A question on multiplying cosets together.

Question

C is the set of complex numbers and Z is the set of integer.

Let $ω$ = $e^{2πi/3} ∈ C$ so that $ω^2 + ω + 1 = 0$.

Let R = Z[ω] = {a + bω : a, b ∈ Z} and let I = $<5>$ be the principal ideal of R generated by 5.

(a) Calculate (i) $[3 + ω]_I · [2 + 2ω]_I$ and

(ii) $[4 + 3ω]_I · [1 + ω]_I$ giving your answers in the form $[a + bω]_I$ with a, b ∈ {0, 1, 2, 3, 4}.

(b) Find a, b ∈ {0, 1, 2, 3, 4} such that $[a + bω]_I$ · $[2 + 3ω]_I$ = $[1]_I$

Where I am at so far:

$[3 + ω]_I · [2 + 2ω]_I$ = $[6 + 8ω + 2ω^2]_I$ = $[6 + 8ω]_I$ + $[2ω^2]_I$

Observe that since $ω^2 + ω + 1 = 0$, $2ω^2 = -2ω -2$.

Thus, $[3 + ω]_I · [2 + 2ω]_I$ = $[6 + 8ω]_I$ + $[-2ω-2]_I$ = $[4 + 6ω]_I$.

Now I'm stuck because I need a and b to be either 0,1, 2, 3, or 4. But I have 6.

Answer 1

For part (a), note that $[6]_{<5>}=[1]_{<5>}$.

For part (b), $[a + bω]_I$ · $[2 + 3ω]_I$ = $[1]_I$

$\implies[2a+(3a+2b)\omega+3b\omega^2]_I$

$=[2a+(3a+2b)\omega-3b(1+\omega)]_I$

$=[(2a-3b)+(3a-b)\omega]_I=[1]_I$

(remembering that $\omega^2=-\omega-1$)

$\implies 2a-3b\equiv1\pmod5$ and $3a-b\equiv0\pmod5$.

Can you solve that system?

I got $a\equiv2$ and $b\equiv1\pmod5$, and you can check that indeed $[2+\omega]_I\cdot[2+3\omega]_I=[1]_I\\$, remembering that $[5]_I=[0]_I$.