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As I understand it discriminants are supposed to easily prove statements like $\sqrt r \not\in \mathbb Q (\sqrt p, \sqrt q)$ for $r,p,q$ distinct odd primes. You can easily calculate a discriminant like $$D_{\mathbb Z[\sqrt p, \sqrt q]/\mathbb Z} (1,\sqrt p, \sqrt q, \sqrt p \sqrt q)=2^8 p^2q^2$$ Let $\mathscr O$ be the ring of integers of $\mathbb Q (\sqrt p, \sqrt q)$ over $\mathbb Q$. Apriori you don't know a basis for the ring of integers, but you do know that the discriminant $ D_{\mathscr O/\mathbf Z}$ generated by $a \in \mathbb Z$ should satisfy $ak^2 = 2^8 p^2q^2$ for some integer $k \in \mathbb Z$. So if $z$ is any prime with $z|a$ then $z|2pq$. If you assume to the contrary that $\sqrt r \in \mathbb Q (\sqrt p, \sqrt q)$ and you can show that $z|2rq$ and $z|2pq$ so it must be that $z=2$ so $a=2^m$ for some $m$. It seems like $p$ should divide $a$ so that this should be a contradiction. Does $p|a$? How do you see that?

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There is a relation for the tower of discriminants. Let $A$ be a Dedekind domain with field of fractions $F$. Let $F''/F'/F$ be a tower of field extensions with $A'$ (resp. $A''$) the integral closure of $A$ in $F'$ (resp. $F''$). Then the discriminants satisfy $$ D_{A''/A} = \mbox{Norm}_{A'/A}(D_{A''/A'}) D_{A'/A}^{[F'':F']} $$ so in particular $D_{A'/A} | D_{A''/A}$. So in the above example we have that the discriminant of the ring of integers in $\mathbb Q(\sqrt p)$ is $4p$ so that $4p|a$.

http://math.stanford.edu/~conrad/676Page/handouts/transdisc.pdf

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