First, recall that I mention in my first post on NCG cohomologies of field extensions that a first order differential calculus gives rise to a differential graded algebra. Let's briefly describe this:

Given our [possibly noncommutative] algebra ${A = \Omega^0}$, we've already described it's first order differential calculus as an ${A}$-bimodule ${\Omega^1}$ together with a linear map ${d: \Omega^0 \rightarrow \Omega^1}$ that obeys the rules 1) ${d(ab) = a \cdot db + da \cdot b}$ (where ${\cdot}$ is the action of the algebra on the bimodule) and 2) ${\Omega^1 = \{a \cdot db : a, b \in \Omega^0\}}$. The differential graded algebra is then defined as graded algebra ${\Omega = \bigoplus_n \Omega^n}$ together with the map ${d : \Omega^n \rightarrow \Omega^{n+1}}$ and the rules: ${\Omega^0 = A}$, ${\Omega^1}$ as before, ${d^2 = 0}$, and ${\Omega}$ generated by ${\Omega^0}$ and ${\Omega^1}$. The product of the differential graded algebra is the wedge product ${\wedge}$, as in classical differential geometry.

Putting the above together, we get a complex ${A = \Omega^0 \xrightarrow{d^0} \Omega^1 \xrightarrow{d^1} \ldots}$. Here ${d^n}$ is the restriction ${d^n = d |_{\Omega^n}}$. The cohomology is then defined as ${H^n(A) = \text{ker}(d^n)/\text{Im}(d^{n-1})}$, with ${H^0(A) = \text{ker}(d^0)}$ as we saw in the last two posts.

In our case of the field extension ${\mathbb{F}_4/\mathbb{F}_2}$, we have as our algebra the polynomial ring ${A = \mathbb{F}_2[x]}$. We've already defined a differential calculus on it, the polynomial ring ${\Omega^1 = \mathbb{F}_4[x]}$, where ${\mathbb{F}_4 = \mathbb{F}_2(\mu)}$, together with the differential map ${df = \frac{f(x + \mu) - f(x)}{\mu}}$. Using this, how can we describe the differential graded algebra ${\Omega = \bigoplus_n \Omega^n}$ ?

First, recall that ${\mathbb{F}_4}$ is a two dimensional vector space over ${\mathbb{F}_2}$ with basis ${\{1, \mu\}}$. This means that the polynomial ring ${\Omega^1}$ can be ``spanned'' by ${\{1, \mu\}}$ over the polynomial ring ${\mathbb{F}_2[x]}$. In other words, every ${f \in \Omega^1}$ can be written as ${f = f_1 \cdot 1 + f_\mu \cdot \mu}$, with ${f_1, f_\mu \in \Omega^0}$. In order for us to get a graded algebra, ${\Omega^2}$ must have a basis ${\{1 \wedge \mu\}}$, hence it's a one dimensional space over ${\Omega^0}$ and thus is isomorphic to ${\Omega^0}$. ${\Omega^n = 0}$ for ${n > 2}$, as we're using a classical anticommutative wedge product.

But what is the map ${d^1: \Omega^1 \rightarrow \Omega^2}$ ? Well, from our description of ${\Omega^1}$, we can easily express the action of ${d}$ on ${\Omega^1}$ as an action on ${\Omega^0}$, which is well defined. Id est, for ${f \in \Omega^1}$,

$\displaystyle d^1 f = d^0(f_1) \wedge 1 + f_1 \wedge d^0(1) + d^0(f_\mu) \wedge \mu + f_\mu \wedge d^0(\mu)$

Now, ${d^0(a) = 0}$ for any constant polynomial, and ${d^0(g) \in \Omega^1}$ for any ${g \in \mathbb{F}_2}$. So ${d^0g = \partial_1 g \cdot 1 + \partial_\mu g \cdot \mu}$. Hence we have linear maps ${\partial_1, \partial_\mu : \mathbb{F}_2[x] \rightarrow \mathbb{F}_2[x]}$. We return to ${d^1}$ using this idea:

$\displaystyle d^1 f = (\partial_1 f_1 \cdot 1 + \partial_\mu f_1 \cdot \mu) \wedge 1 + (\partial_1 f_\mu \cdot 1+ \partial_\mu f_\mu \cdot \mu) \wedge \mu$

The wedge product is anticommutative, so this leaves us with:

$\displaystyle d^1 f = (\partial_1 f_\mu - \partial_\mu f_1) \cdot 1 \wedge \mu$

And ${d^2 f = 0}$, as the rest of the ${\Omega^n}$'s are 0. Now back to the cohomologies: ${H^1(\mathbb{F}_4/\mathbb{F}_2) = \text{ker}(d^1)/\text{Im}(d^0)}$, where ${\text{ker}(d^1) = \{f \in \Omega^1 : \partial_1 f_\mu = \partial_\mu f_1 \}}$ and ${\text{Im}(d^0) = \{f_1 \cdot 1 + f_\mu \cdot \mu \in \Omega^1 : \exists f \in \mathbb{F}_2[x] \; \text{such that} \; df = f_1 \cdot 1 + f_\mu \cdot \mu\}}$. Hence the we need to find a description of the ``partial derivatives'' ${\partial_1}$ and ${\partial_\mu}$. In other words, we need to describe ${d^0f}$ as:

$\displaystyle \frac{f(x + \mu) -f(x)}{\mu} = \partial_1 f + \partial_\mu f \cdot \mu$

By noting that ${\mu^3 = 1}$, I was able to write down a formula for ${\partial_i x^{3n}}$, and then formulas for ${\partial_i x^{3n+1}}$ and ${\partial_i x^{3n+2}}$ in terms of ${\partial_i x^{3n}}$. But the binomial expansion makes these formulas just too messy to work with. I then tried finding a basis under which the maps would have a nice, neat formula, but I had no luck there. Next I tried looking at the action of the Galois group on $d$.

Now, the only non-trivial element of the $\text{Gal}(\mathbb{F}_4/\mathbb{F}_2)$ is the Frobenius automorphism $\sigma(x) = x^2$, exempli gratia, $\sigma\mu=\mu^2=\mu+1$. The action of the Galois group extends to an action on the polynomial ring - one that leaves $\mathbb{F}_2$ fixed. Hence we have

$\displaystyle \sigma d f = \sigma(\partial_1 f) + \sigma(\partial_\mu f \cdot \mu) = \partial_1 f + \partial_\mu f \cdot (\mu + 1) = df + \partial_\mu f$

This allows us to write the ``partial derivatives'' in terms of $d$ and the Frobenius automorphism. Id est,

$\displaystyle $$\partial_\mu f = \sigma (d f) - d f$

and

$\displaystyle \partial_1 f = df - \sigma(d f) \mu + (df) \mu$

It remains to be seen if these formulas for the partials will actually help me calculate the remaining cohomologies.

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