Why the extension $L/K$ is ramified in exactly those primes that divide the relative discriminant

algebraic-number-theorycommutative-algebra

Let $L$ is the finite extension field of the number field $K$ and $p$ is a prime ideal over the ring of integer $O_K$. Then, $pO_L= \prod_{j=1}^{g} P_j^{e_j}$. By CRT, we have $$ O_L/pO_L=\prod_{j=1}^{g} O_L/P_j^{e_j}.$$
My question:

Why the extension $L/K$ is ramified in exactly those primes that divide the relative discriminant ?

How does this imply that the extension $L/K$ is unramified in all but finitely many prime ideals ?

Updated:

Yes I got proof of first question.

But what about the 2nd question?

My approach for 2nd question:

Since only finitely many prime ideals $P_j$ divide the discriminant in the above product, the extension $L/K$ ramifies only at finitely many primes. All others primes do not ramify in the extension.

Is it true ?

Best Answer

Once you know the unique factorization in prime ideals then $P O_L = \prod_j Q_j^{e_j} $

For each $j, O_L/Q_j$ is a finite field, let $a_j$ a generator of its multiplicative group, then $O_L/Q_j = O_K[a_j]/(Q_j \cap O_K[a_j])$.

(CRT) The $Q_j$ are distinct maximal ideals so there is some element which is $\equiv 1 \bmod Q_j$ and $\equiv 0 \bmod Q_i$ for $i \ne j$, which means you can construct some $a \in O_L$ which is $\equiv a_j \bmod Q_j$ for all $j$. Moreover we can construct it such that $L = K(a)$.

Let $f \in O_K[x]$ be the minimal polynomial of $a$ so that we now consider $O_K[a] = O_K(x]/(f)$, a $O_K$-monogenic subring having all the $O_L/Q_j$ in its quotients.

Then $\forall j, e_j=1$ iff $f$ is separable $\bmod P$ iff $Disc(f) \not \in P$

(where $Disc(\prod_l (x-b_l))= \prod_{l\ne m} (b_l-b_m)$)

Proof : the factorization of $f \bmod P$ gives all the maximal ideals of $O_K[x]/(f)$ above $P$ and the degree of the irreducible factors tell the dimension of $O_K[x]/(f)/m$ as $O_K/P$ vector space, if all the $e_j=1$ then $O_L/P \cong \prod_j O_L/Q_j$ so if unramified the dimensions must sum to $\dim(O_L/P)=\deg(f)$.

Only finitely many prime ideals of $O_K$ contain $Disc(f)$, all the others are unramified in $O_K[a]$ thus in $O_L$.

The relative discriminant of $O_L/O_K$ is the ideal generated by all the $Disc(f)$ for each minimal polynomial $f$ of some $a \in O_L$ such that $L = K(a)$.

Being an ideal of $O_K$ only finitely many primes of $O_K$ contain it (and are ramified), the others are unramified in $O_L$.

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