# Showing that $R[X]/(Xf-1) \cong R[1/f]$ [duplicate]

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Let $R$ be an integral domain with quotient field $K$. Let $0 \neq f \in R$. I want to prove

Statement: $R[X]/(Xf-1) \cong R[1/f]$.

Argument: Consider the epimorphism $\phi: R[X] \rightarrow R[1/f]$ which sends $X$ to $1/f$. The goal is to show that the kernel of $\phi$ is the ideal generated by $Xf-1$. If $\phi(p[X])=0$ then $p(1/f)=0$. This shows that $1/f$ is a root of $p(X)$. Viewing $p[X]$ as an element of $K[X]$, this implies that there exists some $g(X) \in K[X]$ such that $p(X) = (X-1/f) g(X) = (Xf-1) \left[g(X) /f\right]$. I will be done if i can show that $g(X)/f \in R[X]$.

Question 1: How to complete the argument?

Question 2 (optional): Is there any other proof to the statement?

## marked as duplicate by user26857, Hippalectryon, apnorton, Pedro Tamaroff♦Feb 13 '15 at 0:39

• It would help to know, how do you define $R[1/f]$? – Thomas Andrews Feb 12 '15 at 21:59
• @ThomasAndrews: It is just the localization of $R$ at $\left\{1,f,f^2,...\right\}$. – Manos Feb 12 '15 at 22:01
• In my opinion, one of the most interesting answers is this: math.stackexchange.com/a/768018/121097 – user26857 Feb 12 '15 at 22:13
• @user26857: Indeed, that is a brilliant answer. – Manos Feb 12 '15 at 22:41
• @user26857 The proof you link to follows from the special case $\,b=1\,$ of the short proof I gave here. – Bill Dubuque Feb 17 '15 at 3:43

To complete the argument, you can observe the following : if $P\in R[X]$ verifies $P(\frac{1}{f}) = 0$, writing $P = \sum_{k=0}^d a_k X^k$, you have that $\sum_{k=0}^d a_k f^{d-k} = 0$ which shows that $f$ is root of the polynomial $Q = \sum_{k=0}^d a_{d-k} X^k$. Note that $Q(X) = X^d P(\frac{1}{X})$ and $P(X) = X^d Q(\frac{1}{X})$. Now, as $f$ is root of $Q(X)$, you can write that $Q(X) = (X-f)R(X)$ for some $R(X) \in K[X]$. Then, you can check by hand (writing $R(X) = \sum_{k=0}^{d-1} \beta_k X^k$ and injecting that in $Q(X) = (X-f)R(X)$ and then expanding and regrouping same powers of $X$) that $R \in R[X]$. Now this implies that $X^d Q(\frac{1}{X}) = X^d(\frac{1}{X}-f)R(\frac{1}{X}) = (1-fX) X^{d-1} R(\frac{1}{X})$. The element $X^{d-1} R(\frac{1}{X})$ is a polynomial $S(X)$ in $R[X]$ and, as $P(X) = X^d Q(\frac{1}{X})$, we have $P(X) = (1-fX)S(X)$.

For your second question : you can show that $R[X]/(fX-1)$ verifies the universal property of the localization of $R$ wrt powers of $f$. This will straightforwardly imply the isomorphism, as $R$ being a domain implies that $R[1/f]$ is the localisation of $R$ wrt powers of $f$. To verify the universal property, you have to show that the obvious morphism $R \to R[X]/(fX-1)$ sends powers of $f$ in $(R[X]/(fX-1))^{\times}$, and that every ring morphism $\varphi : B \to R[X]/(fX-1)$ sending powers of $f$ in $(R[X]/(fX-1))^{\times}$ factor through the morphism $R \to R[X]/(fX-1)$.

• Beware of euclidian division : in $\mathbf{Z}[X]$, you can perform without any problem the euclidian division of $6X^3 - X^2 + 7X-5$ by $2X-1$. But try to do the euclidian by $3X-1$ instead, and you will quickly have problems. So what you have stated if in fact false, my bad. – ujsgeyrr1f0d0d0r0h1h0j0j_juj Feb 12 '15 at 23:22
• I believe the correct statement is: euclidean division holds for univariate polynomials over an integral domain, provided that the leading coefficient of the divisor is a unit. If i remember correctly, that's how it is stated in Lang's Algebra. – Manos Feb 12 '15 at 23:32
• Yes yes, this is obvioulsy true, that's how it's stated everywhere, but the question was : can we say something more in other cases, and the answer is no. Btw I therefore corrected my answer. – ujsgeyrr1f0d0d0r0h1h0j0j_juj Feb 12 '15 at 23:35

Let $S=R[1/f]$ and let $i:R\to S$ be the obvious map. You can easily check that the pair $(S,i)$ has the following property:

whenever $A$ is a ring and $q:R\to A$ is a morphism such that $q(f)$ is invertible in $A$ there is a unique morphism $\bar q:S\to A$ such that $\bar q\circ i=q$.

On the other hand, let $T=R[X]/(fX-1)$ and let $j:R\to T$ be the obvious map. Show that the pair $(T,j)$ has exactly the same property, that is:

whenever $A$ is a ring and $q:R\to A$ is a morphism such that $q(f)$ is invertible in $A$ there is a unique morphism $\bar q:T\to A$ such that $\bar q\circ j=q$.

Finally, show that if $(U_1,k_1)$ and $(U_2,k_2)$ are two pairs which have the above propeerty, then $U_1\cong U_2$.

• Fantastic, thanks. – Manos Feb 12 '15 at 22:13
• That's the second part of my answer, by the way. – ujsgeyrr1f0d0d0r0h1h0j0j_juj Feb 12 '15 at 22:19
• Well, I wanted to avoid terms like «universal property», factors, and such and write everything out explicitly. – Mariano Suárez-Álvarez Feb 12 '15 at 22:21
• Oh, I am a firm believer that students should be exposed to those ideas as quickly as possible, but avoiding the mumbo jumbo. Category theory and its language should only be taught to people when they already know most of it, and then only when they need it. – Mariano Suárez-Álvarez Feb 12 '15 at 22:39
• @RobertGreen There's a difference between avoiding a term and avoiding a concept! – Pedro Tamaroff Feb 13 '15 at 0:40