<<

. 8
( 18 .)



>>

2
√√ √ √ √ √ √√
hence Q( 2, 3) Q( 2 + 3). Since Q( 2 + 3) Q( 2, 3) we must have
√ √ √√
Q( 2 + 3) = Q( 2, 3).
Referring to Example 2.10 we see that
√ √
degQ ( 2 + 3) = 4.
Let us ¬nd a non-zero polynomial in ker µ√2+√3 Q[X].
√ √ √
Referring to Example 2.10 or Proposition 2.12 we see that 2 + 3 ∈ Q( 2), hence
/
√ √
√ ( 2 + 3) = 2.
degQ( 2)

One polynomial in ker µ√2+√3 Q( 2)[X] is
√ √ √ √ √
(X ’ ( 2 + 3))(X ’ ( 2 ’ 3)) = X 2 ’ 2 2X ’ 1.
Since this is monic and of degree 2,

2
√ √√
minpolyQ( 2), 2+ 3 (X) = X ’ 2 2X ’ 1.
Similarly, √
minpolyQ(√2),’√2+√3 (X) = X 2 + 2 2X ’ 1.
Consider
p(X) = minpolyQ(√2),√2+√3 (X) minpolyQ(√2),’√2+√3 (X)
√ √
= (X 2 ’ 2 2X ’ 1)(X 2 + 2 2X ’ 1)
= X 4 ’ 10X 2 + 1.
√ √
Then p( 2 + 3) = 0 so p(X) ∈ ker µt . Since deg p(X) = 4 and p(X) is monic, we have
minpolyQ,√2+√3 (X) = X 4 ’ 10X 2 + 1.
Definition 3.10. Let L/K be a ¬nite extension. An element u ∈ L for which L = K(u) is
called a primitive element for the extension L/K.
Later we will see that if char K = 0 then all ¬nite extensions L/K have primitive elements.
Lemma 3.11. Let L/K be a ¬nite extension and u ∈ L. Then u is a primitive element for
L/K if and only if degK u = [L : K].
3.1. ALGEBRAIC EXTENSIONS 31

Proof. K(u) ⊆ L is a ¬nite dimensional K-vector subspace. Then K(u) = L if and only
dimK K(u) = dimK L. Since degK u = dimK K(u) and [L : K] = dimK L the result follows.
Sometimes the minimal polynomial of an element in an extension is introduced in a di¬erent
but equivalent way.
Proposition 3.12. Let t ∈ L be algebraic over K. Then
I(t) = {f (X) ∈ K[X] : f (t) = 0} ⊆ K[X]
is an ideal which is principal and has an irreducible monic generator q(X) ∈ K[X]. In fact,
q(X) = minpolyK,t (X).
Proof. It is easy to see that I(t) K[X] and therefore I(t) = (q(X)) for some monic
generator q(X). To see that q(X) is irreducible, suppose that q(X) = q1 (X)q2 (X) with
deg qi (X) < deg q(X). Now as q1 (t)q2 (t) = 0, we must have q1 (t) = 0 or q2 (t) = 0, hence
q1 (X) ∈ I(t) or q2 (X) ∈ I(t). These possibilities give q(X) | q1 (X) or q(X) | q2 (X) and
so deg q(X) deg q1 (X) or deg q(X) deg q2 (X), contradicting the above assumption that
deg qi (X) < deg q(X).
The irreducible monic polynomial minpolyK,t (X) is in I(t) so q(X) | minpolyK,t (X) and
therefore q(X) = minpolyK,t (X).
The next Lemma will often be useful.
Lemma 3.13. Let L/K be an extension and suppose that u1 , . . . , un ∈ L are algebraic. Then
K(u1 , . . . , un )/K is a ¬nite extension.
Proof. Use induction on n together with Proposition 2.8 and Theorem 2.6(ii).
We now come to an important notion for extensions.
Definition 3.14. The extension L/K is algebraic or L is algebraic over K if every element
t ∈ L is algebraic over K.
Proposition 3.15. Let L/K be a ¬nite extension. Then L/K is algebraic.
Proof. Let t ∈ L. Since the K-vector space L is ¬nite dimensional the powers 1, t, . . . , tn , . . .
must be linearly dependent over K, hence for suitable coe¬cients cj ∈ K not all zero and some
m 1 we have
c0 + c1 t + · · · + cm tm = 0.
But this means that t is algebraic over K.
Proposition 3.16. Let M/L and L/K be algebraic extensions. Then the extension M/K
is algebraic.
Proof. Let u ∈ M . Then u is algebraic over L, so there is a polynomial
p(X) = p0 + p1 X + · · · + pm X m ∈ L[X]
of positive degree with p(u) = 0. By Lemma 3.13, the extension K(p0 , . . . , pm )/K is ¬nite and
so is K(p0 , . . . , pm , u)/K(p0 , . . . , pm ). By Theorem 2.6(ii), K(p0 , . . . , pm , u)/K is ¬nite, so by
Proposition 3.15, u is algebraic over K.
Definition 3.17. For an extension L/K, let
Lalg = {t ∈ L : t is algebraic over K} ⊆ L.
Proposition 3.18. For an extension L/K, Lalg is a sub¬eld containing K and Lalg /K is
algebraic.
Proof. Clearly K ⊆ Lalg . We must show that Lalg L.
Let u, v ∈ Lalg . Then by Lemma 3.13, K(u, v)/K is a ¬nite dimensional extension, hence
every element of K(u, v) is algebraic over K. In particular, u + v and uv are in K(u, v) and if
u = 0, u’1 is also in K(u, v). Therefore u + v, uv and u’1 are all algebraic over K.
32 3. ALGEBRAIC EXTENSIONS OF FIELDS

Example 3.19. In the extension C/Q we can consider Calg C which is called the sub¬eld
of algebraic numbers. Similarly, in the extension R/Q the sub¬eld
Ralg = Calg © R C
consists of all the real algebraic numbers. Elements of C’Calg are called transcendental complex
numbers; examples are e and π. The sets Calg and Ralg are both countable, whereas C and R
are uncountable, so there are in fact many more transcendental numbers but it can be hard to
determine whether a given number is transcendental or not. A more usual notation for Calg
is Q since this is the algebraic closure of Q which will be discussed later. When dealing with
algebraic extensions of Q we will usually work with sub¬elds of Q = Calg .
We end this section with a technical result.
Proposition 3.20. Let K(u)/K be a ¬nite simple extension. Then there are only ¬nitely
many subextensions F/K K(u)/K.
Proof. Consider the minimal polynomial minpolyK,u (X) ∈ K[X]. Now for any subexten-
sion F/K K(u)/K we can also consider
minpolyF,u (X) = c0 + c1 X + · · · + ck’1 X k’1 + X k ∈ F [X],
which divides minpolyK,u (X) in F [X]. The Unique Factorization Property 1.32 implies that
minpolyK,u (X) has only ¬nitely many monic divisors in K(u)[X], so there are only a ¬nite
number of possibilities for minpolyF,u (X). Now consider F0 = K(c0 , c1 , . . . , ck’1 ), the extension
¬eld of K generated by the coe¬cients of minpolyF,u (X). Then F0 F and so minpolyF,u (X) ∈
F0 [X] is irreducible since it is irreducible in F [X]; hence minpolyF,u (X) = minpolyF0 ,u (X). We
have
[K(u) : F ] = deg minpolyF,u (X) = deg minpolyF0 ,u (X) = [K(u) : F0 ],
hence F = F0 .
This shows that there are only ¬nitely many subextensions F/K K(u)/K, each of which
has the form K(a0 , a1 , . . . , a ’1 ), where
’1
a0 + a1 X + · · · + a ’1 X + X ∈ K(u)[X]
is a factor of minpolyK,u (X) in K(u)[X].

3.2. Splitting ¬elds and Kronecker™s Theorem
We can now answer a basic question. Let K be a ¬eld and p(X) ∈ K[X] be a polynomial
of positive degree.
Question 3.21. Is there an extension ¬eld L/K for which p(X) has a root in L?
A stronger version of this question is the following.
Question 3.22. Is there an extension ¬eld E/K for which p(X) factorizes into linear factors
in E[X]?
Definition 3.23. p(X) ∈ K[X] splits in E/K or over E if it factorizes into linear factors
in E[X].
Of course, if we have such a ¬eld E then the distinct roots u1 , . . . , uk of p(X) in E generate
a sub¬eld K(u1 , . . . , uk ) E which is the smallest sub¬eld of E that answers Question 3.22.
Definition 3.24. Such a minimal extension of K is called a splitting ¬eld of p(X) over K
and we will sometimes denote it by K(p(X)) or Kp .
We already know how to answer Question 3.21.
Theorem 3.25 (Kronecker™s Theorem: ¬rst version). Let K be a ¬eld and p(X) ∈ K[X] be
a polynomial of positive degree. Then there is a ¬nite extension L/K for which p(X) has a root
in L.
3.2. SPLITTING FIELDS AND KRONECKER™S THEOREM 33

Proof. We begin by factorizing p(X) ∈ K[X] into irreducible monic factors qj (X) together
with a constant factor c:
p(X) = cq1 (X) · · · qr (X).
Now for any j we can form the quotient ¬eld K[x]/(qj (X)) which is a ¬nite dimensional (simple)
extension of K and in which the coset X + (qj (X)) satis¬es the equation

qj (X + (qj (X))) = 0 + (qj (X)).

Hence p(X) has a root in K[x]/(qj (X)).
Of course, this construction is only interesting if qj (X) to has degree bigger than 1 since a
linear polynomial already has a root in K.

To answer Question 3.22 we iterate this construction. Namely, having found one root u1 in
an extension L1 /K we discard the linear factor X ’ u1 and consider the polynomial

p(X)
p1 (X) = ∈ L1 [X].
X ’ u1
We can repeat the argument to form a ¬nite extension of L1 (and hence of K) containing a
root of p1 (X) and so on. At each stage we either already have another root in L1 or we need to
enlarge the ¬eld to obtain one.

Theorem 3.26 (Kronecker™s Theorem: second version). Let K be a ¬eld and p(X) ∈ K[X]
be a polynomial of positive degree. Then there is a ¬nite extension E/K which is a splitting
¬eld of p(X) over K.

In practise we often have extension ¬elds ˜lying around in nature™ containing roots and we
can work inside of these. When working over Q (or any other sub¬eld of C) we can always ¬nd
roots in C by the Fundamental Theorem of Algebra. We then refer to a sub¬eld of C which is
a splitting ¬eld as the splitting sub¬eld.

Example 3.27. Find a splitting ¬eld E/Q for p(X) = X 4 ’ 4 over Q and determine [E : Q].

Solution. Notice that
p(X) = (X 2 ’ 2)(X 2 + 2),
√ √ √ √
so ¬rst we adjoin the roots ± 2 of (X 2 ’ 2) to form Q( 2, ’ 2) = Q( 2) which gives an

extension Q( 2)/Q of degree 2. √ √
Next consider the polynomial X 2 + 2 ∈ Q( 2)[X]. The √ complex roots of X 2 + 2 are ± 2i
and these are not √
real, so this polynomial is irreducible √ Q( 2)[X]. Hence we need to consider
in
√√ √
Q( 2, 2i) = Q( 2, i) and the extension Q( 2, i)/Q( 2) which has degree 2.

C


Q( 2, i)
adjoin roots of X 2 + 2 2

Q( 2)
adjoin roots of X 2 ’ 2 2

Q
√ √
Thus the splitting sub¬eld of p(X) over Q in C is Q( 2, i) and [Q( 2, i) : Q] = 4.
34 3. ALGEBRAIC EXTENSIONS OF FIELDS

Of course we could have started by adjoining roots of X 2 + 2 and then of X 2 ’ 2, giving a
tower
C


Q( 2, i)
adjoin roots of X 2 ’ 2 2

Q( 2i)
adjoin roots of X 2 + 2 2

Q

An important point is that if a splitting ¬eld exists inside of a given extension ¬eld F/K, it is
unique.

Proposition 3.28. Let F/K be an extension ¬eld and p(X) ∈ K[X]. If E1 , E2 F are
splitting sub¬elds for p(X) over K then E1 = E2 .

Proof. Let u1 , . . . , uk ∈ F be the distinct roots of p(X) in F . By de¬nition, K(u1 , . . . , uk )
is the smallest sub¬eld containing K and all the uj . But K(u1 , . . . , uk ) must be contained in
any splitting sub¬eld, so E1 = K(u1 , . . . , uk ) = E2 .

Since we will frequently encounter quadratic polynomials we record a useful result on roots
of such polynomials. Recall that p(X) = aX 2 + bX + c ∈ K[X] is quadratic if a = 0 and its
discriminant is
∆ = b2 ’ 4ac ∈ K.
The proof of the next result is the standard one which works provided 2 has an inverse in K,
i.e., when char K = 2.

Proposition 3.29. Let K be a ¬eld of characteristic di¬erent from 2. Then the quadratic
polynomial p(X) = aX 2 + bX + c ∈ K[X] has
• no roots in K if ∆ is not a square in K;
• one root ’b/(2a) = ’(2a)’1 b if ∆ = 0;
• two distinct roots
’b + δ ’b ’ δ
= (2a)’1 (’b + δ), = (2a)’1 (’b ’ δ),
2a 2a
if ∆ = δ 2 for some non-zero δ ∈ K.

In particular, the splitting ¬eld of p(X) over K is K if ∆ is a square in K and K(δ) otherwise,
where δ is one of the two square roots of ∆ in some extension of K such as the algebraic closure
K which we will introduce in Section 3.4.

Example 3.30. Find a splitting ¬eld E/Q for p(X) = X 3 ’ 2 over Q and determine [E : Q].

√ Solution. By the Eisenstein Test 1.37, p(X) is irreducible over Q. One root of p(X) is

3
2 ∈ R so we adjoin this to Q to form an extension Q( 3 2)/Q of degree 3. Now
√ √ √
p(X) = (X ’ 2)(X 2 + 2X + ( 2)2 )
3 3 3


√ √2
and the second factor has the non-real complex roots 3 2 ζ3 and 3 2 ζ3 which lie in the extension
√ √ √
Q( 3 2, ζ3 )/Q( 3 2) of degree 2. So the splitting sub¬eld of X 3 ’ 2 in C over Q is Q( 3 2, ζ3 ) for
3.3. MONOMORPHISMS BETWEEN EXTENSIONS 35

which [Q( 3 2, ζ3 ) : Q] = 6.

C
2


R ∞




Q( 3 2, ζ3 )

iiii q VV

<<

. 8
( 18 .)



>>