I shall prove some properties on a polynomial ring $R[x]$ over a commutative ring $R$, and there are two with which I struggle:
For some $f=a_0+a_1x+\cdots+a_nx^n\in R[x]$,
$f$ is invertible in $R[x] \Leftrightarrow a_0$ is invertible, and $a_1,…,a_n$ are nilpotent.
$fg$ is primitive $\Leftrightarrow f$ and $g$ are primitive, where primitive means: $(a_0,…,a_n)=(1)$.
The first one is quite intuitive and $a_0$ must be invertible since the first coefficient of the unit-polynomial shall be $1$, but I do not follow the nilpotency of the other coefficients.
Can anyone help?
Best Answer
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).