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While solving another problem (specifically Exercise 7.2 in Atiyah & Macdonald's Introduction to Commutative Algebra), I got stuck in the following step:

If $R$ is an integral domain, how I can prove that $R[[x]]$ is an integral domain?

Here $R[[x]]$ is the set of all formal series in $x$ with coefficients in $R$. So typical element of $R[[x]]$ would have the form $a_0+a_1x+a_2x^2+\cdots$ where $a_i\in R$.

So I need to prove that if $$ (a_0+a_1x+a_2x^2+\cdots)(b_0+b_1x+b_2x^2+\cdots)=0 $$ then $a_i=0$ and $b_i=0$ for all $i\ge 0$. Now, I am not particularly fond of opening up those brackets :( Is there any slick way of proving this?

Prism
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  • How about using the well ordering principle on the coefficients?, its easy to get a contradiction that way – ADR Aug 01 '13 at 19:28
  • @ADR: Hmm... Since $R$ is arbitrary integral domain, it may not well-ordered, so I am sure you mean totally something else... – Prism Aug 01 '13 at 19:34
  • @Prism I meant on the coefficients, like let $a_ib_i$ minimum respect to that $a_i \neq 0$ or $b_i \neq 0$ – ADR Aug 01 '13 at 19:39
  • @ADR: Oh yes! Thanks. I think that particular idea is employed below in BenjaLim's argument. – Prism Aug 01 '13 at 19:41

3 Answers3

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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.

  • All the answers I got are nice, but this one seems to be the slickest in my opinion. – Prism Aug 01 '13 at 19:17
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    @Prism I'd just like to add to BenjaLim's excellent answer that the idea here is similar to the proof that the polynomial ring over an integral domain is an integral domain. In that proof, we use the notion of the degree of a polynomial. In this case, we're using the notion of the order of a formal power series. Indeed, BenjaLim proved that $\text{ord}(fg)=\text{ord}(f)+\text{ord}(g)$ for formal power series $f$ and $g$ just as $\text{deg}(fg)=\text{deg}(f)+\text{deg}(g)$ for polynomials $f$ and $g$. – Amitesh Datta Aug 02 '13 at 08:30
  • @AmiteshDatta: Thanks for elucidating on the key point. :) – Prism Aug 02 '13 at 18:41
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    The $(i+j)$-th term is going to be $(a_0b_{i+j}+a_1b_{i+j-1}+\cdots+a_{i+j}b_0)x^{i+j}$. How do we know this will be non-zero? – user5826 Jun 06 '18 at 00:29
  • @AlJebr I know I am two years late to your comment, but we only care about showing that there is at least one non-zero term. Some other coefficients can be zero. The argument shows that there is at least one non-zero term (namely the first one $a_i b_j$), and thus the product is non-zero. Note that in the solution $i$ and $j$ were chosen carefully. – Prism May 21 '20 at 06:58
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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 $n$, $a_n=0$.

PVAL-inactive
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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 $a_{n+1}=0$.

Jared
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