I want to show that $\mathbb Z[i]/(3) \cong \mathbb F_9$, and I know that $\mathbb Z[i]/(3) \cong \mathbb Z[x]/(3, x^2 + 1) \cong \mathbb F_3[x]/(x^2 + 1)$. How do we know that $\mathbb F_3[x]/(x^2 + 1) \cong \mathbb F_9$? I know that $x^2 + 1$ is irreducible, is this important?
$\endgroup$2 Answers
$\begingroup$Since the polynomial x$^{2}$+1 is irreducible, $\mathbb{F}_{3}$[x]/(x)$^{2}$+1) is definitely a field. It is like dividing a ring by its prime element.
Now the mod group is basically remainder mod(x$^{2}$+1), which is basically a linear polynomial (ax+b) where a,b$\in\mathbb{F}_{3}$.
Since, $\mathbb{F}_{3}$ has 3 elements, we have 3 choices for a and b each. Therefore the mod group has 9 elements.
$\endgroup$ 3 $\begingroup$$F_3[X]/(x^2+1)$ is an extension of degree 2 of $F_3$ so its cardinal is $3^2$. Two finite field which have the same order are isomorphic.
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