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

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.2 Exercise 7.2.4

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

Solution: Let $\alpha$ and $\beta$ be two nonzero elements in $R[[x]]$. Let $$\alpha=\sum_{n\geqslant 0}a_nx^n,\quad \beta=\sum_{n\geqslant 0}b_nx^n.$$Suppose $i$ is the smallest nonnegative integer $n$ such that $a_n\ne 0$ and $j$ is the smallest nonnegative integer $m$ such that $b_m\ne 0$. Then $a_i\ne 0$, $b_j\ne 0$ and $$\alpha=\sum_{n\geqslant i}a_nx^n,\quad \beta=\sum_{m\geqslant j}b_mx^m.$$Then it is clear that $$\alpha\beta=a_ib_jx^{i+j}+\text{terms with higher degree}.$$Because $R$ is an integral domain, we have $a_ib_j\ne 0$. So $\alpha\beta\ne 0$. Therefore the ring $R[[x]]$ of formal power series is also an integral domain.