Invertible and primitive polynomials

Solution 1:

Here are indications for 1.

1) Suppose that $f$ is invertible
i) As you already know $a_0$ is invertible.
ii) Consider a prime ideal ${\mathfrak p } \subset R$. If $fg=1$ then $ \bar f\bar g=\bar 1$ in $(R/{\mathfrak p }) [x]$.
But in a domain a polynomial ( like here $\bar f$) can only be invertible if it is constant ( what is the degree of a product of polynomials in a domain? )
So every $\bar a_i, \; i\geq1$ satisfies $\bar a_i=0 $ and thus $ a_i\in {\mathfrak p }$ for all prime ideals in $R$. This shows that $a_i$ is nilpotent.

2) Suppose $a_0$ invertible and the other $a_i$ nilpotent
Use that if $a$ is invertible and $n$ nilpotent then $a+n$ is invertible (and to prove this statement try $a=1$ first).

Solution 2:

For 2: write $f=a_0+\cdots+a_nx^n$, $g=b_0+\cdots+b_mx^m$, $fg=c_0+\cdots+c_d x^d$, with $d\leq n+m$, $a_n\neq 0$, $b_m\neq 0$, $c_d\neq 0$.

Let $\mathfrak{m}$ be a maximal ideal of $R$; if $f$ and $g$ are each primitive, then there exists $i,j$ such that $a_i\notin\mathfrak{m}$ but $a_{0},\ldots,a_{i-1}\in\mathfrak{m}$; and $b_j\notin\mathfrak{m}$, but $b_0,\ldots,b_{j-1}\in\mathfrak{m}$. Then $$\begin{align*} c_{i+j} &= a_0b_{i+j} + \cdots + a_{i-1}b_{j+1}\\ &\quad\mathop{+} a_ib_j\\ &\quad\mathop{+}a_{i+1}b_{j-1}+\cdots + a_{i+j}b_0. \end{align*}$$ Every term on the first line of the right hand side is in $\mathfrak{m}$, and every term on the last line of the right hand side is in $\mathfrak{m}$; so $c_{i+j}\in\mathfrak{m}$ if and only if $a_ib_j\in\mathfrak{m}$. Hence...

Conversely, note that $(c_0,c_1,\ldots,c_d)\subseteq (a_0,\ldots,a_n)(b_0,\ldots,b_m)$. So if $fg$ is primitive, then...