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Let a be an element of a ring R, and let $R' = R[x]/(ax − 1)$ be the ring obtained by adjoining an inverse of a to R. Let α denote the residue of x (the inverse of a in R').

Show that every element of B in R' can be written as $B=α^kb$ with b in R

I have a few questions about this just to make sure I understand the problem. Firstly, can someone briefly explain how ax-1 is the inverse of a in R[x]/(ax − 1)?

Secondly, by residue of x, don't we mean the result of R[x] modulo (ax-1)? or am I misunderstanding?

so basically α = ax-1? And I need to show that every element in this quotient ring can be written as $(ax-1)^kb$? If someone can give a very brief pointer on how to start the proof or how I want to think about this problem, that would be great.

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    You never specified what $B$ is. – 2017-01-30
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    Oh, I see, there was a typo in the title. – 2017-01-30
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    Yea, sorry. Thank you! – 2017-01-30
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    Also, Lana, do we know anything about the ring $R$? If you have a division algorithm in $R[x]$, the proof becomes very simple. – 2017-01-30
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    No, we know nothing about R. – 2017-01-30

3 Answers 3

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In congruence language we have $\,x\equiv 1/a\,$ and the idea is simply to put a sum of fractions over a common denominator, e.g. in the degree $\,k=2\,$ case we have

$$ bx^2+cx+d\ \equiv\, \dfrac{b}{a^2}+\dfrac{c}{a}+d\ \equiv\, \dfrac{\overbrace{b + c a + d a^2}^{\large :=\ r\ \in\ R}}{a^2}\, \equiv\, \dfrac{r}{a^2}\, \equiv\ r\, x^2$$

Obviously the same method works for any degree, e.g. as below

$$\begin{align} &\qquad\ \ \ r_k x^k + \cdots + r_1 x + r_0\\ \equiv &\ \ x^k a^k(r_k x^k + \cdots + r_1 x + r_0)\\ \equiv &\ \ x^k (\underbrace{r_k + \cdots + r_1 a^{k-1}\!+r_0 a^k}_{\large :=\ r\ \in\ R})\\ \end{align}$$

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Let $\alpha$ denote the residue class of $x$ in $R'$. That is, $\alpha = x + (ax-1)$ where $(ax-1)$ is the ideal generated by $ax-1$. Notice that \begin{align*} [a + (ax-1)]\alpha - [1 + (ax-1)] &= [a + (ax-1)][x + (ax-1)] - [1 + (ax-1)] \\ &= [ax-1] + (ax-1) \\ &= 0 + (ax-1). \end{align*}

Now, all that coset notation is cumbersome, which is why we defined $\alpha$ in the first place. Let me rewrite that last computation, dropping the coset notation:

\begin{align*} a\alpha - 1 &= ax-1 \\ &= 0. \end{align*}

All we are saying is that in the ring $R'$, $x + (ax-1)$ acts as the inverse of $a + (ax-1)$. But to write things more succinctly we like to just think of this thing $\alpha$ that acts as the inverse of $a$.

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If $\alpha$ denotes the image of $x$ under the homomorphism $R[x]\to R[x]/(ax-1)$, we have that $a\alpha=1$ (we “identify” elements of $R$ with their image under the obvious homomorphism $R\to R[x]/(ax-1)$).

The image of $f(x)=b_0+b_1x+\dots+b_kx^k\in R[x]$ can be written $$ b_0+b_1\alpha+\dots+b_k\alpha^k= \alpha^k(a^kb_0+a^{k-1}b_1+\dots+b_k) $$

If $R$ is an integral domain, then the homomorphism $R\to R[x]/(ax-1)$ is injective; however it need not be. Precisely, $b\in R$ is mapped to $0$ if there exist $k$ with $a^kb=0$.