$\mathbb{F}_2[\alpha] \cong \mathbb{F}_2[x]/(x^2+1)$? My question involves part (b) of Chapter 11 problem 6.4 in Artin's Algebra textbook.

In each case, describe the ring obtained from $\mathbb{F_2}$ by adjoining an element $α$ satisfying the given relation:
(a) $α^2+α+1=0$
(b) $α^2+1=0$
(c) $α^2+α=0$

Now, I obtained that the ring in part (a) is isomorphic to $\mathbb{F_4}$ and that the ring in part (c) is isomorphic to $\mathbb{F_2}\times\mathbb{F_2}$.
It seems to me that the ring in part (b) would be isomorphic to $\mathbb{F}_2[x]/(x^2+1)$, but my teacher doesn't agree.
He said,

"Be careful: notice the polynomial $x^2+1$ is not irreducible over $\mathbb{F}_2$. Adjoining a root of a reducible polynomial is not the same as taking the quotient $\mathbb{F}_2[x]/(x^2+1)$"

So, is my teacher right, or am I? And why?
 A: The answer depends on how you interpret "adjoining an element $\alpha$ to $F$ satisfying $f(\alpha) = 0$".
The standard interpretation is: consider the algebraic closure $\bar F$ of $F$, take an element $\alpha \in \bar F$ satisfying $f(\alpha) = 0$ and look at the smallest field containing both $F$ and $\alpha$, i.e, $F(\alpha)$. If $f$ is irreducible, this is isomorphic to $F[x]/(f)$. This makes the answer to (a) ${\Bbb F}_4$ and to (b) and (c) both ${\Bbb F}_2$.
Since Artin says "the ring obtained from ${\Bbb F}_2$ $\dots$", you could also interpret it as: "add a new free element $\alpha$ to $F$, subject only to the constraint $f(\alpha) = 0$". That is, by definition, consider $F[x]/(f)$. With this interpretation, the answer to (a) is still ${\Bbb F}_4$, (b) is (isomorphic to) ${\Bbb F}_2[x]/(x^2)$ and (c) is ${\Bbb F}_2^2$.
A: Yes, $x^2+1=(x+1)^2$ in $\mathbb{F_2}[x]$ you deduce that $\mathbb{F_2}[x]/(x^2+1)$ is a $2$ dimensiobal vector space over $\mathbb{F_2}$.
Adjoining an element (of the algebraic closure) which satisfies  $\alpha^2+1=0$ is adjoining $1$ and the result is $\mathbb{F_2}$.
