I am trying to study for an exam and I am not sure of this solution my professor posted to an exercise. I am given the polynomial quotient ring $\mathbb{Z}_6/(x^2+2x)$ and have to find all units and zero divisors.
My initial idea was to do a multiplication table, but with a group this big I am sure there must be another way. For units, I was thinking all polynomials that are coprime to $x$ and $x+2$, but this isn’t what the solutions have.
He lists the units as $1, 5, x+1, 2x+5, 3x+1, 3x+5, 4x+1, 5x+5$, but gives no reason why. I understand these are all coprime to the ideal, but why isn’t $x+3$ an ideal for example?
He then says the zero divisors are everything that isn’t a unit or zero. Why is this? I thought it was possible for an element to be neither a unit nor a zero divisor.
Thank you very much.
Best Answer
Hint: first we can think of elements of $\mathbb{Z}_6[x]/(x^2+2x)$ is of the form $a+bx$ where $a,b\in \mathbb{Z}_6$.
Now $a+bx$ is a unit if and only if there is $c+dx$ such that $(a+bx)(c+dx)=1$. Note that since $x^2=-2x$ in the ring, we have \begin{align}1=(a+bx)(c+dx)&=ac+(ad+bc)x+bdx^2\\&=ac+(ad+bc)x+bd(-2x)\\ &=ac+(ad+bc-2bd)x \end{align}
Then you need to solve system of equation $ac=1$ and $ad+bc-2bd=0$.