# Is there a clever way to find the units of the quotient ring $\mathbb{Z}_6[x]/\langle x^2+2x\rangle$?

I've come to the problem of finding the units of the quotient ring $$R:=\mathbb{Z}_6[x]/\langle x^2+2x\rangle$$. I've no problem doing this using multiplication table -- i.e. working out the multiplication of the $$6^2$$ elements in $$R$$ with the extra rule that $$x^2\equiv 4x$$ over in $$R$$ (this comes from the fact that $$x^2+2x$$ is, obviously, in the principal ideal $$\langle x^2+2x\rangle$$). However, this is pretty tedious and not very instructive. Is there a prettiest way of doing this?

Recall that $$R=\{[a_0+a_1x]|a_0,a_1\in \mathbb{Z}_6\}$$ is just the set of classes of the remainders of the Euclidean division by $$x^2+2x$$.

Thanks a lot for any advice!

[For convenience of notation, I will not distinguish between elements of $$\mathbb{Z}_6[x]$$ and their images in quotient rings. For instance, I will write elements of $$R$$ as just $$a_0+a_1x$$ instead of your $$[a_0+a_1x]$$.]

A unit of $$R$$ is just an element which is not in any maximal ideal, so it is enough to identify the maximal ideals. If $$M\subset R$$ is a maximal ideal, then $$M$$ contains either $$2$$ or $$3$$, since $$2\cdot 3=0$$ in $$R$$. Similarly, either $$x\in M$$ or $$x+2\in M$$.

If $$2\in M$$, then in either case we get $$x\in M$$. But $$M=(2,x)$$ is already a maximal ideal, with quotient $$R/M\cong \mathbb{Z}_2$$ (corresponding to the homomorphism $$\mathbb{Z}_6[x]\to\mathbb{Z}_2$$ sending $$f(x)$$ to $$f(0)$$ mod $$2$$), so $$(2,x)$$ is the only maximal ideal containing $$2$$.

If $$3\in M$$, then we get two different possibilities, either $$M=(3,x)$$ or $$M=(3,x+2)$$. These are both maximal ideals with quotient $$\mathbb{Z}_3$$ (the first corresponds to the homomorphism $$\mathbb{Z}_6[x]\to \mathbb{Z}_3$$ sending $$f(x)$$ to $$f(0)$$ mod $$3$$ and the second corresponds to the homomorphism $$\mathbb{Z}_6[x]\to \mathbb{Z}_3$$ sending $$f(x)$$ to $$f(-2)$$ mod $$3$$).

So, the units are just the elements of $$R$$ that are not in $$(2,x)$$, $$(3,x)$$ or $$(3,x+2)$$. For an element $$a+bx\in R$$ to not be in $$(2,x)$$ or $$(3,x)$$ means that $$a$$ is not divisible by $$2$$ or $$3$$ (so $$a$$ is $$1$$ or $$5$$). For $$a+bx$$ to not be in $$(3,x+2)$$ means that $$a-2b$$ is not divisible by $$3$$.

• What is the meaning of the $M=(2,x)$ notation for example? – DaveWasHere Dec 5 '18 at 21:34
• But, yes I get the idea of working with maximal ideal! – DaveWasHere Dec 5 '18 at 21:35
• By $(2,x)$ I mean the ideal generated by $2$ and $x$. – Eric Wofsey Dec 5 '18 at 21:40
• There was nothing special about $\mathbb{Z}_6$ here really: if you had $\mathbb{Z}_n$ instead, then you would consider the prime factors of $n$ in place of where I considered $2$ and $3$. – Eric Wofsey Dec 5 '18 at 22:18
• No, $x$ is a unit in that ring so it's not in any maximal ideals. To find the maximal ideals you have to factor $x^2+2=x^2-1=(x+1)(x-1)$. Since $x^2+2=0$ in the ring, this means a maximal ideal must contain either $x+1$ or $x-1$. – Eric Wofsey Dec 5 '18 at 22:49