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How would one display the Cayley table for $F_3 [x]/(x^2 +2)$ and show that it is a ring (I have assumed addition and multiplication are associative and that multiplication is distributive over addition).

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  • $\begingroup$ In $\mathbf F_3[x]$ , $x^2 $\endgroup$ – Bernard May 4 '16 at 15:41
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Hint:

In $\mathbf F_3[x]$ , $x^2+2=x^2-1$, hence by the Chinese remainder theorem, the quotient is isomorphic to $$\mathbf F_3[x]/(x-1)\times\mathbf F_3[x]/(x+1)\simeq\mathbf F_3\times\mathbf F_3.$$

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If you are looking for a less powerful approach than Bernard's answer, you can just write down each element of $\frac{\mathbb{F}_3[x]}{x^2+2}$. $$\frac{\mathbb{F}_3[x]}{x^2+2} = \{ a_0 + a_1 \alpha \ | \ \alpha^2=-2 = 1 \textrm{ and } a_0,a_1\in \mathbb{F}_3\}.$$ When you take a quotient of a ring by an ideal, you always get a ring. But you can write down the multiplication table if you wanted to.

For example: $(1 + \alpha)*\alpha = \alpha + \alpha^2 = \alpha + 1$.

Another example: $(1 + 2\alpha)*(2+\alpha) = (2 + \alpha + 4\alpha + 2\alpha^2) = 2 +5\alpha + 2\alpha^2 = 2 + 2\alpha + 2 = 4 + 2\alpha = 1 + 2\alpha.$

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