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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?

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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.

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  • $\begingroup$ For a moment I thought $\mathbb{F}_{25} = \mathbb{Z}_{25}$... So $\mathbb{F}_5 [x] /\langle x^2 + x\rangle \cong \mathbb{F}_{25} \cong \mathbb{F}_5^2$ as rings? $\endgroup$ – 3x89g2 Nov 10 '16 at 5:07
  • $\begingroup$ The last isomorphism is only an isomorphism of groups, not rings. The ring $\mathbb{F}_5^2$ is not a field: notice that $(1,0)\cdot (0,1)=(0,0)$ so it has zero divisors. $\endgroup$ – Eric Wofsey Nov 10 '16 at 5:13
  • $\begingroup$ So the quotient is isomorphic to $\mathbb{F}_{25}$ as field, but isomorphic to $\mathbb{Z}_5^2$ as ring?... $\endgroup$ – 3x89g2 Nov 10 '16 at 5:20
  • $\begingroup$ Er, did you mean $\mathbb{F}[x]/(x^2+2)$ instead of $\mathbb{F}_5[x]/(x^2+x)$ before? (The first is isomorphic to $\mathbb{F}_{25}$, and the second is isomorphic to $\mathbb{F}_5^2$, as rings.) $\endgroup$ – Eric Wofsey Nov 10 '16 at 5:21
  • $\begingroup$ Oh yes. I made a typo! I'm talking about $Q = \mathbb{F}_5 [x] /\langle x^2 + 2\rangle$. So $Q \cong \mathbb{F}_{25}$ as field, but isomorphic to $\mathbb{Z}_5^2$ as ring? $\endgroup$ – 3x89g2 Nov 10 '16 at 5:23

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