ L_1)\dashv(R_1\circ R_2).$$ In the end I managed to find a collection of adjoint functors such that the proof of the identity follows directly from this fact.

Denote $\mathbf{Ring}_\ast$ for the category of pointed rings, i.e. of pairs $(A,f)$, where $f\in A$ such that a morphism $(A,f)\to(B,g)$ is an ordinary ring homorphism $A\to B$ mapping $f$ to $g$. Next, denote $\mathbf{Ring}_\ast^\mathrm{u}$ for the subcategory, where the chosen element $f$ of a pointed ring $(A,f)$ is a unit in $A$. Then, it is a well-known fact that the "localisation functor" $$L\colon\textbf{Ring}_\ast\to\textbf{Ring}_\ast^\mathrm{u}:(A,f)\mapsto(A_f,f)$$ is a left adjoint to the inclusion $R\colon\textbf{Ring}_\ast^\mathrm{u}\hookrightarrow\textbf{Ring}_\ast$.

We start by factorising this inclusion functor through two other categories. Thus, let $\textbf{Ring}_{\ast,\ast}$ denote the category of double pointed rings and $\textbf{Ring}_{\ast,\ast,I}$ a special category of tuples $(R,x,y,I)$ such that $I$ is an ideal of $R$ containing $xy-1$. A morphism $(R,x,y,I)\to(S,z,w,J)$ in this category is a ring homomorphism $R\to S$ mapping $x$ to $z$ and $y$ to $w$ such that the image of $I$ is contained in $J$. Then, the inclusion functor $R$ factorises trivially as $$\textbf{Ring}_\ast^\text{u}\xrightarrow{i}\textbf{Ring}_{\ast\ast,I} \xrightarrow{U_I}\textbf{Ring}_{\ast\ast}\xrightarrow{U_\ast} \textbf{Ring}_\ast,$$ where $i\colon (R,u)\mapsto (R,u,u^{-1},(0))$ and the functors $U_I$ and $U_\ast$ are the obvious forgetful functors.

Now, by the composition of adjunctions, $L$ factorises as the composition of the left adjoints of these three functors. One can check that for the following three functors $$\begin{array}{cccc} Q:&(R,x,y,I)&\mapsto&(R/I,x)\\ E:&(R,x,y)&\mapsto&(R,x,y,(xy-1))\\ P:&(R,x)&\mapsto&(R[y],x,y) \end{array}$$ we have $Q\dashv i$, $E\dashv U_I$, and $P\dashv U_\ast$. In conclusion, this implies that $$\begin{align} (A_f,f)&=L(A,f)\\ &\cong(Q\circ E\circ P)(A,f)\\ &=(Q\circ E)(A[x],f,x)\\ &=Q(A[x],f,x,(xf-1))\\ &=\left(A[x]/(xf-1),f\right). \end{align}$$