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original poster
hbghlyj
posted 2025-8-11 17:06
In a PID, two ideal-quotient equalities $(K:I)=J$ and $(K:J)=I$ force $K=IJ$.
Proof: write $I=⟨a⟩, J=⟨b⟩, K=⟨c⟩$. In a PID,
$$
\big(⟨x⟩:⟨y⟩\big)=\{r: ry\in⟨x⟩\}=⟨x/\gcd(x,y)⟩.
$$
Our hypotheses
$$
(K:I)=J,\qquad (K:J)=I
$$
become
\begin{align*}
(c:a)&=⟨b⟩\Rightarrow⟨c/\gcd(c,a)⟩=⟨b⟩,\\
(c:b)&=⟨a⟩\Rightarrow⟨c/\gcd(c,b)⟩=⟨a⟩.
\end{align*}
Thus, up to a unit,
$$
c=b\gcd(c,a)=a\gcd(c,b).
$$
From $c=b\gcd(c,a)$ we get $\gcd(c,b)=\gcd(b\gcd(c,a),b)=b$. Plugging into the second display gives $c\sim ab$. Hence $⟨c⟩=⟨ab⟩$, i.e.
$$
K=IJ.
$$
(Aside: the same implication holds in any ring where $I$ and $J$ are invertible ideals, e.g. Dedekind domains: from $(K:I)=J$ multiply by $I$ to get $K=IJ$. In a PID every nonzero ideal is invertible.) |
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