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Let $R$ be a principal ideal domain and consider an infinite strictly decreasing chain of ideals in $R$, say $I_1\supset I_2\supset \dots$. Show that $\cap_{i=1}^{\infty} I_i =(0)$

(I took down my first attempt, and heres where I am so far)

My attempt at a proof:

Since $R$ is a prinicpal ideal domain, every ideal in $R$ is a principal ideal. This implies that for an infinite strictly decreasing chain of ideals in $R$, $I_1\supset I_2\supset \dots$. each $I_i$, where $i\in\mathbb{N}$, is generated by a single element of $R$, $a_i$.

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    So $a_i$ must have less irreducible factors than $a_{i+1}$, and a nonzero element contained in $I_i$ must have at least the number of factors needed to construct $a_{i+1}$, so any nonzero element contained in every $I_i$ must have an infinite number of primes!2012-10-10

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(Did not realize that I had to wait for two days before I could close this question off)

Since $(a_1)\supset (a_2)$, it follows that $a_1|a_2$ and $a_2$\not|a_1$, and any element $b$ of $I_2$ and $I_1$ must have at least the number of irreducible factors of $a_2$ since $a_2|b$, which is greater than the number of irreducible factors of $a_1$. If $I_i$ is generated by $a_i$ and the number of irreducible factors of $a_i$ is greater than $a_{i-1}$, and $I_{i}\supset I_{i+1}$, where $I_{i+1}$ generated by $a_{i+1}$, than $a_i|a_{i+1}$ and $a_{i+1}$\not|a_{i}$. This implies that $a_{i+1}$ has more irreducible factors than $a_i$, and that any element of $I_{i+1}$ must contain at least the number of factors of $a_{i+1}$. By induction, any non zero element of $\cap_{i=1}^{\infty}I_i$ must contain an infinite number of factors, contradiction.