## A Book of Abstract Algebra, Second Edition (1982)

### Chapter 29. DEGREES OF FIELD EXTENSIONS

In this chapter we will see how the machinery of vector spaces can be applied to the study of field extensions.

Let *F* and *K* be fields. If *K* is an extension of *F*, we may regard *K* as being a vector space over *F*. We may treat the elements in *K* as “vectors”

and the elements in *F* as “scalars.” That is, when we add elements in *K*, we think of it as vector addition; when we add and multiply elements in *F*, we think of this as addition and multiplication of scalars; and finally, when we multiply an element of *F* by an element of *K*, we think of it as scalar multiplication.

We will be especially interested in the case where the resulting vector space is of finite dimension. If *K*, as a vector space over *F*, is of finite dimension, we call *K* a *finite extension* of *F*. If the dimension of the vector space *K* is *n*, we say that *K* is an *extension of degree n* over *F*. This is symbolized by writing

[*K* : *F*] = *n*

which should be read, “the degree of *K* over *F* is equal to *n*.”

Let us recall that *F*(*c*) denotes the smallest field which contains *F* and *c*. This means that *F*(*c*) contains *F* and *c*, and that any other field *K* containing *F* and *c* must contain *F*(*c*). We saw in __Chapter 27__ that if *c* is algebraic over *F*, then *F*(*c*) consists of all the elements of the form *a*(*c*), for all *a*(*x*) in *F*[*x*]. Since *F*(*c*) is an extension of *F*, we may regard it as a vector space over *F*. Is *F*(*c*) a finite extension of *F*?

Well, let *c* be algebraic over *F*, and let *p*(*x*) be the minimum polynomial of *c* over *F*. [That is, *p*(*x*) is the monic polynomial of lowest degree having casa root.] Let the degree of the polynomial *p*(*x*) be equal to *n*. It turns out, then, that the *n* elements

1, *c*,*c*^{2}, …, *c ^{n}*

^{ }

^{−}

^{ 1}

*are linearly independent and span* *F*(*c*). We will prove this fact in a moment, but meanwhile let us record what it means. It means that the set of *n* “vectors” {1,*c*,*c*^{2}, …, *c ^{n}*

^{ }

^{−}

^{ 1}} is a basis of

*F*(

*c*); hence

*F*(

*c*) is a vector space of dimension

*n*over the field

*F*. This may be summed up concisely as follows:

**Theorem 1** *The degree of F*(*c*) *over F is equal to the degree of the minimum polynomial of c over F*.

PROOF: It remains only to show that the *n* elements 1, *c*, …, *c ^{n}*

^{ }

^{−}

^{ l}span

*F*(

*c*) and are linearly independent. Well, if

*a*(

*c*) is any element of

*F*(

*c*), use the division algorithm to divide

*a*(

*x*) by

*p*(

*x*):

*a*(*x*) = *p*(*x*)*q*(*x*) + *r*(*x*) where deg *r*(*x*)≤*n* − 1

Therefore,

This shows that every element of *F*(*c*) is of the form *r*(*c*) where *r*(*x*) has degree *n* − l or less. Thus, every element of *F*(*c*) can be written in the form

*a*_{0} + *a*_{1}*c* + ⋯ + *a _{n}*

_{ }

_{−}

_{ 1}

*c*

^{n}^{ }

^{−}

^{ 1}

which is a linear combination of 1,*c*,*c*^{2}, …, *c ^{n}*

^{ }

^{−}

^{ 1}.

Finally, to prove that 1,*c*,*c*^{2}, …, *c ^{n}*

^{ }

^{−}

^{ l}are linearly independent, suppose that

*a*

_{0}+

*a*

_{1}

*c*+ ⋯ +

*a*

_{n}_{ }

_{−}

_{ l}

*c*

^{n}^{ }

^{−}

^{ 1}= 0. If the coefficients

*a*

_{0},

*a*

_{1}, …,

*a*

_{n}_{ }

_{−}

_{ 1}were not all zero,

*c*would be the root of a nonzero polynomial of degree

*n*− 1 or less, which is impossible because the minimum polynomial of

*c*over

*F*has degree

*n*. Thus,

*a*

_{0}=

*a*

_{1}= ⋯ =

*a*

_{n}_{ }

_{−}

_{ 1}= 0.■

For example, let us look at (): the number is not a root of any monic polynomial of degree 1 over . For such a polynomial would

have to be ,and the latter is not in [*x*] because is irrational. However, is a root of *x*^{2} − 2, which is therefore the minimum polynomial of over , and which has degree 2. Thus,

In particular, every element in () is therefore a linear combination of 1 and ,that is, a number of the form where *a*,*b* ∈ .

As another example, *i* is a root of the irreducible polynomial of over *x*^{2} + 1 in [*x*].Therefore *x*^{2} + 1 is the minimum polynomial of *i* over *x*^{2} + 1 has degree 2, so [(*i*) : ] = 2. Thus, (*i*) consists of all the linear combinations of 1 and *i* with real coefficients, that is, all the *a* + *bi* where *a*,*b* ∈ .Clearly then, (*i*) = , so the degree of over is equal to 2.

In the sequel we will often encounter the following situation: *E* is a finite extension of *K*, where *K* is a finite extension of *F*. If we know the

degree of *E* over *K* and the degree of *K* over *F*, can we determine the degree of *E* over *F* This is a question of major importance! Fortunately, it has an easy answer, based on the following lemma:

**Lemma** *Let* *a*_{1}*a*_{2}, …, *a _{m}* be a basis of the vector space

*K*over

*F*, and let

*b*

_{1},

*b*

_{2}, …,

*b*be a basis of the vector space

_{n}*E*over

*K*.

*Then the set of mn products*{

*a*}

_{i}b_{j}*is a basis of the vector space*

*E*over the field

*F*.

PROOF: To prove that the set {*a _{i}b_{j}*} spans

*E*, note that each element

*c*in

*E*can be written as a linear combination

*c*=

*k*

_{1}

*b*

_{1}+ ⋯ +

*k*with coefficients

_{n}b_{n}*k*in

_{i}*K*. But each

*k*,

*because it is in*

_{i}*K*, is a linear combination

*k _{i}* =

*l*

_{i}_{1}

*a*

_{1}+ ⋯ +

*l*

_{im}*a*

_{m}with coefficients *l _{ij}* in

*F*. Substituting,

and this is a linear combination of the products *a _{i}b_{j}* with coefficient

*l*in

_{ij}*F*.

To prove that {*a _{i}b_{j}*} is linearly independent, suppose ∑

*l*= 0. This can be written as

_{ij}a_{i}b_{j}(*l*_{11}*a*_{1} + ⋯ + *l*_{1m} *a _{m}*)

*b*

_{1}+ ⋯ + (

*l*

_{n}_{1}

*a*

_{1}+ ⋯ +

*l*)

_{nm}a_{m}*b*= 0

_{n}and since *b*_{1}, …, *b _{n}* are independent,

*l*

_{il}a_{1}+

*· · ·*+

*l*= 0 for each

_{im}a_{m}*i*. But

*a*

_{1}, …,

*a*are also independent, so every

_{m}*l*= 0. ■

_{ij}With this result we can now conclude the following:

**Theorem 2** *Suppose* *F* ⊆ *K* ⊆ *E* *where* *E* *is a finite extension of* *K* and *K* is a finite extension of *F*.*Then ∈ is a finite extension of* *F*, *and*

[*E : F*] = [*E : K*][*K : F*]

This theorem is a powerful tool in our study of fields. It plays a role in field theory analogous to the role of Lagrange’s theorem in group theory. See what it says about any two extensions, *K* and *E* of a fixed “base field” *F* : If *K*is a subfield of *E*_{9} then the degree of *K* (over *F*) divides the degree of *∈* (over *F*).

If *c* is algebraic over *F*, we say that *F*(*c*) is obtained by *adjoining c* to *F*. If *c* and *d* are algebraic over *F*, we may find adjoin *c* to *F*, thereby obtaining *F*(*c*), and then adjoin *d* to *F*(*c*). The resulting field is denoted *F*(*c*, *d*) and is the smallest field containing *F*, *c* and *d*. [Indeed, any field containing *F*, *c* and d must contain F(*c*), hence also *F*(*c*,*d*).] It does not matter whether we first adjoin *c* and then *d* or vice versa.

If *c*_{1}, …, *c _{n}* are algebraic over

*F*, we let

*F*(

*c*

_{1}, …,

*c*) be the smallest field containing

_{n}*F*and

*c*

_{1}, …,

*c*. We call it the field obtained by

_{n}*adjoining*

*c*

_{1},…,

*c*to

_{n}*F*. We may form

*F*(

*c*

_{1}, …,

*c*) step by step, adjoining one

_{n}*c*at a time, and the order of adjoining the

_{i}*c*is irrelevant.

_{i}An extension *F*(*c*) formed by adjoining a single element to *F* is called a *simple extension* of *F*. An extension *F*(*c*_{1}, …, *c _{n}*)formed by adjoining a finite number of elements

*c*

_{1}, …,

*c*is called an

_{n}*iterated extension*. It is called “iterated” because it can be formed step by step, one

*simple*extension at a time:

F ⊆ *F*(*c*_{1}) ⊆ *F*(*c*_{1},*c*_{2}) ⊆ *F*(*c*_{l},*c*_{2},*c*_{3}) ⊆ … ⊆ *F*(*c*_{1}, …, *c _{n}* (1)

If *c*_{1}, …, *c _{n}* are algebraic over F, then by

__Theorem 1__, each extension in Condition (1) is a finite extension. By

__Theorem 2__,

*F*(

*c*

_{1}

*c*

_{2}) is a finite extension of

*F*; applying

__Theorem 2__again,

*F*(

*c*

_{1},

*c*

_{2},

*c*

_{3}) is a finite extension of

*F*; and so on. So finally,

*if*

*c*

_{1}, …,

*c*

_{n}*are algebraic over*F,

*then*

*F*(

*c*

_{1}, …,

*c*) is a finite

_{n}*extension of F*.

Actually, the converse is true too: *every finite extension is an iterated extension*. This is obvious: for if *K* is a finite extension of *F*, say an extension of degree *n*, then *K* has a basis {*a*_{1}, …,*a _{n}*} over

*F*. This means that every element in

*K*is a linear combination of

*a*

_{1}, …,

*a*with coefficients in

_{n}*F*; but any field containing

*F*and

*a*

_{1}, …,

*a*obviously contains all the linear combinations of

_{n}*a*

_{1}, …,

*a*; hence

_{n}*K*is the smallest field containing

*F*and

*a*

_{1}, …,

*a*. That is,

_{n}*K*=

*F*(

*a*

_{1}, …,

*a*).

_{n}In fact, if *K* is a finite extension of *F* and *K* = *F*(*a*_{1}, …, *a _{n}*), then

*a*

_{1}, …,

*a*have to be

_{n}*algebraic*over

*F*. This is a consequence of a simple but important little theorem:

**Theorem 3** *If K is a finite extension of* *F*, *every element of K is algebraic over* *F*

PROOF: Indeed, suppose *K* is of degree *n;* over *F*, and let *c* be any element of *K*. Then the set {1,*c*,*c*^{2}, …, *c ^{n}*) is linearly dependent, because it has

*n*+ 1 elements in a vector space

*K*of dimension

*n*. Consequently, there are scalars

*a*

_{0},…,

*a*∈

_{n}*F*, not all zero, such that

*a*

_{0}+

*a*

_{1}

*c*+ ⋯ +

*a*= 0. Therefore

_{n}c^{n}*c*is a root of the polynomial

*a*(

*x*) =

*a*

_{0}+

*a*

_{1}

*x*+ ⋯ +

*a*in

_{n}x^{n}*F*[

*x*].■

Let us sum up: *Every iterated extension F*(*c*_{1}, …, *c _{n}*),

*where*

*c*

_{1}, …,

*c*are

_{n}*algebraic over*

*F*,

*is finite extension of*

*extension of*

*F*.

*Conversely, every finite extension of F is an iterated extension F*(

*c*

_{1}, …,

*c*),

_{n}*where c*

_{1}, …,

*c*

_{n}*are algebraic over F*.

Here is an example of the concepts presented in this chapter. We have already seen that () is of degree 2 over , and therefore () consists of all the numbers *a* + *b* where *a,b* ∈ . Observe that cannot be in(); for if it were, we would have = *a* + *b* for rational *a* and *b;* squaring both sides and solving for would give us = a rational number, which is impossible.

Since is not in (), cannot be a root of a polynomial of degree 1 over () (such a polynomial would have to be *x* − ). But is a root of *x*^{2} − 3, which is therefore the minimum polynomial of over ().Thus, (,) is of degree 2 over (), and therefore by __Theorem 2__, (,) is of degree 4 over .

By the comments preceding __Theorem 1__, {1,} is a basis of () over ,and {1,} is a basis of (,) over (). Thus, by the lemma of this chapter, {1,,,} is a basis of (,) over . This means that (,) consists of all the numbers *a* + *b*+ *c* + *d* , for all *a b,c*, and d in .

**For later reference.** The technical observation which follows will be needed later.

By the comments immediately preceding __Theorem 1__, every element of *F*(*c*_{1}) is a linear combination of powers of *c*_{1}, with coefficients in *F*. That is, every element of *F*(*c*_{1}) is of the form

where the *k _{i}* are in

*F*. For the same reason, every element of

*F*(

*c*

_{1}

*c*

_{2}) is of the form

where the coefficients *l _{j}* are in

*F*(

*c*

_{1}).Thus, each coefficient

*l*is equal to a sum of the form (2). But then, clearing brackets, it follows that every element of

_{j}*F*(

*c*

_{1}

*c*

_{2}) is of the form

where the coefficients *k _{ij}* are in

*F*.

If we continue this process, it is easy to see that every element of *F*(*c*_{1},*c*_{2},…, *c _{n}*) is a sum of terms of the form

where the coefficient *k* of each term is in *F*.

**EXERCISES**

**A. Examples of Finite Extensions**

**1** Find a basis for (*i*) over , and describe the elements of (*i*).(See the two examples immediately following __Theorem 1__.)

**2** Show that every element of (2 + 3*i*) can be written as *a* + *bi*, where *a,b* ∈ Conclude that (*2* + 3*i*) = .

**# 3** If , show that {1, 2

^{1/3}, 2

^{2/3},

*a*,2

^{1/3}

*a*, 2

^{2/3}

*a*} is a basis of (

*a*) over . Describe the elements of (

*a*).

**# 4** Find a basis of () over , and describe the elements of .

**5** Find a basis of over and describe the elements of ().(See the example at the end of this chapter.)

**6** Find a basis of (,,) over , and describe the elements of (,,.

**7** Name an extension of over which *π* is algebraic of degree 3.

**† B. Further Examples of Finite Extensions**

Let *F* be a field of characteristic ≠ 2. Let *a* ≠ *b* be in *F*.

**1** Prove that any field *F* containing + also contains and .[HINT: Compute ( + )^{2} and show that ∈ *F*. Then compute ( + ), which is also in *F*] Conclude that *F*( + ) = *F*(,).

**2** Prove that if *b* ≠ *x*^{2}*a* for any *x* ∈ *F*, then ∉*F*(). Conclude that *F*(,) is of degree 4 over *F*.

**3** Show that *x* = + satisfies *x*^{4} − 2(*a* + *b*)*x*^{2} + (*a − b*)^{2} = 0. Show that *x* = also satisfies this equation. Conclude that

**4** Using parts 1 to 3, find an uncomplicated basis for (*d*) over , where *d* is a root of *x*^{4} − 14*x*^{2} + 9. Then find a basis for over .

**C. Finite Extensions of Finite Fields**

By the proof of the basic theorem of field extensions, if *p*(*x*) is an irreducible polynomial of degree *n* in *F*[*x*], then *F*[*x*]*/*⟨*p*(*x*)⟩ ≅ *F*(*c*) where c is a root of *p*(*x*). By __Theorem 1__ in this chapter, *F*(*c*) is of degree *η* over F. Using the paragraph preceding __Theorem 1__:

**1** Prove that every element of *F*(*c*) can be written *uniquely* as *a*_{0} + *a _{1}c* + ⋯ +

*a*

_{n}_{ }

_{−}

_{ l}

*c*

^{n}^{ }

^{−}

^{ 1}, for some

*a*

_{0},…,

*a*

_{n}_{ }

_{−}

_{ 1}∈

*F*.

**# 2** Construct a field of four elements. (It is to be an extension of

_{2}.) Describe its elements, and supply its addition and multiplication tables.

**3** Construct a field of eight elements. (It is to be an extension of _{2}).

**4** Prove that if *F* has *q* elements, and *a* is algebraic over *F* of degree *n*, then *F*(*a*) has *q ^{n}* elements.

**5** Prove that for every prime number *p*, there is an irreducible quadratic in * _{p}*[

*x*]. Conclude that for every prime

*p*, there is a field with

*p*

^{2}elements.

**D. Degrees of Extensions (Applications of Theorem 2)**

Let F be a field, and *K* a field extension of *F*. Prove the following:

**1** [*K*:*F*] = *1* iff *K* = *F*.

**# 2** If [

*K*:

*F*] is a prime number, there is no field properly between

*F*and

*K*(that is, there is no field

*L*such that

*F*

*L*

*K*).

**3** If [*K*:*F*] is a prime, then *K* = *F*(*a*) for every *a* ∈ *K* − *F*.

**4** Suppose *a*,*b* ∈ *K* are algebraic over *F* with degrees *m* and η, where *m* and « are relatively prime. Then:

(*a*)*F*(*a*,*b*)is degree *mn* over *F*.

(*b*) *F*(*a*) *F*(*b*)= *F*.

**5** If the degree of *F*(*a*) over *F* is a prime, then *F*(*a*) = *F*(*a ^{n}*) for any

*n*(on the condition that

*a*∉

^{n}*F*).

**6** If an irreducible polynomial *p*(*x*) ∈ *F*[*x*] has a root in *K*, then deg *p*(*x*)*|*[*K*:*F*].

**E. Short Questions Relating to Degrees of Extensions**

Let *F* be a field.

Prove parts 1−3:

**1** The degree of *a* over *F* is the same as the degree of 1/*a* over *F*. It is also the same as the degrees of *a* + *c* and *ac* over *F*, for any *c* ∈ *F*.

*2* *a* is of degree 1 over *F* iff *a* ∈ *F*.

**3** If a real number *c* is a root of an irreducible polynomial of degree >1 in [*x*], then *c* is irrational.

**4** Use part 3 and Eisentein ’ s irreducibility criterion to prove that (where ra, *m* *n* ∈ ) is irrational if there is a prime number which divides ra but not *n*,and whose square does not divide ra.

**5** Show that part 4 remains true for where *q* >1.

**6** If *a* and *b* are algebraic over *F*, prove that F(*a*, *b*) is a finite extension of *F*.

**† F. Further Properties of Degrees of Extensions**

Let *F* be a field, and *K* a finite extension of *F*. Prove each of the following:

**1** Any element algebraic over *K* is algebraic over *F*, and conversely.

**2** If *b* is algebraic over *K*, then [*F*(*b* :*F*] | [*K*(*b*) :*F*].

**3** If fe is algebraic over *K*, then [*K*(*b*) :*K*] ≤ [*F*(*b*) : *F*]. (HINT: The minimum polynomial of *b* over *F* may factor in *K*[*x*], and 6 will then be a root of one of its irreducible factors.)

**# 4** If

*b*is algebraic over

*K*, then [

*K*(

*b*):

*F*(

*b*)] ≤ [

*K*:

*F*]. [HINT: Note that

*F*⊆

*K*⊆

*K*(

*b*) and

*F*⊆

*F*(

*b*) ⊆

*K*(

*b*). Relate the degrees of the four extensions involved here, using part 3.]

**# 5** Let

*p*(

*x*) be irreducible in

*F*[

*x*]. If [

*K*:

*F*] and deg

*p*(

*x*) are relatively prime, then

*p*(

*x*) is irreducible in

*K*[

*x*].

**† G. Fields of Algebraic Elements: Algebraic Numbers**

Let *F* ⊆ *K* and *a*,*b* ∈ *K*. We have seen on page 295 that if *a* and *b* are algebraic over *F*, then *F*(*a*, *b*) is a finite extension of *F*.

Use the above to prove parts 1 and 2.

**1** If *a* and *b* are algebraic over *F*, then *a* + *b*, *a* − *b*, *ab*, and *a* / *b* are algebraic over *F*. (In the last case, assume *b* ≠ 0.)

**2** The set {*x* ∈ *K* : *x* is algebraic over *F*} is a subfield of *K*, containing *F*.

Any complex number which is algebraic over is called an *algebraic number*. By part 2, the set of all the algebraic numbers is a field, which we shall designate by .

Let *a*(*x*) = *a*_{0} + *a _{1}x* + ⋯ +

*a*be in [

_{n}x^{n}*x*], and let

*c*be any root of

*a*(

*x*). We will prove that

*c*∈ . To begin with, all the coefficients of

*a*(

*x*) are in (

*a*

_{0},

*a*

_{1}, …,

*a*).

_{n}**3** Prove: (*a*_{0}, *a*_{1}, …, *a _{n}*) is a finite extension of .

Let (*a*_{0}, …, *a _{n}*) =

_{1}Since

*a*(

*x*) ∈

_{1}[

*x*],

*c*is algebraic over

_{1}Prove parts 4 and 5:

**4** _{1}(*c*) is a finite extension of _{1} hence a finite extension of . (Why?)

**5** *c* ∈.

Conclusion: The roots of any polynomial whose coefficients are algebraic numbers are themselves algebraic numbers.

A field *F* is called *algebraically closed* if the roots of every polynomial in *F*[*x*] are in *F*. We have thus proved that is algebraically closed.