# $R$ commutative ring having a non-zero nilpotent, then $R^{\times}\subsetneq (R[X])^{\times}$

Let $R$ be a commutative ring. Also there is $a\in R$, $a\ne 0$, such that $a^n=0$. Then $R^{\times}\subsetneq(R[X])^{\times}$, so there is an element in $(R[X])^{\times}$ which is not contained in $R^{\times}$

I am struggling in finding such an element in $(R[X])^{\times}$ which is not contained in $R^{\times}$.

Any tips? Thanks :)

Consider $1+ax$. More generally see Chapter 1 Exercises 1-2 of Introduction to Commutative Algebra by Atiyah and Macdonald .

• It is a little bit easier to consider $1-ax$. – Martin Brandenburg Nov 1 '14 at 10:46
• When I first did this problem, I used the difference of squares rule instead of the identity in the other answer. You can also do it by inducting on the degree of nilpotence as $(1+ax)(1-ax)=1-a^2x$ and $a^2$ is strictly less nilpotent then $a$ as long as $a\ne 0$. – PVAL-inactive Nov 1 '14 at 18:34

EDIT: If you just want a hint only read the next two lines.

Consider the general identity

$$A^n - B^n = (A-B)(A^{n-1}+ A ^{n-2}B + ...+AB^{n-2}+B^{n-1})$$

which holds in any commutative algebra. Now let $A=1$ and $B=aX$, where $a^n = 0$ in $R$. Since $(aX)^n = a^n X^n = 0$, we have

$$1 = 1-0 = 1^n - (aX)^n = (1-aX)(1+aX+a^2X^2+...+a^{n-1}X^{n-1}).$$

This shows $1-aX \in R[X]^\times$, so $R[X]^\times$ has elements that are not in $R^\times$.