If a commutative ring with $1$ does not have a unit of order $2$, then $R$ does not have an element of order $2$.

Question

Let $R$ be a commutative ring with $1$. This is a simple question, but I can't see whether it is true or not.

Suppose $R$ does not have a "unit" of (additive) order $2$, i.e. , a unit $u$ such that $u=-u$. Then it is necessarily true that $R$ does not have an element of order $2$?

Answer 1

This is not true. For example, consider the ring $\mathbb{Z}[X]/(2X)$. The units in this ring are only $\pm 1$, but $X+X=0$.

Answer 2

No, consider the example $R = \mathbb{C}[X]/\langle X^2-1\rangle$. If you denote $x = X + \langle X^2-1\rangle$ then $x^2=1$ and hence it has order $2$ in the multiplicative monoid.

On the other hand, $R$ cannot have a non-zero element $p(X)$ such that $2p(X) + \langle X^2-1\rangle = \langle X^2-1\rangle.$ Otherwise, $2p(X) \in \langle X^2-1\rangle$ which implies that $p(X) \in \langle X^2-1\rangle$.