# Let R be UFD. Consider $R[[x]]$

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.

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

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 $$i, so $$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$$.