Residue Rings of Polynomial Rings over $\mathbb F_p$

Let $\mathbb Z$ be the ring of integers, $p$ a prime and $\mathbb F_p = \mathbb Z/p\mathbb Z$ the field with $p$ elements. Let $x$ be an indeterminate. Set $R_1 = \mathbb F_p[x]/(x^2-2)$, $R_2 = \mathbb F_p[x]/(x^2-3)$. Determine whether the rings $R_1$ and $R_2$ are isomorphic in each of the following cases:

(a) $p = 2$

(b) $p = 5$

(c) $p = 11$

I'm pretty sure about my answer to (c), but not very sure about (a) & (b). Any comments would be greatly appreciated for PhD Quals prep. Thank you.

Attempt at Solution:

(c) When $p = 11$, $x^2 - 2$ is irreducible but $x^2 - 3$ is reducible [$2$ is a quadratic nonresidue $\bmod 11$; $3$ is a quadratic residue]. So $(x^2 - 2)$ is a maximal ideal and hence $R_1$ is a field, whereas $R_2$ is not. So they are not isomorphic.

(b) When $p = 5$, both $x^2 - 2$ and $x^2 - 3$ are irreducible, so both $R_1$ and $R_2$ are fields. As any polynomial in $R_1$ or $R_2$ of degree $\ge2$ is equal to some polynomial of degree $0$ or $1$, effectively, the elements of $R_1$ and $R_2$ can be represented by $a_0 + a_1x$, where $a_0, a_1 \in \{0,1,2,3,4\}$. So $R_1 = R_2 =$ finite field with $5^2$ elements, i.e. they are isomorphic.

(a) When $p=2$, both $x^2 - 2 = x^2$ and $x^2 - 3 = x^2-1$ are reducible. So although $R_1$ and $R_2$ can each be represented by $a_0 + a_1x$, where $a_0, a_1 \in \{0,1\}$. I am not sure whether each of them is isomorphic to $\mathbb Z/4\mathbb Z$ or the Klein $4$-group.

• Dear Conan, concerning your last sentence, be very careful: the question is about rings not groups. An $\mathbb F_2$- algebra will never be isomorphic to $\mathbb Z/4\mathbb Z$, but that is irrelevant: $\mathbb F_2/(x^2+x+1)$ and $\mathbb F_2/(x^2)$ both have the Klein group as underlying group but yet they are not isomorphic as rings . – Georges Elencwajg Aug 15 '12 at 9:51
• Thanks Georges - sadly I have to admit that I have made a similar mistake before. Thank you for the reminder. – Conan Wong Aug 15 '12 at 14:08

(a) $R_1 = F_p[t]$, where $t$ is the image of $x$. $R_2 = F_p[s]$, where $s$ is the image of $x + 1$. Since $t^2 = 0$ and $s^2 = 0$, $R_1$ and $R_2$ are isomorphic.