Let $R$ be a Dedekind domain. If $I$,$J$ are $R$-ideals, prove that there exists an $\alpha\in I$ such that $\gcd(\langle\alpha\rangle,IJ)=I$ Let $R$ be a Dedekind domain. If $I$,$J$ are $R$-ideals, prove that there exists an $\alpha\in I$ such that $\gcd(\langle\alpha\rangle,IJ)=I$.
Well if I could prove that there existed an $\alpha\in I$ and $j\in J$ such that $\alpha r_1 +jr_2=1$, then I would have $I\supseteq\gcd(\langle\alpha\rangle,IJ)=\langle\alpha\rangle+IJ\supseteq I\langle\alpha\rangle+IJ = I(\langle\alpha\rangle+J)=IR=I$.  But who knows if that is the case.  Honestly I have no idea what to do here, I tried looking at some examples but ideals in $\mathbb{Z}$ are too trivial and finding $\alpha$ in $\mathbb{Z}[\sqrt{-5}]$ is too complex (no pun intended).  Hopefully someone can help me with this, thanks.
 A: The problem can be rephrased as saying: find $\alpha$ such that $(\alpha, IJ) = I$, or again, showing that $I/IJ$ is principal as an ideal in $R/IJ$.  
To this end, prove that every ideal in $R/IJ$ is principal.  (Incidentally, you will need to assume that $IJ$ (and hence $IJ$) is non-zero.  Note that if $J =0$ then the statement is false unless $R$ is a PID, while if $I = 0$ the statement is trivial.)
There is nothing special about $IJ$ here; prove that for any non-zero ideal $I$ in $R$, every ideal in $R/I$ is principal.  For this, you might want to use the CRT, as Benjamin Lim suggested.
A: For a nonzero prime ideal $\mathfrak{p}$ of a Dedekind domain $R$ and a nonzero ideal $I$ of $R$, let $v_{\mathfrak{p}}(I)$ be the multiplicity to which $\mathfrak{p}$ appears in the prime ideal factorization of $I$.  
Step 1: Show that any finite set $\{ \mathfrak{p}_1,\ldots,\mathfrak{p}_n \}$ of prime ideals in $R$ and non-negative integers $a_1,\ldots,a_n$, there is $\alpha \in R \setminus \{0\}$ such that for all $1 \leq i \leq n$, $v_{\mathfrak{p}}( (\alpha)) = a_i$.  
(Suggestion: use the Chinese Remainder Theorem.)
Step 2: Let $\mathfrak{p}_1,\ldots,\mathfrak{p}_n$ be prime ideals containing all prime divisors of $I$ and $J$.  By Step 1, there is $\alpha \in R \setminus \{0\}$ such that for all $1 \leq i \leq n$, $v_{\mathfrak{p}_i}((\alpha)) = v_{\mathfrak{p}_i}(I)$.  Thus $(\alpha) = I K$ where $K$ is a nonzero ideal which is prime to both $I$ and $J$.  It follows that 
$\operatorname{gcd}((\alpha),IJ) = \operatorname{gcd}(IK,IJ) = I \operatorname{gcd}(K,J) = I$. 
