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.


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 $ad+bc-2bd=0$.


First, note that this quotient ring is finite (since $x^2+2x$ is monic). In fact it must have $36$ elements.

Now in a finite (commutative) ring $R$, every element is either a unit, zero or a zero-divisor.

Why? Well, for any nonzero element $x$ in $R$, consider the sequence $x^n$ for $n\in \Bbb{N}$. By pigeonhole $x^n=x^m$ for some $n,m$ with $n<m$. Then $x^nx^{m-n}=x^n$, or $x^n(x^{m-n}-1)=0$. There are two possibilities. Either $x^{m-n}-1=0$ and $x^{m-n}=1$, so $x$ is a unit, or $x^{m-n}-1\ne 0$. In this case, let $a$ be the least integer such that $x^a(x^{m-n}-1)=0$. Then $c=x^{a-1}(x^{m-n}-1)\ne 0$ but $xc=0$.

For the specifics of finding the actual units, user9077 has already given a good answer, so I won't duplicate that.


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