Let $R$ be a commutative ring with unity, and let $f(X),g(X)\in R[X]$. Assume the ideals generated by the coefficients of $f(X),g(X)$ are both $R$. Prove that the ideal generated by the coefficients of the product $f(X)g(X)$ is also $R$.
Let $$f(X)=a_nx^n+\cdots +a_0, \\g(X)=b_nx^n+\cdots + b_0$$ (I am not assuming that they have the same degree - $a_n$ or $b_n$ can be $0$). We are given $$(a_n,\ldots,a_0)=(b_n\ldots, b_0)=R$$ Now $$f(X)g(X)=\sum_{i=0}^n\Bigg(\sum_{j+k=i}a_jb_k\Bigg)x^i$$ so the ideal generated by the coefficients is $I=(a_0b_0, a_1b_0+a_0b_1,\ldots, \sum_{j+k=n}a_jb_k)$. How can one show $I=R$? My only strategy (Update 1: Thomas' answer shows this is incorrect) is that $I=R$ because $$I=Ra_0b_0+R(a_1b_0+a_0b_1)+\cdots + R\sum_{j+k=n}a_jb_k\\=Ra_0+Ra_1+\cdots + Ra_n\\=R+R+\cdots +R\\=R$$ where in the 3rd expression I am distributing the $R$ through each sum and using the fact that $Rt=R$ for any $t\in R$.
Update 2: Since $(a_1,\ldots, a_n)=(b_1,\ldots, b_n)=R$, there exists $r_0,\ldots , r_n, r_0',\ldots, r_n'\in R$ such that $r_0a_0+\cdots r_na_n =1, r_0'b_0+\cdots + r_n'b_n=1$, so multiplying these two equations gives $$r_0r_0'a_0b_0 + r_0r_1'a_0b_1 + r_1r_0'a_1b_0 + \cdots=1 \tag{$*$}$$ but the left hand side isn't an element of $I$ since $a_1b_0 + a_0b_1$ have distinct coefficients in $(*)$. I am unsure if this can be modified.
Write $f = a_0 + ... + a_mX^m$ and $g = b_0 + ... + b_nX^n$. It suffices to show that for each maximal ideal $M$, there is a coefficient of $fg$ not in $M$. Let $k$ and $l$ be minimal with $a_k,b_l \notin M$. Check that the coefficient of $X^{k+l}$ is not in $M$.