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Let R be a commutative ring with identity element and no zero divisors. Let x∈R be an element with x⋅x = 1. How do I show that either x = 1 or x = -1?

Riccardo Carnevali
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    Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Shaun Mar 12 '22 at 21:24

1 Answers1

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Given that

$x^2 = x \cdot x = 1, \tag 1$

we have

$(x + 1)(x - 1) = x^2 - 1 = 0; \tag 2$

thus if

$x \ne 1, \tag 3$

or

$x - 1 \ne 0, \tag 4$

we have

$x + 1 = 0, \tag 5$

since $R$ has no zero divisors. From (5),

$x = - 1; \tag 6$

the case

$x = 1 \tag 7$

may be treated in a similar manner.

Robert Lewis
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  • More directly $,(x!+!1)(x!-!1)=0\Rightarrow x!+!1=0,$ or $,x!-!1=0\Rightarrow x=-1,$ or $,x = 1.\ $ Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Mar 13 '22 at 00:06