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

nonequilibrium

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

2. Relevant equations
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3. The attempt at a solution

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

- [itex]\mathbb Z_4[/itex] No, as it can obviously not be cyclical

- [itex]\mathbb Z_2 \oplus \mathbb Z_2[/itex] No. Because say there is an isomorphism [itex]\phi: \frac{\mathbb Z_2[X]}{X^2+1} \to \mathbb Z_2 \oplus \mathbb Z_2[/itex], then say [itex]\phi(X) = (a,b)[/itex], then [itex]\phi(1) = \phi(X^2) = \phi(X)* \phi(X) = (a^2,b^2) = (a,b)[/itex] 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?

micromass

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 [itex]\mathbb{Z}_2\oplus\mathbb{Z}_2[/itex] has no nilpotent elements, while [itex]\mathbb{Z}[X]/(X^2+1)[/itex] does.

nonequilibrium

Aha true :) thank you