Factorization of extension is injective Let A be a Dedekind domain with field of fractions K, and let B be the integral closure of A in a finite separable extension L of K. Now I want to show the map from Id(A) to Id(B) is injective. I know B is also a Dedekind domain, which means Id(B) is also an free abelian group.
Now if we have two different ideal from Id(A), then is it true that their extension in B are also different? 
 A: Since we are dealing with Dedekind domains, the map $I \mapsto IB$ from $\textrm{Id } A$ to $\textrm{Id } B$ is a homomorphism of groups, so it suffices to prove that the following result, which is true in much greater generality: $I$ is a fractional ideal of $A$, and $IB = B$, then $I = A$.  This result holds when $A$ is an integrally closed integral domain, and $B$ is the integral closure of $A$ in some field extension of $A$'s quotient field.
If $x \in I$, then $x \cdot 1 \in IB$, so $I \subseteq B$, and hence $I \subseteq A$, so you already know that $I$ is an integral ideal of $A$.  Suppose that $I$ is a proper ideal of $A$.  Then $I$ is contained in some maximal ideal $\mathfrak m$ of $A$, with $\mathfrak m B = B$ as well.  If we localize at $A \setminus \mathfrak m$, we then have $\mathfrak m B_{\mathfrak m} = B_{\mathfrak m}$
Since $B_{\mathfrak m} = \mathfrak m B_{\mathfrak m} = (\mathfrak m A_{\mathfrak m})B_{\mathfrak m}$, we can write $$1 = b_1x_1 + \cdots + b_tx_t$$ for some $b_i \in B_{\mathfrak m}$ and $x_i \in \mathfrak m  A_{\mathfrak m}$.  Since $B$ is the integral closure of $A$ in $L$, $B_{\mathfrak m}$ is the integral closure of $A_{\mathfrak m}$ in $L$, and in particular $b_1, ... , b_t$ are integral over $A_{\mathfrak m}$.  Hence the $A_{\mathfrak m}$-algebra $M := B_{\mathfrak m}[b_1, ... , b_t]$ is finitely generated as a module over $A_{\mathfrak m}$, and we also have $(\mathfrak m A_{\mathfrak m})M = M$ (because $(\mathfrak m A_{\mathfrak m})M$ is an ideal of $M$ containing the identity).
By Nakayama's lemma, $M = 0$, which is absurd.
If you want use the full machinery of Dedekind domains, you can argue much more simply as follows: since $IB = B$, $I$ must be an integral ideal of $A$.  Now since $IB$ is the identity of $\textrm{Id } B$,  $B = (IB)^{-1} = I^{-1}B$ (where here $(IB)^{-1}$ is the inverse operation in $\textrm{Id } B$, and $I^{-1}$ is the inverse operation in $\textrm{Id } A$).  Now again $I^{-1}B = B$, so $I^{-1}$ is an integral ideal of $A$.  The only way this can happen is if $I = A$.
