Is $\mathbf{Z}[X]/(2,X^2+1)$ a field/PID?

  1. $\mathbb{Z}[X]$ is not a PID because the ideal $(2,X)$ is not principal.

  2. $\mathbb{Z}[X]/(X^2+1)$ is the ring of Gaussian integers, which is a PID. It is not a field, because $2$ is not invertible (for instance).

  3. $\mathbb{Z}[X]/(2,X^2+1)$ is not even a domain, because as you noticed $X^2+1=(X+1)^2$ in $\mathbb{F}_2$.

  4. Your answer is correct.


Hint: $(X+1)^2 = X^2+2X+1= X^2+1+2X$, so the class of $(X+1)^2$ in $Z[X]/(2,X^2+1)$ is zero and thus the class of $X+1$ is a divisor of zero.

Tags:

Abstract Algebra

Ring Theory

Finite Fields

Proof Verification

Principal Ideal Domains

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