Trying to find the mistake

Question

I know that if $\alpha$ is algebraic over $F$ then $F(\alpha)=F[\alpha]$. Also i know the fact that $\frac {\mathbb Z_2[X]}{(X^2+1)}\cong \mathbb Z_2[i]$ Also $i$ satisfies the polynomial $X^2+1$ over $F_2$, so $i$ is algebraic over $\mathbb{Z_2}$ hence $\mathbb Z_2[i] =\mathbb Z_2(i)$ which is not true because $\frac {\mathbb Z_2[X]}{(X^2+1)}$ is not a field.

What is the mistake in above argument?

Answer 1

The mistake is that, apparently, you mean $\;\Bbb Z_2=\Bbb F_2:=\;$ the field with two elements, and here we have that

$$x^2+1=(x+1)^2\implies\Bbb F_2[x]/\langle x^2+1\rangle=\Bbb F_2[x]/\langle(x+1)^2\rangle$$

Answer 2

I know that if $\alpha$ is algebraic over $F$ then $F(\alpha)=F[\alpha]$.

This statement is only true if $\alpha$ is an element of some field extension of $F$ (or equivalent, some extension of $F$ that is a domain, since you can then take the field of fractions). Indeed, this assumption is needed to define $F(\alpha)$ at all, since by definition $F(\alpha)$ means "the subfield generated by $F$ and $\alpha$". In your case, the ring $\mathbb{Z}_2[i]$ is not a domain, since $(i+1)^2=0$ but $i+1\neq 0$, so this assumption is not satisfied.