fractional ideals in the localization of a Dedekind I'm reading Janusz, Algebraic number fields, 1973, pag.16-17, where defining a fractional ideal of a Dedekind domain $R$. A fractional ideal of $R$ is a non-zero finitely generated $R$-submodule $\mathfrak{M}$ of $K$, $K$ being the quotient field of $R$. And if $\mathfrak{M}$ is a fractional ideal of $R$, we define $\mathfrak{M}^{-1}:=\{x\in K:x\mathfrak{M}\subseteq R\}$, also a fractional ideal of $R$.
I can't understand this part:

EXERCISE. Multiplication and inversion behave properly with respect to
  localization. That is, if $\mathfrak{p}$ is a prime ideal of $R$ and $\mathfrak{M}$ a fractional ideal of $R$, then $\mathfrak{M}R_{\mathfrak{p}}$ is a fractional ideal of $R_{\mathfrak{p}}$, and $(\mathfrak{M}R_{\mathfrak{p}})^{-1} = \mathfrak{M}^{-1}R_{\mathfrak{p}}$. Also $(\mathfrak{M}\mathfrak{N})R_{\mathfrak{p}}=(\mathfrak{M}R_{\mathfrak{p}})(\mathfrak{M}R_{\mathfrak{p}})$ for $\mathfrak{M},\mathfrak{N}$ fractional ideals of R.

But, what is $\mathfrak{M}R_{\mathfrak{p}}$? There is no definiton in the book, before this exercise. I know what is the extended ideal $\mathfrak{m}R_{\mathfrak{p}}$ for an ideal (in the usual sense) $\mathfrak{m}$ of $R$, but for fractional ideals i think it doesn't make sense.
 A: $\mathfrak{M} R_{\mathfrak{p}}$ is the $R_{\mathfrak{p}}$-submodule of $K$ generated by $\mathfrak{M}$.  This is pretty close to the definition of the extension of an ideal $I$ of $R$ under a homomorphism of rings $f: R \rightarrow S$, which is the $S$-submodule of $S$ generated by $f(I)$.  
As an aside, Janusz's definition of a fractional ideal is "wrong", in the sense that it does not agree with the standard definition of a fractional ideal over an abitrary integral domain (though it is equivalent for Noetherian domains, hence Dedekind domains; still, I don't see why one should internalize the wrong definition for the general case).  For any domain $R$ with fraction field $K$, a fractional $R$-ideal is an $R$-submodule $M$ of $K$ such that there is $a \in R \setminus \{0\}$ with $a M \subset R$.  Then it is easy to see:
$\bullet$ Any finitely generated $R$-submodule of $K$ is a fractional $R$-ideal.  (Take $a$ to be the product of the denominators of a finite generating set.)  
$\bullet$ Any "integral" ideal of $R$ is a fractional $R$-ideal.  (Thus if $R$ is not Noetherian this definition is more permissive.)
Thus fractional ideals are submodules of $K$ of the form $\frac{1}{a}I$ for an integral ideal $I$.  Thus, if $I_{\mathfrak{p}} = I R_{\mathfrak{p}}$ is the extended/localized integral ideal, then the extended/localized fractional ideal is just $(\frac{1}{a}I) R_{\mathfrak{p}} = \frac{1}{a} I_{\mathfrak{p}}$.  
(As a final thought: for any fractional ideal $I$ in a domain $R$ one can define $I^* = \{x \in K \ | \ xI \subset R\}$.  Then tautologously one has $I I^* \subset R$; 
one says $I$ is invertible if there is equality.  This occurs iff there is some fractional ideal $J$ with $IJ = R$.  Dedekind domains are characterized among integral domains by the property that all nonzero fractional ideals are invertible.  In any domain an invertible ideal must be finitely generated.) 
