6

Let $M/L/K$ be a tower of number fields with discriminant of $M/K: d_M$ and of $L/K: d_L$. I would like to find a transitivity theorem for the discriminant and by letting $p_i$ and $q_i$ be integral basis for $M$ and $L$ respectively and $A =[a_{ij}]$ the transition matrix between the basis, a calculation gives:

$$[M:L]^{[L:K]}d_L = \det(A)^2d_M$$

However, these two links give different(even from each other) answers:

Divisibility of discriminants in number field extensions $([M:L]^2 d_L = \det(A)^2 d_M)$

Quadratic subfield of cyclotomic field (discriminant of $M$ is divisible by discriminant of $L$ to the power $[M:L]$

Both of these are given in the accepted answers and use different notation. Which of the three is correct?(The last one is not strictly contradictory but probably often will be...).

Community
  • 1
Asvin
  • 8,229

1 Answers1

7

A couple of definitions are in order to clarify the issue.

Definition 1. Consider a field extension $L / K$ and a basis $\{\alpha_1,\dotsc,\alpha_n\} \subset \mathcal{O}_L$ of $L$ over $K$. Its discriminant is $$ d(\alpha_1,\dotsc,\alpha_n) = \det(\text{Tr}\,(\alpha_i\alpha_j)) $$

Definition 2. The discriminant $\mathfrak{d}_{L/K}$ of a field extension $L/K$ is the ideal of $\mathcal{O}_K$ generated by the discriminants $d(\alpha_1,\dotsc,\alpha_n)$, where $\{\alpha_1,\dotsc,\alpha_n\}$ ranges over all the bases of $L/K$ with elements in $\mathcal{O}_L$.

Note that if $\alpha_1,\dotsc,\alpha_n$ generate $\mathcal{O}_L$ over $\mathcal{O}_K$ as a free module, then $\mathfrak{d}_{L/K} = d(\alpha_1,\dotsc,\alpha_n)\mathcal{O}_K$, because then all other bases can be obtained transforming $\alpha_1,\dotsc,\alpha_n$ by an appropriate matrix with entries in $\mathcal{O}_K$.

Clearly $\mathfrak{d}_{L/K}$ is always principal if $\mathcal{O}_K$ is a PID, e.g. if $K = \mathbb{Q}$, but it may not be in general.

Finally, we have the following:

Theorem. If $K \subseteq L \subseteq M$ is a tower of fields, then $$ \mathfrak{d}_{M/K} = \mathfrak{d}_{L/K}^{[M:L]} N_{L/K}(\mathfrak{d}_{M/L}) $$ Proof. See Neukirch's Algebraic Number Theory, corollary 2.10, chapter III.

A.P.
  • 9,728
  • 2
    The explicit answer to your question is: the third one! – A.P. May 14 '15 at 22:14
  • If I were to define my $d_M$ and $d_L$ as the discriminant for the specified basis, my computation is correct. However, the discriminant of an extension is the minimum such discriminant-of-basis(over $\Bbb Q$) and this causes the difference, is that correct? – Asvin May 14 '15 at 22:23
  • Unless there is a typo, your computation cannot be correct because it contains $d_L$ instead of $d_L^{[M:L]}$. You are right in saying that what you call the "discriminant of an extension" $L/K$ is exactly the generator of $\mathfrak{d}{L/K}$, if it is principal. I would strongly advise you against using that unless $K \neq \Bbb{Q}$: otherwise not only it may not be defined because $\mathfrak{d}{L/K}$ may not be principal, but you cannot define it uniquely if $\mathcal{O}_K$ is not totally ordered. – A.P. May 14 '15 at 22:34
  • (continued) Indeed, if $I = (d)$ is a principal ideal of a ring $R$, then $I = (ud)$ for every unit $u$ of $R$. But if $\mathcal{O}_K \not \subset \Bbb{R}$, then there is no canonical choice of unit. – A.P. May 14 '15 at 22:38
  • Yes, my calculation was wrong. I completely messed up the powers. Thank you for your answer, it helped quite a bit! – Asvin May 14 '15 at 23:45