Question

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

Commutative Algebra
Suppose that R is a ring and that $$\displaystyle{a}^{{2}}={a}$$ for all $$\displaystyle{a}\in{R}{Z}$$. Show that R is commutative.

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.