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Let $K$ be an algebraic extension of the rational numbers and $L$ an algebraic extension of $K$. Let $\mathfrak{a}_K = (a, \alpha )$ be an ideal of the ring of integers $\mathcal{O}_K$ of $K$, with $a \in \mathbb{Z}$ and $\alpha \in \mathcal{O}_K$. Let $\mathfrak{a}_L = (a, \alpha )$ be an ideal of the ring of integers $\mathcal{O}_L$ of $L$. By $\mathfrak{a}_L$, I mean simply $\mathfrak{a}_K$ taken as an ideal of $\mathcal{O}_L$.

What is the relationship between the norms $[\mathcal{O}_L : \mathfrak{a}_L]$ and $[\mathcal{O}_K : \mathfrak{a}_K]$ and what is a reference for this result?

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up vote 5 down vote accepted

The ring $\mathcal O_L$, thought of as an $\mathcal O_K$-module, is locally free of rank equal to $[L:K]$. Thus $$\mathcal O_L/\mathfrak a_L = \mathcal O_L/\mathfrak a_K \mathcal O_L = (\mathcal O_K/\mathfrak a_K) \otimes_{\mathcal O_K} \mathcal O_L$$ is locally free (and hence actually free) over $\mathcal O_K/\mathfrak a_K$ of rank $[L:K]$. Consequently $[\mathcal O_L:\mathfrak a_L] = [\mathcal O_K: \mathfrak a_K]^{[L:K]}.$ (This is an ideal-theoretic analogue of the fact that if $a \in K$, then $N_{L/\mathbb Q}(a) = N_{K/\mathbb Q}(a)^{[L:K]}.$)

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Thank you very much! I expected them to be related by the exponent $[L:K]$ but I didn't know how to show it! Thanks. – Samuel Hambleton Feb 16 '12 at 22:14

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