Throughout, let $K$ be a field of characteristic $p\ne 0$ and $E/K$ a cyclic extension of order $p^{m-1}$ with $m >1$. The algebraic closure $\overline K^\mathrm{a}$, the separable algebraic closure $\overline K^{\mathrm{s}}$ are always fixed. We use $\mathbf{F}_p$ to denote the finite field of $p$ elements.
For proposition 2 in the post, let $G$ be the Galois group of the extension of $\overline K^\mathrm{s}/K$ (which is, the projective limit of $\mathrm{Gal}(K’/K))$, with $K’$ running over all finite and separable extension of $K$; see this post for the definition of projective limit). The reader is expected to know how to induce a long exact sequence from a short exact sequence, for example from this post.
In this post (the reader is urged to make sure that he or she has understood the concept of characters and more importantly Hilbert’s theorem 90), we have shown that if $[E:K]=p$, then $E=K(x)$ where $x$ is the zero of a polynomial of the form $X^p-X-\alpha$ where $\alpha \in K$. In this belated post, we want to show that, whenever it comes to an extension of order $p^{m-1}$, we are running into the a polynomial of the form $X^p-X-\alpha$. The theory behind is called Artin-Schreier theory, which has its own (highly non-trivial) nature.
Definition 1. An Artin-Schreier polynomial $A_\alpha(X) \in K[X]$ is of the form
An immediate property of Artin-Schreier polynomials that one should notice is the equation
To see this, one should notice that for $x,y \in K$ we have $(x+y)^p=x^p+y^p$.
With this equation we can easily show that
Proposition 1. If $A_\alpha(X)$ has a root in $K$, then all roots of $A_\alpha(X)$ is in $K$. Otherwise, $A_\alpha(X)$ is irreducible over $K$. In this case, let $x$ be a root of $A_\alpha(X)$, then $K(x)/K$ is a cyclic extension of degree $p$.
Proof. We suppose that $x \in K$ is a root of $A_\alpha(X)$. Then
Therefore, by induction, we see easily that $x, x+1, \cdots, x+p-1$ are roots of $A_\alpha(X)$, all of which are in $K$.
Now we suppose that $A_\alpha(X)$ has no root in $K$. Let $x \in \overline K$ be a root of $A_\alpha(X)$. Then in $\overline K[X]$, the polynomial will be written in the form
because, again due to the equation $A_\alpha(X+Y)=A_\alpha(X)+A_\alpha(Y)-A_\alpha(0)$, we can see that $x,x+1,\dots,x+p-1$ are roots of $A_\alpha$.
By contradiction we suppose that $A_\alpha$ is reducible, say $A_\alpha(X)=f(X)g(X)$ where $1 \le d=\deg f < p$ and $f,g \in K[X]$. It follows that
where $\{n_1,\dots,n_d\} \subset \{1,2,\cdots,p\}$. If we expand the polynomial above, we see
Therefore $\left(\sum_{j=1}^{d}n_j-dx\right) \in K$ which is absurd because we then have $x \in K$. Therefore we see that $A_\alpha$ is irreducible.
To see that $K(x)/K$ is Galois, we first notice that this extension is normal : $K(x)$ contains all roots of $A_\alpha(X)$. This extension is separable because all roots of $A_\alpha(X)$, namely $x,x+1,\dots,x+p-1$, are pairwise distinct, i.e. $A_\alpha(X)$ has no multiple roots.
Finally, to see why the Galois group of $K(x)/K$ is cyclic, we notice the action of the Galois group $G$ over the roots of $A_\alpha(X)$. Since $A_\alpha(X)$ is irreducible, there exists $\sigma \in G$ such that $\sigma(x)=x+1$. We see easily that $\sigma^j(x)=x+j$ so $\sigma$ generates $G$ which has period $p$. $\square$
The correspondence between extensions of degree $p$ and polynomials of the form $X^p-X-\alpha$ inspires us to consider them in a distinguished manner.
Definition 2. The field extension $E/K$ is called an Artin-Schreier extension if $E=K(x)$ for some $\alpha \in L \setminus K$ such that $x^p-x\in K$.
Consider the map $\wp:\overline K^\mathrm{s} \to \overline K^\mathrm{s}$ defined by $u \mapsto u^p-u$. We certainly want to find the deep relation between Artin-Schreier extensions of a given field $K$ and the map $\wp$. One of the key information can be found through the following correspondence.
Proposition 2. There is an isomorphism $\operatorname{Hom}(G,\mathbf{F}_p) \cong K/\wp(K)$.
Proof. We first notice that $\wp$ is a $G$-homomorphism, that is, it commutes with the action of $G$ on $\overline K^\mathrm{s}$. Indeed, for any $x \in \overline K^\mathrm{s}$ and $g \in G$, we have
On the other hand, $\wp$ is surjective. Indeed, for any $a \in \overline{K}^\mathrm{s}$, the equation $X^p-X=a$ always has a solution in $\overline K^\mathrm{s}$ because the polynomial $X^p-X-a$ is separable.
We can also see that the kernel of $\wp$ is $\mathbf{F}_p$. This is because the splitting field of $X^p-X$ is the field of $p^1$ elements, which has to be $\mathbf{F}_p$ itself. Therefore we have obtained a short exact sequence
where $\iota$ is the embedding. Taking the long exact sequence of cohomology, noticing that, by Hilbert’s Theorem 90, $H^1(G,\overline{K}^\mathrm{s})=0$, we have another exact sequence
where the first arrow is induced by $\wp$ and the second by $\iota$. Therefore we have $\operatorname{Hom}(G,\mathbf{F}_p) \cong K/\wp(K)$. One can explicitly show that there is a surjective map $K \to \operatorname{Hom}(G,\mathbf{F}_q)$ with kernel $\wp(K)$ that defines the isomorphism. For $c \in K$, one solves $x^p-x=c$, then $\varphi_c:g\mapsto g(x)-x$ is the desired map. The key ingredient of the verification involves the (infinite) Galois correspondence, but otherwise the verification is very tedious. We remark that for any $\varphi \in \operatorname{Hom}(G,\mathbf{F}_p)\setminus\{0\}$ and put $H=\ker\varphi$. Then $K^H/K$ is an Artin-Schreier extension with Galois group $G/H$ and on the other hand $H=\mathrm{Gal}(\overline K^\mathrm{s}/K^H)$. $\square$
We conclude this post by showing that, under a certain condition, one can find an Artin-Schreier extension $L/E$ such that $L/K$ is cyclic of order $p^m$.
Lemma 1. Let $\beta \in E$ be an element such that $\operatorname{Tr}_K^E(\beta)=1$, then there exists $\alpha \in K$ such that $\sigma(\alpha)-\alpha = \beta^p-\beta$, where $\sigma$ is the generator of $\operatorname{Gal}(E/K)$.
Proof. Notice that $\operatorname{Tr}_K^E(\beta^p)=\operatorname{Tr}_K^E(\beta)^p=1$, which implies that $\operatorname{Tr}_K^E(\beta^p-\beta)=0$. By Hilbert’s theorem 90, such $\alpha$ exists. $\square$
Lemma 2. The polynomial $f(X)=X^p-X-\alpha$ is irreducible over $E$; that is, let $\theta$ be a root of $f$, then $E(\theta)$ is an Artin-Schreier extension of $E$.
Proof. By contradiction, we suppose that $\theta \in E$. By Artin-Schreier, all roots of $f$ lie in $E$. In particular, $\sigma(\theta)$ is a root of $f$. Therefore
which implies that
It follows that $\sigma\theta-\theta-\beta$ is a root of $g(X)=X^p-X$. This implies that $\sigma\theta-\theta-\beta\in\mathbf{F}_p \subset K$ and therefore
However, by assumption and Artin-Schreier, $\sigma\theta-\theta \in \mathbf{F}_p \subset K$ we therefore have $\operatorname{Tr}_K^E(\sigma\theta-\theta)=0$ and finally
which is absurd. $\square$
Proposition 3. The field extension $K(\theta)/K$ is Galois, cyclic of degree $p^m$ of $f$, whose Galois group is generated by an extension $\sigma^\ast$ of $\sigma$ such that
Proof. First of all we show that $K(\theta)=E(\theta)$. Indeed, since $K \subset E$, we have $K(\theta) \subset E(\theta)$. However, since $\theta \not \in E$, we must have $K \subset E \subsetneq K(\theta)$. Therefore $p=[E(\theta):K(\theta)][K(\theta):E]$, which forces $E(\theta)$ to be exactly $K(\theta)$.
Let $h(X)$ be the minimal polynomial of $\theta$ over $K$ of degree $p^m$. Then we give an explicit expression of $h$. Notice that since $f(X)$ is the polynomial of $\theta$ over $E$ of degree $p$, we must have $f(X)|h(X)$. For any $k$, we see that $f^{\sigma^k}(X)|g^{\sigma^k}(X)$ too. However, since $\sigma$ fixes $K$, we must have $g^{\sigma^k}(X)=g(X)$, from which it follows that $f^{\sigma^k}(X)|g(X)$ for all $0 \le k \le p^{m-1}-1$. Since the degree of each $f^{\sigma^k}(X)$ is $p$, we obtain
Knowing that $\theta$ is a root of $g$, we see that $\theta+\beta$ is a root of $g(X)$ too because
and by induction we see that for $0 \le k \le p^{m-1}-1$, $f^{\sigma^k}(X)$ has a root in the form
By Artin-Schreier, all roots of $f^{\sigma^k}(X)$ lie in $E(\theta)$ and therefore $h(X)$ splits in $E(\theta)$. Since $E(\theta)/E$ is separable, $E/K$ is separable, we see also $E(\theta)/K$ is separable, which means that $E(\theta)=K(\theta)$ is Galois over $K$.
To see why $K(\theta)/K$ is cyclic, we consider an homomorphism $\sigma^\ast$ of $K(\theta)$ such that $\sigma^{\ast}|_E=\sigma$ and that $\sigma^\ast(\theta)=\theta+\beta$. It follows that $\sigma^\ast \in \operatorname{Gal}(K(\theta)/K)$ because its restriction on $K$, which is the restriction of $\sigma$ on $K$, is the identity. We see then for all $0 \le n \le p^{m}$, one has
In particular,
from which it follows that $(\sigma^\ast)^{p^{m-1}}$ has order $p$, which implies that $\sigma^\ast$ has order $p^m$, thus the Galois group is generated by $\sigma^\ast$. $\square$
此内容由惯性聚合(RSS阅读器)自动聚合整理，仅供阅读参考。 原文来自 — 版权归原作者所有。