TITLE: Is there an isomorphism of $F[x]/\langle x^2+a\rangle$ to $F[x]/\langle x^2+x+b\rangle$ just as fields?
QUESTION [5 upvotes]: Question
If $F$ is a field of characteristic $2$, can there be irreducible polynomials $x^2+a$ and $x^2+x+b$ over $F$ such that $F[x]/\langle x^2+a\rangle$ is isomorphic to $F[x]/\langle x^2+x+b\rangle$ just as fields?
My Attempt
I know they cannot be isomorphic over $F$ because, if they were so, let $u$ be the representative of $x$ in $F[x]/\langle x^2+a\rangle$, so that $u^2=a$, and let $v$ be the representative of $x$ in $F[x]/\langle x^2+x+b\rangle$, so that $v^2=v+b$, then we can suppose, without loss of generality, that $F[u]=F[v]$ is the same field with $F$ as common subfield, and there would exist $\alpha,\beta\in F$ such  that $v=\alpha+\beta u$, so:
$$v+b=v^2=(\alpha+\beta u)^2=\alpha^2+2\alpha\beta u+\beta^2a=\alpha^2+\beta^2a,$$
so that $v\in F$, a contradiction.

REPLY [2 votes]: Describing what I think is actually the same example as in Atticus Stonestrom's fine answer, but in different words. I think of this answer as an application of Lüroth's theorem.
Consider the field of rational functions $L=\Bbb{F}_2(u)$. It has the subfields
$$
K_1=\Bbb{F}(u^2)
$$
and
$$
K_2=\Bbb{F}(u^2+u).
$$
Lüroth says that both $K_1$ and $K_2$ are isomorphic to the function field $F=\Bbb{F}_2(t)$. Indeed, this follows from the fact that both $u^2$ and $u^2+u$ are transcendental over $\Bbb{F}_2$ given that $u$ is transcental.
It also follows that $L$ is a quadratic extension $K_1(u)=K_2(u)$ of both $K_1$ and $K_2$, and that the minimal polynomials of $u$ over $K_1$ and $K_2$ have the required forms.
So we get the required example from $L$, when we identify both $K_1$ and $K_2$ with $F$.
First by mapping $t\mapsto u^2$ then by mapping $t\mapsto u^2+u$. The minimal polynomial of $u$ over $K_1$ is then $x^2+t$, but the minimal polynomial of $u$ over $K_2$ is $x^2+x+t$. All because $t$ means different elements of $L$ in the two cases.