Show $L$ can be extended to $M$ with $M/F$ cyclic

Question

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?

Answer 1

Assuming $p=2$ here, because the Witt vector arithmetic is simplest in that case.

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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

1. $(0,0)$ is the additive neutral element of $W_2(F)$.
2. $(1,0)$ is the multiplicative neutral element of $W_2(F)$.
3. The additive inverse of $x=(x_0,x_1)$ is $x'=(x_0,x_1+x_0^2)$ because $$x+x'=(x_0+x_0,x_1+[x_1+x_0^2]+x_0^2)=(0,0).$$
4. The mapping $W_2(F)\to F, (x_0,x_1)\mapsto x_0$ is a homomorphism of rings.
5. 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 $$2+2=(0,1)+(0,1)=(0+0,1+1+0\cdot0)=(0,0).$$
6. 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 $W_2(F)$.

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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.