# Principal Ideal Domain Exercise

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 $R$, $a_i$.

• ...and $(a_i)\supset (a_{i+1})$ if and only if what property holds about $a_i$ and $a_{i+1}$...?
– Matt
Oct 10 '12 at 3:56
• Hint: How does the number of irreducible factors of $a_i$ compare to that of $a_{i + 1}$? How many irreducible factors must a nonzero element of $I_i = (a_i)$ have? How many must a nonzero element contained in every $I_i$ have? Oct 10 '12 at 3:59
• 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! Oct 10 '12 at 4:14

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.