Let $R$ be a commutative ring. Finding the units of $R[X]$ is not too difficult. We’ll do this in this notes.
Define the Nilradical $N$ as ${x\in R:\exists k\in\mathbb{N}, x^k=0}$. To see that this is an ideal, simply note that $x_1,x_2,\cdots,x_n\in N$ implies
\[y:=\sum_{i=1}^na_ix_i,\quad\Longrightarrow y^{nK}=0\]where $K\gg 0$ such that $x_i^K=0$, $\forall i$.
Now,
Theorem. $R[X]^\times=R^\times+N[X]$.
Proof. $\Longleftarrow$: WLOG let $f=1-h$ where $h\in N[X]$. Then $h$ is nilpotent by the same reason above. Clearly
\[f^{-1}=1+h+h^2+\cdots\in R[X].\]$\Longrightarrow$: Suppose $n\geq 1$ and
\[f=1+\cdots+a_{n-1}X^{n-1}+a_nX^n,\] \[g=1+\cdots+b_{m-1}X^{m-1}+b_mX^m,\] \[fg=1.\]Inductively, we can show that $a_n^k$ annihilates the set ${b_{m-k+1},\cdots,b_m}$. Indeed, this follows from
\[0=\sum_{i+j=n+m-k+1}a_ib_j,\quad 1\leq k\leq n+m.\](Coeffecients of negative degree are considered $0$ of course.) In particular, taking $k=m+1$ implies that $a_n^{m+1}$ annihilates $b_0=1$. Thus $a_n\in N$.
Notice that $f-a_nX^n$ is invertible in $R[X]$, because $1-a_n(f^{-1}X^n)\in R^\times+N[X]$. By induction of $\deg(f)$, we win.
Remark: The $\Longrightarrow$ direction has a one line proof by using mod $\mathfrak{p}$ reduction for each $\mathfrak{p}\in\operatorname{Spec}(R)$ and the fact that $N$ is the intersection of all prime ideals.
Post date: 2025/06/02
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