note that the elements of Z_{3}[x]/<x^{2}- 1> will be cosets that we can represent as polynomials of degree < 2, to save on notation instead of writing p(x) + I, i will simply write p(u), and thus u = x + I, and i will write a instead of a + I

(taking advantage of the ring isomorphism Z_{3}-->Z_{3}[x]/<x^{2}- 1> given by: a--> a + <x^{2}+1>).

there are exactly NINE of these cosets:

0,1,2,u,u+1,u+2,2u,2u+1,2u+2. since this is a commutative ring (since Z_{3}is commutative) i will only write "half" the products (a little more than half, actually).

0*0 = 0

0*1 = 0

0*2 = 0

0*u = 0

0*(u+1) = 0

0*(u+2) = 0

0*(2u) = 0

0*(2u+1) = 0

0*(2u+2) = 0

1*1 = 1

1*2 = 2

1*u = u

1*(u+1) = u+1

1*(u+2) = u+2

1*(2u) = 2u

1*(2u+1) = 2u+1

1*(2u+2) = 2u+2 so far, these are all trivial

2*2 = 1 (4 = 1 (mod 3))

2*u = 2u

2*(u+1) = 2u+2 (remember distributivity still works)

2*(u+2) = 2u+1

2*(2u) = u (associativity still applies, as well)

2*(2u+1) = u+2

2*(2u+2) = u+1

u*u = u^{2}= (u^{2}- 1) + 1 = 1 (since u = x + I, we are essentially saying x^{2}- 1 + I = I, so x^{2}+ I = 1 + I, that is u^{2}= 1. u is a "third square root of 1". note this can't happen in a field)

u*(u+1) = u^{2}+ u = 1 + u = u+1

u*(u+2) = 2u+1 (what did i do here?)

u*(2u) = 2

u*(2u+1) = u+2

u*(2u+2) = 2u+2

(u+1)*(u+1) = 2u+2

(u+1)*(u+2) = 0 <---a zero divisor. interesting....

(u+1)*(2u) = 2u+2 <---note that cancellation does NOT hold (u+1)*(u+1) = (u+1)*(2u), but u+1 does not equal 2u.

(u+1)*(2u+1) = 0 <---more zero divisors.

(u+1)*(2u+2) = u+1...but u+1 is NOT the multiplicative identity. very bizarre, hmm?

(u+2)*(u+2) = u+2 <--more strangeness

(u+2)*(2u) = u+2

(u+2)*(2u+1) = 2u+1 <---u+2 behaves VERY strangely. note that 1 is a root of x^{2}-1, and so is u, so u+2 is "sort of like" 1+2 = 0.

(u+2)*(2u+2) = 0

(2u)*(2u) = u

(2u)*(2u+1) = 2u+1

(2u)*(2u+2) = u+1

(2u+1)*(2u+1) = u+2

(2u+1)*(2u+2) = 0

(2u+2)*(2u+2) = 2u+2

all in all, a very strange ring of order 9.