A problem on Algebraic Number Theory, Norm of Ideals - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T23:02:30Z http://mathoverflow.net/feeds/question/9235 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/9235/a-problem-on-algebraic-number-theory-norm-of-ideals A problem on Algebraic Number Theory, Norm of Ideals 7-adic 2009-12-18T03:13:07Z 2012-05-23T04:50:24Z <p>A problem on Algebraic Number Theory</p> <p>K and L are number fields over Q.(Q is rational number filed) K is a subfield of L.</p> <p>O_K is the integers of K. and O_L is the integers of L.</p> <p>P is a prime ideal of O_L. p is a prime ideal of O_K. P is over p.</p> <p>The residue class degree f is defined to be f=[O_L/P:O_K/p]. The norm of P is Norm(P)=p^f</p> <p>This is the usual definition of Norm of an ideal.(See Serre's Local fields and Serge Lang's Algebraic Number Theory)</p> <p>Swinnerton-Dyer's A Brief Guide to Algebraic Number Theory has a different definition on Norm of an ideal.(Page 25)</p> <p>if A is an ideal of O_L, Norm(A)= ideal in O_K generated by elements Norm(a) where a is in A.</p> <p>I dont know why these two definitions are the same. Swinnerton-Dyer claims so in his book. Can anyone here give a hint, an explanation or anything else?</p> http://mathoverflow.net/questions/9235/a-problem-on-algebraic-number-theory-norm-of-ideals/9248#9248 Answer by Rado for A problem on Algebraic Number Theory, Norm of Ideals Rado 2009-12-18T07:16:46Z 2009-12-18T07:16:46Z <p>I thought I should know this but then i ended up looking it up. Its in Lang's Algebraic Nubmer Theory pages 24-26 (at least for A principal, but that should be enough). That is if you want to know the proof. I have no idea where the intuition comes from but I bet it is using lattices somehow.</p> http://mathoverflow.net/questions/9235/a-problem-on-algebraic-number-theory-norm-of-ideals/9251#9251 Answer by H. Hasson for A problem on Algebraic Number Theory, Norm of Ideals H. Hasson 2009-12-18T07:40:01Z 2009-12-18T17:54:26Z <p>Hmm... I can see offhand how to deal with it if L/K is Galois, but I'd have to think about it otherwise... In the Galois case, above p you have r many prime ideals, each with ramification index e, and residue degree f. The rough sketch is to view this as a problem about discrete valuations, rather than prime ideals.</p> <p>N(P) (according to your second definition) = &lt; N(a)| a in P >. We know this is an ideal in O_K, and it only remains to describe its decomposition into primes. Since the ramification index of p in (each) P above it is e, the minimal p-adic valuation of an element in N(P) is f. So if t is a parametrizing element of the p-adic valuation (choose it in O_K), then u*t<sup>f</sup> generates N(P)<sub>p</sub> where u is in O_K - P (check that N(P) isn't divisible by other prime ideals, with similar methods).</p> <p>Hope that helps a bit with the intuition.</p> <p><hr /></p> <p>After reading Adam's solution, I noticed a few things were wrong in my argument. They were corrected in the body.</p> http://mathoverflow.net/questions/9235/a-problem-on-algebraic-number-theory-norm-of-ideals/9286#9286 Answer by Adam Topaz for A problem on Algebraic Number Theory, Norm of Ideals Adam Topaz 2009-12-18T16:48:25Z 2009-12-18T21:23:33Z <p>Ok, here's the argument: First recall that the usual norm for non-zero elements of a field is transitive in towers; thus the same is true for your second definition of the norm of an ideal. In particular, $N_{K|Q}\circ N_{L|K} = N_{L|Q}$. The fact that the norm $N_{L|Q}(\mathfrak{P}) = [\mathcal{O}_L:\mathfrak{P}] \cdot \mathbb{Z}$ is easy to see for a prime $\mathfrak{P}$ in $\mathcal{O}_L$; edit: and thus the same is true for any integral ideal $\mathfrak{a}$. Now let $\mathfrak{p} = \mathcal{O}_K \cap \mathfrak{P}$ and $(p) = \mathbb{Z} \cap \mathfrak{P}$.</p> <p>We have $N_{L|Q}(\mathfrak{P}) = p^{f(\mathfrak{P}|p)} = N_{K|Q}N_{L|K}\mathfrak{P}$. In particular, we deduce that $N_{L|K}\mathfrak{P} = \mathfrak{p}^d$ for some $d$. Moreover, we know that</p> <p>$N_{K|Q}\mathfrak{p}^d = p^{d \cdot f(\mathfrak{p}|p)} = p^{f(\mathfrak{P}|p)}.$</p> <p>But then $d = f(\mathfrak{P}|p) / f(\mathfrak{p}|p) = f(\mathfrak{P}|\mathfrak{p})$ as required. </p> http://mathoverflow.net/questions/9235/a-problem-on-algebraic-number-theory-norm-of-ideals/97720#97720 Answer by KConrad for A problem on Algebraic Number Theory, Norm of Ideals KConrad 2012-05-23T04:42:36Z 2012-05-23T04:50:24Z <p>Here is a proof that the ideal norm as defined in the books by Serre and Lang is equal to the ideal norm as defined in Swinnerton-Dyer's book. We will start from the definition given by Serre and Lang, state some of its properties, and use those to derive the formula as given by Swinnerton-Dyer.</p> <p>Background: Let $A$ be a Dedekind domain with fraction field $K$, $L/K$ be a finite separable extension, and $B$ be the integral closure of $A$ in $L$. For any prime $\mathfrak P$ in $B$ we define ${\rm N}_{B/A}({\mathfrak P}) = \mathfrak p^f$, where $f = f({\mathfrak P}|{\mathfrak p})$ is the residue field degree of $\mathfrak P$ over $\mathfrak p$, and this norm function is extended to all nonzero ideals of $B$ by multiplicativity from its definition on (nonzero) primes in $B$. </p> <p>Properties.</p> <p>1) The map <code>${\rm N}_{B/A}$</code> is multiplcative (immediate from its definition).</p> <p>2) Good behavior under localization: for any (nonzero) prime ${\mathfrak p}$ in $A$, <code>${\rm N}_{B/A}({\mathfrak b})A_{\mathfrak p} = {\rm N}_{B_{\mathfrak p}/A_{\mathfrak p}}({\mathfrak b}B_{\mathfrak p})$</code>. Note that $A_{\mathfrak p}$ is a PID and $B_{\mathfrak p}$ is its integral closure in $L$; the ideal norm on the right side is defined by the definition above for Dedekind domains, but it's more easily <em>computable</em> because $B_{\mathfrak p}$ is a finite free $A_{\mathfrak p}$-module on account of $A_{\mathfrak p}$ being a PID and $L/K$ being separable. The proof of this good behavior under localization is omitted, but you should find it in books like those by Serre or Lang.</p> <p>3) For nonzero $\beta$ in $B$, <code>${\rm N}_{B/A}(\beta{B}) = {\rm N}_{L/K}(\beta)A$</code>, where the norm of $\beta$ on the right is the field-theoretic norm (determinant of multiplication by $\beta$ as a $K$-linear map on $L$). To prove this formula, it is enough to check both sides localize the same way for all (nonzero) primes $\mathfrak p$: <code>${\rm N}_{B_{\mathfrak p}/A_{\mathfrak p}}(\beta{B}_{\mathfrak p}) = N_{L/K}(\beta)A_{\mathfrak p}$</code> for all $\mathfrak p$. If you know how to prove over the integers that <code>$[{\mathcal O}_F:\alpha{\mathcal O}_F] = |{\rm N}_{F/{\mathbf Q}}(\alpha)|$</code> for any number field $F$ then I hope the method you know can be adapted to the case of $B_{\mathfrak p}/A_{\mathfrak p}$, replacing ${\mathbf Z}$ with the PID $A_{\mathfrak p}$. That is all I have time to say now about explaining the equality after localizing.</p> <p>Now we are ready to show <code>${\rm N}_{B/A}({\mathfrak b})$</code> equals the ideal in $A$ generated by all numbers ${\rm N}_{E/F}(\beta)$ as $\beta$ runs over $\mathfrak b$.</p> <p>For any $\beta \in \mathfrak b$, we have $\beta{B} \subset \mathfrak b$, so ${\mathfrak b}|\beta{B}$. Since <code>${\rm N}_{B/A}$</code> is multiplicative, <code>${\rm N}_{B/A}({\mathfrak b})|{\rm N}_{E/F}(\beta)A$</code> as ideals in $A$. In particular, <code>${\rm N}_{E/F}(\beta) \in {\rm N}_{B/A}({\mathfrak b})$</code>. Let $\mathfrak a$ be the ideal in $A$ generated by all numbers <code>${\rm N}_{E/F}(\beta)$</code>, so we have shown <code>$\mathfrak a \subset {\rm N}_{B/A}(\mathfrak b)$</code>, or equivalently <code>${\rm N}_{B/A}(\mathfrak b)|\mathfrak a$</code>. To prove this divisibility is an equality, pick any prime power ${\mathfrak p}^k$ dividing $\mathfrak a$. We will show ${\mathfrak p}^k$ divides <code>${\rm N}_{B/A}(\mathfrak b)$</code>.</p> <p>To prove ${\mathfrak p}^k$ divides <code>${\rm N}_{B/A}(\mathfrak b)$</code> when ${\mathfrak p}^k$ divides $\mathfrak a$, it suffices to look in the localization of $A$ at $\mathfrak p$ and prove ${\mathfrak p}^kA_{\mathfrak p}$ divides <code>${\rm N}_{B/A}(\mathfrak b)A_{\mathfrak p}$</code>, which by the 2nd property of ideal norms is equal to <code>${\rm N}_{B_{\mathfrak p}/A_{\mathfrak p}}(\mathfrak b{B_{\mathfrak p}})$</code>. Since $B_{\mathfrak p}$ is a PID, the ideal <code>${\mathfrak b}B_{\mathfrak p}$</code> is principal: let $x$ be a generator, and we can choose $x$ to come from $\mathfrak b$ itself. By the 3rd property of ideal norms, <code>${\rm N}_{B_{\mathfrak p}/A_{\mathfrak p}}(xB_{\mathfrak p}) = {\rm N}_{E/F}(x)A_{\mathfrak p}$</code>. Showing <code>${\mathfrak p}^kA_{\mathfrak p}$</code> divides <code>${\rm N}_{E/F}(x)A_{\mathfrak p}$</code> is the same as showing <code>${\rm N}_{E/F}(x) \in {\mathfrak p}^kA_{\mathfrak p}$</code>. Since $x$ is in in $\mathfrak b$, <code>${\rm N}_{E/F}(x) \in \mathfrak a \subset {\mathfrak p}^k$</code>, so <code>${\rm N}_{E/F}(x) \in {\mathfrak p}^kA_{\mathfrak p}$</code>. QED</p>