# Is there a redundant assumption in Exercise 2, page 14, from Janusz "Algebraic Number Fields"?

The exercise says:

Let $R$ be an integral domain with quotient field $K$ and let $M$ be an $R$-submodule of a finite dimensional $K$-vector space. Prove $M=\bigcap_{P} R_P M$, where the intersection is taken over all maximal ideal of $R$.

My question is whether the bolded assumption that the vector space is finite dimensional is redundant. If not, please find the mistake in the following solution:

Solution (without the assumption): It is obvious that $M\subseteq R_{P} M$ for every $P$, hence $M\subseteq \bigcap_{P} R_P M$. For the other direction, let $x\in \bigcap_{P} R_P M$. Put $$I = \{ a\in R : ax \in M\}.$$ Let $P$ be maximal. As $x\in R_PM$, it has the form $x = \frac{u}{v} y$, for some $u,v\in R$ with $v\not\in P$ and $y\in M$. Thus, $vx=uy \in M$. Hence, $v\in I$ and in particular, $I\not\subseteq P$. So $I$ is not contained in any maximal ideal, whence $I = R$. We conclude that $1\in I$, so $x=1\cdot x\in M$. QED

• What do you mean by $R_PM$? Jun 27, 2016 at 13:59
• $R_P$ is the localization of $R$ at $P$ and $R_P M$ is the minimal $R_P$ module that contains $M$. It is composed of expressions of the form $\sum_i \frac{r_i}{s_i} x_i$ with $r_i,s_i\in R$ but $s_i\not\in P$ and $x_i\in M$ (note that everything is contained in some $K$-vector space, so all operations are defined). Jun 27, 2016 at 15:01
• Isn't this $M_P$? Why need such an unusual notation? Jun 27, 2016 at 15:03
• Because Janusz uses it. Jul 2, 2016 at 6:12

$x \in R_PM$ yields only $x = \sum_{i} \frac{u_i}{v_i}m_i$ with $m_i \in M, u_i \in R, v_i \notin P$. Nevertheless this does not kill the argument, since we can multiply by $v = \prod_i v_i \notin P$ and get $vx = \sum_{i} u_i\frac{v}{v_i}m_i \in M$, hence $v \in I$.
• This assumption does not affect you, when you consider Algebraic number fields. Think of $R$ being some ring of integers and the vector space being some finite field extension of $K$.