### An Example of an Inseparable Irreducible Polynomial Suppose t is

```An Example of an Inseparable Irreducible Polynomial
Suppose t is an indeterminate and let K = Z3 (t). Then since K/(t) ∼
= Z3 , the ideal (t) is
maximal, hence prime, and therefore t is prime. (Theorems III.2.19 and III.3.4 in [Hun].) As
a result, the polynomial f (x) = x6 + tx3 + 2t is irreducible in K[x] by Eisenstein’s Criterion
(Theorem II.6.15 in [Hun]).
We notice that f (x) = x6 + tx3 + 2t = (x3 )2 + t(x3 ) + 2t = g(x3 ), where g(x) = x2 + tx + 2t—
which is also irreducible in K[x] by Eisenstein. Suppose g(x) = (x − r)(x − s) ∈ F [x] for
r, s ∈ F , some extension of K. Then f (x) = (x3 − r)(x3 − s) ∈ F [x]. Suppose α ∈ F˜ is a
root of f , where F˜ is an extension—possibly trivial—of F . Then r = α3 or s = α3 . In the
former case, x3 − r = x3 − α3 = (x − α)3 , using the fact that the characteristic of K is 3.
(See (2) below.) In any case, the final splitting of f will be f (x) = (x − α)3 (x − β)3 . So
although f is an irreducible polynomial of degree 6 in Z3 (t)[x], f has only two distinct roots
in its splitting field.
So, do we really need to introduce indeterminates to get an example of an inseparable
irreducible polynomial? The answer, unfortunately, is “yes.” Suppose, for example, we took
g(x) = x2 + x + 2. This polynomial is irreducible in Z3 [x]. (Why?) Now take f (x) = g(x3 ) =
x6 + x3 + 2 ∈ Z3 [x]. Now we could try to apply the technique from the previous paragraph.
The problem here is that f (x) is not irreducible:
x6 + x3 + 2 = (x2 )3 + x3 + 2 = (x2 )3 + x3 + 23 = (x2 + x + 2)3 ,
(1)
the last equation following from (2) below.
In fact, we have the following
Theorem. Let K be a field of characteristic p. If f (x) is monic irreducible in K[X] then
the following conditions are equivalent:
m
i. f (xp ) is irreducible in K[x] for all m ≥ 1,
ii. f (x) ∈
/ K p [x].
But for any finite field K of characteristic p, the Corollary below implies that K = K p and
thus that K[x] = K p [x], by (2).
Theorem (Freshman’s Dream). If R is a commutative ring of prime characteristic p, then
ϕ : R → R defined by ϕ(r) = rp is a ring homomorphism.
Proof.
p
ϕ(r1 + r2 ) = (r1 + r2 ) =
p X
p
i=0
i
r1i r2p−i = r1p + r2p = ϕ(r1 ) + ϕ(r2 ),
(2)
since all but the first and last terms of the sum have coefficients divisible by the prime p.
Also,
ϕ(r1 r2 ) = (r1 r2 )p = r1p r2p ,
since R is commutative.
Corollary. If K is a finite field with characteristic p, then ϕ : K → K defined by ϕ(r) = rp
is a field automorphism.
Proof. The theorem says that ϕ is a field homomorphism, which is nonzero since ϕ(1) = 1.
Then since fields have no proper ideals, ϕ must be one-to-one. But one-to-one functions
from finite sets to themselves must also be onto.
References
[Hun] Thomas W. Hungerford, Algebra, Springer-Verlag, New York, 1974.
c R. Kubelka 2014
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