1
$\begingroup$

Let $R$=ring with identity $1$. If $d^2=a$ for all $d \in R$,The only unit in $R$ is $1$...show this.

I was thinking $$1\cdot d^2=d^2\cdot 1$$ which equals $1\cdot d=d\cdot 1$.

I found out a unit means that $uv=vu=1$.

$\endgroup$
9
  • $\begingroup$ Something's up here. If a ring has $1$, then by definition it also has $0$ and $-1$. $\endgroup$
    – Stuck
    May 31, 2021 at 6:29
  • 2
    $\begingroup$ @innerproduct note you can have $1=-1$ $\endgroup$ May 31, 2021 at 6:42
  • $\begingroup$ You need to work with the definition of a unit. $\endgroup$ May 31, 2021 at 6:45
  • 3
    $\begingroup$ @innerproduct We need $0$ but it’s not a unit (except in the zero ring), so no claim about it is made in the question. The equation $1=-1$ holds in some rings, in particular in all rings with the additional property that $a^2=a$ for all $a\in R$. $\endgroup$ May 31, 2021 at 7:11
  • 1
    $\begingroup$ @innerproduct The definition of a ring gives you $0$ of course, but the proof here does not require it. With the idempotent property it is easy to see that $1=(1)^2=(-1)^2=-1$. Think about the ring (even field) with two elements. I said "you can have" not "every ring must have". The rings in this question are Boolean Rings and all have $1=-1$ $\endgroup$ May 31, 2021 at 7:13

2 Answers 2

4
$\begingroup$

If $a$ is invertible, then $1=aa^{-1}=a^2a^{-1}=a$.

$\endgroup$
1
$\begingroup$

Suppose $a \in R$ is a unit. Then there exists $b \in R - \{0\}$ such that $ab = 1.$ By the definition of $R,$ we have $a^2 = a,$ and so $a(a-1) = 0.$ But then $a - 1 = ab(a-1) = ba(a-1) = 0,$ hence $a = 1.$

(Note that $R$ must be commutative, so the order of multiplication does not matter.)

$\endgroup$
2
  • $\begingroup$ where is the a-1 coming from? $\endgroup$
    – shaylene
    May 31, 2021 at 6:52
  • $\begingroup$ @Amanda if $a^2 = a,$ then $0 = a^2 - a = a(a-1).$ $\endgroup$ May 31, 2021 at 6:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .