2
$\begingroup$

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 :)

$\endgroup$
4
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ It is a little bit easier to consider $1-ax$. $\endgroup$ – Martin Brandenburg Nov 1 '14 at 10:46
  • 1
    $\begingroup$ 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 $a\ne 0$. $\endgroup$ – PVAL-inactive Nov 1 '14 at 18:34
3
$\begingroup$

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 $R^\times$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.