fields and galois theory - Neil Strickland - University of Sheffield

Proof. We write α for the image of x in K, so f(α) = 0. As f(x) is irreducible, we see from Corollary 4.25
that K is a field and K = F p (α). If f(x) has degree s we also see from Proposition 5.2 that K ≃ F s p as vector
spaces over F p , so in particular |K| = p s . As f(x) | ϕ q−1 (x) | x q−1 − 1 | x q − x, we see that α q = α. Here
q = p r and so one checks that σ r (t) = t pr = t q , so we see that σ r (α) = α. Now put K ′ = {a ∈ K | σ r (a) = a}.
We see from Proposition 1.31 that K ′ is a subfield of K = F p (α), and it contains α so it must be all of K.
This means that every element in K is a root of the polynomial g(x) = x q − x. However, g(x) has degree q
and so cannot have more than q roots in any field. We must therefore have |K| ≤ q.
We next consider the order of α in K × . As explained above we have f(x) | x q−1 − 1 and so α q−1 = 1,
so the order of α divides q − 1. Write r for this order, and suppose (for a contradiction) that r < q − 1.
It then follows from Proposition 8.3 that x q−1 − 1 is divisible by (x r − 1)f(x), so Lemma 9.6 tells us that
x r − 1 and f(x) are coprime mod p. This means that there exist polynomials a(x), b(x) ∈ F p [x] with
a(x)(x r − 1) + b(x)f(x) = 1. We now put x = α, remembering that f(α) = 0 = α r − 1, to get 0 = 1, which is
impossible. We must therefore have r = q − 1 instead, so the subgroup of K × generated by α is isomorphic
to C q−1 . On the other hand, we have shown that |K| ≤ q so |K × | ≤ q − 1. This can only be consistent if
K × = 〈α〉 ≃ C q−1 as claimed.
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Example 9.8. [eg-F-eight]
Put f(t) = 1 + t + t 3 and g(t) = 1 + t 2 + t 3 , considered as elements of F 2 [t].
Example 8.4 we see that
f(t)g(t) = 1 + t + t 2 + t 3 + t 4 + t 5 + t 6 = ϕ 7 (t) in F 2 [t].
By direct expansion and
We also claim that f(t) is irreducible over F 2 . Indeed, any nontrivial factorisation of f(t) would involve a
factor of degree one, which would give a root of f(t) in F 2 = {0, 1}. However, we have f(0) = f(1) = 1 so
there is no such root. By the same argument we see that g(t) is also irreducible. It follows that there are
fields K = F 2 [α]/f(α) and L = F 2 [β]/g(β) of order 8.
We next claim that in K we have g(α 3 ) = 0. Indeed, by construction we have f(α) = 0, and f(t) divides
t 7 − 1, so α 7 = 1, which implies α 9 = α 2 . The relation f(α) = 0 can also be rewritten as α 3 = 1 + α, which
squares to give α 6 = 1 + α 2 . It follows that
g(α 3 ) = 1 + α 6 + α 9 = 1 + (1 + α 2 ) + α 2 = 0
as claimed. This means that we can define a homomorphism λ: L → K by λ(β) = α 3 . One can check that
this is actually an isomorphism, with λ −1 (α) = β 5 .
Proposition 9.9. [prop-units-cyclic]
Let K be a field, and let U be a finite subgroup of K × . Then U is cyclic.
Proof. Put U[d] = {x ∈ U | x d = 1}. As a polynomial of degree d can have at most d roots, we see that
|U[d]| ≤ d for all d. The claim thus follows from Lemma 9.11 below.
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Lemma 9.10. [lem-cyclic-test-aux]
Let U be a finite abelian group of order n such that |U[d]| ≤ d for all d. Then |U[d]| = d whenever d divides
n.
Proof. We can define a group homomorphism α: U −→ U by α(x) = x n/d . We note that α(x) d = x n = 1,
so α(x) ∈ U[d], so |U[d]| ≥ | image(α)|. On the other hand, the First Isomorphism Theorem tells us
| image(α)| = |U|/| ker(α)|. Here |U| = n, and it is clear from the definitions that ker(α) = U[n/d], so
| ker(α)| ≤ n/d, so | image(α)| ≥ n/(n/d) = d. Putting this together gives |U[d]| ≥ d, but also |U[d]| ≤ d by
assumption, so |U[d]| = d as claimed. The groups and homomorphisms considered can be displayed in the
following diagram:
U[n/d] U
U/U[n/d]
α
1 U U[d]
50
≃
α
image(α)
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