Self inverting Rings

Question

Would it be possible for a ring to have elements that are their own additive inverses? What I mean is, would it be possible to have a ring $K$ of mathematical objects $A$ such that: $$A+A=i,\;\forall A\in K$$

Where $i$ is the additive identity?

Answer 1

Yes, it is possible; consider any ring of characteristic $2$; since $1 + 1 = 0$ in such a ring by definition, we have for all $a$ in the ring $a + a = a(1 + 1) = a0 = 0$; this implies $-a = a$; an example, as Dietrich Burde mentioned in his comment, is $\Bbb Z_2 = \Bbb Z/2\Bbb Z$. A more complex example is $\Bbb Z_2[x]$, the polynomial ring over $\Bbb Z_2$, or $GF(2^n)$, the finite field with $2^n$ elements, or $GF(2^n)[x]$, the polynomials with coefficients in $GF(2^n)$; the list goes on . . .

Note that the characteristic of a commutative unital ring $A$ is the minimum number of times $1_A$ must be added to itself to produce $0$; it is denoted by $\text{char}A$; it is considered infinite of there is no finite number of times $1_A$ may be added to itself to produce $0$; see this widipedia entry.

Note that if $\text{char}A = 2$, then $A[x]$ is an infinite ring of characteristic $2$.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

Answer 2

Yes, consider $K=\Bbb{Z}/2\Bbb{Z}=\{0,1\}$, with addition and multiplication Modulo 2.