Question

Suppose that R is a ring and that a^2 = a for all a in RZ. Show that R is commutative.

Commutative Algebra
ANSWERED
asked 2021-01-31
Suppose that R is a ring and that \(\displaystyle{a}^{{2}}={a}\) for all \(\displaystyle{a}\in{R}{Z}\). Show that R is commutative.

Answers (1)

2021-02-01

For any \(\displaystyle{a}\in{R}\), we have,
\(\displaystyle{\left({a}+{a}\right)}={\left({a}+{a}\right)}^{{2}}\)
\(\displaystyle\Rightarrow{\left({a}+{a}\right)}={\left({2}{a}\right)}^{{2}}\)
\(\displaystyle\Rightarrow{\left({a}+{a}\right)}={4}{a}^{{2}}\)
\(\displaystyle\Rightarrow{\left({a}+{a}\right)}={a}^{{2}}+{a}^{{2}}+{a}^{{2}}+{a}^{{2}}\)
\(\displaystyle\Rightarrow{\left({a}+{a}\right)}={a}+{a}+{a}+{a}\ldots{\left({1}\right)}\)
Now, let \(\displaystyle{a},{b}\in{R}\)
\(\displaystyle{a}+{b}={\left({a}+{b}\right)}^{{2}}\)
\(\displaystyle={a}^{{2}}+{a}{b}+{b}{a}+{b}^{{2}}\)
\(\displaystyle={a}+{a}{b}+{b}{a}+{b}\)
\(\displaystyle\Rightarrow{a}{b}+{b}{a}={0}\)
Adding ba on both sides of above equation, we get,
\(ab+(ba+ba)=ba\)
From equation 1, we can observe that same terms on both sides of equation get cancelled out.
So, applying same observation on equation 2, we get, \(ab=ba\)
Hence, Ring R is commutative.

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