# The different of a Galois extension of an algebraic number field

Since the ramification indices of a prime are the same in a Galois extension, the following proposition is likely to be true. If it is true, how do we prove it?

Proposition Let $K$ be an algebraic number field. Let $L$ be a finite Galois extension of $K$. Let $G$ be the Galois group of $L/K$. Let $\mathfrak{D}_{L/K}$ be the different. Then $\sigma(\mathfrak{D}_{L/K}) = \mathfrak{D}_{L/K}$ for every $\sigma \in G$.

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Isn't it better to use the defn of inverse different as the "dual module" to the integers $\mathfrak o_L$ under trace to $K$? That is, $$\hbox{inverse different}\;=\; \{\alpha\in L\;:\; \hbox{tr}^L_K(\alpha\cdot \mathfrak o_L)\subset \mathfrak o_K\}$$ Trace is certainly Galois-invariant, and taking inverse of a fractional ideal is, also.
The different was first defined by Dedekind. He defined it as $GCD$ of all $f'(\alpha)$, where $\alpha$ is an integer of $L$ generating $L$ over $K$ and $f(X)$ the minimal polynomial of $\alpha$ over $K$. There is the Hilbert's definition. I posted a question about it a while ago. The diffferent can be defined as the annihilator of the Kähler differential module $\Omega_{B/A}$. – Makoto Kato Aug 24 '12 at 22:07