Quadratic reciprocity problem

Question

How can I use quadratic reciprocity to prove that $-3$ is a quadratic residue $\pmod p$ if and only if $p=2$ or $p \equiv 1 \pmod 6$ and deduce that $\mathbb{Z}[\sqrt{-3}]/(p)\cong \mathbb{F}_p \oplus \mathbb{F}_p$ if $p \equiv 1 \pmod6$ and $\mathbb{Z}[\sqrt{-3}]/(p)\cong \mathbb{F}_{p^2}$ if $p \equiv 5 \pmod6?$

Answer 1

For the first part, note first that the statement is clear for $p=2$. So if $p$ is odd, we have $$ \left(\frac{-3}{p}\right) = \left(\frac{-1}{p}\right)\left(\frac{3}{p}\right). $$ If $p\equiv 1\mod 4$, then $$ \left(\frac{-1}{p}\right)\left(\frac{3}{p}\right) = \left(\frac{p}{3}\right), $$ so that $-3$ is a quadratic residue mod $p$ if and only if also $p\equiv 1\mod 3$. Alternatively, if $p\equiv 3\mod 4$, then $$ \left(\frac{-1}{p}\right)\left(\frac{3}{p}\right) = -\left(-\left(\frac{p}{3}\right)\right) = \left(\frac{p}{3}\right), $$ so that again $-3$ is a quadratic residue mod $p$ if and only if $p\equiv 1\mod 3$. Putting these together, we see that $-3$ is a quadratic residue mod $p$ if and only if either $p\equiv 1\mod 12$ or $p\equiv 7\mod 12$; that is, if and only if $p\equiv 1\mod 6$.

As to the second part of your question, we have $$ \mathbb{Z}/(p) \cong \mathbb{Z}[x]/(p,x^2+3) \cong (\mathbb{Z}/(p))[x]/(x^2+3) = \mathbb{F}_p[x]/(x^2+3), $$ where now $x^2+3$ is regarded as an ideal in $\mathbb{Z}/(p) = \mathbb{F}_p$. If $-3$ is a quadratic residue mod $p$, then $p$ splits, so that $x^2+3$ is reducible over $\mathbb{F}_p$, and then $$ \mathbb{F}_p[x]/(x^2+3) \cong \mathbb{F}_p[x]/\mathfrak{p}_1 \times \mathbb{F}_p[x]/\mathfrak{p}_2 \cong \mathbb{F}_p\times\mathbb{F}_p, $$ where the final isomorphism follows since both $\mathfrak{p}_1$ and $\mathfrak{p}_2$ must be generated by linear polynomials. In the other case, where $-3$ is not a quadratic residue, then $p$ is inert, so that $x^2+3$ remains irreducible over $\mathbb{F}_p$, so that $(x^2+3)$ is prime and hence irreducible. So $\mathbb{F}_p[x]/(x^2+3)$ is a field; since it has $p^2$ elements, it is the field $\mathbb{F}_{p^2}$.