Localisation is isomorphic to a quotient of polynomial ring

Question

I am having trouble with the following problem.

Let $R$ be an integral domain, and let $a \in R$ be a non-zero element. Let $D = \{1, a, a^2, ...\}$. I need to show that $R_D \cong R[x]/(ax-1)$.

I just want a hint.

Basically, I've been looking for a surjective homomorphism from $R[x]$ to $R_D$, but everything I've tried has failed. I think the fact that $f(a)$ is a unit, where $f$ is our mapping, is relevant, but I'm not sure. Thanks

Answer 1

In the quotient $R[x]/(ax-1)$, the class of $x$ is invertible. What is its inverse?

This will tell you where $x$ must be mapped under $R[x]\to R_D$ if this must will induce an isomorphism $R[x]/(ax-1)\to R_D$.

Answer 2

As $\,\,ax-1=0\,\,$ in the quotient $\,\,R[x]/(-1+ax)\,\,$, there doesn't seem to be much choice here: the map $$x\to\frac{1}{a}\,\,,\,\,f(x)\to f\left(\frac{1}{a}\right)\,,\,f(x)\in R[x]$$

seems like a reasonable choice: it is almost trivial that it is a ring homom. (remember the usual evaluation map), it also is easy to show it is onto $\,R_D\,$ , and its kernel is the ideal it must be.

Answer 3

You can use the universal properties, namely, the universal property of the localization, of the polynomial ring, and of the quotient.

Recall that if $R$ is a commutative ring and $D$ is a multiplicative subset, there is a homomorphism $\phi\colon R\to R_D$, given by $\phi(r) = \frac{rd}{d}$ (where $d\in D$ is arbitrary); this map is well-defined, and has the universal property:

If $T$ is any ring, and $f\colon R\to T$ is a ring homomorphism with the property that $f(d)$ is a unit in $T$ for each $d\in D$, then there exists a unique ring homomorphism $\mathcal{F}\colon R_D\to T$ such that $\mathcal{F}\circ\phi = f$.

Consider the natural embedding $R\to R[x]$ followed by the quotient map $R[x]\to R[x]/(ax-1)$. Call this $f$. Note that $f(a)$ is a unit in $R[x]/(ax-1)$, hence so are $f(a^n)$ for every $n$. Therefore, every element of $D$ is mapped to a unit in $R[x]/(ax-1)$, which means that there is a unique homomorphism $\mathcal{F}\colon R_D\to R[x]/(ax-1)$ with the property that $\mathcal{F}\circ \phi = f$.

The claim is that $\mathcal{F}$ is an isomorphism.

To establish this, we construct an inverse. The universal property of the polynomial ring $R[x]$ tells us that if we specify how to map $R$ and what element we want $x$ to map to, we get a homomorphism. We define a homomorphism $R[x]\to R_D$ by mapping the coefficient ring $R$ to $R_D$ using $\phi$, and mapping $x$ to $\frac{1}{a}\in R_D$. This gives us a homomorphism $g\colon R[x]\to R_D$.

Now, the polynomial $ax-1$ is mapped to $0$: $g(ax-1) = \phi(a)\frac{1}{a}-\phi(1) = \frac{aa}{a}\frac{1}{a} - \frac{a}{a} = 0_{R_D}$, so by the universal property of the quotient, the map $g$ factors through $R[x]/(ax-1)$. That is, there is a unique homomorphism $\mathcal{G}\colon R[x]/(ax-1)\to R_D$ such that $g = \mathcal{G}\circ \pi$, where $\pi\colon R[x]\to R[x]/(ax-1)$ is the canonical projection.

So now we have homomorphism $\mathcal{F}\colon R_D\to R[x]/(ax-1)$ and $\mathcal{G}\colon R[x]/(ax-1)\to R_D$. I claim that $\mathcal{G}$ is the inverse of $\mathcal{F}$.

First, consider $$\begin{array}{rcccl} &&R&&\\ &{\small\phi}\swarrow & {\small f}\downarrow&\searrow{\small\phi}\\ R_D & \stackrel{\mathcal{F}}{\to} & \frac{R[x]}{(ax-1)} &\stackrel{\mathcal{G}}{\to} & R_D \end{array}$$ Now, notice that $\mathcal{G}f=\phi$, since elements of $R$ in $R[x]$ are mapped to $R_D$ as $\phi$. So this diagram commutes; that is, $\mathcal{GF}\phi =\mathrm{id}_{R_D}\phi$. But the universal property of the localization says that there is a unique map $R_D\to R_D$ that makes the diagram $$\begin{array}{rcl} &R&\\ {\small\phi}\swarrow &&\searrow{\small\phi}\\ R_D & \longrightarrow& R_D \end{array}$$ commute; clearly the identity does, but we just saw that $\mathcal{GF}$ does as well. That means that we must have $\mathcal{GF} = \mathrm{id}_{R_D}$.

On the other hand, consider the composition $\mathcal{FG}\colon R[x]/(ax-1)\to R[x]/(ax-1)$. The restriction to (the image of) $R$ of this map is just $$\mathcal{FG}(\pi(r)) = \mathcal{F}(g(r)) = \mathcal{F}(\phi(r)) = f(r) = r+(ax-1)$$ (where $(ax-1)$ means the ideal of $R$, not the element $ax-1$). And the image of the class of $x$ is $$\mathcal{FG}(\pi(x)) = \mathcal{F}(g(x)) = \mathcal{F}\left(\frac{1}{a}\right) = f(a)^{-1} = x+(ax-1).$$ So the map $\mathcal{FG}$ agrees with the identity on $\pi(R)$ and on $\pi(x)$, hence equals the identity. So $\mathcal{FG}=\mathrm{id}_{R[x]/(ax-1)}$.

Thus, $\mathcal{F}=\mathcal{G}^{-1}$, so $\mathcal{F}$ is an isomorphism.

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Added. As an alternative of the latter part: once we know that $\mathcal{GF}=\mathrm{id}_{R_D}$, we conclude that $\mathcal{F}$ is one-to-one. Now notice that $\mathcal{F}$ is onto: since $\mathcal{F}\circ \phi = f$, the image of $\mathcal{F}$ contains the image of $f$; the image of $f$ includes the image of all scalars under the projection $R[x]\to R[x]/(ax-1)$. The image of $\mathcal{F}$ also includes the image of $x$, since $\mathcal{F}(\frac{1}{a}) = \mathcal{F}(a)^{-1}$, and the inverse of $\mathcal{F}(a)$ is $x+(ax-1)$. Since $x+(ax-1)$ and $r+(ax-1)$, $r\in R$, generate $R[x]/(ax-1)$, it follows that $\mathcal{F}$ is onto. Since it was already one-to-one, $\mathcal{F}$ is an isomorphism.