Is the splitting field of $x^2+1$ over $\Bbb Z_2$ is $\{0,1,i,1+i\}$.

Question

I am not sure about my answer. Since x^2+1 is reducible with respect to Z2 but X^2-1 is irreducible with respect to Z2 so I made this finite field {0,1,i,1+i} nd said it as the splitting field of above polynomial. apologize for typing mistakes.

Answer 1

No. The splitting field of $x^2+1$ over $\mathbf Z_2$ is $\mathbf Z_2$ itself since $x^2+1=(x+1)^2$ has $1$ as a double root in $\mathbf Z_2$.