Why does $\frac{\mathbb{Z}_5[x]}{\langle x^2+x+1\rangle}$ have $25$ elements, but $\frac{\mathbb{Z}_5[x]}{\langle x-1\rangle}$ have $5$ elements? I understand this must come from the degrees of the polynomials generating these ideals, but I don't fully understand how.
Any help appreciated!
Because in the second, every quadratic polynomial can be written as $$ p(x) = q(x) (x - 1) + r(x) $$ by the long-division algorithm, and in this decomposition, $\deg(r) < 1$; that means that $r$ is a constant, and there are only five choices.
A parallel argument for the second shows that every item can be written as a product with a remainder that's a linear polynomial $ax + b$ in $x$, and the 5 choices for each of $a$ and $b$ provide your 25 elements.
Just look at the viable representatives in the quotient, $a\overline{x}+b$ with $a,b\in\Bbb Z_5$ for the former and $a$ with $a\in\Bbb Z_5$ for the latter. That's $25$ choices for the first and $5$ for the second.
$\mathbb Z_5[x]/(x^2+x+1)\cong \mathbb Z_5[\alpha]$ where $\alpha$ root of $f(x)=x^2+x+1$ and degree of extension $\mathbb [\mathbb Z_5[\alpha]:\mathbb Z_5]=2$ then number of element is $5^2=25$ .
But $\mathbb Z_5[x]/(x-1)\cong \mathbb Z_5[1]\cong \mathbb Z_5$