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Let $A$ be a integral domain, $b\in A$ a nonzero element and $S=\{b^n\mid n \in \mathbb{N}\cup\{0\}\}$

Prove: $S^{-1}A \cong A[x]/(bX-1)$

For this we define the homomorphism $\varphi: X\longrightarrow1/b$ and prove that $\ker\varphi=(bX-1)$

Since $\varphi(bX-1)=b·1/b-1=0$ we have that $\ker\varphi \supseteq(bX-1)$

Now we have to see that any $f\in \ker\varphi$ is in the ideal $(bX-1)$, the obvious way to approach this is to divide $f$ by $bX-1$, but we can't apply the division algorithm directly since the main coefficient of $bX-1$ is not necessarily $1$.

Any help to prove $\ker\varphi \subseteq(bX-1)$?

Michael Hardy
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user605734 MBS
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  • I changed $Ker\varphi,$ coded as Ker\varphi, to $\ker\varphi,$ coded as \ker\varphi. The backslash does not only prevent italicization, but also results in the proper amount of space between $\ker$ and $\varphi,$ and it doesn't just add space, but rather the spacing depends on the context, so that, for example, the space after $\ker$ in $\ker\varphi$ differs from that in $\ker(\varphi).$ If you want a capital K, you can write $\operatorname{Ker}\varphi,$ coded as \operatorname{Ker}\varphi. In genuine LaTeX (as opposed to MathJax, which is used here) you can$,\ldots$ – Michael Hardy Nov 23 '20 at 17:39
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    Does this answer your question? Localisation isomorphic to a quotient of polynomial ring – Viktor Vaughn Nov 23 '20 at 17:39
  • $\ldots,$put \newcommand{\Ker}{\operatorname{Ker}} before the \begin{document} line and thereafter just type \Ker each time. Similarly one writes $3\cos\varphi$ rather than $3cos\varphi,$ and $2\log x$ rather than $2log x$ and $c\det A$ rather than $cdetA$ etc. $\qquad$ – Michael Hardy Nov 23 '20 at 17:39

1 Answers1

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Hint: $S^{-1}A = \{\frac{a}{b^n}\mid a\in A, n\geq 0\}$.

The quotient ring $Q = A[x]/(bX-1)$ contains a zero of $bX-1$, namely the residue class of $x$, which you can denote by $b^{-1}$. This gives the ring extension $A[b^{-1}]$.

Wuestenfux
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