# Two fractions in $R[\frac{1}{x}]$ are equal iff $(\exists n\ge 0)(x^n\cdot ax^i=x^n\cdot bx^j)$

Given a ring $R$ and an element $x\in R$, we can adjoint the inverse of $x$ by ring extension $R[\frac{1}{x}]=R[x,y]/\langle xy-1\rangle$, so any elements in $R[\frac{1}{x}]$ is of form $\frac{a}{x^i}$.

I aim to prove that in the ring $R[\frac{1}{x}]$, for any two elements $\frac{a}{x^i},\frac{b}{x^j}$, they are equal iff $(\exists n\ge 0)(x^n\cdot ax^j=x^n\cdot bx^i)$.

It sounds quite counter-intuitive since I expect $ax^j= bx^i$ is enough. Why it doesn't hold, may I please ask some counterexamples?

I tried to prove it but have not found a way yet. Could someone please tell me how to prove the fact: Two fractions in $R[\frac{1}{x}]$ are equal iff $(\exists n\ge 0)(x^n\cdot ax^j=x^n\cdot bx^i)$?

Thanks for any help.

Edit: I have not seen the definition of localization in my algebra course, I googled the construction and wonder why we make the equivalence relation $t(r_1s_2 − r_2s_1) = 0$. And also it does not directly give the desired conclusion because here we require a multiple of $x$, instead of an arbitary element in $R$, playing the role of $t$. If possible, may I please ask for an argument does not involve the usage of definition of localization? Thanks!

• I guess you a have typo in your post: $a/x^i = b/x^j$ iff $\exists n \geq 0$ s.t. $x^n \cdot ax^j = x^n \cdot bx^i$. – Federico Sep 23 '17 at 9:44
• @Federico Yes...Thanks for pointing it out! – PropositionX Sep 23 '17 at 9:49
• @PropositionX there is another answer that you will find useful. – Xam Sep 26 '17 at 20:23
• I added some remarks to my answer that may help you to better understand these matters - which are often puzzling to most who first encounter these topics. Please feel welcome to ask questions if anything is not clear, – Bill Dubuque Sep 27 '17 at 15:38

Something that could go wrong is if you try to invert a nilpotent element. For example, let $R = k[\epsilon] / (\epsilon^2)$ for some field $k$, and attempt to invert the element $\epsilon \in R$. Then the extended ring is $S = k[\epsilon, t] / (\epsilon^2, t \epsilon - 1)$. At this point, we want to say that $1 / \epsilon^0 \neq 1 / \epsilon^1$, since $\epsilon \neq 1$ in $R$.

However, we can take the relation $t \epsilon = 1$ in $S$, and multiply it by $\epsilon$ to find that $t \epsilon^2 = 0 = \epsilon$. At this point the relation $t \epsilon = 1$ turns into $0 = 1$, and so $S$ is in fact the zero ring. So every expression involving fractions in $\epsilon$ are equal! This demonstrates the need for that extra $x^n$ in the criterion you have there: in this case, it's saying that the expressions are equal for some power of $\epsilon$. In particular, higher powers of $\epsilon$ are all $0$ and so everything is equal in the extended ring.

The moral of the story is basically that "bad stuff" happens when we invert nilpotent elements or zero-divisors: the theory developed around localisation gives good tools for dealing with how to do this. The "universal property of localisation" is also a very powerful way to look at how to introduce fractions into commutative rings.

In many prior answers you can find explanations motivating the definition of the equivalence relation in the common pair-based normal form construction of localizations. But this does not provide immediate intuition for the alternative presentation-based construction of the fraction ring (localization) $$\rm\,S^{-1} R\,$$ as, conceptually, the universal way of adjoining inverses of $$\rm\,S\,$$ to $$\rm\,R\,$$ via $$\rm\,S^{-1} R = R[x_i]/(s_i x_i - 1),\,$$ as in your special case. This presentation-based approach allows us to exploit to the hilt the universal properties of quotient rings and polynomial rings to quickly and easily construct and derive basic properties of localizations (avoiding the many tedious verifications always "left for the reader" in the more commonly pair=based normal-forms approach).

In this presentation-based approach there is no need to unintuitively massage an equivalence relation to account for annihilated elements. Rather we can directly compute what is annihilated by a simple (widely-overlooked) trick:  if $$\rm\,r\in R\,$$ lies in the kernel of $$\rm\, R\to S^{-1} R = R[x]/(1\!-\!sx),\,$$ then $$\rm\,r \in (1\!-\!sx)R[x]\,\Rightarrow\, r = (1\!-\!sx)f(x)\,$$ so by comparing coefficients we obtain

$$\quad \begin{eqnarray}\rm n = deg\, f\quad and\quad r &=&\rm (1\!-\!sx)\,f(x) &\Rightarrow&\ \rm f(0) = r\qquad\,\ \ \ via\ \ coef\ x^0 \\ \rm\Rightarrow\ (1\!+\!sx\!+\dots+\!(sx)^n)\, r &=&\rm (1\!-\!(sx)^{n+1})\, f(x) &\Rightarrow&\ \rm f(0)\,s^{n+1}\! = 0\quad via\ \ coef\ x^{n+1} \\ & & &\Rightarrow&\ \rm\ \ \,\bbox[5px,border:1px solid #c00]{r\ s^{n+1} = 0} \end{eqnarray}$$

As Xam wrote in his answer, this immediately yields the result that you seek. and using the above proof requires no prior knowledge of the (pair-based) localization construction.

If you introspect on the construction used in the pair-based approach you will find that it boils down to choosing $$\rm\color{#0a0}{constant}$$ normal-form reps of equivalence classes in the above quotient rings $$\rm f(x) = s^n f(x)/s^n = s^n(f_n x^n+\cdots +f_1 x+ f_0)/s^n = (f_n+ f_1 s^{n-1} + f_0 s^n)/s^n = \color{#0a0}{r/s^n},\ r\in R$$

analogous to how Hamilton's pair-based construction of $$\,\Bbb C = \Bbb R[x]/(x^2+1)$$ chooses $$\rm\color{#0a0}{least\ degree}$$ (= linear) normal form reps $$\rm\,a+bx\,$$ of polynomials $$\rm \bmod\, x^2+1\,$$ with the ring operations and congruence transported to the pair normal form reps, e.g. transported multiplication of pairs is

$$\rm\begin{eqnarray}\rm (a,\ b) &&\rm (c,\ d) &\!\!=&\rm (ac\!\color{#000}{\bf -}\!bd,\quad\ \ ad\!+\!bc)\\[.2em] \rm i.e.\ \bmod{\color{#c00}{\,x^2\!+1}}\!:\ \ \color{#c00}{x^2\equiv -1}\ \,\Rightarrow\,\ (a\! +\! b\color{#c00}x)&&\rm(c\! +\! d\color{#c00}{ x})\, &\!\!\!\rm\,\equiv&\rm (ac\!\color{#c00}{\bf -}\!bd) + (ad\!+\!bc)\, x\\[.3em] \rm i.e.\quad\, (a\! +\! b{\it i}\,)&&\rm(c\! +\! d {\it i}\,)\, &\!\! =\,&\rm (ac\!\color{#000}{\bf -}\!bd) + (ad\!+\!bc)\,{\it i}\end{eqnarray}$$

The pair-based localization construction is precisely analogous, i.e. the construction transports the ring operations and congruence to normal form reps of the equivalence classes of the quotient rings. When viewed in this way, the equivalence relation imposed on the fraction pair reps $$\rm\,(r,s^n)\,$$ arises naturally and no longer seems contrived.

This is a protoypical example of ubiquitous techniques employed in normal form computation in quotient objects, so it is well-worth the effort to study it carefully.

For further discussion (including literature references) see this answer.

• Thanks a lot! This is much easier then my brute-force method. Learnt a lot from this answer! – PropositionX Sep 28 '17 at 4:01

We have the following isomorphisms $$R[x,x^{-1}]\cong R[x,y]/(xy-1)\cong R[x]_x.$$ The last ring is the localization of $R[x]$ at the multiplicatively closed subset $S=\{1,x,x^2,\ldots\}$. Now, by definition of localization we have $\frac{a}{x^i}=\frac{b}{x^j}$ if only if there is an element $x^n\in S$ such that $x^n(ax^j-bx^i)=0$, i.e., $x^n\cdot ax^j=x^n\cdot bx^i$.

• Thanks for your answer. But I have not seen the definition of localization, and I doubt why the definition os like this. May I please ask for some further explaination without the usage of definition of localization? – PropositionX Sep 23 '17 at 21:01
• @PropositionX I suggested to Xam to undelete this answer since I think that - with the help of the alternative approach in my answer - it may prove to be more comprehensible to you (and possibly to other readers too). – Bill Dubuque Sep 26 '17 at 20:30