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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

user26857
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johnq
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  • To be clear, this holds for any commutative ring $R$, and $a \in R$ can be any element. (See the answer of Maxime) – James Sep 16 '21 at 11:16

5 Answers5

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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.

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.

darij grinberg
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Arturo Magidin
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  • Is it correct just to say that since $\mathcal{G}\circ f=\phi$ and $\mathcal{F}\circ \phi=f$, it follows that $\mathcal{G}\circ\mathcal{F}\circ \phi=\phi$ and $\mathcal{F}\circ\mathcal{G}\circ f=f$, and the surjectivity of both $\phi$ and $f$ implies $\mathcal{F}\circ\mathcal{G}=\operatorname{id}$ and $\mathcal{G}\circ\mathcal{F}=\operatorname{id}$? – user557 Jan 11 '17 at 10:53
  • Excellent! Dummit and Foote said "It is not difficult to see that." (Example (2) on page 708 in Abstract Algebra) You are so good at applying these universal properties. Do you know which books I can find a similar proof? – bfhaha Dec 04 '18 at 04:48
  • Note that this nice proof uses neither that $R$ is an integral domain (any commutative ring would do), nor that $a$ is nonzero. – darij grinberg Dec 07 '18 at 02:36
  • @Cary Why is $f$ and $\phi$ surjective? – Babai Aug 28 '20 at 03:27
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Here's another answer using the universal property in another way (I know it's a bit late, but is it ever too late ?)

As for universal properties in general, the ring satisfying the universal property described by Arturo Magidin in his answer is unique up to isomorphism. Thus to show that $R[x]/(ax-1) \simeq R_D$, it suffices to show that $R[x]/(ax-1)$ has the same universal property !

But that is quite easy: let $\phi: R\to T$ be a ring morphism such that $\phi(a) \in T^{\times}$.

Using the universal property of $R[x]$, we get a unique morphism $\overline{\phi}$ extending $\phi$ with $\overline{\phi}(x) = \phi(a)^{-1}$.

Quite obviously, $ax -1 \in \operatorname{Ker}\overline{\phi}$. Thus $\overline{\phi}$ factorizes uniquely through $R[x]/(ax-1)$.

Thus we get a unique morphism $\mathcal{F}: R[x]/(ax-1) \to T$ with $\mathcal{F}\circ \pi = \phi$, where $\pi$ is the canonical map $R\to R[x]/(ax-1)$. This shows that $\pi: R \to R[x]/(ax-1)$ has the universal property of the localization, thus it is isomorphic to the localization.

This is essentially another way of seeing Arturo Magidin's answer

Maxime Ramzi
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    Note that this nice proof uses neither that $R$ is an integral domain (any commutative ring would do), nor that $a$ is nonzero. – darij grinberg Dec 07 '18 at 02:36
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Define $\phi: R[x] \rightarrow R_D$ by sending $x$ to $1/a$. We will prove that the kernel of $\phi$ is the ideal of $R[x]$ generated by $ax -1$.

Let $p\left(x\right) \in \ker \phi$. Note that $p(x)$ can be naturally viewed as an element of $R_D[x]$. Since $\phi(p(x))=0$, we have $p(1/a)=0$ in $R_D$. So $1/a$ is a root of $p(x) \in R_D[x]$. Since the leading coefficient of $ax-1$ is a unit in $R_D$, Euclidean division applies and we get that $p(x) = (ax-1) g(x)$ for some $g(x) \in R_D[x]$. Inspecting coefficients on both sides starting from lowest degree terms, we see that in fact $g(x) \in R[x]$ and we are done.

PS: This answer is inspired by Robert Green's answer (and our subsequent interaction) to an identical question that i recently asked, being ignorant of the existence of this post.

darij grinberg
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Manos
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    Could you elaborate on what you mean by "inspecting coefficients on both sides starting from lowest degree terms, we see that in fact $g(x) \in R[x]$"? When I factor out $(ax-1)$ in $R_D[x]$, the lowest-degree terms are too messy to comprehend. If I start with $c_nx^n + c_{n-1}x^{n-1} + \dotsb + c_1x + c_0$, then my coefficients on $g$ are $(c_n/a)x^{n-1} + ((c_{n-1} + c_n/a)/a)x^{n-2}$, and progressively more complicated after that. – Eric Auld Oct 29 '15 at 02:25
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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 $R[x]/(ax-1)\to R_D$.

Mariano Suárez-Álvarez
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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.

DonAntonio
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    Proving that the kernel is $(ax-1)$ can be tricky. Did you try it? – Bill Dubuque May 31 '12 at 23:39
  • Not really, but seeing that $,R,$ is an integral domain, I think we can slip into is fractions field and then the proof there goes as usual: a pol. vanishes at some element iff it is divided by the minimal polynomial of that element, which is either $,ax-1,$, if there's no demmand of the min. pol. to be monic, or its associate element (in the pol. ring) $,x-1/a,$ – DonAntonio May 31 '12 at 23:59
  • That could be circular, since in many presentations this is a precursor to constructing the fraction field (by localization). It can be done without any knowledge of fraction fields. – Bill Dubuque Jun 01 '12 at 00:12
  • How'd you do it? In may case, I'd go by constructing the fractions field in a more elementary way and not as the localization wrt the prime ideal $,{0},$ , namely: as the set of equivalence classes of elements of the form $,a/b,,,,a,b\in R,,,,b\neq 0,$ under the well-known equiv. relation $,a/b ~ c/d \Longleftrightarrow ad=bc,,in,,R$ and the usual operations (this is the way I first learned this, way before I had a commutative algebra course). – DonAntonio Jun 01 '12 at 02:21
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    Hint $\ $ Since the image is a domain, the kernel $\rm:P:$ is prime. Note $\rm\ g = ax!-!1\in P.:$ Suppose $\rm: f \in P.:$ Hence $\rm: f + f_0:! g =: x:!h \in P,\ x \not\in P\Rightarrow\ h\in P.:$ By induction on degree $\rm:g:|:h\ $ so $\rm\ g:|:xh-f_0g = f.:$ Therefore $\rm: P = (g) = (ax!-!1).\ $ QED $\ \ $ – Bill Dubuque Jun 01 '12 at 03:33
  • Ok, I'm almost there, but I still can't see why you can assume we can factor out $,x,$ in the expression for $,f+f_0g,$...? As far as I see, that expression could have a non-zero free doefficient. I can't also see why $,g\mid h,$... – DonAntonio Jun 01 '12 at 03:46
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    Hint $\ $ Recall $\rm:x:|:p(x)\iff p(0) = p_0 = 0.:$ For $\rm:p = f + f_0 g:$ we have $\rm:p_0 = f_0 + f_0 g_0 = f_0 - f_0 = 0.:$ Thus $\rm:x:|:p,:$ so $\rm:p = x:g,:$ for some $\rm:g\in R[x].:$ Finally $\rm:g:|:h:$ follows from the (implicit) induction hypothesis, namely that $\rm:g:$ divides all elements of $\rm:P:$ of smaller degree than $\rm:f.$ – Bill Dubuque Jun 01 '12 at 03:51
  • Oh, wait: I didn't know $,f_0,$ mean $,f(0),$ ! Ok, so now it is clear. Why from this $,g\mid h,$ though? – DonAntonio Jun 01 '12 at 04:00
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    Since $\rm:h\in P:$ has smaller degree than $\rm:f,:$ our induction hypothesis yields that $\rm:g:|:h.:$ We are proving by induction on degree that every element $\rm:f\in P:$ is divisible by $\rm:h = ax!-!1\in P.:$ – Bill Dubuque Jun 01 '12 at 04:03
  • Oh, thank you. It's nice, yet I think going into the fractions fields is more elementary and easy, but it's always nice to learn new approaches. – DonAntonio Jun 01 '12 at 04:05
  • Yes, but only if one already has available a rigorous development of fraction fields. Often this is not done rigorously until fraction fields are derived as a special case of localizations. In that case it could be circular to employ fraction fields in proofs about localizations. – Bill Dubuque Jun 01 '12 at 04:10