What happened to Emmy Noether's *Zukunftsphantasie* ?

Question

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 $\mathfrak{D}$ ?

Answer 1

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.)