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.

  • 5
    $\begingroup$ 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 . $\endgroup$ – Georges Elencwajg Aug 15 '12 at 9:51
  • $\begingroup$ Thanks Georges - sadly I have to admit that I have made a similar mistake before. Thank you for the reminder. $\endgroup$ – 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.

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.