TITLE: When is $\mathbb{F}_p[x]/(x^2-2)\simeq\mathbb{F}_p[x]/(x^2-3)$ for small primes?
QUESTION [11 upvotes]: 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.

REPLY [2 votes]: 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.