1
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ What do you mean by $R_PM$? $\endgroup$
    – user26857
    Jun 27, 2016 at 13:59
  • 1
    $\begingroup$ $R_P$ 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). $\endgroup$
    – Lior B-S
    Jun 27, 2016 at 15:01
  • $\begingroup$ Isn't this $M_P$? Why need such an unusual notation? $\endgroup$
    – user26857
    Jun 27, 2016 at 15:03
  • 1
    $\begingroup$ Because Janusz uses it. $\endgroup$ Jul 2, 2016 at 6:12

1 Answer 1

2
$\begingroup$

$x \in R_PM$ 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 $v \in I$.

So yes, the assumption is redundant.

$\endgroup$
4
  • $\begingroup$ Thanks for the correction. I will accept this answer in a while; unless someone will find a gap (why Janusz added this assumption in his book?) $\endgroup$
    – Lior B-S
    Jun 27, 2016 at 12:24
  • $\begingroup$ 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 $K$. $\endgroup$
    – MooS
    Jun 27, 2016 at 12:26
  • $\begingroup$ The more interesting question would be: Taking the assumption, is there another easy proof, which essentially uses the assumption? $\endgroup$
    – MooS
    Jun 27, 2016 at 12:30
  • $\begingroup$ I totally agree with your latter comment, I wonder which proof he had in mind. Regarding the former comment, I don't know of any interesting example in which this assumption does not hold, and I have no problem to assume it. But as it is an exercise, I am trying to check that I haven't cheated myself... $\endgroup$
    – Lior B-S
    Jun 27, 2016 at 12:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .