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## Homework Statement

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?

## Homework Equations

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## 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?