let $f=\sum_{n=0}^\infty a_nx^n$ and $g =\sum_{n=0}^\infty b_nx^n$. Suppose the gcd of $a_n$ and $b_n$ are both 1. Then show the gcd of the coefficients of $fg$ is also one.

I know that since R is in particular a domain, then $R[[x]]$ is also domain. However, I'm not really sure how to proceed further.

  • $\begingroup$ The g.c.d. of which $ a_n $ and $b_n$? $\endgroup$ – Bernard Mar 18 at 0:58
  • $\begingroup$ I mean the gcd of all the $a_n$'s is one and the same for all the $b_n$'s $\endgroup$ – davidh Mar 18 at 1:00
  • $\begingroup$ @Bernard He wants to prove that the product of two primitive power series is primitive. $\endgroup$ – user26857 Mar 20 at 10:25

I like the proof of @JohnBrevik fine, but here’s my version, which you may find more visual:

For a prime $p$ of $R$, the ring $R/(p)$ is an integral domain. If we have a series $f\in R[[x]]$ let $\tilde f$ be the corresponding series in $R/(p)$. In case $p$ does not divide all coefficients of $f$, the series $\tilde f$ will have an initial degree $n$ so that $\tilde f(x)=x^n\varphi(x)$ with $\varphi(0)\ne0$, i.e. $\varphi$ has nonzero constant term.

Now under your hypotheses on $f$ and $g$, if $p$ is any prime of $R$, then we have $\tilde f(x)=x^n\varphi(x)$ and $\tilde g=x^m\psi(x)$ where as above, $\varphi(x)$ and $\psi(x)$ have nonzero constant terms. Since $R/(p)$ is an integral domain, $\varphi\psi$ will also have nonzero constant term. And $\widetilde{fg}(x)=\tilde f(x)\tilde g(x)=x^{m+n}\varphi(x)\psi(x)$, whose $x^{m+n}$-coefficient is nonzero. Thus $fg$ has a coefficient indivisible by $p$. Do this for all $p$, and get your result.


Suppose that the coefficients of $fg$ have a common prime factor $p\in R$. One of $a_0$ and $b_0$, WLOG $a_0$, is divisible by $p$. Let $k$ be the smallest index with the property that $a_k$ is not divisible by $p$ (such $k$ must exist). Then every term $a_ib_{k-i}$ contributing to the $x^k$ coefficient of $fg$ except $a_kb_0$ is divisible by $p$. Since the coefficient is divisible by $p$ by assumption, $p\mid b_0$. Now prove by induction that $p\mid b_j$ for all $j$: Supposing $p\mid b_0, p\mid b_1, \dots, p\mid b_{j-1}$, examine the $x^{k+j}$ coefficient of $fg$ as $\sum a_i b_{k+j-i}$. For $i<k, p\mid a_i$, so $p\mid a_i b_{k+j-i}$. For $i>k, p\mid b_k+j-i$ and so again $p\mid a_i b_{k+j-i}$. This leaves only the contributing term when $i=k$. By hypothesis $p\nmid a_k$, and so in order for $p$ to divide the coefficient of $fg$, $p$ must divide $b_{k+j-k} = b_k$.


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