Rings, Modules, Fields
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15 First Applications
Let’s understand finite fields. They are finite extensions of \(\FF _p\), and we have already seen in the proof of the primitive element theorem that every element is the root of \(x^{p^k}x\). Since this polynomial is separable, our
field is the splitting field of \(x^{p^k}x\). Indeed such a field exists, as given the splitting field, the roots of \(x^{p^k}x\) form a subfield (since it is the fixed points of \(Fr^k\)) which cannot be proper, so there is a unique finite
field of size \(p^k\) for each \(p,k\). It is degree \(k\), and \(Fr\) generates its cyclic Galois group. Thus we get the following proposition:

Proposition 15.1. There is a unique finite field of size \(p^k\) for each \(k\) and prime \(p\), which is
the splitting field of the polynomial \(x^{p^k}x\). We can denote the field \(\FF _{p^k}\), and it’s generators also generate the multiplicative group. An extension of finite fields is of the form \(\FF _{p^{nk}}/\FF _{p^k}\),
and is cyclic with generator \(Fr^k\). The union of all these finite extensions is the algebraic closure, and the absolute Galois group is \(\hat {\ZZ }\).
We can use Galois theory to get an algebraic proof of the fundamental theorem of algebra:

Proof. Suppose we have a finite Galois extension of \(\CC \). If its degree is a power of \(2\), there is an index two subgroup of its Galois group, corresponding to a degree \(2\)
extension, which cannot exist as \(\CC \) has all of its square roots. If the degree is not a power of \(2\), a \(2\)Sylow corresponds to an odd extension, which means there is an irreducible odd degree polynomial, but any odd
degree polynomial has a root in \(\RR \). □
We can also get a nice proof of the primitive element theorem.

Proof. We have already dealt with finite fields, so we assume the fields are infinite. If there are finitely many intermediate extensions of \(L/K\), this follows since any element not in the
union of these intermediate extensions is primitive. Conversely if \(\alpha \) is a primitive element over \(K\), \(\alpha \) has a minimal polynomial over \(K\). \(\alpha \)’s minimal polynomial over any intermediate extension
divides that, and there are finitely many factors in the algebraic closure, hence finitely many intermediate fields. □
A good first example of extensions we can study are cyclotomic extensions, where we adjoin roots of unity.
The cyclotomic polynomial \(\phi _n(x)\) is the minimal polynomial over \(\QQ \) of a primitive \(n^{th}\) root of unity, \(\zeta _n\). We will see that it can be constructed explicitly (and depends only on \(n\))
by the equation \(\Pi _{dn}\phi _d(x) = x^n1\). To do this we will need the discriminant of a polynomial \(f \in K[x]\), which is given by \(a^{2\deg (f)2}\prod _{i<j} (\alpha _i\alpha _j)^2\),
where \(\alpha _i\) are the roots of \(f\) in the algebraic closure counted with multiplicity, and \(a\) is the leading term of \(f\). Note that the formula yields \(0\) for an inseparable polynomial, and is invariant under the
Galois action, so gives an element of \(K\) for a separable polynomial.

Proof. For positive characteristic, since we know the multiplicative group is cyclic of order \(p^l1\), if there is a root, \(kp^l1\), and conversely if \(kp^l1\), then
\(x^k1x^{p^l}x\), of which \(\FF _{p^l}\) is the splitting field.
For characteristic \(0\), we will first prove that if \(\phi _n(\zeta _n) = 0\), \(\phi _n(\zeta _n^p) = 0\) for \(p \nmid n\) since primes generate \(\ZZ /n\ZZ ^\times \). To see this we will first compute the discriminant
of \(x^n1\).
\[ \prod _{i<j}(\zeta _n^i\zeta _n^j)^2 = (1)^{\frac {n(n1)}{2}}\prod _{i\neq j}\zeta _n^i(1\zeta _n^{ji}) = (1)^{\frac {n(n1)}{2}}\prod _{i}\zeta _n^i\prod _{j\neq 0}(1\zeta
_n^{j})\]
\[ = (1)^{\frac {n(n1)}{2}}\prod _{i}\zeta _n^in = n^n(1)^{\frac {(n+2)(n1)}{2}} \]
Now if \(\zeta _n^p\) isn’t a root of \(\phi _n\) then \(\phi _n(\zeta _n^p) n^n\) because \(\phi _n(x) = \prod (x\zeta _n^{k_i})\) for some \(k_i\) (the \(\) makes sense in the integral closure of \(\ZZ \)).
However, \(\phi _n(\zeta _n^p) \equiv \phi _n(\zeta _n)^p = 0 \pmod {p}\), so \(p\) divides \(n^n\), a contradiction. □
A classical problem is determining whether or not an irreducible polynomial \(f\) has a root \(\alpha \) that can be expressed in the field \(K\) via radicals. Note that a radical is a extension, so that we would like to know when
\(K(\alpha )/K\) is contained in an extension that is a chain of radical extensions, or extensions of the form \(L(a^{\frac {1}{l}})\). Now that we understand roots of unity a bit, we can use Galois theory to answer
this question. We say that the Galois group of a polynomial \(f\) is the automorphism group of its splitting field.
Unfinished, need Kummer theory

Proof. Note we can assume the polynomial is separable, as the purely inseparable part will not change the automorphism group, and will not change the ability to write a root in terms of
radicals. First suppose the Galois group of the splitting field is solvable. Then we have a chain of extensions \(K_n/K_{n1}/\dots K_0 =K\) such that each extension is abelian. For the converse, suppose that we have some cyclic
extensions \(K_n/K_{n1}/\dots K_0 = K\) containing \(K(\alpha )\). Then if \(K_n\) is degree \(l\), consider the chain \(K_n(\zeta _l)/K_{n1}(\zeta _l)/\dots K_0(\zeta _l)/K_0\). Each of these extensions is still
radical, but now they are all abelian extensions, since the conjugates of \(a^{1/m}\) are products of it and a root of unity, so the Galois group is a subgroup of \(\ZZ /m\ZZ \). Now the Galois group of \(f\) is a quotient of a
solvable group, hence solvable. □
A nice property of finite Galois extensions is that they have a normal basis, that is a basis consisting of conjugates.

Proof. Suppose this were the case, and choose a relation \(\sum _i a_i\chi _i = 0\) with the minimal number of nonzero \(a_i\) Now given any two \(\chi _i,\chi _j\) whose
\(a_i,a_j\) are nonzero, find a \(h\) such that \(\chi _i(h) \neq \chi _j(h)\). Then \(\sum _i a_i\chi _i(gh) = \sum _i a_i\chi _i(g)\chi _i(h) = 0\), so we can subtract a multiple of the original relation from this
to get one with fewer \(\chi _i\), a contradiction. □

Proof. Certainly the condition is necessary. First assume the extension is cyclic, so that we can view \(L/K\) as a \(K[\sigma ]\) module where \(\sigma \) generates the Galois group.
\(\sigma ^n1\) annihilates the module, and by linear independence of characters no smaller polynomial of \(\sigma \) can. Thus \(L \cong K[\sigma ]/(\sigma ^n1)\) as modules, so any generating element produces a
normal basis.
Now assume the field is infinite. It suffices to find an \(\alpha \in L\) that satisfies the property \(\sigma _i\sigma _j(\alpha )\) is an invertible matrix, as then any relation \(\sum c_i\sigma _i = 0\) can be multiplied by
each \(\sigma _j\) to see the \(c_i\) are 0. Choose a basis \(e_i\) of \(L/K\), and consider the polynomial \(\Delta (c_1,\dots ,c_n) = \det (\sigma _i\sigma _j(\sum c_ie_i))=\det (\sum c_i\sigma _i\sigma
_j(e_i))\). If the polynomial \(\Delta \) is nonzero, since \(K\) is infinite, there is a choice of \(c_i \in K\) that \(\Delta \) is nonzero on, providing a normal basis. By linear independence of characters, \((\sigma
_i(e_1),\dots , \sigma _i(e_n))\) are linearly independent in \(L^n\), so \((\sigma _1(e_i),\dots , \sigma _n(e_i))\) are too. Then we can choose some \(c_i \in L\) such that \(\sum _k c_k\sigma _i(e_k) =
\delta _{i,1}\). Hence \(\Delta (c_1,\dots ,c_n) \neq 0\), so \(\Delta \neq 0\).
□
Normal bases can help explicitly find primitive elements of intermediate extensions. We call an element that is part of a normal basis a normal element. We say that the trace of an element \(\alpha \) of a
finite extension \(L/K\) is the trace of multiplication by that element as a \(K\)linear map, and the norm is the determinant of the same endomorphism. Note that when the extension is Galois, we can write the trace as
\(\Tr _{L/K}(\alpha ) = \sum _{\tau \in \Gal (L/K)}\tau (\alpha )\) and the norm as \(\N _{L/K}(\alpha ) = \prod _{\tau \in \Gal (L/K)}\tau (\alpha )\).

Proposition 15.11. If \(\alpha \) is a normal element of \(M/K\), and \(L\) is an intermediate
extension, then \(\Tr _{M/L}(\alpha )\) is a primitive element. If \(L\) is a normal, then this is a normal element.