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Unformatted text preview: 9.4. SPLITTING FIELDS AND AUTOMORPHISMS 427 9.3.7. If K is a ﬁeld of characteristic p Ã a 2 K , then .x C a/p D x p C
p . Hint: The binomial coefﬁcient p is divisible by p if 0 < k < p .
Now, suppose K is a ﬁeld of characteristic p and that f .x/ is an irreducible polynomial in KŒx. If f .x/ has a multiple root in some extension
ﬁeld, then Df .x/ is identically zero, by Exercise 9.3.6. Therefore, there
is a g.x/ 2 KŒx such that f .x/ D g.x p / D a0 C a1 x p C : : : ar x rp .
Suppose that for each ai there is a bi 2 K such that bi D ai . Then
f .x/ D .b0 C b1 x C br x r /p , which contradicts the irreducibility of
f .x/. This proves the following theorem:
Theorem 9.3.2. Suppose K is a ﬁeld of characteristic p in which each
element has a p t h root. Then any irreducible polynomial in KŒx has only
simple roots in any ﬁeld extension. Proposition 9.3.3. Suppose K is a ﬁeld of characteristic p . The map
a 7! ap is a ﬁeld isomorphism of K into itself. If K is a ﬁnite ﬁeld, then
a 7! ap is an automorphism of K . Proof. Clearly, .ab/p D ap b p for a; b 2 K . But also .a C b/p D
ap C b p by Exercise 9.3.7. Therefore, the map is a homomorphism. The
homomorphism is not identically zero, since 1p D 1; since K is simple,
the homomorphism must, therefore, be injective. If K is ﬁnite, an injective
map is bijective.
I Corollary 9.3.4. Suppose K is a ﬁnite ﬁeld. Then any irreducible polynomial in KŒx has only simple roots in any ﬁeld extension. Proof. K must have some prime characteristic p . By Proposition 9.3.3,
any element of K has a p t h root in K and, therefore, the result follows
from Theorem 9.3.2.
I 9.4. Splitting Fields and Automorphisms
Recall that an automorphism of a ﬁeld L is a ﬁeld isomorphism of L onto
L, and that the set of all automorphisms of L forms a group denoted by ...
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