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Math 421 : Abstract Algebra II
- Preliminaries. You should be prepared to present, to the class at the blackboard and to hand in, the proofs for every result except definitions, the simpler examples, and those statements marked with ] (meaning that either the result or the proof takes us beyond the scope of the course, so the result is included only for clearer understanding or enrichment) or with [ (meaning that to do the whole proof is not worth the effort, although you are encouraged to do some part of it, or to try a few examples of your own devising, to help you understand the statement; of course, this is always a good idea!). The proofs of the results marked ∗ are particularly difficult. To the usual taxonomy of mathematical results — lemma, proposition, theorem, corollary, remark — I have chosen to add the medieval “scholium”: a result included more to provide a broader understanding than to apply to the proofs of later results. In general, our terminology and notation follows that of Dan Saracino, Abstract Algebra: A First Course (Waveland Press, Prospect Heights, Illinois, 1992). In particular, we use ⊆ for “is a subset of” and ⊂ for “is a proper subset of.” Absolute value bars around a finite set means the number of elements in that set. A minus sign between sets (rather than a backslash ) denotes set difference. The symbols Z , Q , R and C denote the sets of all integers, rational numbers, real numbers and complex numbers respectively. But throughout our course, the term ring means commutative ring with unity (= multiplicative identity), and a ring homomorphism takes the unity to the unity. If we speak of one ring containing another, it is assumed that the smaller is a subring of the larger, i.e., that the operations on the smaller are the restrictions of those on the larger and that the rings share the same unity (so that, by this definition, a proper ideal of a ring is not a subring). Domain means integral domain, i.e., a ring in which, if a product is zero, then one of the factors must be zero; or equivalently, if ab = ac and a 6 = 0, then b = c. A domain D has a field of fractions (called the “quotient field” in Saracino’s text) consisting of all fractions a/b where a, b ∈ D and b 6 = 0; it is the smallest field containing D, in the sense that any field containing D also contains (an isomorphic copy of) the field of fractions of D. A unit in a ring is an element with a multiplicative inverse, and an ideal is a nonempty subset that is closed under addition and “captures” multiplication. If a 1 , a 2 ,... , an are elements of the ring R, then the smallest ideal of R that contains these elements is the set
a 1 R + a 2 R + · · · + anR = { a 1 r 1 + a 2 r 2 + · · · + anrn : r 1 , r 2 ,... , rn ∈ R } ,
called the ideal (of R) generated by a 1 , a 2 ,... , an. An ideal aR generated by a single element is called a principal ideal. The subset of a ring R consisting of only one element, the zero of R, is an ideal; we will often denote this ideal by 0 rather than by { 0 } (by “abuse of language,” as the group of mathematicians “Nicolas Bourbaki” say); so R − 0 means the set of nonzero elements of R. We will need:
0.1 Fundamental Theorem of Ring Homomorphisms [. (1) Let I be an ideal in the ring R. Then the set R/I, consisting of the cosets a + I as a varies over R, is a ring with the operations induced by the operations in R (i.e., (a+I)+(b+I) = (a+b)+I and (a+I)(b+I) = (ab)+I), called the factor ring of R by I (or just “R mod I”); and the natural (or canonical) map η : R → R/I, defined by η(a) = a + I for each a in R, is a surjective ring homomorphism. (2) Let R, S be rings and ϕ : R → S be a ring homomorphism. Then the kernel Ker(ϕ) = {a ∈ R : ϕ(a) = 0} of ϕ is an ideal of R, and the factor ring R/ Ker(ϕ) is isomorphic (as a ring) to the image ϕ(R) of ϕ, a subring of S, by the isomorphism ϕ : R/ Ker(ϕ) → S induced by ϕ as follows: ϕ(a + Ker(ϕ)) = ϕ(a).
(One often says that ϕ “factors through” R/ Ker(ϕ) or through ϕ, since if η denotes the natural map R → R/ Ker(ϕ), then ϕ = ϕη, where juxtaposition means composition of functions.) A final comment about notation: Just as the symbol Z was inspired by the German “Zahlen” (“numbers”), we will usually use K to denote a field, inspired by the German “Korps”. The word “ring” is the same in both English and German; in the literature a ring may be denoted A, for the French “anneau”.
- Special Types of Rings and Ideals.
1.1 Definition. A proper ideal I in a ring R is called (1) prime iff, for a, b in R, ab ∈ I implies a ∈ I or b ∈ I. (2) maximal iff there is no proper ideal J of R that properly contains I.
1.2 Examples. If p is a prime integer (2, 3, 5, 7, etc.) then p Z is both a prime ideal and a maximal ideal in Z. The zero ideal is prime but not maximal in Z. No other ideals in Z are prime or maximal.
1.3 Fact ]. In any ring, any element that is not a unit is contained in at least one maximal ideal. But if we were to use the definition of ring in Saracino’s text, then there would be rings that have no maximal ideals at all.
If we set x = (0, 1 , 0 , 0 , 0 ,... ), then by identifying the element r of R with (r, 0 , 0 , 0 ,... ) we can write any polynomial in a more familiar way:
(f 0 , f 1 , f 2 ,... ) = f 0 + f 1 x + f 2 x^2 +....
Thus, x is not an “unknown quantity” of some sort (though it is called an indeterminate or “vari- able”); the powers of x just serve in the sum expression as placeholders for the coordinates of the infinite sequence. Though the sum is written as though it had infinitely many terms, it is really finite, since all the “later” fi’s are 0; and when this finiteness is important, we will often write f in the form f = f 0 + f 1 x + f 2 x^2 +... + fnxn^ ,
or in the more familiar descending order of powers of x. If f is written in this way, fn may or may not be nonzero (i.e., the last nonzero coefficient may have come earlier); whether it is so should be made clear in a separate statement. Hereafter we will usually write polynomials in this way and often incorporate the x into the name of the polynomial: f (x) instead of just f (as if a polynomial were a function instead of a sequence of coefficients; see Discussion 1.12 below). The ring of all polynomials over R in the indeterminate x is denoted R[x]. For a given polynomial f (x) = f 0 + f 1 x + f 2 x^2 + · · · , the largest integer n for which fn 6 = 0 is called the degree of f (x) (we will say that the degree of the zero polynomial is −∞), and that coefficient fn is called the leading coefficient of f (x). If the leading coefficient is 1, f (x) is called monic. To include polynomials of more than one indeterminate, we form a polynomial ring over a polynomial ring: x^2 + y^2 ∈ (R[x])[y] = R[x, y].
The definitions of addition and multiplication of polynomials as sequences were chosen to reflect the familiar addition and multiplication of polynomials as sums. See also Discussion 1.12 below.
1.11 Proposition. Let D be a domain and f (x), g(x) ∈ D[x] − 0. Then deg(f (x)g(x)) = deg(f (x)) + deg(g(x)).
1.11a Corollary. If D is a domain, then the only units in the polynomial ring D[x] are the units in D.
1.12 Discussion. Let R ⊆ S be rings, and let a be an element of S and f (x) be a polynomial in R[x]. Then we get an element of S by “substitution”:
f (a) = f 0 + f 1 a + f 2 a^2 +... ,
where x in the sum expression for f (x) is replaced by a to yield a sum in S, a finite sum because the later fn’s are 0. If we hold f (x) fixed and let a vary over the elements of S, we get a “polynomial function” from S to S, i.e., an element of F unc(S, S). It is usually still denoted f (x), though it is possible (see Scholium 1.13; but rare — see Corollary 3.3) that different polynomials will yield the same polynomial function. The definitions of addition and multiplication of polynomials (as sequences or sums) in Definition 1.10 were chosen so that they would make the association of a polynomial in R[x] to the corresponding polynomial function in F unc(S, S) into a ring homomor- phism [; i.e., (f + g)(a) = f (a) + g(a) and (f g)(a) = f (a)g(a) for all a in S. (But most polynomial functions, like most functions, are not ring homomorphisms.) On the other hand, if we hold the element a of S fixed and vary f (x) over the elements of R[x], we get a function from R[x] to S, “evaluation at a,” defined by εa(f (x)) = f (a) for each f (x) in R[x]. The function εa is a ring homomorphism [this statement is the only part of the discussion that requires proof], so its kernel, {f (x) ∈ R[x] : a is a root of f }, is an ideal in R[x].
1.13 Scholium. Let K be a field with only finitely many elements; say |K| = q. Then every element of K is a zero of the polynomial xq^ − x. It follows that the distinct polynomials 0 and xq^ − x give the same polynomial function in F unc(K, K).
(Suggestion: K − 0 is a multiplicative group with q − 1 elements; apply Lagrange’s Theorem.)
1.14 Fact ]. If in the definition of polynomial we remove the restriction that the fi’s in the sequence f = (f 0 , f 1 , f 2 ,... ) are eventually zero, the definitions of addition and multiplication are still meaningful, since they require only a finite number of additions and/or multiplications in each component. These operations make the set of all sequences f = (f 0 , f 1 , f 2 ,... ) of elements of R, i.e., all possibly infinite sums f (x) = f 0 + f 1 x + f 2 x^2 +... ,
into a ring, the ring of formal power series R[[x]] in x over R. But there is no natural way to interpret evaluation, i.e., substituting a value a for x, into a power series — unless a = 0 or f (x) is really a polynomial — since infinite sums are usually not meaningful. (In some cases, an infinite sum can be interpreted as a limit, as in real analysis; but in a general ring, with no sense of “neighborhoods,” a limit is meaningless.)
- Special Types of Domains. Throughout this section, D denotes a domain. We consider three classes of domains: UFD’s, PID’s and Euclidean domains. We will see that each is more restrictive than the last, and then we
2.11 Definition. A domain D is called a unique factorization domain, or UFD (or “factorial ring”), iff every nonzero nonunit a is a product of irreducible elements, and this factorization is “unique up to order and units”, i.e., if a = p 1 · · · pm and a = q 1 · · · qn where all the pi’s and qj ’s are irreducibles, then m = n and the qj ’s can be rearranged so that pi is associate to qi for each i = 1,... , m.
2.12 Proposition. An irreducible element in a UFD is a prime element.
2.13 Discussion [. Let D be a UFD. Pick a set {pλ}λ∈Λ consisting of one element from each associate class of irreducible elements. (E.g., in Z , one might pick all the positive prime numbers; in K[x] where K is a field, one might pick all the monic irreducible polynomials.) Then every nonzero element a of D has a unique expression in the form
(2-1) a = u
λ∈Λ
pe λλ ,
where the eλ’s are nonnegative integers, all but finitely many of which are equal to zero, and u is a unit in D. For example, in the UFD Z , we have
−240 = (−1) · 24 · 31 · 51 · 70 · 110 · 130 ·... and 35 = 1 · 20 · 30 · 51 · 71 · 110 · 130 ·....
Moreover, any element a of the field of fractions K of D has a unique expression in the form (2-1), but the eλ’s now may be negative integers, still all but finitely many equal to zero. For example, in the field of fractions Q of Z , we can write
− 24035 = (−1) · 24 · 31 · 50 · 7 −^1 · 110 · 130 ·.... Thus, for any UFD D, we get a family of functions vλ : (K − 0) → Z , called the essential valuations of D, given by vλ(a) = eλ, the exponent on pλ in the expression (2-1) for a, for each a in K − 0. Note the following useful fact: An element a of K − 0 is in D iff vλ(a) ≥ 0 for every λ in Λ. For convenience, we extend each vλ to all of K by setting vλ(0) = ∞, yielding functions vλ : K → Z ∪ {∞}.
2.14 Proposition. If v : K → Z ∪ {∞} is one of the essential valuations of the UFD D, then v satisfies, for any a, b in K: (0) v(a) = ∞ iff a = 0, (1) v(ab) = v(a) + v(b), and (2) v(a + b) ≥ min{v(a), v(b)}.
2.15 Proposition. If a function v : K → Z ∪{∞} satisfies conditions (0)–(2) in Proposition 2.14, then it also satisfies the following conditions: (3) v(1) = v(−1) = 0. (4) v(−a) = v(a) for all a in K. (5) If v(a) 6 = v(b), then v(a + b) = min{v(a), v(b)}. More generally, if one of the elements a 1 , a 2 ,... , an of K has v-value strictly smaller than that of any of the others, then v(a 1 + a 2 +... + an) = min{v(a 1 ), v(a 2 ),... , v(an)}. (6) The subset v(K − 0) of Z is the set of all multiples of some nonnegative integer e. If e 6 = 0 and we replace v by (1/e)v, then v becomes surjective and still satisfies conditions (0)–(2).
2.16 Definition. A domain D is a principal ideal domain, or PID, iff every ideal in D is principal.
2.17 Lemma. (1) If I 1 ⊆ I 2 ⊆ I 3 ⊆... is a chain of ideals in a ring R, then the union ⋃∞ n=1 In is also an ideal of R. (2) If D is a PID, then any chain a 1 D ⊆ a 2 D ⊆ a 3 D ⊆... of ideals in D must terminate, i.e., for some n we must have anD = an+1D = an+2D =.... (3) An irreducible element in a PID generates a maximal ideal (and hence is prime). (4) If a prime element of a domain divides a product of irreducibles, then it is associate to one of the irreducibles in the product. (5) If an element of a domain is a product of prime elements, then up to order and units that factorization is the only factorization of that element into irreducibles.
2.18 Theorem ∗. Every PID is a UFD.
(Suggestion: Use (2) of the lemma to show by contradiction that every nonzero nonunit element in D has at least one irreducible factor. Then use (2) again to show that every nonzero nonunit is a product of (one or more) irreducibles in at least one way. Then the uniqueness of the factorization follows from (3) and (5).)
2.19 Example. Not every UFD is a PID: Z [x] is a UFD but not a PID; in particular, the ideal 2 Z [x] + x Z [x] is not principal.
2.20 Discussion ]. The (Krull) dimension of a ring R is the length of the longest chain of prime ideals in R, P 0 ⊂ P 1 ⊂ P 2 ⊂... ⊂ Pn (counting inclusions, not prime ideals, so this chain has length n). By Lemma 2.17(3), a PID has dimension 1 (or dimension 0, if it is the trivial case of a PID, a field). In fact, a UFD of dimension 1 (or 0) is a PID. (The UFD Z [x] has dimension 2: A chain of primes of length 2 is 0 ⊂ x Z [x] ⊂ 2 Z [x] + x Z [x].)
assume v is surjective. Then the domain V is called a discrete (rank-one) valuation ring, or DVR, and v : K → Z ∪ {∞} is the (normalized) valuation associated to V. Examples of these rings are the subrings Z (^) (p) of Q for the various prime integers p, defined by Z (^) (p) = {a/b : a, b ∈ Z , p - b}. In fact, a DVR V is a very nice domain: Let p be an element of V for which v(p) = 1; then p and its associates are the only primes in V , i.e., all the nonzero ideals in V have the form pnV for some positive integer n, and they are arranged in a chain:
V ⊃ pV ⊃ p^2 V ⊃ p^3 V ⊃...
Moreover, for each nonzero element a of K, either a or a−^1 is in V.
2.25 Discussion ]. The emphasis in these notes on essential valuations rather than prime factors is somewhat unusual in introductory presentations. The reason it was selected is that an essential family of valuations exists for many domains that are not UFD’s; among them “rings of algebraic integers” like Z [√−5] = {a + b√−5 : a, b ∈ Z }, where (2 + √−5)(2 − √−5) and 3 · 3 are essentially different factorizations of 9 into irreducibles.
- Criteria for Irreducibility of Polynomials. Just as it is often difficult to tell whether a large integer is prime, it is often difficult to tell whether a polynomial f (x) over a field, even over Q , is irreducible. Indeed, since the set of polynomials of degree less than that of f (x) is infinite (except over a finite field), it is in principle even more difficult. There are a few elementary results in this direction:
3.1 Proposition. Let K be a field, r ∈ K, and f (x) ∈ K[x]. Then r is a root of f (x) iff x − r divides f (x) in K[x].
(Suggestion: Apply Lemma 2.23 with g(x) = x − r.)
3.2 Corollary. Let K be a field and f (x) ∈ K[x] with deg(f (x)) > 1. If f (x) has a root in K, then f (x) is reducible in K[x]. If deg f (x) = 2 or 3 , then f (x) is irreducible in K[x] iff f (x) has no root in K.
3.3 Corollary. If K is a field and f (x) ∈ K[x] − 0 , then f (x) has at most deg(f (x)) different roots in K.
Proposition 3.6 below sometimes appears in elementary texts, but the proof is usually incomplete in that it ignores the intervening results. The first lemma (or something like it) is sometimes credited to Kronecker, but it is usually called Gauss’s Lemma.
3.4 Gauss’s Lemma (valuation version). Let K be a field and v : K → Z ∪ {∞} be a function satisfying conditions (0)–(2) of Proposition 2.14. Extend v to K[x] by letting v(g(x)) denote the minimum v-value of the coefficients of g(x). Then for g(x), h(x) in K[x], v(g(x)h(x)) = v(g(x)) + v(h(x)).
(Suggestion: Let i, j be the smallest subscripts for which v(gi) and v(hj ) attain the respective minima. Then the coefficient of xi+j^ in g(x)h(x) has the correct v-value.) Gauss’s Lemma says that the extension of v to K[x] also satisfies condition (1) of Proposi- tion 2.14. Conditions (0) and (2) also hold, but (1) is the hardest to prove and the one we will need below.
3.5 Corollary. Let D be a UFD with field of fractions K, and let f (x), g(x) ∈ D[x] − 0 and h(x) ∈ K[x] for which f (x) = g(x)h(x). If the coefficients of g(x) have no common factor in D (except units), then h(x) ∈ D[x].
(Suggestion: It suffices to show that vλ(h(x)) ≥ 0 for all the essential valuations vλ of D.)
3.6 Proposition. Let D be a UFD with field of fractions K, and let f (x) ∈ D[x]. Then any root of f (x) in K has the form: a factor (in D) of the constant term of f over a factor of the leading coefficient of f.
One of the few general results that yield irreducibility for polynomials of higher degree is:
3.7 Eisenstein’s Criterion. Let D be a UFD with field of fractions K, let p be a prime element of D, and let
f (x) = f 0 + f 1 x + f 2 x^2 +... + fnxn^ ∈ D[x] − 0.
If p - fn (so that fn 6 = 0), if p|fi for each i = 0, 1 , 2 ,... , n− 1 , and if p^2 - f 0 , then f (x) is irreducible in K[x].
(Suggestion: Suppose that f (x) = g(x)h(x) in K[x], and let v be the essential valuation of D associated to p. Then we may assume that v(g(x)) = 0 (so that v(h(x)) = 0 also by Gauss’s Lemma) and that v(g 0 ) = 1 and v(h 0 ) = 0. Let i be smallest so that v(gi) = 0, and argue that h(x) must be a constant, i.e., a unit in K[x].)
3.8 Proposition. For any (positive) prime integer p, the polynomial Φp(x) = 1+x+x^2 +.. .+xp−^1 is irreducible over Q.
4.2 Remark [. For K, L, a 1 , a 2 ,... as in (1) of this definition, the smallest subring of L containing K and all the ai’s is just the result of evaluating all polynomials at the ai’s, i.e., taking all the polyno- mials f (x 1 , x 2 ,... ) in K[x 1 , x 2 ,... ] and considering all the elements of L of the form f (a 1 , a 2 ,... ); so we denote this subring by K[a 1 , a 2 ,... ]. The field of fractions of this ring is K(a 1 , a 2 ,... ); so if K[x 1 , x 2 ,... ] is a field, then it is equal to K(a 1 , a 2 ,... ). But K(a 1 , a 2 ,... ) may be strictly larger than K[a 1 , a 2 ,... ]. In fact, if there are only finitely many ai’s, then we claim that the ring they generate is equal to the field they generate iff all the ai’s are algebraic over K. (A first step toward proving this claim is part of Theorem 4.5.)
4.3 Examples. (1) Since √3 is a zero of the polynomial x^2 −3 in Q [x], √3 is algebraic over Q. Also, the fact that (√3)^2 = 3 means that if we have any polynomial expression in √3 with coefficients in Q , we can rewrite the relation so that it has no squared or higher powers of √3, i.e., it has the form a + b√3 where a, b are in Q. Thus, Q [√3] = {a + b√3 : a, b ∈ Q }. Moreover, if we have a quotient of two such expressions, (a + b√3)/(c + d√3), then we can multiply numerator and denominator by the “conjugate” c − d√3 of the denominator, and again reach the form (rational) + (rational)√3. Thus, Q [√3] is a field, so Q [√3] = Q (√3); and a vector space basis for Q (√3) over Q is { 1 , √ 3 }, so [ Q (√3) : Q ] = 2. (2) Similarly, since √^5 3 is a root of x^5 − 3, we see a Q -basis for Q ( √^5 3) is 1, √^53 , √^532 , √^533 , √^534 , and [ Q ( √^5 3) : Q ] = 5. (“Rationalizing a denominator” is now much harder, but it is still possible.) But the primitive fifth root of unity, ω 5 = cos(2π/5) + i sin(2π/5) = e^2 πi/^5 , one of the complex roots of the polynomial x^5 − 1, is such that [ Q (ω 5 ) : Q ] = 4, since x^5 − 1 factors as (x − 1)(x^4 + x^3 + x^2 + x + 1), and ω 5 is a root of the second factor, which is irreducible by Proposition 3.8. A Q -basis for Q (ω 5 ) is 1, ω 5 , ω^25 , ω^35. (3) Q (3^1 /^2 , 31 /^4 , 31 /^8 ,... ) is an algebraic extension of Q that is not finite. (4) The same extension field can be generated in many different ways. For example,
Q ( 4
2 , i 4
2 , −i 4
2) = Q ( 4
2 , i).
(5) ] It is true, but difficult to prove (cf. Serge Lang, Algebra, Addison-Wesley, Reading, MA, 1965, Appendix), that π (the ratio of the circumference of a circle to its diameter) is transcendental over Q. As a result, all of the powers of π are linearly independent over Q , so Q (π)/ Q is not a finite extension. The same is true of e, the base of the natural logarithms.
4.4 Proposition. A finite field extension is always algebraic.
(Suggestion: If a is an element of the larger field, then the powers of a cannot all be linearly independent over the smaller field.)
4.5 Theorem. Let a be an element of an extension field of K such that a is algebraic over K. Then there is a unique monic irreducible polynomial p(x) in K[x] such that, for each f (x) in K[x], f (a) = 0 iff p(x)|f (x). Moreover, K[x]/(p(x)K[x]) ∼= K[a] = K(a) (the isomorphism being given by a map that takes the coset x + (p(x)K[x]) to a and is the identity on K), a K-basis for K(a) is 1 , a, a^2 ,... , a(deg(p))−^1 , and [K(a) : K] = deg(p(x)).
4.6 Definition. If a is algebraic over K, then the monic irreducible polynomial p(x) in Theorem 4. is called the minimal polynomial of a over K, and denoted Irr(a, K).
4.7 Examples. (1) Irr(√ 3 , Q ) = x^2 − 3 = Irr(−√ 3 , Q ); but Irr(√ 3 , R ) = x − √3. (2) Irr( √^53 , Q ) = x^5 − 3 = Irr(ω 5 √^53 , Q ), where ω 5 is as in Example 4.3(2). (3) Irr(ω 5 , Q ) = x^4 + x^3 + x^2 + x + 1. (4) Irr( √^43 , Q (√3)) = x^2 − √3. (5) Irr((1 + √3)/ 2 , Q ) = (x − (1 + √3)/2)(x − (1 − √3)/2) = x^2 − x − (1/2). (6) Irr(i, R ) = x^2 + 1 = Irr(i, Q ).
4.8 General Example. For a general positive integer n, the cyclotomic polynomial Φn(x) is the minimal polynomial of the “primitive” n-th roots of unity in C ; one of these is
ωn = cos(2π/n) + i sin(2π/n) = e^2 πi/n^ ,
and the others are the powers ωnm where m ∈ { 2 , 3 ,... , n − 1 } and gcd(m, n) = 1. Thus, for example, ω 1 = 1, ω 2 = −1, ω 4 = i, and ω 43 = −i. (The n-th roots of unity form a cyclic group under multiplication, and the primitive ones are the generators of that group — cf. Saracino, Theorem 4.4(ii), page 38). The polynomial Φn(x) is defined by induction: Φ 1 (x) = x − 1, and for n > 1,
Φn(x) = (xn^ − 1)/
d|n,d 6 =n
Φd(x)
(In effect, to obtain Φn(x), we take the polynomial xn^ − 1 whose roots are all the n-th roots of unity, and remove the roots of strictly smaller order.) It follows from Proposition 3.8 that, if p is prime, then [ Q (ωp) : Q ] = p − 1.
First, we consider the tools themselves. The first Euclidean tool is a “collapsing compass”: it can draw a circle with a given center through a given point, but it cannot, a priori, pick up a length from one part of one part of the plane and use it as a radius for a circle with a center elsewhere. But the second proposition in Euclid’s Elements shows how to use a collapsing compass to transfer a length in this way, so we need not worry about this difference between the Euclidean tool and a “real” compass. There is a more important difference in the second Euclidean tool, however: it is not a ruler, to measure distances; it is only a straightedge, to draw the straight line through two points. (More about why this difference is important below.) These tools correspond to three of the “Postulates” set down in Euclid’s Elements: “Let the following be postulated: 1. To draw a straight line [segment] from any point to any point. 2. To produce [i.e., extend] a straight line [segment] continuously in a straight line. 3. To describe a circle with any center and distance [i.e., radius]” Second, it will be convenient to speak of “constructible numbers”. Most construction problems start with a given line segment, like a segment on which to construct an equilateral triangle, or the radius of a given circle to which a tangent from an exterior point is to be constructed; and when a segment is not given, as in the problem of bisecting a given angle, we can choose one at random. We associate points in the plane with complex numbers by making that given line segment the segment joining 0 and 1 in the usual “complex plane” (or “Argand diagram”) representation of the complex numbers: one end of the given segment becomes (0,0) and the other (1,0) to establish a rectangular coordinate system on the plane, and the point (a, b) is associated to the complex number a + bi. We describe recursively what it means to “locate” a point, or equivalently a complex number, in the plane: The numbers 0 and 1 are (automatically) located, as the endpoints of the given segment. To locate another point P means to draw two lines, determined by points that have already been located; or two circles, each with a center and some point on the circumference that have already been located; or such a line and such a circle; so that the two curves (lines and/or circles) meet at P. We will call a complex number constructible if it can be “located” in this way (in a finite number of steps).
5.2 Lemma. The complex number a + bi is constructible iff the nonnegative real numbers |a| and |b| are constructible.
(A proof of this result amounts to a description of how, given a + bi, one could construct |a| and |b|, and vice versa.)
5.3 Proposition [. The set E (for Euclidean) of constructible complex numbers is a subfield of C that contains the square roots of each of its elements.
Proof sketch. For positive real numbers, which can be interpreted as lengths, the field operations can be done geometrically: addition and subtraction of lengths are easy, and multiplication and division are accomplished by using similar triangles with the original given segment, of length 1, as one side of one triangle. To construct a segment of length √a, where a is a positive real, draw a segment of length a + 1, a perpendicular line 1 unit into the segment, and a semicircle of which the segment is the diameter; the part of the perpendicular from the segment to the semicircle has length √a. It is not difficult to see how all this can be generalized to all complex numbers, but there are many cases to consider. In particular, for the square root of a complex number, it is convenient to write it in “polar form”: r(cos θ + i sin θ) where r is the distance from the origin O to the point P in the plane corresponding to this number, and θ is the angle that the ray OP−→ makes with the positive real axis; the square roots of this number are ±√r(cos(θ/2) + i sin(θ/2)). Euclid’s Elements showed how to perform certain constructions, and it has always been a favorite pastime of amateur mathematicians to repeat these constructions and to find ways to do others. (George Martin, The Foundations of Geometry and the Non-Euclidean Plane (Springer-Verlag, New York, 1975) page 479: “The old games are the best games.”) But there are three constructions that the Greek mathematicians were unable to do with Euclidean tools:
- Duplication of the Cube: Given (an edge of) a cube, construct (an edge of) the cube with exactly twice the volume of the given cube. (This is also called the “Delian Problem,” because, according to legend, it was inspired by the instruction from the oracle of Apollo at Delos that, to end a plague, the cubical altar there must be doubled in size.)
- Squaring the Circle: Given a circle, construct a square with exactly the same area.
- Trisecting the Angle: Given an angle, construct an angle exactly one-third the size (in angle measure). It is not claimed that these tasks are utterly impossible; the ancients constructed other tools that would perform them. Indeed, if the straightedge were a ruler (or even if it were allowed to make two marks on the straightedge — cf. Howard Eves, An Introduction to the History of Mathematics, 5th ed., CBS College Publishing, New York, 1983; Problem Study 4.6, page 93), it would be possible to trisect any angle. The challenge is to perform these constructions with Euclidean tools or to show that they are impossible with these tools.
5.11 Proposition. It is impossible to construct an angle of 20 ◦. Hence, it is impossible to Trisect an Angle of 60 ◦.
(Suggestion: Use the formula cos(3θ) = 4 cos^3 θ − 3 cos θ.) Recall that a regular polygon is one in which all the sides have the same length and all the interior angles have the same measure.
5.12 Theorem (Gauss) ]. It is possible to construct a regular polygon of n sides iff n = 2 r^ p 1 p 2... ps where r is a nonnegative integer and each pi is a Fermat prime, i.e., a prime number of the form 22 m + 1. In particular, it is possible to construct a regular polygon of 17 sides.
For a proof, see Ian Stewart, Galois Theory, 2nd ed. (Chapman and Hall, London, 1989; Chap- ter 17).
- The Galois Group.
6.1 Definition. Let R, S be rings containing the field K. A ring homomorphism ϕ : R → S is called a K-homomorphism iff ϕ(a) = a for every element of K, i.e., iff ϕ leaves every element of K fixed. (This amounts to saying that ϕ, in addition to being a ring homomorphism, is also a linear transformation of vector spaces over K.)
6.2 Lemma. Let R, S be rings containing the field K, and let ϕ : R → S be a K-homomorphism. (1) [ Let f (x 1 , x 2 ,... , xn) be a polynomial with coefficients from K and a 1 , a 2 ,... , an be elements of R. Then ϕ(f (a 1 , a 2 ,... , an)) = f (ϕ(a 1 ), ϕ(a 2 ),... , ϕ(an)). (2) If a in R is a root of the polynomial f (x) from K[x], then ϕ(a) in S is also a root of f (x). (For brevity: “Under a K-homomorphism, roots go to roots.”)
6.3 Lemma. (1) If L/K is a field extension, then every K-homomorphism from L to a ring containing K is a monomorphism, i.e., is injective. (2) If L/K is algebraic over K, then every K-homomorphism from L to itself is an automor- phism. (3) If S is a ring containing K, if L = K(a 1 , a 2... ), and if ϕ : L → S is a K-monomorphism, then ϕ is uniquely determined by the images of a 1 , a 2 ,... , in the sense that if ψ : L → S is also a K-monomorphism and ϕ(ai) = ψ(ai) for each i = 1, 2 ,... , then ϕ = ψ, i.e., ϕ(a) = ψ(a) for every a in L.
(Suggestion: For (2), use the fact that, if a ∈ L, then a K-homomorphism from L to itself takes the finite set of roots of Irr(a, K) in L to itself. For (3), note first that ϕ(a) = ψ(a) for every a in
K[a 1 , a 2 ,... ].)
6.4 Definition. If L/K is an algebraic extension, then the set of all K-automorphisms of L is called the Galois group of L over K and denoted Gal(L/K).
6.5 Proposition. Let L/K be an algebraic extension. (1) Gal(L/K) is a group under the operation of composition of functions. (2) Suppose L = K(a 1 , a 2 ,... ), and let X denote the set of all roots in L of the polynomials Irr(ai, K). Then there is a monomorphism of Gal(L/K) into the symmetric group Sym(X) on the set X, i.e., the set of all 1-1 functions from X onto itself, given by restriction of a K-automorphism of L to the set X.
A good understanding of the following examples is essential to this course. Please study them carefully.
6.6 Examples. (1) Since C = R (i), and Irr(i, R ) = x^2 +1, any element of Gal( C / R ) is determined by the image of i; and that image must be one of the roots of x^2 + 1, i.e., either i or −i. Thus, Gal( C / R ) has only two elements, the identity function (which we will denote id (^) C ) and the familiar “complex conjugation,” i.e., the map taking a+bi (where a, b are real numbers) to a−bi. (A complex conjugate is often denoted with an overbar: a + bi = a − bi.) (2) Similarly, for any element a of Q that is not a square of an element of Q (e.g., a = −1 or 2 or 3 or... ), we have Irr(√a, Q ) = x^2 − a, and Gal( Q (√a)/ Q ) again has two elements, the identity function id (^) Q (√a) and the function that takes b + c√a, where b, c ∈ Q , to b − c√a. (The latter automorphism is also often called “conjugation.”) (3) The polynomial x^4 − 2 is irreducible over Q. Its roots are the real numbers 4
2 and − 4
and the complex numbers i √^4 2 and −i √^4 2; consider the field L = Q ( √^42 , − √^42 , i √^42 , −i √^4 2). By Example 4.3(4), L = Q ( √^42 , i), so the images of √^4 2 and i determine the element of Gal(L/ Q ). The restriction γ of complex conjugation to L gives one element of Gal(L/ Q ) of order 2 (i.e., if we apply it twice in succession, we get the identity). But (see Corollary 6.20 below) there is also at least one Q -automorphism of L that takes √^4 2 to each of the other roots of its minimal polynomial. In particular, there is one, say ϕ, for which ϕ( √^4 2) = i √^4 2. Since ϕ is a field automorphism, we get ϕ^2 ( √^4 2) = ϕ(i)ϕ( √^4 2), and we already know what the second factor is. We don’t know whether ϕ(i) is i or −i, but if it is −i, then ϕγ(i) = i, and we still have ϕγ( √^4 2) = i √^4 2, so we may assume (by replacing ϕ with ϕγ if necessary) that ϕ(i) = i. Then ϕ is an element of Gal(L/ Q ) of order 4, corresponding to the 4-cycle permutation ( √^42 , i √^42 , − √^42 , −i √^4 2) of the roots of x^4 − 2. These four
