$\mathbb{Z}[X]/(X^2-1)$ is not isomorphic with $\mathbb{Z}\times \mathbb{Z}$ 
I have to show that the ring $\mathbb{Z}[X]/(X^2-1)$ is not isomorphic with $\mathbb{Z}\times \mathbb{Z}$. 

I know that $(\mathbb{Z}\times\mathbb{Z})^*=\{(\pm1,\pm1)\}$, so I thought I should be looking for elements which have inverses in $\mathbb{Z}[X]/(X^2-1)$ and hopefully find more or less than 4. But I didnt succeed, so I need hints. Thanks.
 A: $\mathbb{Z}[X]/(X^2-1)$ lacks nontrivial idempotents.
A: A homomorphism $\mathbb{Z}[X]\to \mathbb{Z}\times \mathbb{Z}$ is equivalent to specifying an element of $\mathbb{Z}\times \mathbb{Z}$. 
A homomorphism $\mathbb{Z}[X]/(X^2-1)\to \mathbb{Z}\times \mathbb{Z}$ is equivalent to specifying an element of $\mathbb{Z}\times \mathbb{Z}$ whose square equals $(1,1)$.
What are the elements of $\mathbb{Z}\times \mathbb{Z}$ whose square equals $(1,1)$? Hint: You've only got $4$ possibilities. So, there are $4$ homomorphisms $\mathbb{Z}[X]/(X^2-1)\to \mathbb{Z}\times \mathbb{Z}$. 
Exercise: Check that none of these homomorphisms is surjective.
I hope this helps!
A: $\mathbb{Z}[X]/(X^2-1) \cong \mathbb{Z} \times \mathbb{Z}$ would imply (after modding out the ideal $(2)$) that $$\mathbb{F}_2[X]/(X-1)^2 \cong \mathbb{F}_2 \times \mathbb{F}_2.$$The first ring has a nontrivial nilpotent element, whereas the second one does not.
A: Okay, let's start with the polynomial ring and try to define an isomorphism to the product ring. The multiplicative identity needs to go to the multiplicative identity, for starters:
\[1 \mapsto (1,1)\]
The only remaining question is where $X$ goes. Well, we have $X^2 = 1$, so $X$ has to go to an element that squares to the identity. If it's $(1,1)$ or $(-1,-1)$ the homomorphism won't be injective, so it's either $(1,-1)$ or $(-1,1)$, and it doesn't really matter which:
\[X \mapsto (1,-1)\]
Now a general element of the polynomial ring can be written $a + bX$ for integers $a$ and $b$, and it'll be mapped to: \[(a + b, a - b)\]
But the difference between these two entries is $2b$, so we can never make any element of $\mathbb Z \times \mathbb Z$ whose entries differ by an odd number. In particular, the proposed homomorphism does not hit $(1,0)$ and is not surjective.
A: The ring $\mathbb{Z} \times \mathbb{Z} \cong \mathbb{Z}[X]/(X^2-X)$ has the universal property
$$\mathrm{Hom}_{\mathsf{Ring}}(\mathbb{Z} \times \mathbb{Z},R) \cong \{a \in R : a^2=a\},$$
and the ring $\mathbb{Z}[X]/(X^2-1)$ has the universal property
$$\mathrm{Hom}_{\mathsf{Ring}}(\mathbb{Z}[X]/(X^2-1),R) \cong \{a \in R : a^2=1\}.$$
Thus, if $\mathbb{Z} \times \mathbb{Z}$ and $\mathbb{Z}[X]/(X^2-1)$ were isomorphic, this would imply that every ring $R$ has the same numbers of idempotents and of involutions. This is of course wrong, consider $R=\mathbb{Z}/2 \times \mathbb{Z}/2$ for instance; this ring has $4$ idempotents, but only $1$ involution.
A: By the below Lemma $\,\Bbb Z[x]/(x^2\!-\!1)\cong\:\! \Bbb Z\!\times\! \Bbb Z\,$ $\Rightarrow\, \overbrace{(x\!-\!1)\,a(x)\!+\!(x\!+\!1)\,b(x) = 1}^{\large x-1,\ x+1\ \ {\rm are\ \color{#0a0}{comaximal}\ in}^{\phantom{}}\ \Bbb Z[x]}\ \ \smash{\overset{x\,\to\, 1}\Longrightarrow}\ \  \bbox[4px,border:1px solid #0a0]{2\mid 1} $

Idea $ $ proper factorizations of $\,R/f\,$ $\rm\small\color{#f60}{induce}$ coprime $\small\rm\color{#0a0}{(i.e. comaximal)}$ proper
factorizations of $f\,$ by
Lemma $\,  $ If $\ \begin{align}f\in R\,\ \rm  a\ UFD,\\ {\rm rings}\ G,H\!\neq 0\end{align}\,\ $ then $\, \begin{align} R/f\,\ &\cong\,\ G\!\times\! H\\ \color{#f60}{\bf\large  \Rightarrow} f\ \  &\!\!=\ g\ h\end{align}\, $ with $\ \begin{align} &\color{#0a0}{(g)\!+\!(h)=(1)^{\phantom{|^|}}\!\!}\\ &\,(g),\, (h)\neq\:\! (1)\end{align}\,$ for some $\,g,h\in R^{\phantom{|^|}}\!\!\!$
Proof $\ $ See this answer for a simple $\,4\,$ line proof.
