## [abstract algebra] is this ring isomorphic to...

1. The problem statement, all variables and given/known data
Consider $\frac{\mathbb Z_2[X]}{X^2+1}$, is this ring isomorphic to $\mathbb Z_2 \oplus \mathbb Z_2, \mathbb Z_4$ or $\mathbb F_4$ or to none of these?

2. Relevant equations
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3. The attempt at a solution

- $\mathbb F_4$ No, because $\mathbb Z_2[X]$ is a principle ideal domain (Z_2 being a field) and X²+1 is reducible in $\mathbb Z_2[X]$ and in a principle ideal domain an ideal formed by a reducible element is not maximal and thus the quotient is not a field

- $\mathbb Z_4$ No, as it can obviously not be cyclical

- $\mathbb Z_2 \oplus \mathbb Z_2$ No. Because say there is an isomorphism $\phi: \frac{\mathbb Z_2[X]}{X^2+1} \to \mathbb Z_2 \oplus \mathbb Z_2$, then say $\phi(X) = (a,b)$, then $\phi(1) = \phi(X^2) = \phi(X)* \phi(X) = (a^2,b^2) = (a,b)$ since a and b are either zero or one. As a result "1" and "X" would have the same image. Contradiction.

Is this correct? If so, does this also seem like a good way to do it, or did I overlook an easier way to do this?
 PhysOrg.com science news on PhysOrg.com >> Galaxies fed by funnels of fuel>> The better to see you with: Scientists build record-setting metamaterial flat lens>> Google eyes emerging markets networks
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus It seems to be correct!! And you also used a very nice method to show this If you don't want to work with the isomorphism in (3), then you could perhaps say that $\mathbb{Z}_2\oplus\mathbb{Z}_2$ has no nilpotent elements, while $\mathbb{Z}[X]/(X^2+1)$ does.
 Aha true :) thank you