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edit approved Jun 27, 2015 at 18:06
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edit approved Jun 27, 2015 at 18:06
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In case of a prime power it is not so hard. Denote R$R$ the ring of integers of $Q(\zeta_n)$${\mathbf Q}(\zeta_n)$. First note that $X^n -1$ is separable mod p$p$ when p$p$ does not divide n$n$. This implies $R [1/p] = Z[\zeta_n, 1/p]$$R [1/p] = {\mathbf Z}[\zeta_n, 1/p]$. To check that $R = Z[\zeta_n]$$R = {\mathbf Z}[\zeta_n]$ it then suffices to show that the local rings of $Z[\zeta_n]$${\mathbf Z}[\zeta_n]$ at all primes above p$p$ are DVRs. But you have the explicit element $\lambda = 1 - \zeta$ that you defined above;above and its norm equals to p$p$. This implies that there is only one prime above p$p$ and that this ideal is generated by $\lambda$. In particular the local ring at this prime is regular and hence a DVR.

In case of a prime power it is not so hard. Denote R the ring of integers of $Q(\zeta_n)$. First note that $X^n -1$ is separable mod p when p does not divide n. This implies $R [1/p] = Z[\zeta_n, 1/p]$. To check that $R = Z[\zeta_n]$ it then suffices to show that the local rings of $Z[\zeta_n]$ at all primes above p are DVRs. But you have the explicit element $\lambda = 1 - \zeta$ that you defined above; its norm equals to p. This implies that there is only one prime above p and that this ideal is generated by $\lambda$. In particular the local ring at this prime is regular and hence a DVR.

In case of a prime power it is not so hard. Denote $R$ the ring of integers of ${\mathbf Q}(\zeta_n)$. First note that $X^n -1$ is separable mod $p$ when $p$ does not divide $n$. This implies $R [1/p] = {\mathbf Z}[\zeta_n, 1/p]$. To check that $R = {\mathbf Z}[\zeta_n]$ it then suffices to show that the local rings of ${\mathbf Z}[\zeta_n]$ at all primes above $p$ are DVRs. But you have the explicit element $\lambda = 1 - \zeta$ that you defined above and its norm equals $p$. This implies that there is only one prime above $p$ and that this ideal is generated by $\lambda$. In particular the local ring at this prime is regular and hence a DVR.

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mnr
mnr
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In case of a prime power it is not so hard. Denote R the ring of integers of $Q(\zeta_n)$. First note that $X^n -1$ is separable mod p when p does not divide n. This implies $R [1/p] = Z[\zeta_n, 1/p]$. To check that $R = Z[\zeta_n]$ it then suffices to show that the local rings of $Z[\zeta_n]$ at all primes above p are DVRs. But you have the explicit element $\lambda = 1 - \zeta$ that you defined above; isits norm equals to p. This implies that there is only one prime above p and that this ideal is generated by $\lambda$. In particular the local ring at this prime is regular and hence a DVR.

In case of a prime power it is not so hard. Denote R the ring of integers of $Q(\zeta_n)$. First note that $X^n -1$ is separable mod p when p does not divide n. This implies $R [1/p] = Z[\zeta_n, 1/p]$. To check that $R = Z[\zeta_n]$ it then suffices to show that the local rings of $Z[\zeta_n]$ at all primes above p are DVRs. But you have the explicit element $\lambda = 1 - \zeta$ that you defined above; is norm equals to p. This implies that there is only one prime above p and that this ideal is generated by $\lambda$. In particular the local ring at this prime is regular and hence a DVR.

In case of a prime power it is not so hard. Denote R the ring of integers of $Q(\zeta_n)$. First note that $X^n -1$ is separable mod p when p does not divide n. This implies $R [1/p] = Z[\zeta_n, 1/p]$. To check that $R = Z[\zeta_n]$ it then suffices to show that the local rings of $Z[\zeta_n]$ at all primes above p are DVRs. But you have the explicit element $\lambda = 1 - \zeta$ that you defined above; its norm equals to p. This implies that there is only one prime above p and that this ideal is generated by $\lambda$. In particular the local ring at this prime is regular and hence a DVR.

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mnr
mnr
  • 1.2k
  • 5
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In case of a prime power it is not so hard. Denote R the ring of integers of $Q(\zeta_n)$. First note that $X^n -1$ is separable mod p when p does not divide n. This implies $R [1/p] = Z[\zeta_n, 1/p]$. To check that $R = Z[\zeta_n]$ it then suffices to show that the local rings of $Z[\zeta_n]$ at all primes above p are DVRs. But you have the explicit element $\lambda = 1 - \zeta$ that you defined above; is norm equals to p. This implies that there is only one prime above p and that this ideal is generated by $\lambda$. In particular the local ring at this prime is regular and hence a DVR.