How many elements does the factor ring $\mathbb{Z}_5[x]/(x^2-2)$ have? Find the multiplicative inverses of $x, x+1$ and $3x+2$ in this ring.
Total novice here - how would you be able to count the number of elements in the factor ring? And in this case, if someone could find one multiplicative inverse, e.g $3x+2$, I'm sure can solve the rest. Many thanks.
$\mathbb{Z}_5[x]/(x^2-2)$ means that you look at the ring $\mathbb{Z}_5[x]$ with the relation $x^2-2=0$, or equivalently, $x^2=2$. This means that whenever you have a polynomial of degree $\geq 2$ you can reduce it to a linear polynomial (since you can replace every $x^2$ with $2$). For example, $$\begin{align}3x^4+2x^3+x^2+1=\\3(x^2)^2+2x(x^2)+(x^2)+1=\\3(2)^2+2x(2)^2+(2)+1=\\8x+15=\\3x\end{align}$$
working modulo $5$. This means that every element in $\mathbb{Z}_5 [x]/(x^2-2)$ is on the form $a+bx$, where $a,b\in\mathbb{Z}_5$. Now, how many possibilities are there for $a,b$? This will give you the number of elements in $\mathbb{Z}_5 [x]/(x^2-2)$.
Now for inverses. $$(x+1)(a+bx)=(a+b)x+a+2b=1$$
So $a+2b\equiv_5 1$ and $a+b\equiv_5 0$, using that $x^2=2$. Can you continue from here?