I'm trying to understand the proof of lemma 4.12 on modules over Dedekind domains from Frohlich and Taylor's book 'Algebraic Number Theory' page 94.
I have a Dedekind domain $\mathcal o$, a non-zero prime ideal $\mathfrak p$, its field of fractions $K$ and a finite dimensional $K$-vector space $V$.
The Lemma states
Let $L$ and $M$ be $\mathcal o$-lattices in $V$. Then for almost all $\mathfrak p$ $L_\mathfrak p=M_\mathfrak p$.
$Proof$. Choose free $\mathfrak o$-lattices $F'$ and $F''$ in $V$ with $F''\subset L,M\subset F'$. Let $l$ be an automorphism of the $K$-vector space $V$, mapping a basis of $F'$ onto one of $F''$. For almost all $\mathfrak p$ the matrix representing $l$ has $\mathfrak p$-integral entries and a $\mathfrak p$-unit determinant. For all such $\mathfrak p$, $F_\mathfrak p'=F_\mathfrak p''$, hence $M_\mathfrak p\subset L_\mathfrak p$. Similarly for almost all $\mathfrak p$, $L_\mathfrak p\subset M_\mathfrak p$. This gives the result.
1) I know that I can choose free $\mathfrak o$-lattices $F'$ and $F''$ such that $F'\subset L\subset F''$ and similarly for $M.$ But how can we choose two such lattices which bound both $L$ and $M$?
2) How come the matrix representing $l$ have entries in the localization $\mathcal o_\mathfrak p$? I thought $l$ was $\mathfrak o$-linear. Does the book mean the tensored mapping $1\otimes l$ on $V_\mathfrak p=\mathcal o_\mathfrak p \otimes_\mathcal o V$?
3) Why is the above true for almost all $\mathfrak p$?
I'll be very grateful for answers to any of the questions.
1) Since both are lattices in the same $K$-vector space $V$ one can find nonzero elements $a,b\in \mathcal O$ such that $aM\subseteq L\subseteq b^{-1}M$: take a basis $(e_i)_{i=1,\dots,n}$ of $V$ contained in $M$; since $KL=V$ there is $x\in K$, $x\ne 0$ such that $xe_i\in L$ for all $i$, so $xM\subseteq L$. Now write $x=a/a'$ with $a,a'\in\mathcal O$ and conclude that $aM\subseteq L$. (For the other containment change $M$ by $L$.) Now should be easy to find two free modules $F'$ and $F''$ of the same rank with $F''\subset L,M\subset F'$.
2) $l$ is a matrix whose entries are in $K$. If localize $\mathcal O$ at a prime ideal $\mathfrak p$, then some denominators of the entries of $l$ are in $\mathfrak p$ and some others don't. But there are only finitely many denominators, so for almost all primes $\mathfrak p$ these are outside of $\mathfrak p$. The same holds for the determinant of $l$. Furthermore, we also have that the determinant has the numerator outside of $\mathfrak p$ for almost all $\mathfrak p$. Thus $l$ is an invertible matrix with entries in $\mathcal O_{\mathfrak p}$ for almost all $\mathfrak p$. It follows that the basis of $F'_{\mathfrak p}$ belongs to $F''_{\mathfrak p}$ for almost all primes $\mathfrak p$, that is, $F'_{\mathfrak p}=F''_{\mathfrak p}$ for almost all primes $\mathfrak p$.