Find concise necessary and sufficient conditions on $a_0,a_1,....a_n$ for $\alpha$ to be a unit.

Question

Let $F$ be a field, let $f(X)$ be a polynomial with coefficient in $F$, let $R=F[X]/f(X)$.

Let $F$ is the rational numbers and $f(X)=X^2-1$. Let $\alpha$ be the image of $p(x)=a_0+a_1X+....+a_nX^n$ in $R$. Find concise necessary and sufficient conditions on $a_0,a_1,....a_n$ for $\alpha$ to be a unit.

My attempt: By Chinese remainder theorem, $R=\frac{\mathbb{Q}[X]}{(X^2-1)}\simeq\frac{\mathbb{Q}[X]}{(X-1)} \times \frac{\mathbb{Q}[X]}{(X+1)}$.

Let $\alpha$ be the unit in $R$, and $\frac{\mathbb{Q}[X]}{(X-1)}$ is isomorphic to $\mathbb{Q}$, and units in $\mathbb{Q}$ iff $p(1)\neq 0$. Similarly, $\frac{\mathbb{Q}[X]}{(X+1)}$ is isomorphic to $\mathbb{Q}$, and units in $\mathbb{Q}$ iff $p(-1)\neq 0$.

This implies that $\alpha$ should be unit in $R$ iff $p(1)\neq 0$ and $p(-1) \neq 0$.

Is this idea correct for this question?

Answer 1

Here is a simpler, more general proof. It works for all $f$.

$p(X)$ is a unit in $R$ iff there is $u \in F[X]$ such that $p(X)u(X)\equiv 1 \bmod f(X)$ iff there are $u,v \in F[X]$ such that $p(X)u(X)=1+f(X)v(X)$ iff $\gcd(p(X),f(X))=1$.

In your case, $\gcd(p(X),f(X))=1$ iff $\gcd(p(X),X-1)=1$ and $\gcd(p(X),X+1)=1$ iff $p(1)\neq 0$ and $p(-1)\neq 0$.