# Consequence of Chinese Remainder Theorem

The following proof comes from J.P. May's online notes on Dedekind Rings. The question is as follows.

Let $I$ be a non-zero ideal of a Dedekind domain $R$. Prove that there is an ideal $J$ such that $IJ$ is principal.

The proof given there goes as follows: Let $I=P_i^{r_i}\cdots P_n^{r_n}$, where $P_i$ are distinct maximal ideals and $r_i>0$. By Chinese remainder theorem, we have that $R/I$ is the product of the $R/P_i^{r_i}$. I understand till this point.

I do not follow how the CRT implies that if $b_i\in R-P_i^{r_i+1}$, then there exists an $a\in R$ such that $a-b_i\in P_i^{r_i+1}$ for all $i$.

I know Chinese Remainder Theorem as given in Proposition 1.10 of Atiyah-Macdonald.

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If it helps, this is called the approximation lemma. Pete Clark has a proof in his answer to this math.stackexchange.com/questions/143616/… –  Jack Schmidt Jun 3 '13 at 5:21
Isn't it just that you note that $R/(P_{1}\ldots P_{n}.I)$ is the direct product of the rings $R/P_{i}^{1+n_{i}}$ by what you already know –  Geoff Robinson Jun 3 '13 at 5:24
Let $f:\bigoplus_i R/P_i^{r_i+1}\to R/(\prod_i P_i^{1+r_i})$ be the isomorphism promised by CRT. Then $$f(\overline{b_1},\ldots,\overline{b_n})=\overline{a}$$ determines $a$ modulo $\prod_i P_i^{1+r_i}$. You do know that the inverse of $f$ is $$f^{-1}(\overline{a})=(\overline{a},\overline{a},\ldots,\overline{a}),$$ right? –  Jyrki Lahtonen Jun 3 '13 at 5:38
$R\mapsto R/P_1^{r_1+1}\times \cdots R/P_n^{r_n+1}$ is surjective by CRT. The element $(b_1,\cdots , b_n)$ is a non-zero element of the codomain. Thus we can find an element $a\in R$ with the given properties. –  messi Jun 3 '13 at 5:42

For me it's hard to understand May's point of view in this proof.

He defines a fractional ideal (of an integral domain $R$ with the field of fractions $K$) as being an $R$-submodule $J$ of $K$ for which there is a non-zero element $d\in R$ such that $dJ\subseteq R$. Furthermore, $J$ is called invertible if $JJ^{-1}=R$, where $J^{-1}=\{x\in K:xJ\subseteq R\}$. $R$ is a Dedekind domain if every non-zero ideal of $R$ is invertible.

Now let $I$ be a non-zero ideal in a Dedekind domain $R$. We have $II^{-1}=R$. By choosing $d\in I$, $d\neq 0$, we have $dI^{-1}\subseteq R$. Obviously $I(dI^{-1})=dR$ and setting $J=dI^{-1}$ we find an ideal $J$ of $R$ such that $IJ$ is principal.

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Yes this is what i observed and that is why i asked a question related to the above link. Thanks –  messi Jun 5 '13 at 9:23