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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?

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Yes, $\mathbb{Z}/2\mathbb{Z}$. Then $1+1=0$ and also $0+0=0$. – Dietrich Burde Apr 15 '14 at 18:59
A direct product of arbitrarily many copies of the two-element field $\mathbb{F}_2$ satisfies this. In fact, these rings are precisely all $\mathbb{F}_2$-algebras (and additionally, the one-element ring). – Pavel Čoupek Apr 15 '14 at 19:02
But can such rings have infinite sets? – Disousa Apr 15 '14 at 19:03
up vote 3 down vote accepted

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!!!

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

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What about Rings with infinite sets? – Disousa Apr 15 '14 at 19:08
If we want in infinite example, we can take $K[X]$ with $K=\Bbb{Z}/2\Bbb{Z}$, Or we can take $({\cal P}(\Omega),\Delta,\cap)$ for an infinite set $\Omega$ – Omran Kouba Apr 15 '14 at 19:13

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