32
$\begingroup$

A smooth curve $X$ in $\mathbb{P}^n$ is strange if there is a point $p$ which lies on all the tangent lines of $X$.

Examples are $\mathbb{P}^1$ is strange and so is $y=x^2$ in characteristic $2$. These are in fact all (see Hartshorne IV.3.9).

For this reason, in any characteristic we do not use lines to get embeddings in projective space.

But this actually has consequences. Many classical theorems in algebraic geometry do not use 'transcendental methods', i.e. the only result they use is that the base field is algebraically closed, and so they can be applied in finite characteristic. Or they cannot?

Here is where characteristic $2$ breaks down the standard results. For instance, when embedding blow-ups of $\mathbb{P}^2$ in $\mathbb{P^n}$ we use linear systems of conics and cubics in $\mathbb{P}^2$ to separate the points, but this is not possible in characteristic $2$ (have a look at Beauville IV.4 if you do not know how linear systems can embed spaces). This means that in characteristic 2 we cannot interpret cubic surfaces ni $\mathbb{P}^3$ in terms of blow-ups of the projective plane in 6 points in general position and viceversa.

OK, enough intro. My question is: "Are there other examples of results which do not apply in characteristic 2 due to other reasons not involving embeddings in projective space?"

Also, I am looking for results that do not hold in low characteristic. i.e. I know that vanishing theorems do not hold in characteristic p in general, but I am looking for pathologies for some but not all finite characteristic (usually 2, or 3).

I suspect 'probably yes but not many', since I cannot come up with any, but if it turns out there are lots, maybe I'll make this question community wiki.

$\endgroup$
3
  • 1
    $\begingroup$ There are singular strange curves in other characteristics (e.g. $y=x^p$ in char $p$). It's only for smooth projective curves that you get the classification you stated. $\endgroup$ Jul 6, 2011 at 16:15
  • $\begingroup$ You are right, I meant smooth, projective I assumed it. Fixed it. $\endgroup$ Jul 7, 2011 at 8:56
  • $\begingroup$ Please, if you give an example of a result which does not hold, or needs to be treated apart explain the reason. Francesco's answer is a good example of how to do this. $\endgroup$ Jul 7, 2011 at 8:59

10 Answers 10

25
$\begingroup$

I am not going to add any new examples but suggest a systematic way of looking at examples. If one looks at special phenomena in characteristic $2$ one can classify them as follows (though this division is far from clear cut):

  1. They are really special to positive characteristic and not only characteristic $2$.
  2. They are still really really positive characteristic phenomena but they only appear for some numerical invariants that depend on $p$ (normally increasing with $p$) and as $2$ is the smallest prime they appear "earlier" in characteristic $2$ and hence are encountered there first.
  3. They are really special to characteristic $2$.

Some examples and their classification:

  1. Here one can look at failure of strong versions of Bertini, for instance a base point free linear system all of whose members are singular (take the $p$'th power of a very ample linear system). This is uniform in $p$ (though if one starts with a characteristic free ample system, the degree will grow as $p$ grows so in that sense it could also be classified under 2).
  2. The existence of quasi-elliptic fibrations in characteristic $2$ (and $3$) is an example as the same phenomemena of a regular but non-smooth curve over a non-perfect field exists in all positive characteristics. However, by a result of Tate the genus of such an example is bounded from below by a linear function of $p$ so they appear later and later. However, there is one further complication in that the quasi-elliptic case is of Kodaira dimension $0$ which makes $2$ and $3$ special as all other examples are of general type. This gives an example of overlap between 2) and 3).

    Another such example is that of Enriques surfaces. On the one hand the Godeaux construction gives examples of smooth surfaces whose fundamental group scheme is any group scheme of order $p$ with various numerical invariants depending on $p$. However, only in characteristic $2$ (and $3$ I think) is it of Kodaira dimension $0$.

  3. Here the examples that come to my mind are mostly somehow related to quadratic forms. They themselves of course really behave differently in characteristic $2$ (even purely geometrically) as does the orthogonal group. However, their influence goes further, for instance that theta characteristics behave differently in characteristic $2$ can be traced back to quadratic forms.

Addendum: To test my claim I went through the answers given so far and tried to classify as per above. Most of them are already mentioned above but two are not. First there is Jeremy's comment on torsion in $1+p\mathbb Z_p$ which on the face of it belongs to category 3). However, it is clearly related to $p$-adic radius of convergence of the logarithm and exponential series and that radius grows as $p$ grows. Hence, for absolutely ramified rings you can get the same phenomenon in all characteristics, what is special with $2$ is that it happens in the absolutely unramified. Note also that though the consequence mentioned by Jeremy is more arithmetic than algebro-geometric there are consequences of the latter type. Certainly, mixed characteristic ones such as the structure of finite group schemes but also for crystalline issues (technically the divided power structure on $2\mathbb Z_2$ is not nilpotent).

Sándor's example of failure of Kodaira vanishing is mostly of type 1) but the examples usually have $p$ as a parameter in numerical characters making it partly of type 2). There is even the fact that there are a few minimal surfaces of general type in characteristic $2$ (but in no other characteristic) for which $H^1(X,\omega_X^{-1})\ne 0$ which technically is of type 3).

$\endgroup$
3
  • 1
    $\begingroup$ Torsten: I did not mess with your post. At least I tried. I only edited it in a minimal way to move the misplaced p. Please feel to roll it back or re-edit it if this is contrary to what you wanted. [The actual edit: In 1) I changed "grows with $p$" to "grows as $p$ grows".This moved the symbol $p$ to the next line and out of misplacement.] $\endgroup$ Jul 6, 2011 at 21:37
  • $\begingroup$ Great post! I find those in type 1) quite obvious. I was wondering about 2) and 3) $\endgroup$ Jul 7, 2011 at 9:20
  • $\begingroup$ I choose this answer since it somehow includes all others and classifies them. It is also quite complete. $\endgroup$ Jul 11, 2011 at 15:19
14
$\begingroup$

In characteristic 2 (and 3) a lot of familiar result about elliptic curves are false (usually not essentially false, but quite different). For example the Weierstrass equation is more complicated.

In Silverman's book "The arithmetic of elliptic curves" there is a whole appendix dedicated to this subject.

$\endgroup$
3
  • 3
    $\begingroup$ Likewise for elliptic surfaces: only in characteristic 2 and 3 are there "quasi-elliptic surfaces" where the generic fiber is singular. $\endgroup$ Jul 6, 2011 at 17:29
  • $\begingroup$ Can you explain what's the reason behind this? $\endgroup$ Jul 7, 2011 at 8:58
  • 1
    $\begingroup$ I guess one way to think about it is that the quasi-elliptic surfaces are inseparable covers of ${\bf P}^1$ bundles, and once $p>3$ the corresponding curves $y^p = f(x)$ in characteristic zero cannot have genus as small as 1 (though such a curve can be rational for all $p$). $\endgroup$ Jul 7, 2011 at 21:31
9
$\begingroup$

In all characteristic different from $2$ there exists just one family of Enriques surfaces, obtained by the quotient of a smooth $K3$ surface by a base-point free involution.

In characteristic $2$, however, there are some new families of Enriques surfaces, sometimes called quasi Enriques surfaces or non-classical Enriques surfaces or (super)singular Enriques surfaces.

This is because in characteristic $2$ one can consider not only $\mathbb{Z}/2 \mathbb{Z}$, but also the group schemes $\alpha_2$ and $\mu_2$.

See this link or the book by Cossec and Dolgachev "Enriques surfaces" for more details.

$\endgroup$
8
$\begingroup$

Bertini's theorem says that in characteristic $0$ a general member of a base point free linear system on a smooth variety is smooth. Serre gave an example (see [Hartshorne, Ex.III.10.7] of a linear system with moving singularities in characteristic $2$.

$\endgroup$
1
8
$\begingroup$

The multiplicative group $1+p\mathbb{Z}_p$ is torsion free for all $p>2$, but not for $p=2$.

This, for example, causes $p=2$ to require special treatment throughout Iwasawa theory.

$\endgroup$
2
  • $\begingroup$ ...though there's probably only a small segment of algebraic geometry that Iwasawa theory bears on. $\endgroup$ Jul 7, 2011 at 21:25
  • $\begingroup$ You're right, of course; I kind of forgot about the "algebraic geometry" part of the question. $\endgroup$ Jul 11, 2011 at 15:14
7
$\begingroup$
  • In characteristic 2 and 3 there are quasi-elliptic fibrations, i.e., a smooth projective surface S together with a morphism to a smooth projective curve C such that each fiber is a cuspidal rational curve.

  • The Hurwitz bound for the maximal order of the automorphism group of a smooth projective curve of genus at least 2 does not hold in any positive characteristic.

  • In characteristic 2 there exists unirational surfaces that are not rational.

For the first two you use in char 0 at some point that if $f(x)$ is a polynomial in $x$ and $f'(x)$ is zero then the degree of $f$ is zero. To construct actual examples of curves such that the automorphism group has more that 84(g-1) elements you use the existence of Frobenius. For the third example you might argue that the characteristic zero proof uses transcendental methods.

$\endgroup$
1
  • $\begingroup$ "For the first two you use in char 0 at some point that if f(x) is a polynomial in x and f′(x) is zero then the degree of f is zero." But this is not true for any finite characteristic, right? Why the first result holds only in char 2 and 3? $\endgroup$ Jul 7, 2011 at 9:07
4
$\begingroup$

Theta characteristics of algebraic curves behave differently in characteristic two, basically because they are closely related to the two-torsion in the Jacobian.

$\endgroup$
4
$\begingroup$

I've just realized that no one mentioned Kodaira vanishing. Of course it is not particular to char $2$ and there are versions that work, but the original statement fails in char $p$.

$\endgroup$
1
$\begingroup$

there are probably more results that are simply not known in characteristic 2. I am guilty of not deciding some results in that case, such as the paper Comp.Math. 1990. pp.367 ff. As I recall in this case it was a matter of not having a Taylor series in degree two, because we could not divide by 2. I apologize, but after all that work, I was satisfied with a result valid in all characteristics ≥ 3.

$\endgroup$
1
$\begingroup$

If $E$ is an elliptic curve over a finite field $k$ of characteristic $p>3$, then the group homomorphism ${\rm Aut}_k(E)\rightarrow k^*$ sending an automorphism of $E$ to its effect on the tangent space at $O_E$ is injective.

This can be seen (quite easily) using that the degree map $d:{\rm End}_k(E)\rightarrow \mathbf{Z}$ is a positive definite quadratic form (for a precise definition see Silverman p. 88), thus satisfies $d(a+b)\leq d(a)+d(b)+2\sqrt{d(a)d(b)}$, for all $a,b\in {\rm End}_k(E)$.

If now $u\in {\rm Aut}_k(E)$ acts as the identity on the tangent space at $O_E$, then the endomorphism $1-u$ is either zero or it is an inseparable isogeny. In both cases $p$ divides $d(1-u)$. Setting $a=1$, $b=u$ in the previous inequality we see that $d(1-u)\leq 4$ and hence $d(1-u)=0$, and $u=1$ (since $p>3$).

In char $2$ and $3$ there are elliptic curves with automorphism groups that are not cyclic (even not abelian), thus they cannot be embedded in $k^*$. This different behavior of the primes $2$ and $3$ is perhaps due to the fact that they are "archimedeanly" small.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.