Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is Theorem 6.6 from Janusz, Algebraic Number Fields, it says:

Let $R\subseteq R'$ be Dedekind domains with $R'$ integral over $R$ and $\mathfrak{p}$ a nonzero prime ideal of $R$. Suppose $$\mathfrak{p}R'=\mathfrak{P}_1^{e_1}\cdots\mathfrak{P}_g^{e_g}$$ with the $\mathfrak{P}_i$ distinct prime ideals of $R'$ and $e_i>0$. Let $f_i=f(\mathfrak{P}_i|R)$ be the relative degree of $\mathfrak{P}_i$. Then:

1) $\sum_ie_if_i$ is the dimension of $R'/\mathfrak{p}R'$ over $R/\mathfrak{p}$

2) if $K$ and $L$ are the quotient fields of $R$ and $R'$ respectively and the dimension $[L:K]$ is finite then $\sum_ie_if_i\leq[L:K]$

3) if $S$ is the complement of $\mathfrak{p}$ in $R$ and if $R'_S$ is finitely generated over $R_S$ then $\sum_ie_if_i=[L:K]$


My problem is with point 3). It says: suppose $R'_S$ finitely generated module over $R_S$ and take $x_1,\ldots,x_n$ as a minimal generating set, for $R'_S$ over $R_S$. It shows that such $x_i$'s are elements of $L$ linearly indipendent over $K$, and all is ok for me. Then it that we know that the $x_i$'s are linearly indipendent, we prove that they gives a basis for $L$ over $K$. Suppose not. Then $\sum Kx_i$ is a $K$-vector space properly contained in $L$, thus there exists an element $y\in L$ such that $$Ky\cap\sum Kx_i=0$$ But $y$ satisfies a monic polynomial over $K$, hence.........and from now on i can understand. So my question is: why do $y$ satisfy a polynomial over $K$? We don't know a priori that $L$ is algebraic, or finite degree, over $K$. So why Janusz state it as an obvious fact? Do this follow from $R'$ integral over $R$? Or from $R'_S$ finitely generated over $R_S$?

Could someone help me please?

share|cite|improve this question
up vote 3 down vote accepted

Your title question has an affirmative answer for any integral extension $R' \supset R$ of domains: see e.g. Proposition 14.10 of these notes.

share|cite|improve this answer
What is $I_S(K)^{\bullet}$? – Federica Maggioni May 21 '13 at 21:17
Sorry, there appears to be a typo. It should be $I_M(K)^{\bullet}$. $I_M(K)$ are the elements of $M$ which are integral over $K$, and for any ring $R$, $R^{\bullet}$ denotes $R \setminus \{0\}$. – Pete L. Clark May 21 '13 at 23:06
yes, now it's clear, thanks – Federica Maggioni May 22 '13 at 8:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.