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Let $R$ be a Dedekind Domain and $I=P^{a_{1}}_{1}\cdot\cdot\cdot P^{a_{n}}_{n}$ an ideal of $R$. I'm trying to understand the proof that every nonzero ideal in $R/I$ is principal. In particular, why it suffices by the Chinese Remainder Theorem to prove that every Ideal $R/P^{m}$ is principal for powers of prime ideals. If this is the case and $J/I$ is an ideal in $R/I$, then $J/I \cong J/P^{a_{1}}_{1}\times\cdot\cdot\cdot\times J/P^{a_{n}}_{n}$ where each factor is principal. But how does that imply that $J/I$ is principal?

SihOASHoihd
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1 Answers1

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If one has a direct product of commutative rings $$S=S_1\times\cdots\times S_n$$ then the ideals of $S$ are precisely the sets $$I=I_1\times\cdots\times I_n$$ where each $I_j$ is an ideal of $S_j$. If all the $I_j$ are principal: $I_j=Rb_j$, then $I$ is principal, generated by $(b_1,\ldots,b_n)\in S_1\times \cdots\times S_n$.

Here $S_j=R/P_j^{a_j}$, and all the ideals of $S_j$ are principal, so all ideals of $S$ are too.

Angina Seng
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  • I guess I'm just not seeing how $(b_{1},...,b_{n})$ is a generator. There must be something missing at the moment in my understanding of ideals. Suppose $n=2$ and $r$ and $s$ are distinct ring elements. Then how can $(rb_{1},sb_{2})$ be in $(b_{1},b_{2})$? – SihOASHoihd Jun 24 '18 at 16:12
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    $(rb_1,sb2)=(r,s)(b_1,b_2)$ – Angina Seng Jun 24 '18 at 16:25