Wednesday, April 11, 2012

Cohomology of Finite Fields Part II

Last time we described a theory of cohomology for finite field extensions, but we fell short of calculating ${H^0(\mathbb{F}_4 / \mathbb{F}_2)}$ as I had discovered a mistake in my proof shortly after I had posted the blog. Thanks to the kind folks on math.stackoverflow I have a correct proof (and result!). Let's go through it.

Recall that ${H^0(\mathbb{F}_{4} / \mathbb{F}_2) = \text{ker}(d)}$, where ${\mathbb{F}_{4} = \mathbb{F}_2(\mu)}$, ${\mu}$ a root of a monic irreducible polynomial of degree ${2}$ in ${\mathbb{F}_2}$, and

$\displaystyle d: \mathbb{F}_2[x] \rightarrow \mathbb{F}_{4}[x] \; \text{by} \; df = \frac{f(x + \mu) - f(x)}{\mu}$

Hence ${H^0(\mathbb{F}_{4}/\mathbb{F}_2)}$ is the set of all polynomials ${f \in \mathbb{F}_2[x]}$ such that ${f(x + \mu) = f(x)}$. To find it, first note that if ${f(x) \in H^0}$, then so is ${g(f(x))}$ for any ${g \in \mathbb{F}_2[x]}$, as ${g(f(x + \mu)) = g(f(x))}$. If we can show that some polynomial ${f}$ is the only polynomial in ${H^0}$ with smallest degree, then we know that every other ${g(f(x)) \in H^0}$ is spanned by ${\{f(x)^n : n\in \mathbb{N}\}}$ and further show that every other ${h \in H^0}$ has degree divisible by the degree of ${f}$, then we have that ${\{f(x)^n : n\in \mathbb{N}\}}$ is the basis for ${H^0}$.

Notice also that ${f(x) = x^{4} + x \in H^0}$. This follows from the fact that ${\mathbb{F}_{4}^{\ast}}$ is the cyclic multiplicative group of order ${3}$, hence ${\mu^{3} = 1}$. So we have

$\displaystyle f(x + \mu) = (x + \mu)^{4} + x + \mu = x^{4} + \mu^{4} + x + \mu = x^{4} + x + \mu + \mu = f(x)$

It's not hard to show (exempli gratia, by exhaustion) that ${x^4 + x}$ is the smallest degree polynomial in ${H^0}$, and that it's the only polynomial in ${H^0}$ of degree ${4}$.

Now let ${f \in H^0}$, and let ${\text{deg}f = n}$, say ${f(x) = x^n + a_1 x^{n-1} + \ldots + a_n }$. Clearly ${n}$ cannot be odd, as then ${\binom{n}{1}}$ is also odd, so in the expansion

$\displaystyle f(x + \mu) = (x+\mu)^n + \ldots + a_n = x^n + \mu x^{n-1} + \ldots + \text{lower order terms}$

has a ${\mu}$ as a coefficient of the ${x^{n-1}}$ term, so it can't possibly equal ${f(x)}$.

Now, we want that ${4 | \text{deg}f}$. Say ${\text{deg}f = n = 4k + 2}$. Let's look at the coefficient of ${x^{4k}}$ in ${f(x + \mu)}$. We have ${(x+\mu)^{4k + 2} = (x+\mu)^{4k}(x+\mu)^2 = (x^4 + \mu^4)^k(x^2 + \mu^2) = x^{4k+2} + \mu^2 x^{4k} + \text{lower order terms...}}$. Also ${(x+\mu)^{4k + 1} = (x^4 + \mu^4)^k (x + \mu) = x^{4k+1} + \mu x^{4k} + \text{lower order terms...}}$. Hence in ${f(x + \mu) - f(x)}$, ${x^{4k}}$ has a coefficient ${(\mu^2 + \mu)}$. Thus, if ${f}$ is not divisble by 4, it's not in ${H^0}$.

So say ${f \in H^0}$ has degree ${4k}$. Then ${f(x) - (x^4 + x)^k}$ is also in ${H^0}$ and divisible by 4. In this way we see that polynomials in ${t = x^4 + x}$ form ${H^0}$.

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