Recent content by facepalmer

great, thanks for the assistance

Thanks, so the constant is required but along with the coefficient of the linear polynomial when determining units and zero-divisors then.

Hi all, I would just like to get some clarity on units and zero-divisors in rings of polynomials. If I take a ring of Integers, Z4, (integers modulo 4) then I believe the units are 1 & 3. And the zero-divisor is 2. Units 1*1 = 1 3*3 = 9 = 1 Zero divisor 2*2 = 4 = 0 Now, If I...

Excellent, many thanks!

I should also add that for (α)3 = ααα = α3 (α2)3 = α2α2α2 = α6 = 1 (α3)3 = α3α3α3 = α9 = α6α3 = α3 (α4)3 = α4α4α4 = α12 = α6α6 = 1 (α5)3 = α5α5α5 = α15 = α12α3 = α3

in reply to micromass... (σβ)3 I did: αβαβαβ = αααβββ = α3β3 = α3β (as ββ = 1) Now, if I follow what conquest has said I should have (and recalling βα = α-1β) (σβ)3 = αβαβαβ = αα-1α-1β = 1α-1β = βα. and for some of the other elements I get (α2β)3 = α2βα2βα2β = α2α-2ββα2β = α2β (α3β)3 =...

Homework Statement Taking the Dih(12) = {α,β :α6 = 1, β2 = 1, βα = α-1β} and a function Nr = {gr: g element of Dih(12)} Homework Equations Taking the above I have to find the elements of N3. And then prove that N3 is not a subgroup of Dih(12). The Attempt at a Solution For N3 I...