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I am given the task of showing that $\mathbb{Z}[i]/p\mathbb{Z}[i]\cong\mathbb{F}_{p^2}$ for $p\equiv3$ mod $4$ prime.

I understand that there exists only one field of order $p^2$ up to isomorphism, so I thought it'd be a good idea to instead show that $$\mathbb{Z}[i]/p\mathbb{Z}[i]\cong\mathbb{F}_{p}[X]/(X^2+1).$$ As far as I'm aware this would ascertain that $\mathbb{Z}[i]/p\mathbb{Z}[i]$ has order $p^2$ and so the result would follow. To do this I used the map $$\psi:\mathbb{Z}[i]/p\mathbb{Z}[i]\rightarrow\mathbb{F}_{p}[X]/(X^2+1)$$ $$a+bi\mapsto a+bX$$

After some quick calculations it's clear that this is a surjective homomorphism of the given fields, and injectivity can be checked by showing that the kernel of this map is generated by $0$. The map is an isomorphism and so the required result follows.

The fact that I haven't used that $p\equiv3$ mod $4$ leads me to think that there is a mistake in my workings. My only idea is that $X^2+1$ is only irreducible (and so $\mathbb{F}_{p}[X]/(X^2+1)$ is only a field) when $p\equiv3$ mod $4$, though this isn't a statement I'm confident in.

smoking_big_ole_doinks
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1 Answers1

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Your idea is fine. According to the first supplement to quadratic reciprocity, for odd primes $p$, the congruence $x^2\equiv-1\bmod p$ has solutions if and only if $p\equiv1\bmod4$.

J. W. Tanner
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