September 23rd
Today I learned that the relative different is kind of multiplicative in towers, which unsurprisingly follows roughly from the behavior of trace in towers, with some fudging around.
Outlining somewhat, fix extensions $M$ over $L$ over $K,$ and we'll show that\[\diff(\mathcal O_M/\mathcal O_K)=\diff(\mathcal O_M/\mathcal O_L)(\diff(\mathcal O_L/\mathcal O_K)\mathcal O_M).\]For convenience, we'll fix $T=\mathcal O_M,$ $S=\mathcal O_L,$ and $R=\mathcal O_K.$ Then let $T^*_R$ be the dual of $T$ with respect to $R,$ and define $T_S^*$ and $S_R^*$ similarly. After taking inverses, it suffices to show that\[T_R^*=T_S^*(S_R^*T).\]This is done in two steps.
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In one direction, we show $T_S^*(S_R^*T)\subseteq T_R^*$ by taking traces directly: roughly, we have \[\op T_K^M\left(T_S^*(S_R^*T)T\right)=\op T_K^L\op T_L^M\left(S_R^*\op T_L^M(T^*_ST)\right)=\op T_K^L\left(S_R^*\op T_L^M(T^*_ST)\right)\subseteq\op T_K^L(S_R^*S)\subseteq R\] from the transitivity of trace.
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In the other direction, we show $T_R^*(S_R^*)^{-1}T\subseteq T_S^*$ with some finagling. Roughly, this needs \[\op T_L^M\left(T_R^*(S_R^*)^{-1}T\right)\subseteq S.\] The left-hand side is \[\op T_L^M\left(T_R^*(S_R^*)^{-1}T\right)=(S_R^*)^{-1}\op T_L^M\left(T_R^*T\right),\] which is a subset of $S$ if and only if $\op T_L^M\left(T_R^*T\right)\subseteq S_R^*.$ But this follows from taking $\op T_K^L$ and then using transitivity of trace.
As promised, it's not super exciting.