To show a particular element is the unit of a ring

Question

$Problem$

Suppose that a and b belong to a commutative ring $R$ with unity. If $a$ is the unit of $R$ and $b^2 = 0$; Show that $a+b$ is a unit of $R$.

$ Attempt$

$(a+b)(a-b) a^{-2}= (a^2-b^2)a^{-2}= 1$.

How to show that $a+b$ is a unit if we replace $b^2$ with $b^n$

I have just started ring theory after finishing Group theory. Any hint or suggestion will be appreciated.

$(a+b)(a-b)=a^2-b^2$ (note: this uses the assumption that $R$ is commutative. It is not necessarily true in general rings) — JMoravitz, Jul 10 '18 at 00:42
If you feel the duplicate is too general, check out any of the other 18 questions linked to it. And please try the search feature first next time. Thanks. — rschwieb, Jul 10 '18 at 02:53

score 4 · Answer 1 · answered Jul 10 '18 at 00:48

In general, even if you just have $b^n=0$, let $c=-a^{-1}$ and write $a+b$ as $a(1-cb)$ and $1=(1-cb)(1+cb+c^2b^2+\cdot\cdot\cdot)$, where the sum is finite because eventually the terms all vanish, so $1-cb$ as an inverse and thus so does $a+b=a(1-cb)$.

In general, when you have nilpotent elements floating around, remembering power series representations like $\frac{1}{1-x}=1+x+x^2+\cdot\cdot\cdot$ is very useful.

To show a particular element is the unit of a ring

1 Answers1