When is $\mathbb{F}_p[x]/(x^2-2)\simeq\mathbb{F}_p[x]/(x^2-3)$ for small primes?

Solution 1:

The question is, as you observed, about whether the two polynomials are irreducible or not. If both of them factor, then the two rings are isomorphic, both isomorphic to a direct product of two copies of $\mathbb{F}_p$. This is seen as follows. The ring $\mathbb{F}_p[x]/\langle q(x)\rangle$ is isomorphic to $\mathbb{F}_p$, when $q(x)$ is linear. If $x^2-2$ (resp. $x^2-3$) factors, then the two factors $q_1(x)$ and $q_2(x)$ are both linear and coprime (assume $p>3$), so the claim follows from the Chinese remainder theorem: $$ \mathbb{F}_p[x]/\langle q_1(x)q_2(x)\rangle\simeq\mathbb{F}_p[x]/\langle q_1(x)\rangle\oplus\mathbb{F}_p[x]/\langle q_2(x)\rangle. $$ If both polynomials are irreducible, then both rings are isomorphic to the field $\mathbb{F}_{p^2}.$ If one polynomial factors, but the other one does not, then $R_1$ and $R_2$ are not isomoprhic, because the other has zero divisors but the other has not.

The way to test factorizability in this case is by the theory of quadratic residues. $R_1$ is a field, iff $2$ is not a quadratic residue modulo $p$. Similarly $R_2$ is a field, iff $3$ is not a quadratic residue modulo $p$. As you hopefully remember, $2$ is a special case, and we simply use the result that $2$ is a quadratic residue, iff $p\equiv \pm 1\pmod{8}$. With $3$ we need to use the law of quadratic reciprocity once. The law states that (using the Legendre symbol) $$ \left(\frac3p\right)=(-1)^{\frac{p-1}2}\left(\frac{p}3\right). $$ The prime $p$ is a quadratic residue modulo $3$, iff $p\equiv1\pmod3$, so we can conclude that $3$ is a quadratic residue modulo $p$, iff $p\equiv (-1)^{\frac{p-1}2}\pmod3$. This translates to (unless I made a mistake) the result that $3$ is a quadratic residue modulo $p>3$, iff $p$ is congruent to either $\pm1\pmod{12}$.

Therefore with $p=5$ we see that both $2$ and $3$ are quadratic non-residues modulo $5$ (it would be easier to check this by applying the definition), so both $R_1$ and $R_2$ are fields of $25$ elements and thus isomorphic.

You asked about a simpler way of showing that the two fields of order $p^2$ are isomorphic. This also follows from the theory of quadratic residues. We get two fields, when both $2$ and $3$ are quadratic non-residues. But the ratio of two non-squares in a finite field is a square. This means that there exists an integer $m$, $0<m<p$ such that $$ 2\equiv 3m^2\pmod{p}. $$ Therefore, for the purposes of extending the field $\mathbb{F}_p$ $$ \sqrt{2}=\pm m\sqrt{3}. $$ Using this observation it is easy to see that in this case $$ \mathbb{F}_p[\sqrt2]=\mathbb{F}_p[\sqrt3]. $$

A final note. From the above we see that (for primes $p>3$) the isomorphism type of $R_1$ depends on the residue class of $p$ modulo $8$, and the isomorphism type of $R_2$ depends on the residue class of $p$ modulo $12$. Therefore the answer to the question whether $R_1\simeq R_2$ or not can be given in terms of residue classes of $p$ modulo $24$. I leave that to you, though :-)

Solution 2:

As you say, it really depends how much theory you are prepared to use. If $2$ and $3$ are both quadratic non-residues (mod $p$), then the polynomials $x^{2}-2$ and $x^{2}-3$ are both irreducible in $\mathbb{F}_{p}[x],$ and both quotient rings are fields with $p^{2}$ elements. When $p$ is a prime and $n$ is a positive integer, the unique field up to isomorphism with $p^{n}$ elements is the splitting field for $x^{p^{n}}-x$ over $\mathbb{F}_{p}$, which is not so difficult to verify. As a matter of interest, the odd primes $p$ for which $2$ is a quadratic residue are those congruent to $\pm 1$ (mod 8), and the odd primes for which $3$ is a quadratic residue are primes congruent to $\pm 1$ (mod 12) (and 3 itself).

As you have observed, if one of $2,3$ is a quadratic residue (mod $p$) and the other is not, then one factor ring is a field and the other is not an integral domain, so they are certainly not isomorphic rings.

If both $2$ and $3$ are quadratic residues (mod $p$) and $p \not \in \{2,3\},$ then the rings are again isomorphic. Both are isomorphic to a ring direct sum of two copies of $\mathbb{F}_{p}$. For note that if $c^{2} = 2,$ then the image of $\frac{c-x}{2c}$ is an idempotent element of the quotient ring, and a similar argument works with 3 when 3 is a non-zero square. I leave the case $p \in \{2, 3\}$ for you to consider.

Solution 3:

By my earlier comment $(a+b\sqrt3)^2=a^2+3b^2+2ab\sqrt 3=2$

So, if $p\ne2$ then clearly $ab\cong0\mod p$. If $a\cong0$ then $3b^2\cong2\mod p$. If $b\cong0$ then $a^2\cong2$. But since $a,b\in\mathbb F_p$ and $\sqrt2$ generates $R_1$, we get in either case (either one has to be the case since $\mathbb F_p$ is a domain) $\text{Im }\phi\subsetneq R_2$ contrary to the fact that $\phi$ is an isomorphism of fields.

Hence $p=2$ and $R_1=\frac{\mathbb F_2(x)}{(x^2)}\quad R_2=\frac{\mathbb F_2(x)}{(x^2+1)}$

Here $\overline x^2=0$ in $R_1$ so that $\phi(\overline x)^2=0$ in $R_2$. Hence $\phi(\overline x)$ is a square root of $x^2+1$, and the only one is $\overline x+1$

What remains is to check is whether there exists such a ring homomorphism of $R_1$ and $R_2$ that $\phi(\overline x)=\overline x+1$. Here $\phi(1)=1$ $\phi(0)=0$ $\phi(\overline x+1)=\overline x$ and one easily checks this is a ring isomorphism.

If $2$ is a square and $3$ is not, modulo $p$, then $R_1$ is a domain while $R_2$ is not. If $3$ is a square and $2$ is not, modulo $p$, then $R_2$ is a domain while $R_1$ is not.

If both $2$ and $3$ are squares, then I they're both isomorphic, but I don't know how to put forth my proof.