1
$\begingroup$

I have a problem on different and discriminant.

For simplicity, assume $E/F$ is a quadratic extension of number fields with different $\mathfrak D_{E/F}$ and discrimiant $\mathfrak d_{E/F}$. They are integral ideals in $\mathfrak o_{E}$ and $\mathfrak o_{F}$, respectively. We have the relation $$N_{E/F}(\mathfrak D_{E/F})=\mathfrak d_{E/F}.$$

For $\mathfrak p\mid \mathfrak d_{E/F}$, we know $\mathfrak p$ is ramified, $$\mathfrak p=\mathfrak P_1\mathfrak P_2,$$ and $\mathfrak P_1,\mathfrak P_2\mid \mathfrak D_{E/F}$.

Here is my problem. Assume $\mathfrak P_i^{s_i}\mid\mid \mathfrak D_{E/F}$. By the relation $N_{E/F}(\mathfrak D_{E/F})=\mathfrak d_{E/F}$, we have $$ N_{E/F}(\mathfrak P_1^{s_1}\mathfrak P_2^{s_2})\mid\mid \mathfrak d_{E/F}, $$ and thus one has $\mathfrak p^{s_1+s_2}\mid\mid \mathfrak d_{E/F}$. It is contradiction with the fact that if $\mathfrak p$ is tamely ramfied, then $\mathfrak p\mid\mid \mathfrak d_{E/F}$.

What's wrong with the above argument?

  • 1
    If $ \mathfrak p $ is ramified in a quadratic extension, then you have $ \mathfrak P_1 = \mathfrak P_2 $.2017-02-10

0 Answers 0