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

I've been considering the rings $R_1=\mathbb{F}_p[x]/(x^2-2)$ and $R_2=\mathbb{F}_p[x]/(x^2-3)$, where $\mathbb{F}_p=\mathbb{Z}/(p)$.

I'm trying to figure out if they're isomorphic (as rings I suppose) or not for primes $p=2,5,11$.

I don't think they are for $p=11$, since $x^2-2$ is irreducible over $\mathbb{F}_{11}$, so $R_1$ is a field. But $x^2-3$ has $x=5,6$ as solution in $\mathbb{F}_{11}$, so $R_2$ is not even a domain.

For $p=5$, both polynomials are irreducible, so both rings are fields with $25$ elements. I know from my previous studies that any two finite fields of the same order are isomorphic, but I'm curious if there is a simpler way to show the isomorphism in this case, without resorting to that theorem.

For $p=2$, neither ring is even a domain as $x=0$ is a solution of $x^2-2$ and $x=1$ is a solution for $x^2-3$, but I'm not sure how to proceed after that. Thank you for any help.

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If there is an isomorphism $\phi$, then $\phi(1)=1$, $1\in\mathbb F_p$ and hence $\phi(k)=k$ for all $k\in\mathbb F_p$. So, one can show that if $\alpha$ satisfies a certain polynomial in $R_1$ then $\phi(\alpha)$ satisfies the same polynomial in $R_2$. Now one can check that $R_1=F_p[\sqrt 2]$ and $R_2=F_p[\sqrt 3]$. So $\sqrt 2$ goes to a square root of 2 in $R_2$ using the aforementioned result for polynomials. Determine the $p$ for which there exists $a+b\sqrt3\in R_2$ whose square is 2 and then do further elimination. – Host-website-on-iPage Jul 11 '12 at 7:20

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 :-)

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Thanks! I somewhat follow this, but why is the ratio of two non-squares in a finite field a square? – Noomi Holloway Jul 11 '12 at 7:39
@Noomi: The multiplicative group of the field $\mathbb{F}_p$, $p>2$ is cyclic of order $p-1$. The squares form a subgroup of index two, because squaring is a two-to-one mapping. Equivalently, the quotient group $G=\mathbb{F}_p*/(\mathbb{F}_p*)^2$ is cyclic of order two. The non-squares form the non-trivial element of $G$, so the ratio (or the product) of two non-squares must be the neutral element of $G$, i.e. a square. – Jyrki Lahtonen Jul 11 '12 at 7:44
Thanks, this is quite an interesting answer to me. – Noomi Holloway Jul 11 '12 at 7:57

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.

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Thanks Geoff ${}$! – Noomi Holloway Jul 11 '12 at 7:59

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.

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Just check it's a homomorphism. You only need to check $\phi(ab)=\phi(a)\phi(b)$ for $a,b\in\{\overline x,\overline x+1\}\subset R_1$ – Host-website-on-iPage Jul 11 '12 at 7:50
Thank you Aneesh. – Noomi Holloway Jul 11 '12 at 7:54
I have to remark that $R_1\cong\mathbb F_p[\sqrt2]$ only where $x^2-2$ is irreducible. similarly for $R_2$ iff $x^2-3$ is irreducible. So my argument fails for $p$ such that $2$ or $3$ is a square modulo $p$. Sorry about that. Let me edit the answer. – Host-website-on-iPage Jul 11 '12 at 8:39