The ring of formal power series over an integral domain is an integral domain

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

Suppose R is an integral domain. In the previous exercise, we saw that R[[x]] is a ring with 1. Let \alpha = \sum_{n \geq 0} a_n x^n be in R[[x]], with \alpha \neq 0. Then the set \{ i \in \mathbb{Z} \ |\ i \geq 0, a_i \neq 0 \} is nonempty; let k be minimal in this set. Now suppose \beta = \sum_{n \geq 0} b_n x^n is in R[[x]] and that \alpha\beta = 0. Recall that \alpha\beta = \sum_{n \geq 0} (\sum_{i+j=n} a_ib_j) x^n, so that for all n we have \sum_{i+j=n} a_ib_j = 0.

We will now show by induction that b_j = 0 for all j.

For the base case, let n = k. Since \sum_{i+j=k} a_ib_j = 0 and a_i = 0 when i < k, we have a_kb_0 = 0. Since R is an integral domain, b_0 = 0.

For the inductive step, suppose that for some t \geq 0, b_j = 0 whenever 0 \leq j \leq t. Let n = k+t+1. Since \sum_{i+j=k+t+1} a_ib_j = 0, and a_i = 0 when 0 \leq i < k and b_j = 0 when 0 \leq j \leq t, we have a_kb_{t+1} = 0. Since R has no zero divisors and a_k \neq 0, b_{t+1} = 0.

Thus b_n = 0 for all n, and we have \beta = 0. Thus \alpha has no right zero divisors. Similarly, it has no left zero divisors, and thus R[[x]] is an integral domain.

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  • Gobi Ree  On January 17, 2012 at 8:14 pm

    By contrapositive: Let f(x)g(x)=0 with a_nx^n, b_mx^m be minimal nonzero term, each. Then the coefficient of x^{n+m} is a_nb_m=0 which are zero divisors in R.

    • nbloomf  On August 18, 2012 at 9:39 am

      Very nice- I like this much better.

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