## 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.

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 $R$.