# Witt Cohomology Caution

Hopefully I’ll start posting more now that last week is over. Today we’ll look at a counterexample to see that the Witt cohomology we’ve been looking at is not always a a finite type ${\Lambda}$-module. Just to recall a bit, we’re working over a perfect field of characteristic ${p}$, and ${\Lambda=W_{p^\infty}(k)}$. Given a variety ${X}$ over ${k}$ we can use the structure sheaf ${\mathcal{O}_X}$ to form ${\mathcal{W}_n}$, which is the sheaf of length ${n}$ Witt vectors over ${\mathcal{O}_X}$. This is just ${\mathcal{O}_X^n}$ with a special ring structure that on stalks has the property of being a complete DVR with residue field ${k}$ and fraction field of characteristic ${0}$.

The restriction map given by chopping off the last coordinate ${R: \mathcal{W}_n\rightarrow \mathcal{W}_{n-1}}$ gives us a projective system of sheaves and using standard abelian sheaf cohomology we can define ${H^q(X, \mathcal{W})=\lim H^q(X, \mathcal{W}_n)}$.

This brings us to the purpose of today. It is possible that in very nice (projective even) cases we have ${H^q(X, \mathcal{W}_n)}$ a finite type ${\Lambda}$-module, yet have that ${H^q(X, \mathcal{W})}$ is not. Let ${X}$ be a genus zero cuspidal curve with cusp ${P}$. Let ${X'\rightarrow X}$ be the normalization of ${X}$. We will shorthand ${\mathcal{O}}$ and ${\mathcal{O}'}$ as the structure sheaves of ${X}$ and ${X'}$ respectively.

We have that ${\mathcal{O}_x=\mathcal{O}_x'}$ when ${x\neq P}$. We have ${\mathcal{O}_P}$ is the subring of ${\mathcal{O}_P'}$ formed from functions ${f}$ where the differential ${df}$ vanishes at ${P}$.

Let’s use the standard exact sequence we get from normalizing a curve: ${0\rightarrow \mathcal{O}\rightarrow \mathcal{O}'\rightarrow \mathcal{F}\rightarrow 0}$ where ${\mathcal{F}}$ is concentrated at ${P}$ with the property ${\mathcal{F}_P=k}$. If we take the long exact sequence in cohomology we see that ${ H^0(X, \mathcal{F})\hookrightarrow H^1(X, \mathcal{O})\rightarrow H^1(X, \mathcal{O}')\rightarrow}$. Note that ${X'}$ is non-singular of genus ${0}$, so ${H^1(X', \mathcal{O}')=H^1(X, \mathcal{O})=0}$. Also, ${H^0(X, \mathcal{F})=\mathcal{F}_p=k}$. So ${\mathrm{dim}_k H^1(X, \mathcal{O})=1}$.

Now we can use the standard sequence of restriction ${0\rightarrow \mathcal{O}\rightarrow \mathcal{W}_n\rightarrow \mathcal{W}_{n-1}\rightarrow 0}$ and induction to get that the length of the module ${H^1(X, \mathcal{W}_n)}$ is ${n}$. Now let’s use the normalization sequence above and take Witt sheaves associated to all of them. We’ll denote this by ${0\rightarrow \mathcal{W}_n\rightarrow \mathcal{W}_n'\rightarrow \mathcal{F}_n\rightarrow 0}$.

Note that we still have a bijection with the coboundary map ${\delta: H^0(X, \mathcal{F}_n)\rightarrow H^1(X, \mathcal{W}_n)}$. Let’s now think about the Frobenius map ${F}$. Since our field is perfect, we get a bijection ${\mathcal{W}_n'\rightarrow \mathcal{W}_n'}$ and also between ${\mathcal{W}_n\rightarrow \mathcal{W}_n}$. On ${\mathcal{O}_P'}$ we get that ${F(f)=f^p}$ and hence the differential is ${0}$, which means it is in ${\mathcal{O}_P}$.

Applying Frobenius to our exact sequence we get the square

$\displaystyle \begin{matrix} H^0(X, \mathcal{F}_n) & \rightarrow & H^1(X, \mathcal{W}_n) \\ F \downarrow & & \downarrow F \\ H^0(X, \mathcal{F}_n) & \rightarrow & H^1(X, \mathcal{W}_n) \end{matrix}$

Here we see that ${F: H^1(X, \mathcal{W}_n)\rightarrow H^1(X, \mathcal{W}_n)}$ is identically ${0}$. This means that ${p}$ annihilates ${H^1(X, \mathcal{W}_n)}$ which means that it is not only a length ${n}$ ${\Lambda}$-module, but is a vector space over ${k}$ of dimension ${n}$. Thus the projective limit ${H^1(X, \mathcal{W})}$ is an infinite dimensional vector space over ${k}$ and hence is not a finite type ${\Lambda}$-module.