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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.

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Have you tried using the Chinese Remainder Theorem? – user38268 Dec 22 '12 at 0:22
I have not, I'm not understanding how it applies. – esproff Dec 22 '12 at 0:28
Maybe I was thinking of something else. Though what are $j_1,j_2$ above? – user38268 Dec 22 '12 at 0:31
You mean $r_1,r_2$? They are just arbitrary elements of $R$. – esproff Dec 22 '12 at 0:33
up vote 2 down vote accepted

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

$\operatorname{gcd}((\alpha),IJ) = \operatorname{gcd}(IK,IJ) = I \operatorname{gcd}(K,J) = I$.

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Of course Matt E's answer is perfectly correct. I decided to add a different one. – Pete L. Clark Dec 22 '12 at 0:57
+1 For your nice answer. – user38268 Dec 22 '12 at 4:37

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.

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