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

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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.


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.


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


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.”


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.

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