Let a field $K$ of characteristic $2$, and $F/K$ an hyperelliptic curve. Then, $F$ is defined by an equation of the form : $$ Y^2+Y=w(x), \quad w(x) \in K(X)$$
My question is : My question is : Why we can take $w(x)$ such that all irreducible polynomials in the denominator of $w(x)$ occur to an odd power and $w$ is either of odd positive or of non-positive degree ?
(actually, what I don't figure out is mainly how we can take $w$ such as it is is either of odd positive or of non-positive degree)
Thank you !
Presumably the field of constants $K$ is perfect (for example a finite field of characteristic two). In that case we can adjust $Y$ to meet the requirements you listed.
If $$w(X)=a_{2k}X^{2k}+\cdots + a_1X+a_0$$ is a polynomial of an even degree, then by perfectness there exists $b\in K$ such that $b^2=a_{2k}$. Substitution $\tilde{Y}=Y+bX^k$ into $$ Y^2+Y=w(X) $$ gives $$ \tilde{Y}^2+\tilde{Y}=w(X)+b^2X^{2k}+bX^k, $$ and you see that the terms of degree $2k$ cancel on the right hand side. Rinse - repeat until you are left with an odd degree polynomial.
Observe that using $\tilde{Y}$ instead of $Y$ does not change the function field of the curve (or amounts to a birational morphism).