MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Recenly I came across Peter Roquette's article On the history of Artin's $L$-functions and conductors (23 July 2003) in which he talks about some letters from Emil Artin and Emmy Noether to Helmut Hasse in the early 1930s.

Artin is trying to give the definitive form to the definition of his $L$-functions (to include ramified and archimedean places), and has proved what Hasse calls the Führerdiskriminantenproduktformel : for a finite galoisian extension $L|K$ of number fields with group $G=\mathrm{Gal}(L|K)$, the discriminant $\mathfrak{d}$ of $L|K$ can be decomposed as the product

$$ \prod_{\chi}\mathfrak{f}(\chi,L|K)^{\chi(1)} $$

extending over all characters $\chi$ of $G$, where $\mathfrak{f}(\chi,L|K)$ denotes the conductor of $\chi$ (as defined by Artin).

Emmy Noether writes to Hasse that she is looking for a decomposition formula for the different $\mathfrak{D}$ of $L|K$ which would yield Artin’s product formula for the discriminant $\mathfrak{d}$ after applying the norm map $N_{L|K}$. Perhaps this is what she calls her Zukunftsphantasie (a fantasy for the future).

Question. Is there such a decomposition of the different $\mathfrak{D}$ ?

share|cite|improve this question
up vote 11 down vote accepted

I am not sure if I am interpreting the question correctly: is it whether it is possible to assign, naturally, an ideal $I_\chi\subset O_L$ to each $\chi$ in such a way that $N_{L|K} I_\chi=f(\chi,L|K)^{\chi(1)}$? Then the answer is No, even without the word "naturally", even for cyclotomic fields and even locally:

If $K={\mathbb Q}$ and $L={\mathbb Q}(\zeta_{12})={\mathbb Q}(i,\sqrt{3})$, then ${\rm Gal}(L/K)=C_2\times C_2$ has four 1-dimensional characters, of conductor $1$, $3$, $4$ and $12$. However, $3$ and $12$ are not norms of ideals from $L/K$, because there is a unique prime ${\mathfrak p}|3$ of $O_L$ with $e=f=2$, so $N_{L|K}{\mathfrak p}=3^2$, and the norm of any ideal has even valuation at $3$.

Another way of saying this is that the different here is ${\mathfrak D}={\mathfrak q}^2{\mathfrak p}$ (with ${\mathfrak q}|2, {\mathfrak p}|3$) and the discriminant is ${\mathfrak d}=2^43^2$, and while it is possible to split the 3-part of the discriminant into two parts for the two ramified characters, it is not possible for the different. So perhaps Noether's question means something else?

(The observation is joint with my brother.)

share|cite|improve this answer
That's a very nice example, Tim. Yes, while your interpretation is quite natural and is the one I had in mind, she must have had something else in mind because Peter Roquette goes on to say that the situation is not quite clarified yet. – Chandan Singh Dalawat Dec 13 '12 at 2:16
I agree! I just wanted to point out that it cannot be the "obvious" interpretation. Looking at Roquette's paper more carefully (p. 30), it appears that the algebraic structure that Noether was looking it was the cross product algebra of $L$ by $K[G]$. She is asking whether its decomposition into minimal left and right ideals says something about the decomposition of the different, and Artin backs her point of view (p. 32) that the connection of this ring to the discriminant must be studied in the future. – Tim Dokchitser Dec 13 '12 at 10:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.