# Prove that $(\mathbb{Z}[x]/(x^2))^\times \cong \mathbb{Z} \times \mathbb{Z}_2.$

Prove that $(\mathbb{Z}[x]/(x^2))^\times \cong \mathbb{Z} \times \mathbb{Z}_2,$ where $R^\times$ is the multiplicative group of units of the ring $R.$

Using the fact that $x^2 \equiv 0$ in $\mathbb{Z}[x]/(x^2),$ we can easily show that $(\mathbb{Z}[x]/(x^2))^\times = \{ax \pm 1 \,|\, a \in \mathbb{Z} \}.$ Of course, we have the obvious bijection $\varphi : (\mathbb{Z}[x]/(x^2))^\times \to \mathbb{Z} \times \mathbb{Z}_2$ given by $\varphi(ax + 1) = (a,0)$ and $\varphi(ax - 1) = (a,1).$ Unfortunately, this is not a homomorphism. Indeed, we have that $$\varphi((ax+1)(bx-1)) = \varphi((b-a)x -1) = (b-a, 1) = (-a,0) + (b,1) = -\varphi(ax+1) + \varphi(bx-1).$$ It would be interesting to find an explicit isomorphism between these groups, but I realize that we can also use the Fundamental Theorem of Finitely Generated Abelian Groups to establish the existence of an isomorphism without actually finding one; however, I am not sure how to find a set of generators for $(\mathbb{Z}[x]/(x^2))^\times.$ Could someone assist me in finishing this problem?

• It looks like a semi-direct product. Jun 24, 2018 at 22:11
• Since the result is abelian, it is a fully direct product! Jun 24, 2018 at 22:21

You need instead to define $\phi(ax+1)=(a,0)$ and $\phi(ax-1)=(-a,1)$, and that should work fine.
By the way, it was easier to start with isomorphisms in the other direction from $\Bbb{Z}$ (i.e., $(a,0)\mapsto ax+1$) and $\Bbb{Z}/(2)$ (i.e., $(0,1)\mapsto 0x-1$ and then invert the maps and infer the consequences for elements of the form $ax-1$.
• They are one-to-one, just not onto. So they have inverses with domains that are subgroups of $(\Bbb{Z}[x]/(x^2))^{\times}$. Then you just glue the maps together since in abelian groups a map $A\rightarrow C$ and a map $B\rightarrow C$ yield a map $A\times B\rightarrow C$ by multiplying the images. Jun 24, 2018 at 23:16