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What do I do if I have to solve the usual quadratic equation $X^2+bX+c=0$ where $b,c$ are in a field of characteristic 2? As pointed in the comments, it can be reduced to $X^2+X+c=0$ with $c\neq 0$.

Usual completion of square breaks. For a finite field there is Chen Formula that roughly looks like $X=\sum_{m} c^{4^m}$. I am more interested in the local field $F((z))$ or actually an arbitrary field of characteristic 2.

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  • $\begingroup$ What's your method that works for finite fields? $\endgroup$
    – André Henriques
    Commented Nov 18, 2010 at 17:03
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    $\begingroup$ I think you should add "in characteristic 2" to the title $\endgroup$
    – Ewan Delanoy
    Commented Nov 18, 2010 at 17:05
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    $\begingroup$ Actually, over any field, either $b=0$ or the quadratic reduces to an Artin-Schreier equation $x^2+x+c=0$. If the field is finite, it is soluble if and only the trace of $c$ to ${\mathbb F}_2$ is 0, if I remember correctly. $\endgroup$
    – Tim Dokchitser
    Commented Nov 18, 2010 at 17:44
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    $\begingroup$ @BB: Sorry, I didn't want to mislead you: I don't know the answer to that (I was just commenting on Andre's question). The image of $a: x\mapsto x^2+x$ is a linear subspace of the field $K$ in question, and the 'Artin-Schreier' group $K/a(K)$ is the analogue of the 'Kummer group' $K^\times/K^{\times 2}$ for fields of characteristic $\ne 2$, in the sense that it classifies quadratic extensions of $K$. But your question is how to actually determine whether the class of $c/b^2$ is trivial in this group, and I only know how to answer that for finite fields... $\endgroup$
    – Tim Dokchitser
    Commented Nov 18, 2010 at 22:46
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    $\begingroup$ Dear Bugs: Is your aim a criterion (for interesting $K$ of characteristic $p > 0$, at least for $p = 2$) which detects when $x^p - x - a \in K[x]$ is irreducible or not? For local function fields there's a criterion using residues of 1-forms (in Artin-Tate). For global function fields there's an adelic analogue (in Artin-Tate). If $Y$ is a normal affine variety (geom. conn'd) over a finite field $k$ and $a \in k[Y]$, then $x^p - x - a$ is reducible over $k(Y)$ iff has solution in $k(y)$ for closed pts $y$ off a proper closed subset (prove using Lang-Weil). Is it useful for you? $\endgroup$
    – BCnrd
    Commented Nov 19, 2010 at 1:00

2 Answers 2

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I think this solves $X^2+X+c=0$ over $F((t))$:

I want to assume that $c\in F[[t]]$. If not, say $c=at^{-m}+...$, then the quadratic has no solutions when $m$ is odd or $a$ is not a square, and otherwise the substitution $X\mapsto X+\sqrt{a}t^{-m/2}$ gives a new equation with smaller $m$. So, after finitely many steps $c=c_0+c_1t+...$ is integral.

Because $X^2+X+c$ has derivative $1$, by Hensel's lemma the equation has a solution if and only the constant term $c_0$ is of the form $d^2+d$ for some $d$ in $F$. And if it is, Hensel's approximations are obtained by starting with an approximate solution $x_0=d$ and recursively computing $x_{m+1}=x_m-f(x_m)/f'(x_m)=x_m^2+c$. This gives $$ x = d + \sum_{n=0}^\infty (c-c_0)^{2^n} $$ as the solution (the partial sums are the $x_m$). Actually, the approach seems to work over any complete field, reducing the problem to the residue field. Hope this helps.

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    $\begingroup$ Isn't a complete valued field of characteristic $2$ necessarily of the form $F((t))$ for some field $F$? $\endgroup$
    – LSpice
    Commented Apr 17, 2017 at 16:28
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Here is a paper that might help

  • Jorgen Cherly, Luis Gallardo, Leonid Vaserstein, Ethel Wheland, Solving quadratic equations over polynomial rings of characteristic two, Publicacions Matemàtiques, Vol 42 (1998), 131–142, EuDML, Wayback Machine.
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