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Let $R$ be a commutative ring and $f\in R$ such that $f$ is not a zero divisor (thanks Darij for correction). How can I rigorously show that

The canonical map $R\mapsto R[t]/(ft-1)$ is injective.

Context: This step shows up in Rabinowitch trick (when proving Nullstellensatz) where $R=k[x_1, x_2, …, x_n]$.

Prism
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    Surely not true, say, for $f = 0$. You need to require that $f$ is not a zero-divisor. – darij grinberg Jun 21 '15 at 21:19
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    Hint: $R\left[t\right] / \left(ft-1\right)$ is isomorphic to the localization $f^{\star} R$, where $f^{\star}$ is the multiplicative system $\left{1,f,f^2,f^3,\ldots\right}$. – darij grinberg Jun 21 '15 at 21:20
  • @darij Ah makes sense! So this really follows from the fact that the natural map $R\to S^{-1}R$ is injective if $S$ has no zero divisors. – Prism Jun 21 '15 at 22:31

2 Answers2

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If $\,r\in R\,$ lies in the kernel, then $\,r = (1-ft)\,g\,$ for some $\,g\in R[t],\, \deg g =: n.\,$ Then

$$\begin{eqnarray} r &=& (1-ft)\,g &\Rightarrow&\ g(0) = r\qquad\ \, {\rm via\ \ coef}\ t^0 \\[.2em] \Rightarrow\ (1\!+\!ft\!+\dots+\!(ft)^n)\, r &=& (\color{#c00}{1\!-\!(ft)^{n+1}})\, g &\Rightarrow&\ g(0)\,f^{n+1}\! = 0\ \ {\rm via\ \ coef}\ t^{n+1} \\[.2em] & & &\Rightarrow&\ \ \ \ {\color{#0a0}{r\ f^{n+1} = 0}}\\[-1em] \!\!\text{by scaling by}\,\ 1\!+\!ft\!+\dots+\!(ft)^n \,\ &\!\!=& \dfrac{\color{#c00}{1\!-\!(ft)^{n+1}}}{\!\!\!\!\!\!\!\!\!1-ft} \end{eqnarray}$$

$\color{#0a0}{\rm Therefore}\,$ if $\,\color{#0a0}f\,$ is not a zero-divisor then $\,\color{#0a0}{r = 0}\,$ and the map is injective. $\ $ QED

Remark $\ $ This is used in the computation of the kernel of localization maps. As I mentioned here, this simple proof seems to have been missed by most authors (e.g. Rotmans' Advanced Modern Algebra and all others I've seen give a proof like that in egreg's answer).

Edit $ $ Since the point of this approach seems to not be clear to some readers (cf. comments and downvotes), I will emphasize it explicitly. We do not use any recursion or telescopy above. Rather, we simply invoke the well-known identity $\,(1−x)(1+\cdots +x^n)=1−x^{n+1}\,$ for $\,x=ft.\,$ A common proof in many textbooks (and egreg's answer) amounts to repeating a telescopic proof of this identity in the proof at hand, vs. invoking it as a lemma as we do above (i.e. calling it by value vs. name in CS language). Inlining the proof (vs abstracting it out) only serves to complicate matters. Of course instances of the Factor Theorem (like the identity above) are ubiquitous in ring theory. We should strive to exploit this by appropriately abstracting out their application, esp. in instances like this where this structure is slightly hidden so it may not be immediately clear.

See also this answer for motivational conceptual background.

Bill Dubuque
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  • The proof that egreg gives is the same as you do. In fact the telescoping is made explicit in his answer. The remark seems superfluous. – Pedro Jun 21 '15 at 22:48
  • @Pedro The point is that the (standard) proof is much simpler to comprehend when presented like the above. There is no need for that explicit computation. – Bill Dubuque Jun 21 '15 at 22:50
  • Well, that's simply subjective. – Pedro Jun 21 '15 at 22:51
  • I think both presentations are of value :) So thanks for your contribution! – Prism Jun 21 '15 at 22:53
  • @Pedro Of course it goes without saying that judgements of simplicity are subjective. But I don't recall anyone who hasn't preferred this form when I showed it to them (including Cohn, Kaplansky, Rota, and many others that I've long forgotten). – Bill Dubuque Jun 21 '15 at 23:26
  • In any case, one of the motivations for my interest in this and related topics is that I have long been interested in the historical development of these matters, esp, in the strange bias against the more natural, universal presentation-based construction of localizations or fraction fields (see the linked answer). – Bill Dubuque Jun 23 '15 at 14:36
  • @All To clarify, I do not (explicitly) use any telescopy above. Rather, I use the well-known identity$,(1!-!x)(1+x+\cdots+x^n) = 1-x^{n+1},$ for $,x = ft.,$ The common proof in egreg's answer amounts to repeating a telescopic proof of this identity into the proof at hand, vs. invoking it as a lemma as I do (i.e. calling it by value vs. name in CS language). Inlining the proof (vs abstracting it out) only serves to complicate matters. – Bill Dubuque Aug 20 '17 at 22:19
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An element $a$ belongs to the kernel of the canonical morphism if and only if $a=(ft-1)P(t)$, where $P(t)\in R[t]$. Suppose $a\ne0$ and $P(t)=r_0+r_1t+\dots+r_nt^n$ exists, with $r_n\ne0$. Then $$ (ft-1)P(t)= -r_0+(fr_0-r_1)t+(fr_1-r_2)t^2+\dots+(fr_{n-1}-r_n)t^n+fr_nt^{n+1} $$ so \begin{align} r_0&=-a\\ r_1&=-fa\\ r_2&=-f^2a\\ &\;\;\vdots\\ r_n&=-f^na\\ 0&=f^{n+1}a \end{align} If $f$ is not a zero divisor, we have a contradiction.

Note that if $f$ is a zero divisor, the statement is false: if $fb=0$, then taking $P(t)=-b$, we have $$ (ft-1)(-b)=fbt+b=b $$ so $b$ belongs to the ideal generated by $ft-1$.

egreg
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  • See my answer for a simpler way to present the proof. – Bill Dubuque Jun 21 '15 at 22:38
  • Thanks both to you and Bill for your answers. The explicit computation done here was very helpful. – Prism Jun 21 '15 at 22:50
  • @Prism But there is no need for that (unenlightening) explicit computation. That's the point of my answer! – Bill Dubuque Jun 21 '15 at 22:55
  • @BillDubuque: Yes, there is no need, per se. And that does make your answer shorter! I like this approach because expanding that quantity and equating coefficients is the first thing that would come to my mind, so it feels a bit more natural. Your approach is slicker but slightly harder to think of. Just my personal observations :) – Prism Jun 21 '15 at 23:01
  • @BillDubuque Unenlightening? It explicitly shows what $P$ is, which is probably the reason why Rotman used it. Tricks like yours are slick, but as enlightening as my direct proof. If you want to play “who's got the shortest proof”, then I can agree. – egreg Jun 21 '15 at 23:03
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    @egreg There's no need to explicitly compute $P(t)$ or my $g(t).,$ But if one desires to do so then that is also clearer when done my way, $ $ i.e. $,g (t) = (1!+!ft!+\cdots+!(ft)^n)r,$ arises clearly by simply scaling the equation $\ r = (1!-!ft)g(t)\ $ by $\ (1!+!ft!+\cdots+(ft)^n), =, \dfrac{1!-!(ft)^{n+1}!!!!}{1- ft\quad }$ $\qquad$ – Bill Dubuque Jun 22 '15 at 15:20