K-isomorphic splitting fields in field with characteristic 2

Question

The problem is as follows:

Let $K$ be a field of characteristic $2$, let $a$ be an element of $K$ which is not of the form $b^2 + b$ for any $b \in K$, let $f_a(X) = X^2 + X + a$, and let $L_a = K(\alpha)$ be a splitting field over $K$ of $f_a$, with $\alpha \in L_a$ a root of $f_a$.

Prove that if $a,a'$ are two elements of $K$ as above, then $L_a$ is $K$- isomorphic to $L_{a'}$ iff $a−a'$ is of the form $b^2 +b$ for some $b \in K$.

What I have so far:

$"\Leftarrow":$ If $a-a' = b^2+b$ for some $b \in K$, noting that $X^2+X+a = (X+\alpha)(X+\alpha+1)$ over $K(\alpha)$, we can write $X^2+X+a' = X^2+x+a+b^2+b = (X+\alpha+b)(X+\alpha+b+1)$, using the fact that the characteristic of $K$ equals $2.$ We have that $\alpha+b, \alpha+b+1 \in K(\alpha)$ and so $K(\alpha)$ is a splitting field of $f_{a'}$. Since $L_{a'}$ also is a splitting field of this polynomial, the two fields are $K$- isomorphic.

I can't seem to figure out the $"\Rightarrow"$ implication. I would be grateful for hints or solutions.

Thank you very much in advance!

Answer 1

A roadmap to "$\Rightarrow$". Justify the following claims:

1. If $L_a\simeq_KL_{a'}$ then there exists an element $z\in L_{a'}\setminus K$ such that $z^2+z=a$.
2. If $\alpha'\in L_{a'}$ satisfies the equation $\alpha'^2+\alpha'=a'$ then there exist elements $k_0,k_1\in K$ such that $z=k_0+k_1\alpha'$.
3. $z^2+z=k_0^2+k_0+(k_1^2+k_1)\alpha'+k_1^2a'$. And this implies that we must have $k_1=1$.
4. $b=k_0$ works.

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The idea is pretty much that an element $z=k_0+k_1\alpha\in L_a\setminus K$ has a minimal polynomial (over $K$) of the form $x^2+x+m$ if and only if $k_1=1$. This is somewhat analogous to the more familiar case that if $n$ is a square-free integer, the element $z=q_0+q_1\sqrt{n}\in\Bbb{Q}(\sqrt{n})\setminus\Bbb{Q}$ has a minimal polynomial of the form $x^2+q$ if and only if $q_0=0$. We want to "normalize" the quadratic minimal polynomials by adjusting the coefficient of the linear term. Basically because that gives us control of the Galois conjugate ($\overline{z}=z+1$ in the former case and $\overline{z}=-z$ in the latter case). The quirks of characteristic two lead us to this. An analogue exists for other positive characteristics $p$. Look up Artin-Schreier extensions.