# Why $\mathcal O_L/\mathfrak P_i$ is a finite extension of $\mathcal O_K/\mathfrak p$?

$$\newcommand{\p}{\mathfrak p}$$ $$\newcommand{\P}{\mathfrak P}$$ $$\newcommand{\O}{\mathcal O}$$ 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.

• $\mathcal{O}_L$ 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. – Jyrki Lahtonen Oct 7 '18 at 12:33
• @JyrkiLahtonen Thanks for your answer, it is really helpful. – aeei.w.1995 Oct 7 '18 at 13:10