Prove that if is an integral domain then the ring of formal power series is also an integral domain.

Suppose is an integral domain. In the previous exercise, we saw that is a ring with 1. Let be in , with . Then the set is nonempty; let be minimal in this set. Now suppose is in and that . Recall that , so that for all we have .

We will now show by induction that for all .

For the base case, let . Since and when , we have . Since is an integral domain, .

For the inductive step, suppose that for some , whenever . Let . Since , and when and when , we have . Since has no zero divisors and , .

Thus for all , and we have . Thus has no right zero divisors. Similarly, it has no left zero divisors, and thus is an integral domain.

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By contrapositive: Let with be minimal nonzero term, each. Then the coefficient of is which are zero divisors in .

Very nice- I like this much better.