If $R$ is an integral domain, then $R[[x]]$ is an integral domain

Solution 1:

Choose $f,g \in R[[x]]$ that are non-zero. With $f = a_0 + a_1x + \ldots$ and $g = b_0 + b_1x + \ldots$ let $a_i$ and $b_j$ be respectively the coefficients of the smallest non-zero terms in $f$ and $g$. Then $fg = a_ib_jx^{i+j} + \text{(higher order terms)}$ and thus is non-zero. Done.

Solution 2:

Since $(a_0+a_1x+a_2x^2+\cdots)(b_0+b_1x+b_2x^2+\cdots)=0$, the term of degree $0$ must be zero. Therefore the constant coefficient $a_0b_0=0$, assume $b_0 \ne 0$. Then $a_0=0$ and the $x$ coefficient is $a_1b_0$ so $a_1=0$. Now try to find some strong induction to show for any $n$, $a_n=0$.

Solution 3:

Notice we only have to prove that if the $b_i$ are not all $0$, then the $a_i$ are all $0$.

Just use induction. First, let $m$ be minimal with $b_m\ne 0$, so that $a_0=0$ by looking at the coefficient of $x^m$ in the product. Next, suppose that $a_k=0$ for all $k=0,\ldots,n$, and notice that the coefficient of $x^{m+n+1}$ in the product is $a_{n+1}b_m$. Since $b_m\ne 0$, it follows that $a_{n+1}=0$.