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Let $k$ be an algebraically closed field of characteristic $p>0$. Let $L$ be the extension of $k(t)$ obtained by attaching a root of an irreducible polynomial $f\in k(t)[x]$. Is there a way to tell from the form of $f$ when the extension $L/k(t)$ is unramified at all the finite places of $k(t)$? (i.e. the places generated by $t-a$ for some $a\in k$).

Are there criteria for when this is not the case, for example does $f$ have to involve only powers of $x$ divisible by $p$ except for the linear and constant term?

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    $\begingroup$ Note that there must be examples involving other powers of $x$, since the extension gotten by adjoining a root of $f(x)$ is the same as the extension gotten by adjoining a root of $x^{\text{deg}(f)} f(1/x)$. $\endgroup$
    – Michael Zieve
    Commented Jul 31, 2014 at 15:06
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    $\begingroup$ @MichaelZieve: Those extensions are equivalent as field extensions. But since $x\mapsto 1/x$ does not quite preserve the OP's notion of "finite places", i.e., all places in the affine line, I believe that this does not quite work. $\endgroup$
    – Jason Starr
    Commented Jul 31, 2014 at 16:26
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    $\begingroup$ @JasonStarr: What's the problem? The two polynomials I wrote down define the same extension of $k(t)$, and hence are ramified over the same places of $k(t)$. For example, if $q$ is a power of $p$ then $f(x)=x^q+x+t$ defines an extension of $k(t)$ unramified over finite places, so also $tx^q+x^{q-1}+1$ defines the same extension, and visibly the latter polynomial has a term of degree not $1$ or divisible by $q$ (unless $q=2$). $\endgroup$
    – Michael Zieve
    Commented Aug 1, 2014 at 1:06
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    $\begingroup$ My point is that any answer to this question must take account of the fact that there are several different generators for a given field extension. For instance, evaluating my polynomial $tx^q+x^{q-1}+1$ at $x+1$ gives $tx^q+x^{q-1}+x^{q-2}+...+x+(t+2)$, which is another polynomial defining an extension of $k(t)$ unramified over finite places, and this polynomial has terms of every degree up to $q$. $\endgroup$
    – Michael Zieve
    Commented Aug 1, 2014 at 1:13
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    $\begingroup$ @MichaelZieve: You are correct. I was misreading $x\mapsto 1/x$ as $t\mapsto 1/t$, which does not preserve the OPs notion of finite places. $\endgroup$
    – Jason Starr
    Commented Aug 1, 2014 at 2:37

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Presumably you know that $f$ is separable over $k(t)$ or else you wouldn't pose the question. Scale $x$ by $k(t)^{\times}$ so that $f$ becomes $x$-monic in $k[x,t]$ with $df/dx \ne 0$. Now the irreducible plane curve $C := \{f=0\}$ inside the $(x,t)$-plane over $k$ has projection to the affine $t$-line that is finite flat and generically etale (as $f$ is $k(t)$-separable), with non-etale locus supported exactly over the zeros of the $x$-resultant $R \in k[t]$ of $f$ and $df/dx$ (with the resultant nonzero since $f$ is $k(t)$-separable).

Thus, in the "generic" case that $C$ is smooth it follows that $k[C]$ is the integral closure of $k[t]$ in $k(C)$, so the absence of ramification in $k(C)$ over finite places of $k(t)$ is exactly the condition that the resultant $R = {\rm{Res}}_x(f, df/dx) \in k[t] - \{0\}$ is constant. This is an extremely difficult condition to arrange (apart from cases where $df/dx$ doesn't involve $x$ at all, such as $f = x^{p^n} - x + h(t)$ corresponding to an Artin-Schreier covering).

The smoothness of $C$ can be determined algorithmically from $f$ (without having to compute actual points on $C$) by applying the effective Nullstellensatz to test if 1 lies in the ideal $(f,df/dx,df/dt)$ in $k[x,t]$, and the $x$-resultant $R \in k[t]$ can be computed very directly from $f$ as well. But perhaps this is not really a condition on "the form of $f$" (whatever that may mean).

Aside from Artin-Schreier covers, it is a very hard problem to build finite etale covers of the affine line by purely algebraic methods without some serious geometric machinery. In general there's probably no way to detect by simple means from staring at "the form of $f$" and avoiding genuine calculations if it gives an example (apart from $f$ arising by the Artin-Schreier construction).

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  • $\begingroup$ In my application I actually want to prove that some polynomial does not have this property, i.e. the extension it defines does ramify at some finite place. Is there some simple criterion that must be satisfied by a polynomial defining an extension unramified at the finite places? $\endgroup$
    – Alex
    Commented Aug 1, 2014 at 7:08
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    $\begingroup$ Is your polynomial (and $k$) explicit, or is it an abstract situation? Please say more about your motivating situation. There's unlikely to be a simple criterion without doing some genuine computation with the polynomial, but the more that you say about your setup the more likely it is that someone can say something which will be useful to you. $\endgroup$
    – user27920
    Commented Aug 1, 2014 at 13:01
  • $\begingroup$ In my situation I have a field $k=\overline{\ell(a_1,...,a_n)}$ ($n$ free variables over $l$). The polynomial I am interested in is of the form $F(x,u(x)+tv(x))$, where $F\in \ell[t,x]$ is irreducible and separable over $\ell(t)$, $u$ has degree $n$ and coefficients which are linear forms in $a_1,...,a_n$ and $v\in\ell[x]$ also has degree $n$. $n$ can be assumed to be as large as necessary. $\endgroup$
    – Alex
    Commented Aug 3, 2014 at 6:00
  • $\begingroup$ Correction: $F\in\ell[x,y]$ is irreducible and separable over $\ell(x)$, i.e. $F\not\in\ell[x,y^p]$. $\endgroup$
    – Alex
    Commented Aug 3, 2014 at 6:21
  • $\begingroup$ Write $u(x) = ax^n + \dots$ and $v(x) = bx^n + \dots$ with $a, b \in \ell^{\times}$. Consider where $t' := a + tb$ vanishes on the normalization. Take $n$ big enough so $f(x,t)$ has the expected top-degree $x$-part. If $t$ is unramified along $t=-a/b$ on the normalization then there are $d=\deg_y(F)$ distinct points over the point $[1,0,0]$ on the projective closure $C$ of $f=0$ (projective coordinates $[X,T',Z]$). Unramfiedness then gives $d$ unit power series $h_i(T')$ making $(1,T',T'h_i(T'))$ lie on $C$. Can that be ruled out by "the form" of $f$? This may lead nowhere; just a thought. $\endgroup$
    – user27920
    Commented Aug 6, 2014 at 4:26

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