ramification of prime in Normal closure Let $K$ be an algebraic number field and let $p$ be a prime in $\mathbb{Q}$ such that $p$ ramifies in $L$, the Galois closure of $K$. How can I show that $p$ ramifies in $K$ itself?
 A: 
Important fact: If $K, K'$ are extensions of $\mathbb Q$, and $p$ is a prime which is unramified in both $K$ and $K'$, then $p$ is unramified in the compositum $K\cdot K'$. 

One can prove this, for example, by considering the inertia group $I$ of the Galois closure $L$ of $K\cdot K'$: if $K, K'\subset L^I$, then $K\cdot K'\subset L^I$.

We can use this in the following way: let $K = \mathbb Q(\alpha)$, and let $\alpha_2,\ldots,\alpha_n$ be the Galois conjugates of $\alpha$. Then for each $i$, $$\mathbb Q(\alpha)\cong\mathbb Q(\alpha_i),$$so in particular, $p$ is unramified in $\mathbb Q(\alpha)$ if and only if it is unramified in $\mathbb Q(\alpha_i)$ for each $i$.
But the Galois closure
$$L=\mathbb Q(\alpha)\cdot\mathbb Q(\alpha_2)\cdot\ldots\cdot\mathbb Q(\alpha_n),$$
so if $p$ ramifies in $L$, it cannot be unramified in $K$.
A: Here is an idea. Let $A$ be the ring of integers of $K$, and $B$ the ring of integers of $L$. 
Let $\sigma_1,\ldots,\sigma_n$ be the embeddings of the field $K$ into $L$. Consider the
map
$$
f\colon \bigotimes_{\mathbf Z}^n A\rightarrow L
$$
defined by $f(a_1\otimes\ldots\otimes a_n)=\sigma_1(a_1)\cdots\sigma_n(a_n)$. Let $C$ be the image of $f$. Then $C$ is an order in $B$. It follows that the discriminant of $B$ over $\mathbf Z$ divides the discriminant of $C$ over $\mathbf Z$. The latter discriminant is a divisor of some power of the discriminant of $A$ over $\mathbf Z$. Therefore, if $p$ is ramified in $L$, then $p$ divides the discriminant of $B$, and then also the discriminant of $C$, and then also the discriminant of some power of the discriminant of $A$, and hence the discriminant of $A$. In particular, $p$ is ramified in $K$.
