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Let $Z$ be the ring of integers, $p$ a prime and $F_p = Z/pZ$ the field with $p$ elements. Let $x$ be an indeterminate. Set $R_1 = F_p[x]/(x^2-2)$, $R_2 = F_p[x]/(x^2-3)$. 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 mod 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 $Z/4Z$ or the Klein 4-group.

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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

1 Answer 1

up vote 4 down vote accepted

(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.

(b) and (c) were solved by you.

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Thank you Makoto. –  Conan Wong Aug 15 '12 at 14:21

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