Can a ring be both commutative and anticommutative?

Question

As the title says, can a ring ever be both commutative and anti-commutative?

By anti-commutivity, $a \cdot b+b \cdot a =0$.

Edit: A ring must have infinitely many numbers.

Answer 1

By definition, a ring is commutative if and only if $$a b = b a$$ and a ring is anticommutative if and only if $$a b = - b a$$ so a ring that is both commutative and anticommutative must satisfy $$b a = - b a$$ for all $a$ and $b$. In particular, taking $b = 1$, we find that in such a ring, $2 a = a + a = 0$. Conversely, if $2 a = 0$ for all $a$, then a ring is commutative if and only if it is anticommutative.

There are very many examples of such rings. For instance, the field of two elements $\mathbb{F}_2$ is both commutative and anticommutative, and the same is true of its algebraic closure $\overline{\mathbb{F}_2}$ (which is infinite).