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

Question

I've been asked to determined whether the following are fields, PIDs, UFDs, integral domains: $$\mathbf{Z}[X],\quad \mathbf{Z}[X]/(X^2+1),\quad \mathbf{Z}[X]/(2,X^2+1)\quad \mathbf{Z}[X]/(2,X^2+X+1)$$

1. The first is a UFD since $\mathbf{Z}$ is, but not a PID since $\mathbf{Z}[X]/(X)\simeq \mathbf{Z}$ is not a field and $X$ is irreducible.

2. The second is a field since $X^2+1$ is irreducible in $\mathbf{Z}$.

3. I am stuck on this one. I don't think it is a field because: $$\mathbf{Z}[X]/(2,X^2+1)\simeq \mathbb{F}_2[X]/(X^2+1)$$ But $X^2+1$ is not irreducible in $\mathbb{F}_2[X]$ since $(X+1)(X+1)=X^2+1$, so it is not a field. But I can't see how to go further.

4. As in 3. this is $\mathbb{F}_2[X]/(X^2+X+1)$, and $X^2+X+1$ is irreducible in $\mathbb{F}_2$ so it is field.

Are 1, 2, and 4 correct? How can I go further with 3? Thanks for any help.

Answer 1

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.

Answer 2

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.