Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 74 / 5

For $R={\mathcal{O}}_k$ and $S={\mathcal{O}}_K$, define the inverse different, or codifferent, by the trace-dual latticeThe trace pairing is nondegenerate in this separable field extension, so this is a full fractional ideal of $S$. The relative different ideal is its inverse:Every integral $x$ has integral traces against $S$, so $S\subseteq {\mathfrak{D}}^{-1}$ and $\mathfrak{D}\subseteq S$.

For $k=\mathbb{Q}$, take an integral basis $b_1,\dots ,b_n$ and its trace pairing matrix $T=(\mathrm{Tr}(b_i{b}_j))$. In these coordinates the codifferent is $T^{-1}{\mathbb{Z}}^n$, andThe ideal norm of $\mathfrak{D}$ equals this index, by inversion of fractional ideal norms. Therefore the norm of the different ideal relation isHere $r_2$ is the number of conjugate pairs of complex embeddings. The different is an ideal of the upper field, whereas the relative discriminant ideal is its ideal norm in the base field; over $\mathbb{Q}$ the ordinary signed field discriminant also records the signature.

Use $R={\mathcal{O}}_k$, $S={\mathcal{O}}_L$, $T={\mathcal{O}}_K$, and denote their relative codifferents by $C_{L/k}$, $C_{K/L}$ and $C_{K/k}$. Transitivity of the field trace givesFor the forward direction, testing an element $t\in T$ against every $s\in S$ is legitimate because $st\in T$. For the reverse direction, take $s=1$ and use trace transitivity.

Localize at a nonzero base prime ideal. The intermediate ring is semilocal Dedekind domain, so its invertible fractional ideal $C_{L/k}$ is principal there, say $C_{L/k}=d^{-1}S$ locally. This principality follows by choosing, with the Chinese remainder theorem for ideals, a generator at each of its finitely many maximal ideals. The displayed criterion now becomesInvert these fractional ideals using Solution 1. Equality at every base localization gives the global different in a tower formulaThe second factor is extended to the upper ring of integers; that extension is implicit in the printed formula. At a local prime the exponents consequently satisfy $d(K/k)=d(K/L)+e(K/L)d(L/k)$.

For number fields, a prime ideal divides the different ideal exactly when it is ramified. More precisely, the local different exponent and tame ramification relation is $d_{\mathfrak{P}}\ge e_{\mathfrak{P}}-1$, with equality for tame ramification; if $e=1$, the finite residue extension is separable and the exponent is zero. One can see the ramification criterion directly from the trace pairing modulo the base prime: when $e>1$ its quotient ring has a nonzero nilpotent ideal, whose elements pair to zero since multiplication by a nilpotent has trace zero. When $e=1$ the quotient is a product of finite separable fields and its trace pairing is nondegenerate. Its determinant is a unit exactly in the second case.

Let $M=K_1\cap K_2$. Any rational prime ramifying in $M$ must ramify in each $K_i$, because ramification indices multiply in a tower. It would then divide both field discriminants, contrary to their coprimality. Thus $M$ is unramified at every finite prime, so $|d_M|=1$. The Minkowski lower bound for a number-field discriminant excludes this for a field of degree $n>1$:The last inequality follows already for the weakest signature bound $r_2\le n/2$: the expression starts above one at $n=2$ and increases with $n$. Consequently $M=\mathbb{Q}$.

For completeness, normality converts this intersection statement into the desired degree statement. Put $E=K_1{K}_2$. Restriction injects $\mathrm{Gal}(E/K_2)$ into $\mathrm{Gal}(K_1/\mathbb{Q})$. The fixed field of its image in $K_1$ is exactly $K_1\cap K_2$, because $E/K_2$ is Galois. The Galois correspondence therefore gives $[E:K_2]=[K_1:M]=[K_1:\mathbb{Q}]$. This proves coprime discriminants imply linear disjointness: