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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

3 Answers 3

up vote 12 down vote accepted

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
2  
@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

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