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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!

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[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 $3$.

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  • $\begingroup$ What is the meaning of the $M=(2,x)$ notation for example? $\endgroup$ – DaveWasHere Dec 5 '18 at 21:34
  • $\begingroup$ But, yes I get the idea of working with maximal ideal! $\endgroup$ – DaveWasHere Dec 5 '18 at 21:35
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    $\begingroup$ By $(2,x)$ I mean the ideal generated by $2$ and $x$. $\endgroup$ – Eric Wofsey Dec 5 '18 at 21:40
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    $\begingroup$ 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 $3$. $\endgroup$ – Eric Wofsey Dec 5 '18 at 22:18
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    $\begingroup$ 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 $x-1$. $\endgroup$ – Eric Wofsey Dec 5 '18 at 22:49

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