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Show that an integral domain $A$ is a unique factorization domain if and only if every ascending chain of principal ideals terminates, and every irreducible element of $A$ is prime.

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What have you tried? –  Alex Becker May 9 '12 at 7:25
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Dear Mariam, it is best if you tell us what you have tried and where you are stuck. Otherwise, this feels too much like we are solving problems from a book! –  Mariano Suárez-Alvarez May 9 '12 at 8:34
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ACCP gives you existence. The other condition gives you uniqueness. –  lhf May 9 '12 at 18:55
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Hint: To show the "if" part, observe that if $a\in A$ is not irreducible, then $(a)\subset (b)$ for some $b\in A$, and continue in this manner to get an ascending chain of principle ideals. Do you see how this gives you a factorization of $a$ into irreducible elements (assuming the chain terminates)? Showing uniqueness is then fairly easy. Suppose $a_1\cdots a_n=b_1\cdots b_m$ where $a_1,\ldots,a_n,b_1,\ldots,b_m$ are prime. Then each $a_i$ divides one of the $b_j$, but these are irreducible so the two must be equal. For the "only if" part, note that every irreducible is prime and factoring an element $a$ into primes gives you all the principle ideals containing $(a)$, of which there are only finitely many.

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