How Many Solutions for $x+x=1$ in a Ring?

Problem 204

Is there a (not necessarily commutative) ring $R$ with $1$ such that the equation
$x+x=1$ has more than one solutions $x\in R$?

Solution.

We claim that there is at most one solution $x$ in the ring $R$.
Suppose that we have two solutions $r, s \in R$. That is, we have
$r+r=1 \text{ and } s+s=1.$

Then we have $r+r=s+s$. Putting $a=r-s\in R$, we have
$a+a=0.$ Now we compute
\begin{align*}
0&=1\cdot 0 =(r+r)(a+a)\\
&=ra+ra+ra+ra\\
&=(r+r)a+r(a+a)\\
&=1\cdot a+r\cdot 0\\
&=a.
\end{align*}

Therefore we obtain $a=0$ and thus $r=s$.
It follows that the equation $x+x=1$ has only one solution (at most).

More from my site

• Ring is a Filed if and only if the Zero Ideal is a Maximal Ideal Let $R$ be a commutative ring. Then prove that $R$ is a field if and only if $\{0\}$ is a maximal ideal of $R$.   Proof. $(\implies)$: If $R$ is a field, then $\{0\}$ is a maximal ideal Suppose that $R$ is a field and let $I$ be a non zero ideal: $\{0\} […] • Nilpotent Element a in a Ring and Unit Element 1-ab Let R be a commutative ring with 1 \neq 0. An element a\in R is called nilpotent if a^n=0 for some positive integer n. Then prove that if a is a nilpotent element of R, then 1-ab is a unit for all b \in R. We give two proofs. Proof 1. Since a […] • Ring of Gaussian Integers and Determine its Unit Elements Denote by i the square root of -1. Let \[R=\Z[i]=\{a+ib \mid a, b \in \Z \}$ be the ring of Gaussian integers. We define the norm $N:\Z[i] \to \Z$ by sending $\alpha=a+ib$ to $N(\alpha)=\alpha \bar{\alpha}=a^2+b^2.$ Here $\bar{\alpha}$ is the complex conjugate of […]
• A ring is Local if and only if the set of Non-Units is an Ideal A ring is called local if it has a unique maximal ideal. (a) Prove that a ring $R$ with $1$ is local if and only if the set of non-unit elements of $R$ is an ideal of $R$. (b) Let $R$ be a ring with $1$ and suppose that $M$ is a maximal ideal of $R$. Prove that if every […]
• The Polynomial Rings $\Z[x]$ and $\Q[x]$ are Not Isomorphic Prove that the rings $\Z[x]$ and $\Q[x]$ are not isomoprhic.   Proof. We give three proofs. The first two proofs use only the properties of ring homomorphism. The third proof resort to the units of rings. If you are familiar with units of $\Z[x]$, then the […]
• Finite Integral Domain is a Field Show that any finite integral domain $R$ is a field.   Definition. A commutative ring $R$ with $1\neq 0$ is called an integral domain if it has no zero divisors. That is, if $ab=0$ for $a, b \in R$, then either $a=0$ or $b=0$. Proof. We give two proofs. Proof […]
• A Ring is Commutative if Whenever $ab=ca$, then $b=c$ Let $R$ be a ring and assume that whenever $ab=ca$ for some elements $a, b, c\in R$, we have $b=c$. Then prove that $R$ is a commutative ring.   Proof. Let $x, y$ be arbitrary elements in $R$. We want to show that $xy=yx$. Consider the […]
• Rings $2\Z$ and $3\Z$ are Not Isomorphic Prove that the rings $2\Z$ and $3\Z$ are not isomorphic.   Definition of a ring homomorphism. Let $R$ and $S$ be rings. A homomorphism is a map $f:R\to S$ satisfying $f(a+b)=f(a)+f(b)$ for all $a, b \in R$, and $f(ab)=f(a)f(b)$ for all $a, b \in R$. A […]

You may also like...

Ideal Quotient (Colon Ideal) is an Ideal

Let $R$ be a commutative ring. Let $S$ be a subset of $R$ and let $I$ be an ideal of...

Close