Apologize if this is a trivial question. Here is the original question:
Let $\mathbb{F}_5$ be the finite field of order 5, and $\mathbb{F}_5 [x]$ the polynomial ring in one variable over $\mathbb{F}_5$. Let $I$ be the ideal of $\mathbb{F}_5 [x]$ generated by the polynomial $x^4 + 2x^2$. Determine the number of invertible elements of the quotient ring $\mathbb{F}_5 [x]/I$.
Notice that \begin{equation} x^4 + 2x^2 = x^2 (x^2 + 2) \end{equation} Since $\gcd(x^2 , x^2 + 2) = 1$, $\langle x^2 \rangle \langle x^2 + 2\rangle = \mathbb{F}_5$. Thus, by Chinese Remainder Theorem, we have \begin{equation} \mathbb{F}_5 [x] / \langle x^4 + 2x^2 \rangle \cong \mathbb{F}_5 [x] / \langle x^2\rangle \times \mathbb{F}_5 [x] / \langle x^2 + 2\rangle \end{equation}
Then I got stucked. I am confident that since $x^2 + 2$ is irreducible, $\mathbb{F}_5 [x] / \langle x^2 + 2\rangle$ is just $\mathbb{F}_{25}$ (if I got my finite field thing right). But, thinking about possible remainders, then isn't $\mathbb{F}_5 [x] / \langle x^2 + 2\rangle$ just $\mathbb{F}_5^2$? But this is impossible since one is cyclic and the other one is not?
The field $\mathbb{F}_{25}$ is not cyclic as an additive group, and indeed is additively isomorphic to $\mathbb{F}_5^2$. Other than that, what you've said is correct.