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.