Show $L$ can be extended to $M$ with $M/F$ cyclic Suppose that $F$ has characteristic $p$ and $L/F$ is a cyclic extension of degree $p$. I'm trying to show that $L$ can be extended to $M$ where $F\subset L\subset M$ with $M/F$ cyclic of degree $p^2.$ The hint says we can use the theory of Witt 2-vectors, but I searched everything I have and find nothing about the theory of Witt 2-vectors. What is that and how that can prove this statement?
 A: Assuming $p=2$ here, because the Witt vector arithmetic is simplest in that case.

Reviewing the key properties of Witt vectors of length two.
So if $F$ is any field of characteristic two, the ring $W_2(F)$ of Witt vectors of length $2$ over $F$ is the set $F^2$ equipped with the addition
$$
(x_0,x_1)+(y_0,y_1)=(x_0+y_0,x_1+y_1+x_0y_0)
$$
and multiplication
$$
(x_0,x_1)\cdot(y_0,y_1)=(x_0y_0,x_1y_0^2+y_1x_0^2).
$$
The operations on the r.h.s. involving individual elements of $F$ are those of $F$.
So we see that 


*

*$(0,0)$ is the additive neutral element of $W_2(F)$.

*$(1,0)$ is the multiplicative neutral element of $W_2(F)$.

*The additive inverse of $x=(x_0,x_1)$ is $x'=(x_0,x_1+x_0^2)$ because
$$x+x'=(x_0+x_0,x_1+[x_1+x_0^2]+x_0^2)=(0,0).$$

*The mapping $W_2(F)\to F, (x_0,x_1)\mapsto x_0$ is a homomorphism of rings.

*The ring $W_2(F)$ is of characteristic four, because
$$1+1=(1,0)+(1,0)=(1+1,0+0+1\cdot1)=(0,1)=2$$ and
$$2+2=(0,1)+(0,1)=(0+0,1+1+0\cdot0)=(0,0).$$

*The construction of Witt vectors of length two is a functor from the category of fields of characteristic two to the category of rings. In particular componentwise operation of an endomorphism of $F$ gives a ring endomorphism of $W_2(F)$.



Recall that in characteristic two a cyclic extension such as $L/F$ of degree two is always an Artin-Schreier equation. In other words, there exists an element $z_0\in F$ such that $z_0$ is not of the form $x^2-x$ for any $x\in F$, and $L=F(x_0)$ where 
$$x_0^2-x_0=z_0.\qquad(*)$$
The non-trivial $F$-automorphism $\sigma$ of $L$ is fully determined by $\sigma(x_0)=x_0+1$.
Observe that the left side of $(*)$ is really the difference $\phi(x_0)-x_0$, where $\phi$ is the Frobenius automorphism. Now let's replace $(*)$ with a Witt vector equation
$$
\phi(x_0,x_1)-(x_0,x_1)=(z_0,z_1),
$$
or, writing out the Frobenius
$$
(x_0^2,x_1^2)-(x_0,x_1)=(z_0,z_1).\qquad(**)
$$
In the l.h.s. the Witt vector is $(x_0^2+x_0,x_1^2+x_1+x_0^2+x_0^3)$, so this
is equivalent to the pair of equations over $F$
$$
\left\{\begin{array}{lcl}
x_0^2+x_0&=&z_0\\
x_1^2+x_1&=&z_1+x_0^2+x_0^3\end{array}\right.
$$
So we see that with $L=F(x_0)$ and $M=L(x_1)$ both $L/F$ and $M/L$ are Artin-Schreier extensions provided that $z_0$ is not of the form $x^2+x, x\in F$ (given) and that $z_1$ is not of the form $x^2+x+x_0^2+x_0^3$ for any $x\in L$
(need to check that $z_1\in F$ can always be chosen so that this is the case).
Given all this we see that $M/F$ is then cyclic of order four. This is because the Witt vector $(1,0)$ is a fixed point of the Frobenius automorphism, and consequently
$$
(x_0,x_1)+(1,0)=(x_0+1,x_1+x_0)
$$
is a solution of $(**)$ whenever $(x_0,x_1)$ is.
In other words the above $F$-automorphism $\sigma$ of $L$ extends to an $F$-automorphism of $M$ by declaring $\sigma(x_1)=x_1+x_0$. Because $W_2(F)$ has characteristic four, we see that $\sigma$ is of order four as well. Not unexpectedly $\sigma^2$ then fixes the intermediate field $L$ pointwise.
