## Tag Archives: field of characteristic p

### One example of inseparable polynomial

$\mathbf{Theorem:}$ $F$ is a field. If $F[x]$ contains an inseparable polynomial, then $F$ can not have characteristic 0 or $F$ is some finite field with characteristic $p$.

In general, we do not need to worry about the separability of the extension field, because we often deal with finite fields or the one of characteristic 0.

An example of inseparable polynomial
Let $K$ be a field of characteristic $p$, and $F=K(x)$ is the field of rational polynomials, $x$ is the indeterminate. Consider $F[y]$, y is the indeterminate, $x$ is an irreducible element in this UFD. So $f(y)=y^p-x$ is irreducible in $E$ by the Einsenstein criterion. There exists a splitting field of $f(y)$ over $F$. Suppose $\sigma$ is root of $f(y)$ then $\sigma^p=x$. By freshman’s dream $f(y)=(y-\sigma)^p$. So $f$ has multiple roots. It is not separable.

$\mathbf{Remark:}$ Isaacs, p281

### Automorphism of Z_p(t) and its invariant field

$\mathbf{Problem:}$ Let $E=\mathbb{Z}_p(t)$ where $t$ is transcedental over $\mathbb{Z}_p$. Let $G$ be the group of automorphism generated by the automorphism of $E$ such that $t\to t+1$. Determine $F=\text{Inv}G$ and $[E:F]$.

$\mathbf{Proof:}$ Suppose $\sigma$ is the automorphism of $E$ that $t\to t+1$.  Group $G$ is generated by $\sigma$. It is easy to verify that $G$ is the cyclic group $\{1,\sigma,\cdots,\sigma^{p-1}\}$. So If $F=\text{Inv}G$, then $[E:F]=|G|=p$.

Let’s determine $F$. Note that $\mathbb{Z}_p(u)$ where $u=t^p-t$ is invariant under $\sigma$, because $a^p=a$, for $\forall \,a\in \mathbb{Z}_p$.  So $\mathbb{Z}_p(u)\subset F$.

On the other hand, $\mathbb{Z}_p(u)$ is a field of characteristic $p$ and $E$ is a splitting field of $x^p-x-u$ over $\mathbb{Z}_p(u)$. $x^p-x-u$ has no root in $\mathbb{Z}_p(u)$, thus it is irreducible on $\mathbb{Z}_p(u)[x]$ by a previous conclusion. So $[E:\mathbb{Z}_p(u)]=p=[E:F]$, which means $F=\mathbb{Z}_p(u)$ where $u=t^p-t$.

$\text{Q.E.D}\hfill \square$

$\mathbf{Problem:}$ What if $G$ is replaced by the group of automorphism such that $t\to at+b$, $a,b\in\mathbb{Z}_p$, $a\neq 0$

If $a=1$, $b=0$, then $F=E$ and $[E:F]=1$

If $a=1$, $b\neq 0$, then the conclusion is the same with previous problem.

If $a\neq 1,0$, $|G|=p-1=[E:F]$, $F=\mathbb{Z}_p(u)$ where $\displaystyle u=\prod\limits_{i=1}^{p-1}\sigma^{i}(t)$.

$\mathbf{Remark:}$ Jacobson, Algebra I p243.

### f=x^p-x-a in a field of characteritic p

$\mathbf{Problem:}$ Let $F$ be of characteritic $p$ and let $a\in F$. Show that $f(x)=x^p-x-a$ has no multiple roots and that $f(x)$ is irreducible in $F[x]$ if and only if $a\neq c^p-c$ for any $c\in F$.
$\mathbf{Proof:}$ $f'(x)=-1$, so $(f(x),f'(x))=1$ which means $f(x)$ is separable.
For the second part. If $f$ is irreducible then immediately it has no roots in $F$. Conversely, suppose $c$ is a root of $f$ in some splitting field, $c$ is not in $F$. Then by Freshman’s dream, one can verify $c+1,c+2,\cdots, c+p-1$ are also roots of $f$.

$\displaystyle f(x)=(x-c)(x-c-1)\cdots(x-p+1)$

The irreducible polynomial $g(x)$ of $c$ is a factor of $f(x)$. $g(x)$ must contain all the roots of $f(x)$, otherwise the sum of these roots will results in $c\in F$. So $f(x)=g(x)$ is irreducible.
$\mathbf{Remark:}$ Jacoboson, Algebra I, p234.