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

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.

Answer 1

It depends very much on the field $F$. For instance, every field homomorphism must preserve the prime subfield, so if $F=\mathbb{F}_2$ then the answer is no, by your argument above. (Edit: as Daniel Schepler says below, it can be argued much more straightforwardly in that case, since there are of course no irreducible polynomials of the form $x^2+a$ in $\mathbb{F}_2[x]$. So, in particular, there are vacuously no isomorphic fields of the form you want. The same argument can be used to show that the answer is "no" whenever $F$ is a finite field extension of $\mathbb{F}_2$.)

On the other hand, let $F=\mathbb{F}_2(t)$, the field of rational functions in $t$ with coefficients from $\mathbb{F}_2$, and let $a=b=t$. For convenience denote $$F_1:=F[x]\big/\langle x^2+a\rangle=\mathbb{F}_2(t)[x]\big/\langle x^2+t\rangle$$ and $$F_2:=F[y]\big/\langle y^2+y+b\rangle=\mathbb{F}_2(t)[y]\big/\langle y^2+y+t\rangle.$$ Now, we claim that $\overline{y+t}$ is transcendental over $\mathbb{F}_2\subseteq F_2$. To see this, first note that, since $\mathbb{F}_2(t)$ is a field, the canonical map $\mathbb{F}_2(t)\to F_2$ taking $t\mapsto\overline{t}$ is an injection. In particular, since $t$ is transcendental over $\mathbb{F}_2\subseteq\mathbb{F}_2(t)$, this means $\overline{t}$ is transcendental over $\mathbb{F}_2\subseteq F_2$ as well. Now, suppose for contradiction that $\overline{y+t}$ is algebraic over $\mathbb{F}_2$. Then, since $\overline{y}$ is algebraic over $\mathbb{F}_2(\overline{y+t})$ and algebraic extensions of algebraic extensions are algebraic, $\overline{y}$ would also be algebraic over $\mathbb{F}_2$. But this would mean $\overline{t}=\overline{y+t}+\overline{y}$ is algebraic over $\mathbb{F}_2$ too, a contradiction. Thus $\overline{y+t}$ is transcendental over $\mathbb{F}_2$, and so we may define a map $\mathbb{F}_2(t)\to F_2$ by sending $t\mapsto \overline{y+t}$.

By the universal property of polynomial algebras, we may extend this to a map $f:\mathbb{F}_2(t)[x]\to F_2$ that sends $x\mapsto\overline{y}$. We claim $\ker f=\langle x^2+t\rangle$, whence $f$ will descend to an isomorphism $F_1\cong F_2$. First note that $f(x^2+t)=f(x)^2+f(t)=\overline{y}^2+\overline{y+t}=\overline{0}$, so indeed $\ker f\supseteq \langle x^2+t\rangle$. On the other hand, $\langle x^2+t\rangle$ is a maximal ideal of $\mathbb{F}_2(t)[x]$, and $f$ is a non-zero map, so we have $\ker f=\langle x^2+t\rangle$, as desired. Thus this gives an example of two isomorphic fields of the form you are looking for.

Answer 2

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.