Sunday, August 19, 2012

Progress (or lack thereof) on NCG de Rham Cohomology of Finite Field Extensions

I've been working on calculations of the aforementioned entity, in particular, I want $H^0$, id est, the kernel of the exterior derivative, to be $\mathbb{F}_p$ when I pass to the limit leading to the algebraic closure.

I'm going to make that more precise, but first let me clear up some notation: since $H^0$ does not capture the isomorphism class of field extensions, and since the differential calculi of a field $K$ are in one-to-one correspondence with the monic irreducible polynomials in $K[x]$, we shall write $H^0(m(\mu), K)$ to mean the 0th cohomology of the field extension $K[\mu] (m(\mu))$, or simply $H^0(m(\mu))$ when the field is understood. Also, I shall write $\langle m(x) \rangle_K$ to mean the $K$-span of the set $\{ m(x)^n : n\in \mathbb{N}\} = \{ f(m(x)) : f\in K[x]\}$.

More precisely, we know that the algebraic closure of $\mathbb{F}_2$ is the colimit of fields $\mathbb{F}_{2^k}$, $n\in\mathbb{N}$, with field morphisms $\varphi_{kj} : \mathbb{F}_{2^k} \rightarrow \mathbb{F}_{2^j}$ whenever $k$ divides $j$. It is hoped that $H^0$ is a contravariant functor, yielding a morphism $H^0 (\varphi_{kj}) : H^0(m_j(\mu)) \rightarrow H^0(m_k(\mu))$ for monic irreducible polynomials $m_j(\mu)$ and $m_k(\mu)$ of degrees $j$ and $k$, respectively. This will give us a projective system and thus a projective limit:

$ \displaystyle H^0(\varinjlim_{k\in\mathbb{N}} m_k(\mu) ) = \varprojlim_{k\in\mathbb{N}} H^0(m_k(\mu))$

So our goal is to answer two questions:

  1. What conditions are needed on the polynomials $m_k(x)$, $k\in\mathbb{N}$ so that $H^0(\varphi_{jk})$ is well defined?
  2. When it is defined, when does the above limit equal $\mathbb{F}_2$?
(I'm starting to think the first question is vacuous, but I didn't until recently.)

1. Calculating $H^0(m_k(\mu), \mathbb{F}_2)$

So first I tried to calculate the 0th cohomology for the simplest case: $H^0(\mu^2 + \mu + 1)$. Finding the cohomology amounts to finding which polynomials satisfy $f(x +\mu) = f(x)$. After staring at the equation for awhile, it's pretty easy to see that if $f(x) \in H^0$, then $\text{deg} f = 2^k$ for some $k$.

After some effort (and help), I was able to prove that $H^0(\mu^2 + \mu + 1) = \langle x^4 - x \rangle_{\mathbb{F}_2}$. The proof went like this:

  1. Find the smallest polynomial $f(x)$ in $H^0$, in this case, $f(x)=x^4 - x$.
  2. Let $g(x) \in H^0$, prove that $\text{deg} f$ divides $ \text{deg} g$. (This requires some effort - I've only done it in specific cases by exhaustion).
  3. Hence $\text{deg} \, g = r \, \text{deg}\, f$. Then $g(x) - f(x)^r$ has degree divisible by $\text{deg}f$, so it's also in $H^0$, and it's also of smaller degree, so continuing in this fashion we have to end back at $f(x)$, and we're done.

NOTE: I don't know that there is always only one polynomial of smallest degree in $H^0$, but I've not yet found a case where this isn't true, either.

It's not hard to prove that $x^{p^k} - x$ is always in $H^0(m_k(\mu), \mathbb{F}_p)$. However, $H^0$ can contain a lot more than just $x^{2^k} - x$. But since this is a particularly nice polynomial (its splitting field is $\mathbb{F}_{p^k}$) and $\varprojlim \langle x^{p^k} - x \rangle_{\mathbb{F}_p}$ looks like it's just $\mathbb{F}_p$ (I think, I really have no idea how to calculate that limit, I've not thought about it much yet), one might desire to choose polynomials for the algebraic closure such that $H^0(m_k(\mu)) = \langle x^{p^k} - x\rangle$ .

Note that in this case ($k$ divides $j$, so $j = kq$), we have:

$ \displaystyle x^{p^j} - x = - \sum_{i=0}^{q-1} \left( x^{p^{j - k(i+1)}} -x^{p^{j - ik}} \right) = - \sum_{i=0}^{q-1} (x^{p^k} - x)^{p^{k(q-i)}} $

So that $H^0(\varphi_{kj})$ is simply the identity on $H^0(m_j(\mu))$ embedding it into $H^0(m_k(\mu))$

So, as a guess, we're going to try to find a specific set of polynomials such that $H^0(m_k(\mu)) = \langle x^{p^k} - x\rangle$. Also as a guess, our first candidates are...

2. Conway's polynomials and fields

Conway polynomials of degree $k$ for $\mathbb{F}_p$ are the least monic irreducible polynomial in $\mathbb{F}_p[x]$ under a specific lexicographical ordering. To the best of my knowledge, their primarily use is to provide a consistent standard for the arithmetic of $\mathbb{F}_{p^k}$ for portability across different computer algebra systems. Since these are the ``standard convention'' for Galois Fields, we try them first.

Sadly, they let us down. The Conway polynomial of degree 4 for $\mathbb{F}_2$ is $\mu^4 + \mu + 1$. Using a computer algebra system (sage), I found that $H^0( \mu^4 + \mu + 1) = \langle x^8 + x^4 + x^2 + x^1 \rangle_{\mathbb{F}_2}$.

In addition to those polynomials, John Conway has also described the algebraic closure of $\mathbb{F}_2$ as a subfield of $\text{On}_2$, the set of all ordinals with a field structure imposed by his ``nim-arithmetic''. The description of this field, from his ``On Numbers and Games'' does not treat $\overline{\mathbb{F}_2}$ as a direct limit over various simple extensions. Instead, he shows that the ``quadratic closure'' of $\mathbb{F}_2$ lies in $\text{On}_2$, and then extends the quadratic closure to a cubic closure, also in $\text{On}_2$, et cetera.

This means that for us to use his description, we first have calculate things like the minimal polynomial for $\omega$ and $\omega^\omega$ (where $\omega$ is the least infinite ordinal) over $\mathbb{F}_2$, these calculations are quite difficult (I haven't been able to do a single one), in fact, it's quite a bit of work just figuring out which ordinal corresponds to which $\mathbb{F}_{2^k}$.

But fortunately for us, the quadratic closure only relies on finite ordinals, namely $2^k$. With some help from the internet, we have polynomials $m_{2^k}(\mu)$ describing Conway's quadratic closure. They are given by the recursive relations $m_{2^k}(\mu) = m_{2^{k-1}} ( \mu^2 + \mu)$ with $m_2(\mu) = \mu^2 + \mu + 1$, obviously.

So $m_4(\mu)$ corresponds to the Conway polynomial (this is not the case in general), and we've already used Sage to show that this polynomial doesn't have the cohomology we're looking for.

3. So what next?

Well, for one, we could just define the polynomials $m_k(\mu)$ to be such that $H^0 = \langle x^{2^k} -x \rangle$. But is this choice unique?

Turns out no: again using Sage, I have that both $H^0(1 + \mu + \mu^2 + \mu^3 + \mu^4)$ and $H^0(1+\mu^3+\mu^4)$ are $\langle x^{16} -x \rangle$.

So let's revisit the requirement on $H^0$. I was probably being a dunce insisting on it in the first place. After all, if $H^0$ really is a functor (as it should be, though I've not tried to prove it, or even defined the categories), then the projective system is always well-defined, and we really just need to calculate the limit.

In fact, what we really want is that the chain of polynomial subalgebras $H^0(m_k(x))$ becomes coarser as $k$ tends towards larger integers (thus $m_k(x)$ tends towards larger degrees), and that we have sane maps between $H^0(m_j(x))$ and $H^0(m_k(x))$ whenever $k$ divides $j$.

If my assumption that $H^0$ is always ``spanned'' by a single polynomial is correct, and if my sketch of a proof of $H^0 = \langle f(x) \rangle$ always works, then we might be able to find said map. Say that $H^0(m_k(x)) = \langle \tilde{m}_k(x) \rangle$. We know that $x^{2^k} - x \in H^0(m_k(x))$, so that $\langle x^{2^k} - x \rangle \subset \langle \tilde{m}_k(x) \rangle$, we also know that $\langle x^{2^j} - x \rangle \subset \langle x^{2^k} -x \rangle$ as, from before:

$ \displaystyle x^{p^j} - x = - \sum_{i=0}^{q-1} (x^{p^k} - x)^{p^{k(q-i)}} $

Since $x^{2^j} -x \in H^0(m_j(x))$, we know that $f(m_j(x)) = x^{2^j} - x$ for some polynomial $f \in \mathbb{F}_2[x]$. Let $g(x) \in H^0(m_j(x))$, then

$ \displaystyle g(x) = \tilde{m}_j(x)^s + g_{s-1} \cdot \tilde{m}_j(x)^{s-1} + \ldots + g_0$

by definition. Define a map from $H^0(m_j(x)) \rightarrow H^0(m_k(x))$ by

$ \displaystyle g(x) \mapsto f(\tilde{m}_j(x))^s + g_{s-1} \cdot f(\tilde{m}_j(x))^{s-1} + \ldots + g_0$

Assuming that makes sense, and I haven't made an other stupid errors (a big if!), then I guess that leaves the following for a TODO to show that $H^0$ goes to $\mathbb{F}_2$ for any algebraic closure.:

  1. Check that $H^0$ is indeed ``spanned'' by a single polynomial, id est, that $H^0(m(x)) = \langle f(x) \rangle$ for all $m$.
  2. Define the categories for which $H^0$ is a functor, check that the above map works (or find a new one?)
  3. Prove that the subalgebras $H^0$ become coarser, as stated above. (Intuitively, it makes sense that they do, but I haven't thought about a proof yet. Also, if they do become coarser, then it seems ``obvious'' that the limit goes is $\mathbb{F}_2$, as those are the only elements left in an series of smaller and smaller subalgebras.