Proving that $D^{-1}\mathbb{Z}\cong \mathbb{Z}[x]/(px-1)$

Asked Apr 03 '18 at 04:34

Active Apr 03 '18 at 04:34

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Let $p$ be prime, and $D=\{1,p,p^2,p^3,\dots\}$. The ring $D^{-1}\mathbb{Z}$ is defined as the set of fractions $\{\frac{n}{d}:n\in\mathbb{Z},d\in D\}$. How would I show that the ring $D^{-1}\mathbb{Z}\cong \mathbb{Z}[x]/(px-1)$? I know by the first isomorphism theorem that the ideal $(px-1)$ is the kernel of a homomorphism $\varphi :\mathbb{Z}[x]\rightarrow D^{-1}\mathbb{Z}$.

My issue is that I'm not sure how to actually construct this homomorphism. Any thoughts?

asked Apr 03 '18 at 04:34

JB071098

Construct homomorphisms in the two directions and check that they are mutually inverse. This is easier than computing the kernel. – Mariano Suárez-Álvarez Apr 03 '18 at 04:49
Do you mean take some $p\in\mathbb{Z}[x]$ and check that $\varphi(p)=r$ and then check that $\varphi^{-1}(r)=p$? @MarianoSuárez-Álvarez – JB071098 Apr 03 '18 at 05:13
No, define ring homomorphisms $f:D^{-1}Z\to Z[x]/(px-1)$ and $g:Z[x]/(px-1)\to D^{-1}Z$ in some reasonable way, and then check that $f\circ g$ and $g\circ f$ are identity maps. – Mariano Suárez-Álvarez Apr 03 '18 at 05:14
Okay thanks I'll give it a go. @MarianoSuárez-Álvarez – JB071098 Apr 03 '18 at 05:16
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There have been quite a few posts on this topic, for instance: 1, 2, 3, 4. Do any of these help? – Viktor Vaughn Apr 03 '18 at 16:39
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Very much so. Thank you. @Quasicoherent – JB071098 Apr 03 '18 at 16:51
Okay, great. Unless you object, I'm going to vote to close this as a duplicate. – Viktor Vaughn Apr 06 '18 at 02:15

Proving that $D^{-1}\mathbb{Z}\cong \mathbb{Z}[x]/(px-1)$

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