Quotient group element is a unit

Question

I'm studying up for my algebra exam, and I'm not exactly sure how to solve a problem like the following

Let $f = X^2 + 1 \in \mathbb{F}_5[X]$, $R = \mathbb{F}_5[X]/\langle f \rangle$ and $\alpha = X + \langle f \rangle \in R$. Show that $\alpha \in R^*$ and that $\vert \alpha \vert = 4$ in $R^*$.

I have already proven that R is not a field by proving that $\langle f \rangle$ is not a maximal ideal, but I'm unsure of how I should proceed.

Answer 1

Working in $R$ means working with polynomials in $\mathbb{F}_5[X]$ modulo the polynomial $f = X^{2} + 1$.

So you have $\alpha^{2} = X^{2} \equiv -1 \pmod{f}$, and $\alpha^{4} \equiv (-1)^{2} = 1 \pmod{f}$. This shows that $\alpha$ is invertible, with inverse $\alpha^{3} \equiv -X \pmod{f}$. (This is because $1 \equiv \alpha^{4} \equiv \alpha \alpha^{3} \pmod{f}$.

Also, since $\alpha^{4} \equiv 1 \pmod{f}$, the order of $\alpha$ is a divisor of $4$. But since $\alpha^{2} \equiv -1 \not\equiv 1 \pmod{f}$, the order does not divide $2$, and thus it is indeed $4$.

Answer 2

$\alpha^{2}=-1_R$ in $R$ so $\alpha^{4}=\left(-1_R\right)^{2}=1_R$ showing that the order of $\alpha$ is $4$ and that $\alpha$ is invertible.

Explaining:

$\alpha^{2}=\left(X+\left(f\right)\right)^{2}=X^{2}+\left(f\right)=-1+\left(f\right)$ since $f\mid X^{2}-\left(-1\right)=X^{2}+1$