Ring of sequences on integral domain must be integral domain

Question

Let $R$ be a non-trivial commutative unital ring and $S$ be defined as $S:=\{\{a_n\}_0^\infty:a_n\in R\}$ under the operations of addition and sequence convolution. Then if $R$ is an integral domain then $S$ is an integral domain.

Here's how I proceed:

Since $R$ is an integral domain, it has no zero divisors, which means that if $ab=ac$, then $b=c$, for $a,b,c\in R$. Suppose $\{a_n\}*\{b_n\}=\{a_n\}*\{c_n\}$, then

$$\left\{\sum_{i+j=n}a_i b_j\right\}_{n=0}^\infty=\left\{\sum_{i+j=n}a_i c_j\right\}_{n=0}^\infty$$

I'm then trying to proceed by induction:

$\left\{\sum_{i+j=n}a_i b_j\right\}_{n=0}^\infty=\left\{\sum_{i+j=n}a_i b_j\right\}_{n=0}^0+\left\{\sum_{i+j=n}a_i b_j\right\}_{n=1}^N+\left\{\sum_{i+j=n}a_i b_j\right\}_{n=N+1}^\infty$.

Base case: $\left\{\sum_{i+j=n}a_i b_j\right\}_{n=0}^0 = a_0b_0=a_0c_0 =\left\{\sum_{i+j=n}a_i c_j\right\}_{n=0}^0$, so $b_0 = c_0$. Assume that $\left\{\sum_{i+j=n}a_i b_j\right\}_{n=1}^N=\left\{\sum_{i+j=n}a_i c_j\right\}_{n=1}^N$, then:

$$\left\{\sum_{i+j=n}a_i b_j\right\}_{n=1}^{N+1}=\left\{\sum_{i+j=n}a_i b_j\right\}_{n=1}^N+\left\{\sum_{i+j=n}a_i b_j\right\}_{n=N+1}^{N+1}=\left\{\sum_{i+j=n}a_i c_j\right\}_{n=1}^N+a_0b_{N+1}+a_1b_N+...+a_{N+1}b_0.$$

This is where I'm stuck. How do I prove that $\left\{\sum_{i+j=n}a_i b_j\right\}_{n=N+1}^{N+1}$ satisfies the property of the integral domain?

Or am I moving on a wrong path?

Answer 1

Your proof is set up wrong in several ways. First, you need to assume that $a\neq 0$, and you will need to find a way to make use of this assumption (if $a=0$, then $ab=ac$ does not imply $b=c$, even in a domain). In your base case, you cannot conclude that $b_0=c_0$, since $a_0$ could be $0$. Your induction hypothesis is wrong. You are proving that $b_n=c_n$ for all $n$, so your induction hypothesis should be that $b_n=c_n$ for all $n\leq N$.

This question is considerably easier to think about if instead you directly use the definition of a ring having no zero divisors: namely, that $ab=0$ implies $a=0$ or $b=0$. Supposing $a\neq 0$ and $b\neq 0$, you can prove $ab\neq 0$ by considering the least $m$ such that $a_m\neq 0$ and the least $n$ such that $b_n\neq0$, and showing that the $(m+n)$th term of $ab$ will be nonzero.

(Incidentally, your ring $S$ is usually called $R[[x]]$, the ring of formal power series with coefficients in $R$.)