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Notes on Galois theory
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GALOIS THEORY
1. Basic Ring Theory: definitions and examples
In this section we recall some basic definitions from rings and discuss some examples.
Recall: A group is a nonempty set G with a binary operation · : G×G→ G satisfying
(1) · is associative, i.e. a · (b · c) = (a · b) · c, for any a, b, c ∈ G.
(2) there exists 1 ∈ G such that 1 · a = a = a · 1 for all a ∈ G.
(3) for any a ∈ G, there exists a−1 such that a · a−1 = a−1 · a = 1.
Furthermore, (G, ·) is said to be abelian if a · b = b · a for any a and b.
Remark:
(1) In the definition of a group above, the element 1 is unique and is called the
identity of G. Given g ∈ G, the element g−1 is also unique and is called the
inverse of g.
(2) We often use + to denote the group operation for an abelian gorup and in this
case we denote the identity by 0 and the inverse of a by −a.
Definition 1.1. A ring is a set R with two binary operations + (addition) and ·
(multiplication) satisfying
(1) (R,+) is an abelian group.
(2) (R, ·) is a semigroup with an identity 1, i.e.,
– the operation is associative: a · (b · c) = (a · b) · c, for any a, b, c ∈ R;
– d · 1 = 1 · d = d, for any d ∈ R.
(3) (R,+, ·) satisfies the distributive laws, i.e. for any a, b, c ∈ R
– a · (b+ c) = a · b+ a · c;
– (b+ c) · a = b · a+ c · a.
A ring is said to be commutative if rs = sr, for any r, s ∈ R.
1
2 GALOIS THEORY
Remark:
(1) In Definition 1.1, the identity 1 is unique and is called the mutiplicative identity
of R, and 0 is called the additive identity (or the zero) of R.
(2) In this course, all rings, unless otherwise specified, are commutative and dif-
ferent from the trivial ring {0}.
(3) The symbol · for the multiplication operation is often omitted.
Definition 1.2. (1) A nonempty subset I ⊆ R is an ideal if −a, a+ b, ra ∈ I for
any r ∈ R, a, b ∈ I.
(2) An ideal I is called a principal ideal if I is generated by a single element a, i.e.
I = Ra = {ra|r ∈ R}.
Definition 1.3. A commutative ring R is
(1) an integral domain (ID), if R 6= {0}, and rs = 0 implies r = 0 or s = 0;
(2) a principal ideal domain (PID), if R is an ID, and any ideal of R is a principal
ideal.
(3) a field, if (R\{0}, ·) is a group.
Example 1.4. Z,Q,R,C,Z[x],C[x],C[x, y],Z2,Z4.
(1) All but Z4 are IDs.
(2) PIDs: Z,Q,R,C,C[x],Z2.
(3) Fields: Q,R,C,Z2.
Definition 1.5. (1) A subset S ⊆ R is a subring of R, if (S,+) is a subgroup of
(R,+), S is closed under the mulitplication · and S contains the identity 1.
(2) A nonzero subring S of a field F is a subfield of F if S is closed under taking
multiplicative inverses of nonzero elements.
Example 1.6. Among the rings in Example 1.4, we have
(1) a sequence of subrings: Z ⊆ Q ⊆ R ⊆ C ⊆ C[x] ⊆ C[x, y].
(2) a sequence of subfields: Q ⊆ R ⊆ C.
GALOIS THEORY 3
2. The polynomial ring F[x]
Let F be a field. In this section we recall some basic definitions on polynomials and
discuss the main properties of the polynomial ring F[x]. We write a polynomial f in
F[X] as f = anxn + an−1x
n−1 + · · ·+ a1x+ a0, where ai ∈ F.
2.1. Definitions and main properties. The degree of a polynomial f = anxn +
an−1xn−1 + · · ·+ a1x+ a0 is defined as
deg f =
m if f 6= 0 and m = max{i|ai 6= 0}
−∞ otherwise.
For any two polynomials f and g ,
deg fg = deg f + deg g.
Recall: In an integral domain R, a nonzero element r is called a unit if there exists
s ∈ R such that rs = 1 and a nonzero nonunit element r is called an irreducible
element if r = st implies that s or t is a unit. If it is in the polynomial ring F[x] over
a field F, an irreducible polynomial can be understood as follows.
Definition 2.1. Let f = anxn + an−1x
n−1 + · · · + a1x + a0 ∈ F[x] be a polynomial
with an 6= 0.
(1) f is irreducible over F (or in F[X]) if degf ≥ 1 and f = gh implies that either
f or g has degree 0.
(2) f is monic if an = 1.
For any f, g, h ∈ F[x], if f = gh, then we say that g divides f or f is divisible by g,
denoted by g|f , and call g a factor of f .
Definition 2.2. A polynomial d is called the greatest common divisor (gcd) of two
given polynomials f and g if d is monic, d|f, d|g, and that c|f and c|g implies c|d.
Theorem 2.3. The polynomial ring F[X] has the following important properties.
(1) (Division with remainder) For f, 0 6= g ∈ F[X], there exist r, s ∈ F[X] such
that f = gs+ r and deg r < deg g.
4 GALOIS THEORY
(2) F[X] is a PID and thus a unique factorisation domain (UFD), i.e., any 0 6=
f ∈ K[X] can be uniquely factorised as f = uf1 · · · fn up to the reordering of
f1, · · · , fn, where u has degree 0, and f1, · · · , fn are monic irreducible over F.
(3) Suppose that d, f, g ∈ F[x] and d is the gcd of f and g, then there exist
polynomials h, l ∈ F[x] such that
d = fh+ gl.
Definition 2.4. An element a ∈ F is called a root of f ∈ F[X] if f(a) = 0.
Lemma 2.5. Let f ∈ F[x]. An element a ∈ F is a root of f ∈ F[X] if and only if
(X − a) divides f .
Proof. By Theorem 2.3, there exists g, r ∈ F[x] with degr ≤ 0, such that
f = (x− a)g + r.
Hence
a is a root of f , i.e. , 0 = f(a) ⇔ r = 0, as f(a) = r(a) and r is constant
⇔ x− a divides f , as f = g(x− a) + r.
�
2.2. A construction of fields. We first recall the construction of quotient rings.
Let R be a ring and I an ideal of R. Define a relation ∼ on R by
a ∼ b if a− b ∈ I.
Then ∼ is an equivalence relation (check!). Denote by a the equivalence class of a
and let
R/I = {a|a ∈ R}.
Define operations + : R/I ×R/I → R/I and · : R/I ×R/I → R/I by
a+ b = a+ b and a · b = ab
Theorem 2.6. (1) + and · are well-defined.
(2) (R/I,+, ·) is a ring with 0 the zero and 1 the multiplicative identity.
GALOIS THEORY 5
The ring (R/I,+, ·) is called a quotient ring of R. For instance, Zn = Z/nZ is a
quotient ring of Z.
Let f ∈ F[x] and (f) the ideal generated by f . Then
(f) = {gh | g ∈ F[x]}
and any 0 6= h ∈ (f) has degree bigger than or equal to deg f .
Theorem 2.7. Suppose that f is a nonzero polynomial in F[x]. Then the quotient
ring F[x]/(f) is a field if and only if f ∈ F[x] is irreducible over F.
Proof. (“⇒”). Assume for contradiction that f is not irreducible. Then f is either
(i) a nonzero constant or (ii) f = gh for some polynomials g and h of positive degree.
Case (i): (f) = F[x] and so F[X]/(f) = F[X]/F[X] = {0} is not a field.
Case (ii): as deg g > 0, deg h > 0 and deg f = deg g+ deg h, the polynomials g and
h are not zero and their degrees are smaller than deg f . So g and h are not in (f).
Thus
g 6= 0 and h 6= 0
in F[x]/(f). On the other hand,
0 = f = gh = g · h
So F[X]/(f) is not an ID and thus not a field.
Either case leads to a contradiction and so f is irreducible.
(“⇐”). As F[x]/(f) is a commutative ring, we need only to prove that any nonzero
element x ∈ F[x]/(f) has a multplicative inverse. AS
0 6= a,
f does not divide a and so
gcd(f, a) = 1.
By Theorem 2.3, there exist r, s ∈ F[X] such that
1 = ra+ sf.
6 GALOIS THEORY
Then
1 = ra+ sf = ra+ s0 = ra.
This shows that a has a multiplicative inverse r, as required. �
2.3. Criteria for irreducibility of polynomials.
Lemma 2.8 (Gauss). Let p be a prime number and let f, g ∈ Z[x]. Then p | fg
implies p | f or p | g.
Proof. Suppose that
f = b0 + b1x+ · · ·+ brxr, g = c0 + c1x+ · · ·+ csx
s, fg = a0 + a1x+ · · ·+ anxn.
Assume for contradiction that p divides neither f nor g. Then there exist coefficient
bi and cj with i, j minimal and not divisible by p. Note that
ai+j = b0ci+j + b1ci+j−1 + · · ·+ bi−1cj+1 + bicj + · · ·+ bi+jc0,
where we let ct and bl be 0, if t > r and l > s. Since p divides ai+j, b0, . . . , bi−1,
c0, . . . , cj−1, p divides bicj. As p is a prime number, p divides bi or cj, contradicting
that bi and cj are not divisible by p. So p divides f or g. �
Proposition 2.9. Let f ∈ Z[x] ⊆ Q[x] with degf ≥ 1. If f is irreducible over Z,
then it is irreducible over Q.
Proof. Assume for contradiction that f is not irreducible Q. As deg f ≥ 1, f is
not a unit. So f = gh for some polynomials g and h of positive degree in Q[x].
Let m = p1 . . . pr and n = pr+1 . . . ps with each pi a prime number such that mg
and nh are polynomials in Z[x]. We have p1 . . . psf = (p1 . . . prg)(pr+1 . . . psh) with
(p1 . . . prg) and (pr+1 . . . psh) in Z[x], denoted by g0 and h0, respectively. As p1 divides
p1 . . . psf = g0h0 and p1 is a prime number, p1 divides either g0 or h0. So
p2 . . . psf = g1h1
for some g1 and h1 in Z[x], of the same degree as g and f , respectively. Continue in
this manner for s times,
f = gshs
GALOIS THEORY 7
for some gs and hs in Z[x], of the same degree as g and f , respectively. This contradicts
that f is irreducible over Z. Therefore f is irreducible over Q. �
Note: the converse is not true, e.g. f = 2x is irreducible in Q[x], but it is reducible
in Z[x], as 2 is not a unit in Z.
Deducing from the proof of Proposition 2.9, we have the following.
Corollary 2.10. Let f ∈ Z[x] and f = g1g2 with g1, g2 ∈ Q[x]. Then there exist
polynomials h1, h2 ∈ Z[x], of the same degree as g1 and g2, respectively, such that
f = h1h2.
Theorem 2.11 (Eisenstein’s irreducibility criterion). Let f = a0 + a1x+ · · ·+ anxn
be a polynomial in Z[x] with 0 6= an and n ≥ 1. Suppose that there is a prime p such
that p 6 | an, p|a0, p|a1 · · · , p|an−1 and p2 6 | a0. Then f is irreducible over Q.
Proof. Assume for contradiction that f is not irreducible over Q. As deg f = n ≥ 1,
deg f is at least 2 and f has a factorization
a0 + a1x+ · · ·+ anxn = (b0 + · · ·+ brx
r)(c0 + · · ·+ csxs),
where brcs 6= 0 and 1 ≤ r, s < n. By Corollary 2.10, we may assume that all the
coefficients bi and cj are integers. As a0 = b0c0 is divisible by p but not by p2, exactly
one of b0 and c0 is divisible by p. WLOG, suppose
(*): p divides b0 and p does not divide c0.
As p does not divide an = brcs, p does not divide br. Let i be the smallest positive
integer such that p does not divide bi. Notice that 0 < i ≤ r < n and
ai = b0ci + b1ci−1 + · · ·+ bi−1c1 + bic0.
Since p divides ai, b0, . . . , bi−1 and p does not divide bi, p divides c0, contradicting (*).
So f is irreducible over Q. �
Example 2.12. Determine the irreducibility of f = 29x5 + 5
3x4 + x3 + 1
3over Q.
8 GALOIS THEORY
First, note that f is irreducible over Q if and only if 9f is irreducible over Q. Now,
9f = 2x5 + 15x4 + 9x3 + 3, and 3 6 | 2, 3 | 15, 3 | 9, 3 | 3 and 32 6 | 3, so by Eisenstein’s
criterion, f is irreducible over over Q.
Proposition 2.13. Let p be a prime number, f = a0 + a1x + · · ·+ anxn ∈ Z[x] and
f = a0 + a1x+ · · ·+ anxn ∈ Zp[x] with an 6= 0 and n ≥ 1. If f is irreducible over Zp,
then f is irreducible over Q.
Proof. As an 6= 0, an 6= 0 and deg f = deg f = n ≥ 1. Assume (for contradiction)
that f is not irreducible. As deg f ≥ 1, f is not a constant. So f can be written as
f = g1g2
with gi ∈ Q[x] of positive degree, ∀i. By Corollary 2.10, we may assume that gi ∈ Z[x],
∀i. Note that
f = g1g2, deg f = deg f and deg gi ≤ deg gi.
So
deg gi = deg gi ≥ 1.
This shows that f is reducible over Zp, a contradiction. So f is irreducible over Q. �
Example 2.14. Determine the irreducibility of f = x3 + 11x+ 100 over Q.
Consider f = x3 + 2x+ 1 ∈ Z3[x], which is of degree 3. So if f is reducible, then it
has a linear factor and so it has a root in Z3. We can check that in Z3
f(0) = 1, f(1) = 1, f(2) = 1.
Therefore f has no roots in Z3 and so is irreducible over Z3. By Proposition 2.13, f
is irreducible over Q.
GALOIS THEORY 9
3. Field extensions
3.1. Prime fields.
Definition 3.1. The characteristic of a field a ring R, denoted by charR, is the
smallest natural number p such that p · 1 = 0, if such a number exists, and it is zero
otherwise.
For example, the characteristics of Q, R, and C are 0, the characteristic of Zp is p.
Remark: the only ring (up to isomorphism) of characteristic 1 is the trivial ring {0}.
In this course, unless otherwise specified, a ring is different from the trivial ring.
Proposition 3.2. The characteristic of a field is always zero or a prime number.
Proof. Assume for a contradiction that the characteristic n is neither zero nor a prime
number. Then n > 1 and there exist 1 < r, s < n such that n = rs. Thus 0 = n · 1 =
(r · 1)(s · 1), and so r · 1 = 0 or s · 1 = 0, contradicting the minimality of n. �
Definition 3.3. The prime field K of a field F is the intersection of all subfields of F.
For instance the prime field of Q is itself and the prime field of R is Q.
Remark:
(1) a prime field is a field.
(2) the prime field of a field is unique.
(3) a field F has no proper subfield if and only if the prime field of F is itself, e.g.
Zp (p is a prime) and Q .
Recall: a ring homomorphism is a map φ : R→ S such that
φ(a+ b) = φ(a) + φ(b), φ(ab) = φ(a)φ(b), φ(1R) = φ(1S).
Furthermore, φ is called an isomorphism if φ is bijective.
Theorem 3.4. The prime field of a field F is isomorphic to Q if charF = 0, and to
Zp if charF = p > 0 for some prime p ∈ Z.
Proof. Let K be the prime field of F. Define
φ : Z→ K, n 7→ n · 1.
10 GALOIS THEORY
As K is a subfield, any n · 1 ∈ K and so φ is well-defined. Further, φ is a ring
homomorphism (check!).
First consider the case where charF = p > 0. Then p is a prime, Kerφ ∼= pZ and
Imφ ∼= Z/pZ
is a subfield of K. As K is the prime field of F,
K ∼= Zp.
Next suppose that charK = 0. Then Kerφ = 0, i.e. φ is injective, and φ induces a
morphism
φ : Q→ K,n
m7→ φ(n)(φ(m)−1).
Again, φ is injective, and so Q ∼= Imφ is a subfield of K. As K is prime,
K ∼= Q.
This finishes the proof. �
3.2. Field extensions and extension degrees. Recall that the definition of a vec-
tor space over a field K is a set V with an addition
+ : V × V → V, (u, v) 7→ u+ v
and a scalar multiplication
· : K× V → V, (k, v) 7→ k · v
such that for all a, b ∈ K, v, w ∈ V , the following hold.
(i) (V,+) is an abelian group.
(ii) 1 · v = v.
(iii) a · (b · v) = (ab) · v.
(iv) a · (v + w) = a · v + a · w and (a+ b) · v = a · v + b · v.
Suppose that K is a subfield of F. Then we can view F as a vector space over K.
The addition in F is the usual field addition in F and the scalar multiplication is the
GALOIS THEORY 11
multiplication of elements in K with elements in F. For example: R ⊂ C and C is a
vector space over R and R is a vector space over Q.
Definition 3.5. Let F be a field and let K be a subfield of F. Then
(1) F is called an extension field of K.
(2) The pair K ⊆ F is called a field extension.
(3) The dimension of F as a vector space over K, denoted by [F : K], is called the
degree of the field extension K ⊆ F.
For example, R ⊆ C = R ⊕ iR and [C : R] = 2. Every field is an extension field
of its prime field, and so it is a vector space over its prime field. Thus any field is a
vector space over Q or Zp for some prime p.
Theorem 3.6. Let f be an irreducible polynomial of degree d in F[x].
(1) The map φ : F→ F[x]/(f), a 7→ a, is an injective ring homomorphism. Thus
F can be viewed as a subfield of the field F[x]/(f).
(2) The set {1, x, · · · , xd−1} is a basis of F[x]/(f) over F and so the extension
degree [F[x]/(f) : F] = d.
(3) The element x is a root of f in F[x]/(f).
Proof. (1) As the addition and multiplication in F[x]/(f) is induced from those in F[x]
and F is a subring of F[x], φ is a ring homomorphism. We show that it is injective.
Suppose that φ(a) = a = 0. Then a = gf for some g ∈ F[x]. Comparing the degrees
of a and gh gives a = g = 0. So φ is injective and we can view F as a subfield of
F ⊆ F[x]/(f), which by Theorem 2.7 is a field.
(2) Note for any g ∈ F[x], there exist h, r ∈ F[x] with deg r < deg f such that
g = fh+ r.
So g = r ∈ span{1, x, · · · , xd−1}.
Further, 1, x, · · · , xd−1 are linearly independent. Indeed,∑d−1
i=0 aixi = 0 implies that
h =∑d−1
i=0 aixi = gf for some g ∈ F[x]. Comparing the degrees of h and gf gives
h = g = 0. Thus ai = 0 for all i. Therefore {1, x, · · · , xd−1} is a basis of F[x]/(f) over
F and [F[x]/(f) : F] = d.
12 GALOIS THEORY
(3) f(x) = f = 0 and so x is a root of f in F[x]/(f). �
For any irreducible polynomial f ∈ F[x] of degree ≥ 2, it has no roots in F, but
Theorem 3.6 tells us that there is a natural extension field F ⊆ F[x]/(f) such that f
has a root in the extension field F[x]/(f).
Example 3.7. (1) K[x]/(x− a) ∼= K as fields, for any a ∈ K.
(2) x2 + 1 is irreducible in R[x]. R[x]/(x2 + 1) = R ⊕ Rx as vector spaces and
R[x]/(x2 + 1) ∼= C as fields.
Lemma 3.8 (The tower law). Let K ⊆ L ⊆ F be a tower of fields. Suppose that
{xi}i∈I is a basis of L over K, {yj}j∈J is a basis of F over L. Then {xiyj}i∈I,j∈J is a
basis of F over K. In particular, [F : K] = [F : L][L : K].
Proof. Denote by X the linear space span{xiyj|i ∈ I, j ∈ J}, i.e. the space of finite
linear combinations of xiyj over K. Then F ⊇ X . Next for any λ ∈ F,
λ =∑j∈J
ajyj and aj =∑i∈I
aijxi ∈ L,
where aij ∈ K. Then
λ =∑
i∈I,j∈J
aijxiyj ∈ X and so F = X.
It remains to prove the linear independence of xiyj (i ∈ I, j ∈ J). As (yj)j∈J is a
basis of F over L and (xi)i∈I is a basis of L over K,
0 =∑
i∈I,j∈J
aijxiyj =∑j∈J
(∑i∈I
aijxi)yj
implies ∑i∈I
aijxi = 0 and so aij = 0
for all j ∈ J, i ∈ I. So xiyj (i ∈ I, j ∈ J) are linearly independent and thus they form
a basis of F over K. Consequently,
[F : K] = |I||J | = [F : L][L : K].
�
GALOIS THEORY 13
4. Finitely generated field extensions
Let K ⊆ F be a field extension and let {s1, · · · , sn} ⊆ F. Denote by K[x1, · · · , xn]
the polynomial ring over K in n variables. Define
φ : K[x1, · · · , xn]→ F, by f 7→ f(s1, · · · , sn).
Then φ is a ring homomorphism (check). Define
K[s1, · · · , sn] = Imφ
and
K(s1, · · · , sn) = {ab−1 | a, b ∈ K[s1, · · · , sn], b 6= 0}.
By construction, K[s1, . . . , sn] is a subring of F and K(s1, . . . , sn) is a subfield of F.
Moreover, K[s1, · · · , sn] is the smallest subring of F containing K and s1, · · · , sn and
K(s1, · · · , sn) is the smallest subfield of F containing K and s1, · · · , sn.
Example 4.1. Q(√
2) = Q[√
2] = Q⊕Q√
2. Indeed, first note that Q[√
2] is a field,
which we can check directly or show that it it is isomorphic to the field Q[x]/(x2− 2)
and it contains Q and√
2. By definition Q[√
2] ⊆ Q(√
2) and so by the minimality
of Q(√
2),
Q(√
2) = Q[√
2].
The second identity follows from the fact that any power of√
2 is either a power of 2
or an integral multiple of√
2.
By definition, K[s1, · · · , sn] ⊆ K(s1, · · · , sn). Example 4.1 shows that K[s1, · · · , sn]
and K(s1, · · · , sn) can be the same. We will discuss when they are the same in next
section.
Definition 4.2. A field extension K ⊆ F is said to be
(1) finite if the extension degree [L : F] <∞.
(2) finitely generated if there exist finitely many elements s1, · · · , sn ∈ F such that
F = K(s1, · · · , sn).
(3) a simple extension if F = K(s) for some s ∈ F.
14 GALOIS THEORY
Remark:
(1) By definition, a simple field extension is finitely generated.
(2) A finite field extension K ⊆ F is also finitely generated.
Indeed, if F = span{s1, . . . , sn} with s1, . . . , sn a basis of F over K, then
F ⊆ K[s1, . . . , sn] ⊆ K(s1, . . . , sn) ⊆ F,
thus F = K(s1, . . . , sn) and so K ⊆ F is finitely generated.
(3) Both converses of the statements in (1) and (2) are not necessarily true.
Example 4.3. Consider the tower of fields: Q ⊆ Q(√
2) ⊆ Q(√
2,√
3) ⊆ R. What is
the extension degree [Q(√
2,√
3) : Q]?
By Example 4.1,
Q(√
2) = Q[√
2] = Q + Q√
2
and so [Q(√
2) : Q] = 2 and similarly
[Q(√
2)(√
3) : Q(√
2)] = 2.
Note that Q(√
2,√
3) = Q(√
2)(√
3). So by the Tower Law
[Q(√
2,√
3) : Q] = [Q(√
2)(√
3) : Q(√
2)][Q(√
2) : Q] = 4.
Question: Can Q ⊆ Q(√
2,√
3) be a simple extension?
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