√√ √ √ √ √ √√

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