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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 $\sum_ie_if_i=[L:K]$

Proof.

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 says:....now 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?

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1 Answer 1

up vote 2 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.

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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

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