- The associativity and distributivity axioms follow from the diagrams
The zero element is the empty set and each element is its own additive inverse.

The set*S*itself is the multiplicative identity and is the only element with a multiplicative inverse.With

*A*+*B*=*A*∪*B*we do not get a ring. The only thing that goes wrong is*additive inverses*. - Let
**u**,**v**be the vectors (*x*_{1},*y*_{1},*z*_{1}) and (*x*_{2},*y*_{2},*z*_{2}).

The product of the corresponding "pure quaternions"

*x*_{1}*i*+*y*_{1}*j*+*z*_{1}*k*and*x*_{2}*i*+*y*_{2}*j*+*z*_{2}*k*is

(*x*_{1}*x*_{2}+*y*_{1}*y*_{2}+*z*_{1}*z*_{2}+ (*y*_{1}*z*_{2}-*z*_{1}*y*_{2})*i*+ (*z*_{1}*x*_{2}-*x*_{1}*z*_{2})*j*+ (*x*_{1}*y*_{2}-*y*_{1}*x*_{2})*k*

and this corresponds to**u.v + u × v**.The only thing that stops (

**R**^{3}, +, ×) being a ring is that × is not associative.

To see this observe that (**u**×**v**) ×**w**is perpendicular to**u**×**v**and so in the plane spanned by**u**and**v**, while**u**× (**v**×**w**) is perpendicular to**v**×**w**and so in the plane spanned by**v**and**w**. It is then easy to find a counterexample. - This is an ideal (and hence a subring) since it is generated by
*x*^{2}.Check it directly [using the fact that if

*p*(1) =*q*(1) = 0 then (*p*-*q*)(1) = 0 and*p*(1)*r*(1) = 0 for any polynomial*r*] or (cleverer!) observe that this is the ideal <*x*- 1 > generated by the polynomial*x*- 1. - It is easy to check the axioms. The multiplicative inverse of
*a*+*b*√2 is*a*/(*a*^{2}- 2*b*^{2}) -*b*/(*a*^{2}- 2*b*^{2})√2. - This is a ring of order 4 but (
*x*+ 1)^{2}= 0 and so it has a zero divisor and is not a field.As usual, this is a commutative ring with identity and so we need only find multiplicative inverses.

1

^{-1}= 1 and 2^{-1}= 2 since -1 = 2 the "rule" gives x(x + 2) = 1 and we have inverses for two more elements. Squaring this gives (*x*+ 1)(2*x*+ 2) = 1 and so we have two more. Also -*x*(-*x*- 2) = 1 and this gives 2*x*(2*x*+ 1) = 1 and this is the last pair.A cleverer way to work is to observe that powers of

*x*generate the eight non-zero elements of the ring and then (*x*^{m})^{-1}=*x*^{8-m}. - If the additive order of an element
*a*is*m*=*pq*then (writing 1 + 1 = 2, etc) we get*ma*= 0 =*p*.*qa*and so*p*and*qa*are zero divisors. The only way of avoiding this is for*m*to be a prime number.The order of any non-zero element is then the same as the order of 1.

- There is a choice of
*p*elements for each coefficient and hence there are*p*^{k}elements. Since the multiplication rule tells us*p*(*x*) = 0 in this ring, if we can split*p*(*x*) as a product of two lower degree polynomials, these elements of the ring will be zero-divisors and so we will not have a field.In fact, if the polynomial

*p*(*x*) cannot be factorised ("is irreducible") then using the "rule"*p*(*x*) = 0 will give a field.If the polynomial

*x*^{2}-*x*- 1 ∈**Z**_{3}[*x*] could be factored it would have to have linear factors and hence would have a root. But it is easy to check that substituting*x*= 0, 1, 2 will not give 0.