What are the prime ideals in rings of cyclotomic integers? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T15:58:59Z http://mathoverflow.net/feeds/question/10457 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/10457/what-are-the-prime-ideals-in-rings-of-cyclotomic-integers What are the prime ideals in rings of cyclotomic integers? Jon Yard 2010-01-02T03:58:26Z 2010-01-04T11:20:57Z <p>Is a good characterization of Spec $\mathbb{Z}[\zeta_n]$ known? Same question for its unit group.</p> http://mathoverflow.net/questions/10457/what-are-the-prime-ideals-in-rings-of-cyclotomic-integers/10458#10458 Answer by Chandan Singh Dalawat for What are the prime ideals in rings of cyclotomic integers? Chandan Singh Dalawat 2010-01-02T04:19:09Z 2010-01-02T05:17:33Z <p>The extension $\mathbb{Q}(\zeta_n)|\mathbb{Q}$ is abelian of group $(\mathbb{Z}/n\mathbb{Z})^\times$ so class field theory tells you everything about the prime ideals in $\mathbb{Z}[\zeta_n]$, the ring of integers of $\mathbb{Q}(\zeta_n)$.</p> <p>You should try to do the cases $n=3,4$ by hand.</p> <p>As for the group $\mathbb{Z}[\zeta_n]^\times$, an explicit subgroup of "cyclotomic units" can be constructed which has finite index. </p> <p>Any book on Cyclotomic Fields (Lang, Washington) should help. For a start, you can look up Chapter VI of Fröhlich-Taylor.</p> http://mathoverflow.net/questions/10457/what-are-the-prime-ideals-in-rings-of-cyclotomic-integers/10473#10473 Answer by Chandan Singh Dalawat for What are the prime ideals in rings of cyclotomic integers? Chandan Singh Dalawat 2010-01-02T06:48:54Z 2010-01-03T09:54:52Z <p>Let me summarise what Hilbert says in his <em>Zahlbericht</em> about the behaviour of rational primes in the cyclotomic field $\mathbb{Q}(\zeta)$, where $\zeta$ is a primitive $l$-th root of $1$ and $l$ is an odd prime. You can read the original at the <a href="http://gdz.sub.uni-goettingen.de/" rel="nofollow">Göttingen</a> site or a French translation at the <a href="http://www.numdam.org/" rel="nofollow">Grenoble</a> site.</p> <p><strong>Satz 117.</strong> <em>The ideal $\mathfrak{l}=(1-\zeta)\mathbb{Z}[\zeta]$ is prime of residual degree $1$, and $l\mathbb{Z}[\zeta]=\mathfrak{l}^{l-1}$.</em></p> <p><strong>Satz 118.</strong> <em>The discriminant of the field $\;\mathbb{Q}(\zeta)$ is $(-1)^{(l-1)/2}l^{l-2}$.</em></p> <p><strong>Satz 119.</strong> <em>If $p\neq l$ is a rational prime, $f>0$ is the smallest exponent such that $p^f\equiv1\pmod l$, and $e$ is defined by $ef=l-1$, then</em> $$p\mathbb{Z}[\zeta]=\mathfrak{p}_1\ldots\mathfrak{p}_e,$$ <em>where the</em> $\mathfrak{p}_i$ <em>are distinct prime ideals of residual degree $f$.</em></p> <p>These results go back to Kummer (1847). All this was much before anyone dreamt of Class Field Theory.</p> http://mathoverflow.net/questions/10457/what-are-the-prime-ideals-in-rings-of-cyclotomic-integers/10555#10555 Answer by Qiaochu Yuan for What are the prime ideals in rings of cyclotomic integers? Qiaochu Yuan 2010-01-03T01:29:34Z 2010-01-04T11:20:57Z <p><strong>Theorem:</strong> Let $\alpha$ be an algebraic integer such that $\mathbb{Z}[\alpha]$ is integrally closed, and let its minimal polynomial be $f(x)$. Let $p$ be a prime, and let </p> <p>$\displaystyle f(x) \equiv \prod_{i=1}^{k} f_i(x)^{e_i} \bmod p$</p> <p>in $\mathbb{F}_p[x]$. Then the prime ideals lying above $p$ in $\mathbb{Z}[\alpha]$ are precisely the maximal ideals $(p, f_i(\alpha))$, and the product of these ideals (with the multipicities $e_i$) is $(p)$. (<a href="http://modular.math.washington.edu/papers/ant/" rel="nofollow">Theorem 8.1.3</a>.)</p> <p>In this particular case we have $f(x) = \Phi_n(x)$. When $(p, n) = 1$, its factorization in $\mathbb{F}_p[x]$ is determined by the action of the Frobenius map on the elements of order $n$ in the multiplicative group of $\overline{ \mathbb{F}_p }$, which is in turn determined by the minimal $f$ such that $p^f - 1 \equiv 0 \bmod n$ as described in Chandan's answer. (This $f$ is the size of every orbit, hence the degree of every irreducible factor.) When $p | n$ write $n = p^k m$ where $(m, p) = 1$, hence $x^n - 1 \equiv (x^m - 1)^{p^k} \bmod p$. Then I believe that $\Phi_n(x) \equiv \Phi_m(x)^{p^k - p^{k-1}} \bmod p$ and you can repeat the above, but you'd have to check with a real number theorist on that. (<strong>Edit:</strong> Indeed, it's true over $\mathbb{Z}$ that $\Phi_n(x) = \frac{ \Phi_m(x^{p^k}) }{ \Phi_m(x^{p^{k-1}}) }$.)</p>