# Witt vectors and flat liftings of (non)perfect fields

My motivating question is as follows: Why does Theorem 2.1. in Deligne-Illusie's classical work on the Hodge degeneration (EUDML link), i.e. the decomposition theorem, does $$k$$ need to be a perfect field?

As far as I see is what they actually need that $$W_2(k) \rightarrow \mathbb{Z}/p^2$$ is flat, so my more precise question is: How does flatness of $$W_2(k) \rightarrow W_2(\mathbb{F}_p)$$ relate to perfectness of $$k$$? Is there any good reference for (flat) liftings over $$\mathbb{Z}/p^2$$?

This answer is a response to your first question about relating Witt vector flatness with perfectness. The second question (about flat liftings to $$\mathbf{Z}/p^2$$) has a rich literature, see papers that cite Deligne-Illusie (e.g., on mathscinet).

Theorem: For any $$\mathbf{F}_p$$-algebra $$R$$, the flatness of $$\mathbf{Z}/p^2 \to W_2(R)$$ is equivalent to the perfectness of $$R$$.

So Deligne-Illusie is in optimal generality.

Let me explain the non-trivial implication. Assume $$W_2(R)$$ is flat over $$\mathbf{Z}/p^2$$. This implies that the image of multiplication by $$p$$ on $$W_2(R)$$ coincides with the kernel of multiplication by $$p$$. In Witt vector notation, we are saying the following:

$$(*)$$ If $$x \in W(R)$$ satisfies $$px \in V^2W(R)$$, then $$x \in pW(R) + V^2W(R)$$.

As $$R$$ is an $$\mathbf{F}_p$$-algebra, we have $$p = V(1)$$, so $$px = V(1)x = VF(x)$$. So the condition that $$px \in V^2W(R)$$ appearing above is equivalent to $$VF(x) \in V^2 W(R)$$. As $$V$$ is always injective, this just means $$F(x) \in VW(R)$$. We have also seen here that $$pW(R) = VFW(R)$$, so $$(*)$$ simplifies to

$$(**)$$ If $$x \in W(R)$$ satisfies $$F(x) \in VW(R)$$, then $$x \in VFW(R) + V^2 W(R) \subset VW(R)$$.

Now let's use $$(**)$$ to see that $$R$$ must be perfect, i.e., its Frobenius is bijective.

Injectivitity of Frobenius: If $$x \in R$$ satisfies $$x^p = 0$$, then $$[x] \in W(R)$$ satisfies $$F([x]) = 0$$. Using $$(**)$$, we get $$[x] \in VW(R)$$. But $$VW(R)$$ is the kernel of the restriction map $$W(R) \to R$$, so the image $$x \in R$$ of $$[x] \in W(R)$$ must be $$0$$.

Surjectivity of Frobenius: Given $$x \in R$$, we have $$pV([x]) = V(1)V([x]) = VFV([x]) = V^2F([x]) \in V^2 W(R)$$ since $$VF = FV$$ as $$R$$ has characteristic $$p$$. Applying $$(**)$$ gives $$y_1,y_2 \in W(R)$$ such that $$V([x]) = VF(y_1) + V^2(y_2)$$. As $$V$$ is injective, this gives $$[x] = F(y_1) + V(y_2)$$. Applying the restriction map down to $$R$$ (which has kernel $$VW(R)$$) shows that $$x$$ is a $$p$$-th power.