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 $\mathbb{Z}/p^2$?
 A: 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.
