Find all units of $\Bbb Q[x] / (x^2-1)$ Find all units of $\Bbb Q[x] / (x^2-1)$.
I just tried to find them by simply putting $(ax+b)(cx+d)=1$. But it was too complicated to find the undetermined coefficients. I cannot come up with anything related in case the ideal is not irreducible
Could you give me a few hints...?
 A: First, $b+(x^2-1)$ is a unit for all $0\ne b\in\mathbb{Q}$.
Also note that if for $a\ne0$,  $ax+b+(x^2-1)$ is a unit, then so is $x+\frac{b}{a}+(x^2-1)$. So we need to check elements of the form $x+b+(x^2-1)$.
Now:
$$(x+b)(cx+d)=cx^2+dx+bcx+bd=c+bd+(d+bc)x=1$$
i.e. $d=-bc$ and $c+bd=1$ which gives
$$c-b^2c=1\implies c=\frac{1}{1-b^2}\quad\text{and}\quad d=\frac{-b}{1-b^2}$$
Therefore units of $\mathbb{Q}[x]/(x^2-1)$ are of the form $a(x+b)+(x^2-1)$ with $a\ne0$ and $b\ne\pm1$. The inverse is given by
$$(a(x+b)+(x^2-1))^{-1}=\frac{1}{a}\frac{1}{1-b^2}(x-b)+(x^2-1)$$
A: The map $f: \mathbb{Q}[x] \to \mathbb{Q}[x]/ (x-1) \times \mathbb{Q}[x]/(x+1) $ defined by 
$$p(x) \mapsto \bigg( p(x) \ \text{mod} (x-1) \ , \ p(x) \ \text{mod} (x+1) \bigg)$$
is a ring homomorphism with kernel $(x^2-1)$. This induces an isomorphism from $\mathbb{Q}[x]/(x^2 - 1) $ and the range of the homomorphism.  The range is isomorphic to $\mathbb{Q} \times \mathbb{Q}$ which has as its units $(a,b) \in \mathbb{Q}^2$ with $a,b \neq 0$.  Therefore, the units are the elements which have neither $x-1$ or $x+1$ as factors. In other words anything of the form $ax+b$ with $|a| \neq |b|$.            
A: In $\mathbb{Q}[x]/(x^2-1)$, we have $(ax+b)(cx+d)=acx^2 + (ad+bc)x + bd = (ad+bc)x+(ac+bd)$. 
Thus $ax+b$ is a unit if and only if there exist $c, d \in \mathbb{Q}$ such that $ad+bc=0$, $ac+bd=1$. Equivalently, $$ \begin{bmatrix} b & a \\ a & b \end{bmatrix} \begin{bmatrix} c \\ d \end{bmatrix}= \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$ has a unique (since if inverse element exists, then it is unique) solution. Recall the fact from linear algebra:

For square matrix $A$, the equation $Ax=b$ has a unique solution if $\det A \neq 0$. Otherwise, there is no solution or infinitely many solutions.

Therefore $ax+b$ is a unit if and only if $a^2 \neq b^2 $.
