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:

- What conditions are needed on the polynomials $m_k(x)$, $k\in\mathbb{N}$ so that $H^0(\varphi_{jk})$ is well defined?
- When it is defined, when does the above limit equal $\mathbb{F}_2$?

### 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:

- Find the smallest polynomial $f(x)$ in $H^0$, in this case, $f(x)=x^4 - x$.
- 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).
- 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.:

- Check that $H^0$ is indeed ``spanned'' by a single polynomial, id est, that $H^0(m(x)) = \langle f(x) \rangle$ for all $m$.
- Define the categories for which $H^0$ is a functor, check that the above map works (or find a new one?)
- 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.

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