# «SAMUEL MOY Abstract. Assuming some basic knowledge of groups, rings, and ﬁelds, the following investigation will introduce the reader to the theory ...»

## AN INTRODUCTION TO THE THEORY OF FIELD

## EXTENSIONS

## SAMUEL MOY

Abstract. Assuming some basic knowledge of groups, rings, and ﬁelds, the

following investigation will introduce the reader to the theory of rings before

proceeding to elaborate, in greater depth, on the theory of ﬁeld extensions.

Finally, a few consequences of the subject will be examined by solving classical straightedge and compass problems in a manner that eﬀectively utilizes the material.

Contents

1. The Basics 1

2. Ring Theory 1

3. Fields and Field Extensions 4

4. Algebraic Field Extensions 9

5. Classical Straightedege and Compass Constructions 12

6. Acknowledgements 15

7. References 15

1. The Basics Deﬁnition 1.1. : A ring R is a set together with two binary operations + and × (addition and multiplication, respectively) satisyﬁng the following axioms:

(i) (R, +) is an abelian group, (ii) × is associative: (a × b) × c = a × (b × c) for all a, b, c ∈ R, (iii) the distributive laws hold in R for all a, b, c ∈ R:

(a + b) × c = (a × c) + (b × c) and a × (b + c) = (a × b) + (a × c).

Deﬁnition 1.2. The ring R is commutative if multiplication is commutative.

Deﬁnition 1.3. The ring R is said to have an identity (or contain a 1) if there is an element 1 ∈ R with 1×a=a×1=a a∈R for all Deﬁnition 1.4. A ring R with identity 1, where 1 = 0, is called a division ring (or skew ﬁeld ) if ∀ nonzero element a ∈ R, ∃ b ∈ R such that ab = ba = 1.

Deﬁnition 1.5. A commutative division ring is called a ﬁeld.

** Example 1.6.**

Z is a commutative ring with 1(identity). Q, R, C, and Z/pZ (the integers modulo p, where p is prime) are all ﬁelds.

1 2 SAMUEL MOY

2. Ring Theory Before beginning further study of ﬁelds, additional knowledge pertaining to rings is necessary. Most notably, the topics of ideals, ring homomorphisms and isomor- phisms, and quotient rings must ﬁrst be approached.

Deﬁnition 2.1. A subring of the ring R is a subgroup of R that is closed under multiplication (i.e. A subset S of a ring R is a subring if the operations of addition and multiplication in R when restricted to S give S the structure of a ring).

Deﬁnition 2.2. Let R and S be rings.

(1) A ring homomorphism is a map ϕ : R → S satisfying (a) ϕ(a + b) = ϕ(a) + ϕ(b) for all a, b ∈ R, and (b) ϕ(ab) = ϕ(a)ϕ(b) for all a, b ∈ R.

(2) The kernel of the ring homomorphism ϕ, denoted kerϕ, is the set of elements of R that map to 0 in S.

(3) A bijective ring homomorphism is called an isomorphism.

Deﬁnition 2.3. Let R be a ring, let I be a subset of R and let r ∈ R.

(1) rI = {ra | a ∈ I} and Ir = {ar | a ∈ I}.

(2) A subset I of R is a left ideal (respectively right ideal ) of R if (a) I is a subring of R, and (b) I is closed under left multiplication (respectively right multiplication) by elements from R.

(3) A subset I that is both a left ideal and a right ideal is called an ideal (or, a two-sided ideal ) of R.

Deﬁnition 2.4. Let R be a ring and I an ideal of R. Then the quotient ring of R

**by I, denoted R/I is the ring deﬁned by the following binary operations:**

R → R/I deﬁned by r → r + I is a surjective ring homomorphism with kernel I (this homomorphism is called the natural projection of R onto R/I). Thus, every ideal is the kernel of a ring homomorphism and vice versa.

Proof. Let I be the kernel of ϕ. Then the cosets under addition of I are exactly the ﬁbers of ϕ (the sets of elements of R that map to a single element of S). In particular, the cosets r +I, s+I, and rs+I are the ﬁbers of ϕ over ϕ(r), ϕ(s), ϕ(rs), respectively. Since ϕ is a ring homomorphism, ϕ(r)ϕ(s) = ϕ(rs), hence (r + I)(s + I) = rs + I. Multiplication of cosets is well deﬁned and so I is an ideal and R/I is a ring. The correspondance r + I → ϕ(r) is a bijection between the rings R/I and ϕ(R) which respects addition and multiplication. Hence, it is a ring isomorphism.

If I is any ideal, then R/I is a ring (in particular is an abelian group) and the

## AN INTRODUCTION TO THE THEORY OF FIELD EXTENSIONS 3

map π : r → r + I is a group homomorphism with kernel I (natural projection for groups). It remains to check that π is a ring homomorphism. This is immediate**from the deﬁnition of multiplication in R/I:**

** Theorem 2.6 (The Lattice Isomorphism Theorem for Rings).**

Let R be a ring and let I be an ideal of R. The correspondance A ↔ A/I is an inclusion preserving bijection between the set of subrings A of R that contain I and the set of subrings of R/I. Furthermore, A (a subring containing I) is an ideal of R if and only if A/I is an ideal of R/I.

The proof for this theorem will not be provided. However, it follows almost immediately from the Lattice Isomorphism Theorem for Groups. (In addition, rather than considering subgroups, we must consider ideals).

Deﬁnition 2.7. Let I and J be ideals of R.

(1) Deﬁne the sum of I and J by I + J = {a + b | a ∈ I, b ∈ J}.

(2) Deﬁne the product of I and J, denoted by IJ, to be the set of all ﬁnite sums of elements of the form ab with a ∈ I and b ∈ J.

Deﬁnition 2.8. Let A be any subset of the ring R. Let (A) denote the smallest ideal of R containing A, called the ideal generated by A, (A) = I.

I⊇A Deﬁnition 2.9. An ideal M in an arbitrary ring R is called a maximal ideal if M = R and the only ideals containing M are M and R.

Deﬁnition 2.10. Assume R is a commutative ring. An ideal P is called a prime ideal if P = R and whenever the product ab of two elements a, b ∈ R is an element of P, then at least one of a and b is an element of P.

The following two propositions will be useful for later theorems regarding ﬁelds.

Only Proposition 2.12 will be proved now (Proposition 2.11 will appear as a Lemma for a later theorem and will be proved then).

Proposition 2.11. Assume R is a commutative ring. Then R is a ﬁeld if and only if its only ideals are 0 and R.

Proposition 2.12. Assume R is a commutative ring. The ideal M is a maximal ideal if and only if the quotient ring R/M is a ﬁeld.

Proof. This follows from the Lattice Isomorphism Theorem for Rings along with Proposition 2.11. The ideal M is maximal if and only if there are no ideals I with M ⊂ I ⊂ R. By the Lattice Isomorphism Theorem the ideals of R containing M correspond bijectively with the ideals of R/M, so M is maximal if and only if the ideals of R/M are 0 and R/M. By Proposition 2.11 we see that M is a maximal ideal if and only if R/M is a ﬁeld.

4 SAMUEL MOY Example 2.13. To assist in the understanding of what an ideal is, we will consider ideals in the commutative ring with 1, Z. An example of an ideal in this ring is (2) = {2a | a ∈ Z} = multiples of 2 (the ideal generated by the element 2 ∈ Z).

Now, it should not be diﬃcult to see that the only prime ideals in Z are those

**generated by prime numbers:**

Suppose we have the ideal (n), n composite (and positive) (i.e. n = ab, 0 a, b = 1).

WLOG, assume a b n. Clearly a · n ∈ (n) ⇒ a · a · b = a2 · b ∈ (n). However, 0 a2 n and 0 b n ⇒ neither a2 nor b ∈ (n). Therefore, (n) is not a prime ideal. Now take the ideal (p), p prime. Suppose ab ∈ (p). Then p | ab ⇒ p | a or p | b (in other words, a = n · p, n ∈ N or b = n · p, n ∈ N) ⇒ a ∈ (p) or b ∈ (p) ⇒ (p) must be a prime ideal.

In addition, we see that the only maximal ideals in Z generated by a single element are, again, those generated by prime numbers. To see why this is true, choose some ideal (p), p prime. The only ideals that contain the element p ∈ Z are (1) and (p). But (1) = Z! Thus, (p) must then be a maximal ideal. And if we have some maximal ideal (p ) generated by a single element, then it is contained only in Z = (1) and (p ). But this implies that p ∈ Z is divisible by only 1, p ∈ Z ⇒ p is prime.

Finally, if we take Proposition 2.12 into consideration, we obtain the result that, of the ideals for Z generated by a single element, those that are ﬁelds are exactly the quotients: Z/pZ, p a prime. That these quotients are ﬁelds is a result from ** Example 1.6.**

With these theorems and propositions regarding rings at our disposal, we may now proceed to study the more speciﬁc case of ﬁelds and the ﬁeld extensions that arise from them.

3. Fields and Field Extensions Deﬁnition 3.1. The characteristic of a ﬁeld, F, denoted ch(F ), is deﬁned to be the smallest positive integer p such that p · 1F = 0 if such a p exists and is deﬁned to be 0 otherwise.

** Remark 3.2 (Deﬁnition 3.**

1). The characteristic of a ﬁeld F, ch(F ), is either 0 or a prime p. If ch(F ) = p, then for any α ∈ F,

Proof. Suppose ch(F ) = n ∈ N, n not prime (i.e. n composite). Then ∃ a, b ∈ N such that n = ab a, b = 1.

0 = n · 1 = (ab) · 1 = (a · 1) · (b · 1) ⇒ a · 1 = 0 or b · 1 = 0.

But a, b n ⇒ ch(F ) = n. Contradiction!

Deﬁnition 3.3. The prime subﬁeld of a ﬁeld F is the subﬁeld of F generated by the multiplicative identity 1F of F. It is isomorphic to either Q (if ch(F ) = 0) or Fp (if ch(F ) = p).

Deﬁnition 3.4. If K is a ﬁeld containing the subﬁeld F, then K is said to be an extension ﬁeld (or simply an extension) of F, denoted K/F. This notation is shorthand for “K over F.”

## AN INTRODUCTION TO THE THEORY OF FIELD EXTENSIONS 5

Deﬁnition 3.5. The degree of a ﬁeld extension K/F, denoted [K : F ], is the dimension of K as a vector space over F. The extension is said to be ﬁnite if [K : F ] is ﬁnite and is said to be inﬁnite otherwise.** Example 3.6.**

The concept of ﬁeld extensions can soon lead to very interesting

**and peculiar results. The following examples will illustrate this:**

(1) Take the ﬁeld Q. Now, clearly, we have the polynomial p(x) = x2 −2 ∈ Q[x];

√ however, it should be evident that its roots, ± 2 ∈ Q. This polynomial is / then said to be irreducible over Q.

Thus, by considering the quotient ring Q[x]/(x2 −2), we ﬁnd that we obtain √ √ another ﬁeld, denoted Q( 2) (or Q(− 2), which just so happens to be √ isomorphic to Q( 2) | this, of course, is no coincidence).

(2) Take the ﬁeld R. Again, we may easily ﬁnd a polynomial, which is irreducible over our ﬁeld. Choosing p(x) = x2 + 1 ∈ R[x], it is obvious that the roots, ±i ∈ R. Thus, if we consider the quotient ring, R[x]/(x2 + 1), / we obtain the ﬁeld R(i) (∼ C!).

= Since both of the given examples are of polynomials that are irreducible over the particular ﬁelds, it will be of great beneﬁt to examine the subject of irreducible polynomials (and the criteria to label them as irreducible) more closely.

Deﬁnition 3.7. A polynomial p(x) ∈ R[x], a polynomial ring, is said to be irreducible if it cannot be factored as the product of two other polynomials of smaller degrees, both in the polynomial ring, R[x].

Proposition 3.8. Let F be a ﬁeld and let p(x) ∈ F [x]. Then p(x) has a factor of degree one if and only if p(x) has a root in F, i.e. there is an α ∈ F with p(α) = 0.

Proposition 3.9. A polynomial of degree two or three over a ﬁeld F is reducible if and only if it has a root in F.

Notation 3.10. If the polynomial is of degree ≥4, then it may still be reducible without necessarily having roots in the ﬁeld (i.e. the polynomial may have factors of degree ≥2, yet still have no factors of degree 1). Fortunately, the explicit examples that will be shown will only require testing for the irreducibility of polynomials of degrees 2 and 3. In addition, the polynomials that will be tested for irreducibility will be elements of Q[x]; therefore, the actual procedure of checking for irreducibility will be trivial.

** Theorem 3.11.**

Let ϕ : F → F be a homomorphism of ﬁelds. Then ϕ is either identically 0 or is injective, so that the image of ϕ is either 0 or isomorphic to F.

** Lemma 3.12.**

Let F be a ﬁeld. Then its only ideals are 0 and F.

Proof. Let F be a ﬁeld. Suppose ∃ a nonzero ideal I for F. Let 0 = a ∈ I.

F a ﬁeld ⇒ ∃ a−1 such that a · a−1 = a−1 · a = 1. Thus, ∀ r ∈ F, we have r = r · 1 = r · (a−1 · a) = (r · a−1 ) · a ∈ I (because r · a−1 ∈ F ). Hence, I = F.

This leaves I = {0} as the only other possible ideal for F (it is very easy to check that this is ideal).

** Lemma 3.13.**

If F is a ﬁeld, then any nonzero ring homomorphism from F into another ring is injective.

6 SAMUEL MOY Proof. Let ϕ : F → F be a nonzero ring homomorphism. Since kerϕ is an ideal, and because ϕ is a nonzero ring homomorphism, then kerϕ is a proper ideal. As F is a ﬁeld, then kerϕ = 0 (because 0 is the only proper ideal of a ﬁeld F ).

Therefore, ϕ : F → F is injective.

Proof of Theorem 3.11. Proving this theorem requires only combining the results of Lemmas 3.12 and 3.13. As F is a ﬁeld, then its only ideals are 0 and F. Now, if ϕ : F → F is any nonzero ring homomorphism, by Lemma 3.13, it is injective, implying that the image of ϕ is isomorphic to F (i.e. ϕ(F ) ∼ F ). And if we take ϕ to = be a homomorphism that is not nonzero, then, it must be the zero homomorphism and therefore must be identically 0.

** Theorem 3.14.**

Let F be a ﬁeld and let p(x) be an irreducible polynomial. Then there exists a ﬁeld K containing an isomorphic copy of F in which p(x) has a root.

Thus, there exists an extension of F in which p(x) has a root.