$\newcommand{\p}{\mathfrak p}$ $\newcommand{\P}{\mathfrak P}$ $\newcommand{\O}{\mathcal O}$ Let $L/K$ be a number field extension and $\p\neq 0$ a prime ideal of $\O_K$. Since $\O_L$ is Dedekind, we have a decomposition $$\p\O_L=\P_1^{e_1}\cdots\P_g^{e_g}$$ with $e_i\neq 0$. In this note of algebraic number theory, the residue degree of $\P_i$ above $\p$ is defined by the degree of extension $[\O_L/\P_i\colon\O_K/\p]$ (page 28).

My question is, that the author said $\O_L/\P_i$ is a finite extension of $\O_K/\p$ but I cannot see it clearly. Indeed since $\p\subset \p\O_L\subset\P_i$, a ring homomorphism $\varphi\colon\O_K/\p\to\O_L/\P_i$ can be induced and both of them are fields, as $\O_K$, $\O_L$ are Dedekind. But how do we exclude the case that $\varphi$ is a zero homomorphism, that is, the case $\O_K\subset\P_i$? Moreover, why the extension is finite?

Thank you very much.

  • 3
    $\begingroup$ $\mathcal{O}_L$ is a finitely generated $\Bbb{Z}$-module, hence a finitely generated $\mathcal{O}_K$-module. The cosets modulo $\mathfrak{P}_i$ of those generators span $\mathcal{O}_L/\mathfrak{P}_i$ as a vector space over the smaller quotient field. Furthermore, $1\notin\mathfrak{P}_i$ so $\phi$ cannot be the zero homomorphism. $\endgroup$ – Jyrki Lahtonen Oct 7 '18 at 12:33
  • $\begingroup$ @JyrkiLahtonen Thanks for your answer, it is really helpful. $\endgroup$ – aeei.w.1995 Oct 7 '18 at 13:10

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