Let $$R=\mathbb{Z}_3[x]/\langle x^2+x+1 \rangle.$$ Find the number of units in $R$.
Solution: By definition of factor ring we write $$R=\{ax+b+\langle x^2+x+1 \rangle \mid a,b \in \mathbb{Z_3}\}.$$ Here $R$ contain $9$ elements as both $a$ and $b$ have 3 choices.
We can easily verify that the ideal is not prime in the ring, so $R$ is not an integral domain.
Now my doubt is when the ideal is not prime in the ring how do we examine the ring writing all the elements and checking one by one certain properties for instant no. of units? or else we have some short trick to solve it.
That is $$R=\mathbb{Z_3[x]/\langle x^2+x+1 \rangle}=\mathbb{Z_3[x]}/\langle x+2 \rangle \langle x+2 \rangle$$
I'm stuck here
please give any hint is appreciated.
Best Answer
$R=\mathbb Z_3[x]/\langle(x+2)^2\rangle\cong\mathbb Z_3[y]/\langle y^2\rangle,$ the isomorphism between these two quotients being induced by the isomorphism $\mathbb Z_3[y]\to\mathbb Z_3[x], \;P(y)\mapsto P(x+2).$
In $\mathbb Z_3[y]/\langle y^2\rangle,$ $(ay+b)(cy+d)=(ad+bc)y+bd$ hence $ay+b$ is a unit iff $b\ne0.$ So, there are $6$ units $\pmod{y^2}$: $0y+1,0y+2,1y+1,1y+2,2y+1,2y+2$.