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Suppose R is an integral domain and $0\neq f\in R$. Show that $$R[x]/(xf-1)\cong R[\frac{1}{f}]$$ I am able to get the LHS from the isomorphism theorem by using following homomorphism $\phi(x)=\frac{1}{f}$. The kernel will then be $(xf-1)$ and hence the isomorphism.

This question is from Cutkosky $(1.7)$ the notation on RHS means the smallest subring containing $R$ and $\frac{1}{f}$. Obvioulsy R is an integral domain so it may not contain $f$ inverse but Cutkosky uses an homomorphism from $R[x]$ to a quotient field of $R$ denoted by $K$. Using quotient field will only make sense if $f$ inverse is contained in R otherwise it is not a well defined statement.

How to make sense of RHS of above isomorphism?

aitfel
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  • What do you mean you were able to get the LHS from the FIT? The FIT gives you an isomorphism between the two? – Noah Solomon Dec 25 '20 at 05:28
  • See here for a generalization to non-domains (which lends insight). – Bill Dubuque Dec 25 '20 at 06:21
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    @NoahSolomon What I meant by LHS was the form of ideal or ker of $\phi$ which goes into the isomorphism. BTW Joshua cleared my doubt, I made a mistake of taking $f^{-1}$ to be same as $\frac{1}{f}$ – aitfel Dec 25 '20 at 08:16

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"the notation on RHS means the smallest subring containing $R$ and $\frac{1}{f}$"

This is not clear. Perhaps you mean the smallest subring of the field of fractions $K$ of $R$ containing $R$ and $\frac{1}{f}$. The homomorphism $\phi \colon R[x] \to K$ given by $x \mapsto \frac{1}{f}$ is well-defined and has kernel $(fx-1)$. The image is a ring which contains $R$ and $\frac{1}{f}$, so $\phi(R[x]) \supset R[1/f]$. Clearly $\phi(R[x]) = \{a/f^k : a \in R, k \in \mathbb{Z}_{\geq 0}\}$. Just as clearly, $\{a/f^k : a \in R, k \in \mathbb{Z}_{\geq 0}\} \subset R[1/f]$. The result now follows from the first isomorphism theorem.

Joshua P. Swanson
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  • Yeah it meant that. I made a stupid mistake of taking $f^{-1}$ to be same as $\frac{1}{f}$ since I forgot the field of fraction of $R$. Thanks a lot. – aitfel Dec 25 '20 at 08:10