Can someone please help with the following problem taken from Beachy and Blair: Abstract Algebra ch 5 Commutative Rings
Give the multiplication table for the ring $\displaystyle Z_3 [x] / <x^2 - 1>$
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.
that's true, alex, but it's not the "usual" (field) multiplication on Z_{3}xZ_{3}.
but let's use this:
say we want to multiply (u+1)(u+2).
first, which element of Z_{3}xZ_{3} does u correspond to? a little thought shows it must be (1,2) (a root of x-1 in the first factor, and a root of x+1 in the second factor).
since we want to preserve the identity, we have to send 1-->(1,1), the identity of Z_{3}xZ_{3}.
so u+1 = (1,2) + (1,1) = (2,0) <---this explains why u+1 is a 0-divisor.
similarly, u+2 = (1,2) + (2,2) = (0,1) <--explains why u+2 is a 0-divisor.
so (u+1)(u+2) = (2,0)(0,1) = (0,0) = 0
continuing, we can now create a complete correspondence between the 2 versions:
0<-->(0,0) (duh!)
1<-->(1,1)
2<-->(2,2)
u<-->(2,1)
u+1<-->(0,2)
u+2<-->(1,0)
2u<-->(1,2)
2u+1<-->(2,0) wait...was 2u+1 a 0-divisor? yep, sure was.
2u+2 <-->(0,1)
proving this IS an isomorphism is tedious (additively it's clear, but multiplication is a chore) perhaps the distributive law might help....
i think this is a great example of how the additive structure of a ring can have very little to do with the multiplicative structure. and that different multiplicative structures on the very same abelian group can produce VERY different results. put another way: the cardinality of a finite ring tells us a lot less than we might hope.