I read this in Algebraic Number Theory by A. Fröhlich & M. J. Taylor on p94:
$\mathfrak o$ is a Dedekind domain with field of fraction $K$. Let $L, M$ be finitely generated torsion free $\mathfrak o$-modules, $V$ a finite dimensional vector space over $K$, we say $M$ (same for $L$) is a $\mathfrak o$-lattice in $V$ if $M$ is contained in $V$ with $M$ containing a basis of $V$,
My questions are:
Why there is a fractional ideal $[L:M]$ and why it's unique?
And is the "equality" actually means $[L:M]_{\mathfrak p} = [L_{\mathfrak p}: M_{\mathfrak p}]\otimes_{\mathfrak o} \mathfrak o _{\mathfrak p}$? Since he left is a tensor but the right is not.
$\mathfrak o$ is a Dedekind domain and the other notations are defined below:
The fractional ideals of a discrete valuation ring $A$ are not very interesting. They are just the powers $P^k : k \in \mathbb{Z}$ of the unique maximal ideal $P$ of $A$. A Dedekind domain is an integral domain for which the localization at any maximal ideal is a DVR.
Let $I$ be a fractional ideal of $\mathfrak o$. You can identify $I_{\mathfrak p}$ with $I \otimes_{\mathfrak o} \mathfrak o_{\mathfrak p}$ if you want, but in any case $I_{\mathfrak p}$ is equal to (or can be identified with, whatever) a power $\mathfrak p^{\nu_{\mathfrak p}(I)} \mathfrak o_{\mathfrak p}$ of the unique maximal ideal $\mathfrak p \mathfrak o_{\mathfrak p}$ of the localized ring $\mathfrak o_{\mathfrak p}$, where $\nu_{\mathfrak p}(I)$ is some integer. You always have $\nu_{\mathfrak p}(I) = 0$ for all but finitely many primes $\mathfrak p$, can you can recover the ideal $I$ as a product of prime ideals that way: $$I = \prod\limits_{\mathfrak p} \mathfrak p^{\nu_{\mathfrak p}(I)}$$ Conversely, given a sequence of integers $\{ m_{\mathfrak p} : \mathfrak p$ is a prime of $\mathfrak o \}$, all but finitely many of which are zero, there is a unique fractional ideal $J$ of $\mathfrak o$ for which $$J = \prod\limits_{\mathfrak p} \mathfrak p^{m_{\mathfrak p}}$$ i.e. for which $J_{\mathfrak p} = \mathfrak p^{m_{\mathfrak p}} \mathfrak o_{\mathfrak p}$. That's exactly what they're doing here, each fractional ideal $[M_{\mathfrak p} : L_{\mathfrak p}]$ is equal to $\mathfrak p^{m_{\mathfrak p}} \mathfrak o_{\mathfrak p}$ for some integer $m_{\mathfrak p}$, almost all of which are zero. So they're letting $[M : L]$ just be the product $$\prod\limits_{\mathfrak p} \mathfrak p^{m_{\mathfrak p}}$$ Let me know if any of this is still unclear to you.