25 Radical extensions This section contains two classic applications of Galois theory. Thefirst application, the problem of the constructibility of points in the plane raised by Greek mathematicians, shows that the underlying questions do not need to refer tofield extensions or auto- morphism groups in any way. The second application, solving polynomial equations by extracting roots, is the problem that, in a way, gave rise to Galois theory.

� Construction problems In Greek mathematics,13 people used to constructfigures with a straightedge and com- pass. Here one repeatedly enlarges a given setX of at least two points in the plane by adding to it those points that can be constructed as intersections of lines and circles defined in terms of the given points. The question is, for givenX, whether certain points in the plane can be obtained fromX infinitely many construction steps. More formally, a construction step starting out from a subsetX of the plane consists of replacing the setX with the set (X) of all pointsP obtained by applying F the following algorithm: 1. Pick points a, b, c, d X witha=b andc=d. ∈ � � 2. Let� ab be either the line througha andb or the circle througha with centerb. Likewise, let� cd be either the line throughc andd or the circle throughc with centerd. 3. If� and� do not coincide, pickP � � . ab cd ∈ ab ∩ cd b a c a b d b c c d a d In order to perform step 1 of the algorithm,X needs to contain at least two points. In this case, pickingc=a andd=b shows that we haveX (X). As the permissible ⊂F intersections� � consist of 0, 1, or 2 points, (X) isfinite ifX isfinite. ab ∩ cd F

25.1. Definition. LetX=X 0 be a subset of the plane containing at least two points, and define, recursively fori Z 1, the sets ∈ ≥

Xi = (X i 1) X i 1. F − ⊃ − Then (X)= X is called the set of constructible points starting fromX. C i=0 i In addition to� constructing points in the plane, we can also speak of constructing lines and circles in the plane. A line is called constructible if it is possible to construct two

63 Algebra III– 25 § distinct points on it, and a circle is called constructible if it is possible to construct its center and a point on it. As early as thefifth century B.C., Greek mathematics gave rise to several construc- tion problems that the Greeks could not solve and which, for more than two thousand years, defeated many mathematicians, “professional” or not.14 25.2. Squaring the circle. Construct a square whose area is equal to that of a circle with given radius. Hippocrates of Chios, who lived around 430 B.C., showed that 25.2 is solvable if instead of the circle, we take certainfigures bounded by circular arcs, the so-called lunes of Hippocrates15 (see Exercise 14). 25.3. Doubling the cube. Construct a line segment that is √3 2 times as long as a given line segment. This problem is also known as the Delian problem, after the legend in which, through his oracle on the island Delos, the god Apollo decreed that the plague-ridden Athenians should “double” their cubic altar to Apollo. 25.4. Trisecting the angle. Use a straightedge and compass to divide a given angle into three equal parts. For a few angles, such as the right angle, this problem is easy to solve. In most other cases, angle trisection does not seem possible. 25.5. Constructing then-gon. Forn 3, construct a regularn-gon in a given ≥ circle. Strictly speaking, this problem does not belong to the classical “corpus” of the three unsolved Greek problems. But, as it corresponds to the division of the full angle of 2π radials inton equal parts, it is closely related. Since bisecting an angle using a straightedge and compass is easy, the interesting question in 25.5 is for which oddn the problem is solvable. The Greek found solutions forn = 3 andn = 5 but not, for example, forn = 7 orn = 9. To formulate construction problems in terms offield extensions, we identify the plane in the usual way with thefieldC of complex numbers. Using 25.1, the problems mentioned above can easily be reformulated in terms of numbers constructible from a subsetX C. After scaling, we may assume thatX contains the two points 0 and 1. ⊂ ForX equal to 0,1 , the set = (X) is simply called the set of constructible { } C C numbers. Problems 25.2, 25.3, and 25.5 then correspond to the questions of whether the numbers √π and √3 2 and the primitiventh root of unityζ =e 2πi/n C are con- n ∈ structible. The question in 25.4 is whether forα C with α = 1, the set ( 0,1,α ) ∈ | | C { } contains a third root √3 α. 25.6. Proposition. LetX C be a set that contains0 and1. Then (X) is a subfield ⊂ C ofC that containsX. It is closed under complex conjugation and under extracting square roots.

64 Algebra III– 25 § Proof. The proof is an exercise in carrying out elementary constructions. We assume the standard constructions from Exercise 13 known—an exercise we recommend for anyone who has never carried out a construction. It is clear thatX, and in particular 0 and 1, are contained in (X). It therefore suffices to show that for x, y (X), the C ∈C differencex y, the product xy, the inverse 1/x, the complex conjugate x, and the − square roots √x are constructible fromX. ± Forx y, we mark off the distance x y on the line through 0 parallel to the − | − | line throughx andy. For x, wefirst intersect the circle throughx with center 0 with the line through 0 and 1—this gives x —and then intersect it with the circle through | | x with center x . Forx/ R, the point x /x is the intersection of the circle through | | ∈ | | 1 with center 0 with the line through 0 and x; the line through 1 parallel to the line through x /x and x now cuts the line through 0 and x in 1/x. A similarfigure shows | | | | how to multiplyx by a real number y . The productx y is then rotated over the | | | | angle∠y01 to obtain xy. Exercise 1. How should the constructions be adjusted forx R? ∈ y x x y x | |

x 0 1 0 1/x x 0 1 y x /x | | | | | |

x y x − Taking the square root corresponds to constructing √x forx R . After all, for ∈ >0 non-realx, we simply mark off x on the bisector of the angle x01. Forx R , | | ∠ ∈ >0 we intersect the perpendicular in 0 to the line through 0 and 1 with the circle that has � the line segment from x to 1 as diameter. Lets be an intersection point. − s=i √x

u

x 0 1 − The angle ( x)s1 is a right angle (Thales’s theorem), so the triangles ( x)s0 and ∠ − − s10 are similar. The equality of the ratiosx:u=u : 1 shows thats is equal toi √x, and we are done.

Exercise 2. (For whoever did not know this yet . . . ) Formulate and prove Thales’s theorem.

The main result on constructible numbers is that 25.6 in fact characterizes thefield (X): it is the smallestfield that satisfies the conditions of 25.6. To see this, we must C show that the construction steps can only lead to quadraticfield extensions.

65 Algebra III– 25 § 25.7. Proposition. LetX C be given, and letK=Q(X, X) be the subfield ofC ⊂ generated by the elements inX and their complex conjugates. Then every pointz C ∈ that can be obtained fromX through a construction step is algebraic overK of degree [K(z):K] 2. ≤ Proof. Suppose that there exist points a, b, c, d X witha=b andc=d. We must ∈ � � show that an intersection point of the line or circle determined by ab and the line or circle determined by cd has degree at most 2 overK. We distinguish between the three possibilities given by thefigures before 25.1. The line through the pointsa andb consists of the pointsz C for which ∈ (z a)/(a b) is real. Expanding the identity (z a)/(a b) = ( z a)/(a b) gives − − − − − − the following equation for this line:

� :(a b)z (a b) z= ab a b. ab − − − −

Intersecting the lines� ab and� cd amounts to solving two linear equations inz and z with coefficients inQ(a, b, c, d, a, b, c, d) K. If� and� are not parallel, this system ⊂ ab cd has a unique solutionz K, and we haveK(z)=K. If the lines are parallel, then ∈ because� ab and� cd do not coincide, the system is inconsistent and has no solution. For the circle throughc with centerd, the equation is z d = c d ; we can | − | | − | rewrite this as (z d)( z d) = (c d)( c d) or − − − − zz dz d z=c c c d cd. − − − −

For a pointz on the line� ab that lies on this circle, we can use the equation for� ab to write z as a linear expression inz with coefficients inK. Substituting this in the equation for the circle gives a quadratic relation with coefficients inK satisfied byz. Wefind [K(z):K] 2. ≤ In the event thatz is a point that lies both on the circle througha with centerb and on the circle throughc with centerd, taking the difference of the equations for the two circles gives a linear relation betweenz and z with coefficients inK. Since the circles do not coincide, this is not the zero relation, and we are back in the previous case. This proves [K(z):K] 2 for all construction steps. ≤ � Quadratic closure The main result for constructible numbers can be formulated concisely in terms of quadratic closures. Wefirst introduce the “maximum square root extension” of a fieldK (inside an algebraic closure K) as the subfieldK(S) K generated by the set ⊂ S= w K:w 2 K { ∈ ∈ } of square roots of elements ofK. ForK of characteristic different from 2, this extension, which we symbolically denote byK( √K), is an algebraic extension ofK that is normal and separable. In many cases, however, it is of infinite degree overK.

Exercise 3. Show thatQ Q( √Q) is an infinite extension. ⊂ 66 Algebra III– 25 § 25.8. Definition. LetKbeafield of characteristic char(K)= 2 with algebraic clo- � sure K. Then the quadratic closureK quad ofK in K is thefield

quad ∞ K = Ki, whereK 0 =K andK i =K i 1( Ki 1) fori 1. i=0 − − ≥ � �

The similarity between 25.8 and 25.1 is more than superficial.

25.9. Theorem. The set (X) of points constructible from a subsetX C that C ⊂ contains0 and1 is equal to the quadratic closure of thefieldQ(X, X) inC.

Proof. By 25.6, the set (X) is afield that containsX and is closed under complex C conjugation and extracting square roots, so it is clear that the quadratic closure of K=Q(X, X) is contained in (X). C Conversely, it follows from 25.7 that the points that can be formed fromX through a construction step are contained either inK=Q(X, X) itself or in a quadratic extension ofK. Every quadratic extension ofK is of the formK K( √x) and ⊂ therefore contained inK 1 =K( √K). Note that along withK, thefieldK( √K) also maps to itself under complex conjugation. Repeating the previous argument shows that, more generally, the points that can be constructed fromX ini 1 construction ≥ steps are contained inK , withK as in 25.8. This shows that (X) is contained in i i C the quadratic closure ofK=Q(X, X).

Even in the simplest caseX= 0,1 , thefield (X) =Q quad is an infinitefield { } C extension ofQ(X, X)=Q. The following theorem is therefore very useful to determine whether a complex number is in the quadratic closure ofK C. ⊂ 25.10. Theorem. LetK Cbeafield. Then for an elementx C, the following ⊂ ∈ statements are equivalent: 1. The elementx is contained in the quadratic closure ofK.

2. There exist ann Z 0 and a chain ∈ ≥

K=E 0 E 1 E 2 ... E n 1 E n C ⊂ ⊂ ⊂ ⊂ − ⊂ ⊂

of intermediatefields ofK C with[E i :E i 1] = 2 for1 i n andx E n. ⊂ − ≤ ≤ ∈ x 3. The elementx is algebraic overK, and the Galois group of the polynomialf K overK is afinite2-group.

Proof. (2) (1). LetV C be the set of elements that satisfy (2). Then forx V ⇒ ⊂ ∈ with associated chainK=E E E ... E , every quadratic extension 0 ⊂ 1 ⊂ 2 ⊂ ⊂ n E E can be obtained through the adjunction of a square root. It follows that i ⊂ i+1 we haveE K forK as in 25.8, and thereforex E K K quad. This proves i ⊂ i i ∈ n ⊂ n ⊂ V K quad. ⊂ (1) (2). For the inclusionK quad V withV as above, we show thatV is a ⇒ ⊂ subfield ofC that containsK and is closed under the adjunction of square roots. It is obvious thatK is contained inV . It is also clear that along withx V , we also ∈ 67 Algebra III– 25 § have √x V ; in the situation of statement (2), consider the extensionE E (√x) ∈ n ⊂ n of the chain for √x / E . Finally, to see thatV is a subfield ofC, we use the chain ∈ n K=F F F ... F fory V to extend the chainK=E E E 0 ⊂ 1 ⊂ 2 ⊂ ⊂ s ∈ 0 ⊂ 1 ⊂ 2 ⊂ ... E forx V to ⊂ r ∈ K=E E E ... E =E F E F E F ... E F . 0 ⊂ 1 ⊂ 2 ⊂ ⊂ r r 0 ⊂ r 1 ⊂ r 2 ⊂ ⊂ r s This tower consist of successive extensions of degree at most 2, so all elements of Q(x, y) E F are contained inV . This shows thatV is afield. ⊂ r s (2) (3). If forx C, there exists a chainK=E E E ... E as ⇒ ∈ 0 ⊂ 1 ⊂ 2 ⊂ ⊂ n in (2), thenx is certainly algebraic overK. We now extend the chain forx so that the last extension is normal overK. For this, take the normal closureM C ofE overK ⊂ n andσ Gal(M/K) arbitrary. Then the chainK=σ[E ] σ[E ] ... σ[E ] is a ∈ 0 ⊂ 1 ⊂ ⊂ n chain as in (2) forσ(x). By combining these chains for allσ Gal(M/K) as explained ∈ above to one long chain, we obtain a tower of quadratic extensions ending in the

compositumM of the collection offields σ[E n] σ Gal(M/K) . It follows that Gal(M/K) { } ∈ f x is a 2-group, and the Galois group Gal(f x ) of the subextensionΩ K M overK is K K ⊂ therefore also one. (3) (2). It follows from 10.17 (“solvability ofp-groups”) that in the 2-group ⇒ x G = Gal(fK ), there exists a chainG=H 0 H 1 ... H n = 1 of subgroups for ⊃ ⊃ ⊃ f x which all indices [H :H ] are equal to 2. The corresponding subfieldsE Ω K give i i+1 i ⊂ K a chain that satisfies the conditions in (2). Exercise 4. Show that 25.10 is correct for anyfieldK of characteristic different from 2 if, everywhere, we replaceC with an algebraically closed extension ofK.

We now return to the problems 25.2–25.5. If we use the result of Lindemann mentioned before 21.5, which says that the numberπ is transcendental, then it follows from 25.9 and 25.10.3 thatπ and √π are not constructible, so we cannot square the circle with a straightedge and compass. The number √3 2 is algebraic, but of degree 3 overQ and therefore not contained in any extension ofQ whose degree is a power of 2. Doubling the cube with a straightedge and compass is therefore also not possible. Forα C with α 2 =α α = 1, trisecting the angle 10α corresponding toα with ∈ | | ∠ a straightedge and compass is not possible ifX 3 α is irreducible overQ(α, α)=Q(α). − This is the case for “most”α, including all transcendental values ofα; see Exercises 20 and 21. In the rare reducible case, which occurs, for example, forα= 1 andα=i, X3 α ± a splittingfieldΩ Q(α−) has degree at most 2 overK and trisection is possible. In the case of the regularn-gon, we need to determine for whatn the degree of

the cyclotomic extensionQ(ζ n) overQ is a power of 2. This is more of an arithmetic than a geometric problem: for whatn isϕ(n) a power of 2? If such ann has prime ep ep 1 decompositionn= p n p , then 6.16 gives the valueϕ(n) = p n(p 1)p − . This | | − shows that apart from the primep = 2, only primes of the formp=2 m + 1 occur in � � the decomposition ofn and that these primes moreover have exponente p = 1. Note thatp=2 m + 1 can only be prime ifm is a power of 2. After all, ifm has a proper divisoru for which m/u is odd, then 2 m + 1 is divisible by 2u + 1.

68 Algebra III– 25 § 25.11. Definition. A Fermat prime is a prime of the formp=2 2k + 1.

If we writeF = 22k + 1 fork 0, thenF = 3,F = 5,F = 17,F = 257, and k ≥ 0 1 2 3 F4 = 65537 are prime. Fermat’s rash conjecture that all numbersF k are prime explains the name in 25.11. The numberF k is also called thekth Fermat number. Thefifth Fermat numberF = 641 6700417 is not prime, and neither are the numbersF with 5 · k 6 k 32. It is not known 16 whether there exist valuesk 5 for whichF is prime— ≤ ≤ ≥ k it is conjectured that this is not the case. Apart from this open question, there is the following complete solution of 25.5.

25.12. Theorem. Letn 3 be an integer, and writen=2 k n withn odd. Then ≥ · 0 0 the regularn-gon is constructible if and only ifn 0 is a product of distinct Fermat primes.

The constructibility forn of the given form was already proved by Gauss in 1801 using the cyclotomic periods from 24.9.2. A seventeen-pointed star adorns the Gauss monument in Brunswijk, because a regular 17-gon can only be distinguished from a circle by looking very closely.

� Radical closure A characterization as that given in 25.10 for the quadratic closure ofQ inC can also be given for the radical closureQ rad ofQ inC. The definition closely resembles that in 25.8.

25.13. Definition. LetK be of characteristic0 with algebraic closure K. Then the rad rad radical closureK ofK in K is thefieldK = i∞=0 K(i), whereK (0) =K and

n K(i) =K (i 1)( w K:w K (i 1)�for somen 1 ) − { ∈ ∈ − ≥ } for alli 1. ≥ ThefieldK rad is the subfield of K consisting of the elements that can be obtained fromK by applying thefield operations (addition, subtraction, multiplication, and di- vision) and “extracting roots” of arbitrarily high degree. This is the smallest extension ofK that is closed under all root extractions. To avoid separability problems, from now on in this section, we assume thatK is of characteristic 0. Forfields of positive characteristicp, all results concerning roots of degreen remain valid whenn is not divisible byp. Forn=p, we obtain the “correct” generalization by replacing thepth roots of unity, which are zeros of polynomialsX p − a K[X], everywhere by zeros of Artin–Schreier polynomialsX p X a K[X]. See ∈ − − ∈ Exercises 32–34. Inside the algebraic closure K ofK, we have a tower of extensions

K K quad K rad K. ⊂ ⊂ ⊂ By definition, K consists of all elements that are zeros of a monic polynomialf K[X]; ∈ it is a classical question whether the zeros of a polynomialf can be expressed in the

69 Algebra III– 25 § elements ofK “using radicals.” Forf K[X] of degreen 4, there exist explicit ∈ ≤ “radical formulas” to express the zeros off in the coefficients off; we will return to this at the end of the section. For polynomials of degreen 5, the search for similar ≥ formulas continued until into the 19th century. The famous application of Galois theory in this section shows that forn 5, such a general formula does not exist. Thefield ≥ Qrad is not equal to Q, and in 25.17, we will construct polynomials inQ[X] whose zeros are in Q but not inQ rad. A word of warning is in order when using the notation √n a to indicate annth root of an elementa. The ambiguity in this notation, which forn = 2 is restricted to a choice of sign, easily leads to mistakes for generaln. For example, because of 16 = 2 4, it seems logical to expect that adjoining a zeroα ofX 8 16 toQ would lead to the − fieldQ(α) =Q( √2). However, we also have 16 = ( 2)4, so thatQ(α) =Q( √ 2) is − − anotherfield that is as “justifiable.” Similar problems occur when extracting “an”nth root of 1. Thus, it is better to avoid the notation √3 1 because not all zeros ofX 3 1 − generate the same extension ofQ. In this last case, even the degree of the extension depends on the choice of the root.

Exercise 5. Show that 1 +i and 1 i are also eighth roots of 16. − We can limit the ambiguity in the notationK K( √n a) forfield extensions obtained ⊂ by adjoining annth root ofa K toK by only looking at irreducible radicals. In this ∈ situation, the additional condition is imposed that only zeros of irreducible polynomials Xn a K[X] are adjoined. In many cases, it is easy to avoid the problematic radical − ∈ notation.

25.14. Definition.Afinitefield extensionK L is called a radical extension if there ⊂ exists a primitive elementx forK L withx n K for somen 1. Ifx andn can be ⊂ ∈ ≥ chosen such thatX n x n is irreducible inK[X], thenK L is called an irreducible − ⊂ radical extension.

Exercise 6. Show that the cyclotomicfieldQ(ζ 5) is a radical extension ofQ but not an irreducible radical extension.

Although not every radical extension is irreducible, thefieldK rad does not depend on the type of radical extension under consideration (Exercise 31). The main result we are going to prove in this section is the following analog of 25.10.

25.15. Theorem. LetKbeafield of characteristic0 with algebraic closure K. Then for an elementx K, the following statements are equivalent: ∈ 1. The elementx is contained in the radical closure ofK in K.

2. There exist ann Z 0 and a chain offields ∈ ≥

K=E 0 E 1 E 2 ... E n 1 E n K ⊂ ⊂ ⊂ ⊂ − ⊂ ⊂

withE i 1 E i for1 i n a radical extension andx E n. − ⊂ ≤ ≤ x ∈ 3. The Galois group of the polynomialf K overK is a solvable group.

70 Algebra III– 25 § Afinite extensionK E is called solvable if the Galois group Gal(M/K) of a normal ⊂ closureM ofE overK is a solvable group. By 10.14, this means that there exists a chain of subgroups

Gal(M/K)=H H H ... H = 1 0 ⊃ 1 ⊃ 2 ⊃ ⊃ k

in Gal(M/K) for which everyH i+1 is normal inH i andH i/Hi+1 is cyclic of prime order. It follows from 25.15 that afinite extensionK K(x) of afieldK of characteristic 0 x ⊂ is solvable if and only if the equationf K (X) = 0 can be “solved” using radicals. This explains the use of the word “solvable” in group theory.

Exercise 7. Show that a subextension of a solvable extension is solvable and that the compositum of two solvable extensions is also solvable.

Before proving 25.15, let us take a closer look at radical extensions. It turns out that such extensions admit an elegant description when the basefield contains sufficiently many roots of unity.

25.16. Theorem. Letn 2 be an integer andKafield that contains a primitiventh ≥ root of unity. 1. Every cyclic extensionK L of degreen is of the formL=K( √n a), witha K ⊂ ∈ and √n a a zero ofX n a K[X]. − ∈ Xn a 2. Fora K, thefieldL=Ω − is a cyclic extension ofK of degree n/d, withd ∈ K the largest divisor ofn for whicha is adth power inK.

Proof. (1) Letσ be a generator of Gal(L/K) andζ K a primitiventh root of unity. ∈ We construct an elementα L ∗ withσ(α)=ζα by looking at the so-called Lagrange ∈ resolvent 1 2 2 1 n n 1 α=x+ζ − σ(x) +ζ − σ (x) +...+ζ − σ − (x), wherex L is chosen such that we haveα= 0. Note that such anx exists by the ∈ � Artin–Dedekind lemma 23.15. A simple verification givesσ(α) = ζα. The element a=α n is now inK becauseσ leavesa invariant:

σ(a) =σ(α n) =σ(α) n =ζ nαn =α n = a.

Since we haveσ(ζ) =ζ, repeatedly applyingσ to the identityσ(α) = ζα gives the relationσ i(α)=ζ iα. It follows that the subgroup Gal(L/K(α)) Gal(L/K) of powers ⊂ ofσ thatfixα is the trivial subgroup σ n = 1. Consequently, we haveL=K(α), � � withα= √n a a zero ofX n a. n − n n 1 i Xn a (2) Ifα is a zero ofX a, then we haveX a= − (X ζ α) andL=Ω − = − − i=0 − K K(α). Forα=a = 0, the statement of the theorem is clear, so from now on we assume n� n n a K ∗. Forτ Gal(L/K), we then have (τ(α)/α) = (τ(α )/α ) =τ(a)/a = 1, so ∈ ∈ the map ψ : Gal(L/K) ζ −→ � � τ(α) τ �−→ α

71 Algebra III– 25 § i j is well defined. It is also a homomorphism; after all, fromψ(τ)=ζ andψ(τ �) =ζ , we obtain j j i+j (ττ �)(α)=τ(ζ α)=ζ τ(α)=ζ (α), i+j soψ(ττ �) =ζ =ψ(τ)ψ(τ �). The homomorphismψ is injective because aK- automorphism thatfixesα is the identity onL=K(α). We conclude that Gal(L/K) is cyclic of degree n/d, whered is the index of ψ[Gal(L/K)] in ζ . For allτ Gal(L/K), we then have � � ∈ 1 =ψ(τ) n/d = (τ(α)/α) n/d =τ(α n/d)/αn/d, sob=α n/d is an element ofK anda=b d is adth power inK. If, conversely, we have a=b t witht n andb K, then thetth root of unityα n/t/b is inK, and therefore so | ∈ isα n/t itself. The identity above (withd=t) shows thatψ[Gal(L/K)] is annihilated by n/t and lies in the subgroup of indext in ζ . Consequently, we havet d, sod is � � ≤ the largest divisor ofn for whicha is adth power inK. Proof of 25.15. The equivalence of (1) and (2) is proved just as in 25.10: the setV of elements that satisfy (2) forms an extension ofK that is closed under extracting roots and is contained inK rad; it follows that we haveV=K rad. For (2) (3), wefirst note that as in the proof of 25.10, we may assume—after ⇒ if necessary extending the chain forx—that the lastfieldE n of the chain is normal overK. SupposeE =E (x ), withx ni E . Letζ be a primitive root of unity of i+1 i i i ∈ i orderN in K, whereN is a common multiple of alln i, and takeL=E n(ζ). Then K L is a normal extension that admits a chain ⊂

K=E 0 E 0(ζ) E 1(ζ) E 2(ζ) ... E n 1(ζ) E n(ζ)=L. ⊂ ⊂ ⊂ ⊂ ⊂ − ⊂ In this chain, thefirst stepE E (ζ) is an abelian extension (Exercise 24.46), 0 ⊂ 0 and the radical extensionsE (ζ) E (ζ) are cyclic by 25.16.2. By looking at the i ⊂ i+1 corresponding chain of subgroups in Gal(L/K), we conclude that Gal(L/K) is solvable. f x It then follows fromΩ K L that Gal(f), as a quotient of Gal(L/K), is also solvable. K ⊂ For (3) (2), letf=f x have a solvable Galois group Gal(f) = Gal(Ω f /K) of ⇒ K K orderN. Ifζ is again a primitive root of unity of orderN, then thefirst step in the towerK=E E =K(ζ) Ω f (ζ) is a radical extension. By (3), the extension 0 ⊂ 1 ⊂ K E =K(ζ) Ω f (ζ) can be written as a chain of cyclic extensions of degree a divisor 1 ⊂ K ofN. By 25.16, these cyclic extensions are radical extensions; this leads to the desired chain.

� Unsolvable polynomials To see thatQ rad is a proper subfield of Q, we show that there exist polynomials in

Q[X] whose Galois group is not solvable. SinceS n and its subgroups are solvable for n < 5, such a polynomial has degree at least 5. The groupS 5 is not solvable, and there exist polynomials inQ[X] with groupS 5. 25.17. Theorem. Letf Q[X] be an irreducible polynomial of degree5 with exactly ∈ rad three real zeros. Then we have Gal(f) ∼= S5, andf has no zeros inQ .

72 Algebra III– 25 §

Proof. By 24.6, the group Gal(f) is a subgroup ofS 5 of order divisible by 5. This means that Gal(f) contains an element of order 5 (Cauchy’s theorem); such an element inS 5 is necessarily a 5-cycle. If we view Q as a subfield ofC, then complex conjugation f gives an automorphism ofΩ Q that, because of the assumption, interchanges two zeros off andfixes the other three. Now, Gal(f) is a subgroup ofS 5 that contains a 5-cycle and a 2-cycle, and such a subgroup ofS 5 is equal toS 5 (Exercise 2.54). In particular, Gal(f) is not solvable, and by 25.15, the polynomialf has no zeros inQ rad. Making an irreducible polynomial of degree 5 with exactly three real zeros is not that difficult. For example, if we choose f=(X 2 + 2) (X + 2) X (X 2) + 2 =X 5 2X 3 8X+2, · · · − − − then this is an Eisenstein polynomial at 2 that is made by slightly shifting a polynomial with exactly three real zeros. The polynomialf certainly has three real zeros because off(0) = 2 andf(1) = 7 and the limit values forx . There are not more 4 −2 →±∞ becausef � = 5X 6X 8 only changes sign twice. − − Exercise 8. Show thatf=(X 2 + 3)(X2 9)X + 3 also has groupS . − 5 The construction of an unsolvable polynomial of degree 5 given here can be general- ized to any prime degreep 5 (Exercise 27). There also exist several families17 of ≥ polynomials f ∞ with Gal(f ) = S . { n}n=1 n ∼ n � Radical formulas It follows from the fact thatS and its subgroups are solvable forn 4 that the zeros n ≤ of polynomials of degree at most 4 can be expressed in radicals. The most famous example of a “radical formula” is undoubtedly the “quadratic formula” b √b2 4ac x , x = − ± − 1 2 2a for the zerosx andx of a quadratic polynomial aX2 + bX+c R[X]. The formula 1 2 ∈ holds not only overR but also over everyfieldK of characteristic different from 2. Note that the formula in fact expresses the zeros in b/a and c/a and that we may assumea = 1 without loss of generality. The proof of the formula by “completing the b square” amounts to noticing that when expressed in the shifted variableX+ 2 , the polynomial 2 b 2 b2 4c X + bX+c= X+ − K[X] 2 − 4 ∈ loses its linear term and can be “solved”� by extracting� the square root of the discrim- inantb 2 4c of the polynomialX 2 + bX+c. − Exercise 9. Show that this discriminant is equal to that from Exercise 24.39.

For polynomials of degree 3, the situation is more complicated. If we write the polyno- mial asX 3 + aX2 + bX+c K[X], then ifK is not of characteristic 3, the polynomial ∈ a can be expressed in the variableY=X+ 3 , giving Y 3 + pY+q,

73 Algebra III– 25 § wherep andq are polynomial expressions ina,b, andc.

Exercise 10. Expressp andq ina,b, andc.

Next, we can use a trick found around 1500 by the Italian Scipione del Ferro ( 1465– ± 1526). For this, writeY=u+v, and note that the polynomial can now be written as (u+v) 3 +p(u+v)+q=u 3 +v 3 +q + (3uv+p)(u+v). This expression is equal to 0 if we letu andv satisfy u3 +v 3 = q (25.18) − uv= p/3. − Apparently,u 3 andv 3 are zeros of the quadratic polynomial (X u 3)(X v 3) =X 2 +qX (p/3) 3. − − − If we moreover assume thatK does not have characteristic 2, then we can expressu 3 andv 3 in radicals using the quadratic formula given above, and wefind

(25.19)Y=u+v= 3 q/2 + (q/2)2 + (p/3)3 + 3 q/2 (q/2)2 + (p/3)3. − − − � � We have three choices for each of the� third rootsu andv ofu 3 and�v 3, but by (25.18), the rootv= p/(3u) isfixed by the choice ofu, and, as expected, wefind only three − zeros and not nine. See Exercise 37 for notation in terms of third roots that avoids this problem of choice. Del Ferro did not announce his method to the world, and its publication in 1545 by his compatriot Girolamo Cardano (1501–1576) is dominated by intrigue and priority disputes.18 There were also mathematical problems with the solution. If we take a polynomial with three real roots, such asY 3 7Y+6=(Y 1)(Y 2)(Y +3), then the Cardano–Del − − − Ferro formula above leads to an expression for the zeros in terms of complex conjugate numbers: 1 3 1 3 81 + 30√ 3 + 81 30 √ 3. 3 − − 3 − − − � � Once we realize that complex numbers were unknown in the sixteenth century and only lost their mysterious halo late in the eighteenth century with Euler, the initial confusion about this casus irreducibilis becomes understandable.

Exercise 11. Show how the zeros 1, 2, and 3 follow from the radical representation. − [Hint: (3 + 2√ 3)3 = 81 + 30 √ 3.] − − − The zeros of a general polynomial of degree 4 over afield of characteristic different from 2 and 3 can also be expressed in radicals using a trick. The method, which can be found in Cardano’s Ars Magna, goes back to Cardano’s student and son-in-law Ludovici Ferrari (1522–1565). Again, we write the general polynomialX 4 + aX3 + bX2 + cX+d a in terms ofY=X+ 4 and solve an equation of the form Y 4 = pY 2 +qY+r

74 Algebra III– 25 § by rewriting it as

(25.20) (Y 2 +s) 2 = (p+2s)Y 2 +qY+(s 2 +r).

Here, we chooses such that the quadratic polynomial on the right-hand side of ( 25.20) is the square of a polynomial of degree 1:

q 2 (Y 2 +s) 2 = p+2s Y+ . 2√p+2s � � � In order to have the constant term equal tos 2 +r as in ( 25.20), we must choose ans that satisfies the cubic equation

(25.21) (p+2s)(s 2 +r)=q 2/4,

and the Cardano–Del Ferro formula gives us a radical expression fors in terms ofp,q, andr. Wefind four values ofY by solving both quadratic equations q Y 2 +s= p+2sY+ ± 2√p+2s � � � for a solutions of ( 25.21). The resulting radical formula is more impressive than practical.

Exercise 12. How should the method be adjusted in the casep+2s = 0?

Galois theory teaches us that the clever tricks in the Ars Magna cannot be extended to the case of degree 5 or more—not something that is immediately apparent from the manipulations given above. In the next section, we interpret the deduced radical formulas in terms of Galois theory. See Exercise 37 for a method for the degree 4 equation that is more similar to the trick we applied in the cubic case but also fails to clarify why this method works.

Exercises.

13. (Construction steps) Show that the following objects are constructible from three non- collinear pointsx,y, andz in the complex plane: a. the bisector of the line segment xy b. the line throughx perpendicular to the line throughx andy c. the line throughz perpendicular to the line throughx andy d. the line throughz parallel to the line throughx andy e. the circle with centerz and radius x y | − | f. the bisector of the angle∠xyz g. the circle throughx,y, andz h. the rotation of a point aroundy by the angle∠xyz. 14. LetAB be a diagonal of a square andC a third vertex. The lune of Hippocrates on ABC is the shaded area in thefigure below bounded by the half-circle onAB and the quarter circle tangent toAC andBC.

75 Algebra III– 25 § C

A B

Prove that the area of this lune is equal to that of the triangleABC. 15. Afigure known in Islamic architecture consists of a small square that, as shown in the figure below, lies in a large square with the same center. The length of the side of the small square is equal to the distance from a vertex of the square to the closest side of the large square.

× ×

z

b

a × × 0 1 Determine whether the small square is constructible from the large square. [Hint: choose coordinates 0,1 C as in thefigure, and determinez=a+ bi.] ∈ 16. LetX C be a subset, and suppose thatz is constructible fromX. Prove thatx is ⊂ constructible from afinite subsetX X. 0 ⊂ 17. LetK Cbeafield that maps to itself under complex conjugation. Prove that ⊂ Kquad C also maps to itself under complex conjugation. ⊂ 18. Show that the quadratic closureK quad of afieldK of characteristic char(K)= 2 has � no quadratic extensions. 19. Give extensionsQ quad L of degree 3 and degree 5. *Does there also exist an extension ⊂ of degree 4? 20. Show that trisecting the angle associated withα C with α = 1 with a straightedge ∈ | | and compass is not possible ifα is transcendental. 21. Is trisecting the angles of a triangleABC with a straightedge and compass possible a. whenABC is equilateral? b. if the lengths of the sides ofABC are equal to 44, 117, and 125? [Hint: 442 + 1172 = 1252.]

22. LetKbeafield of characteristic different from 2, and defineK 1 =K( √K) as in 25.8. Prove:x K is inK K if and only ifK K(x) is afinite Galois extension for which ∈ 1 \ ⊂ the groupG = Gal(K(x)/K) is abelian of exponent 2.

76 Algebra III– 25 § 23. Letx C be a constructible element. DefineK as in 25.8 forK=K =Q. Then the ∈ i 0 root depth ofx is the smallest numberi 0 for whichx is contained inK . Determine ≥ i the root depth of the following elements:

1 + √2, 3 2 √2,ζ ,ζ , 1 + 2√ 6. − 5 12 − � � � Does the answer depend on the choice of the various roots (of unity) inC?

24. Define real numbersx i R 0 recursively byx 0 = 0 andx i+1 = √2 +x i fori 0. ∈ ≥ ≥ a. Prove:Q Q(x ) is cyclic of degree 2i. ⊂ i b. Prove: the root depth ofx is equal toi for alli 0. i ≥ quad 25. LetK be a numberfield. Prove: for thefieldsK i in the definition 25.8 ofK , we haveK =K for alli 0. i � i+1 ≥ 26. Analogously to the root depth, define the radical depth of a numberx Q rad, and x ∈ show that the radical depth ofx depends only on Gal(f Q). *Do there exist elements of arbitrarily large radical depth? 27. Letp=2k+3> 3 be a prime, and define

2 k f=(X + 2) i= k(X 2i) + 2 Q[X]. − − ∈ a. Prove thatf is an irreducible polynomial� of degreep. b. Prove thatf does not havep 1 real zeros. − 2 [Hint:f � is a polynomial inX , and the chain off �(0) is known.] c. Prove thatf has exactlyp 2 real zeros. Conclude thatf is not solvable by − radicals overQ. 28. Give a chainQ=E E E of irreducible radical extensions withζ E . 0 ⊂ 1 ⊂···⊂ n 47 ∈ n Here,ζ 47 is a primitive 47th root of unity. 29. Express the real numberx = cos(2π/7) in irreducible radicals overQ. 30. Show that every extensionQ Q(ζ ) is solvable by irreducible radicals. ⊂ n [Hint: induction onn.] 31. LetKbeafield of characteristic 0 andx an element ofK rad K. Prove: there exist ⊂ ann Z 0 and a chain offields ∈ ≥

K=E 0 E 1 E 2 ... E n 1 E n K ⊂ ⊂ ⊂ ⊂ − ⊂ ⊂

withx E n andE i 1 E i for 1 i n an irreducible radical extension. ∈ − ⊂ ≤ ≤ 32. (Artin–Schreier radicals) LetKbeafield of characteristicp> 0 andK L a cyclic ⊂ extension of degreep. Prove: we haveL=K(α) for a zeroα of an Artin–Schreier polynomialf=X p X a K[X]. − − ∈ [This is the analog of 25.16.1 forn=p = char(K). Hint: look at the resolvent p 1 i − iσ (x) for an elementx L with trace 1.] i=0 ∈ 33. Let� Kbeafield of characteristicp> 0 andK L the extension obtained by adjoining ⊂ the zeros of the Artin–Schreier polynomialf=X p X a K[X] toK. Prove: − − ∈ K L is a cyclic extension of degree 1 orp. ⊂ [This is the analog of 25.16.2 forn=p = char(K). Hint: Exercise 22.30.]

77 Algebra III– 25 § 34. LetKbeafield of characteristicp> 0. The radical closure ofK is defined as in 25.13, except thatK i is now obtained fromK i 1 by adjoining allw K that satisfy exactly − ∈ one of the following conditions: n 1. We havew K i 1 for somen 1 withp�n. p ∈ − ≥ 2. We havew w K i 1 (informally:w is an Artin–Schreier radical overK). − ∈ − Formulate and prove the analog of 25.15 forK. 35. Determine the real solutions of the equationX 3 = 15X + 4 using the Cardano–Del Ferro formula. [This feat was accomplished by Bombelli in 1572.] 36. LetKbeafield with char(K)= 2, 3. Show thatY=ξη(ξ+η) is a zero ofY 3 +pY+q � ∈ K[Y ] ifξ andη satisfy 3q 3q 2 p ξ3,η 3 = + . 2p ± 2p 3 �� � Why does this not give nine distinct zeros? [Hint: Writeu=ξ 2η andv=ξη 2 in 25.19. This is Cayley’s version of the Cardano–Del Ferro formula.] 37. LetK be as in the previous exercise. Show that the degree 4 equationY 4 = pY 2+qY+r 1 2 2 2 overK has solutionY= 2 (u+v+w) ifu ,v , andw are zeros of the cubic resolvent X3 2pX 2 + (p2 + 4r)X q 2 − − and the signs ofu,v, andw are chosen such that we have uvw=q.

38. Show that thefieldE n in 25.10.2 can be chosen such that we haveE n =K(x). 2k 39. LetF k = 2 + 1 fork Z 0 be thekth Fermat number. ∈ ≥ a. Prove: fork Z 0, we haveF k = 2 + i

40. Determine alln Z 3 for which a regularn-gon, a regularn + 1-gon, and a regular ∈ ≥ n + 2-gon are all constructible.

41. Letk Z 0, and letp be a prime factor ofF k. ∈ ≥ k+1 a. Prove: the order of (2 modp) in the groupF p∗ is equal to 2 , and fork 2, the k+2 ≥ order of (Fk 1 modp) inF p∗ is equal to 2 . − b. Prove: fork 2, we havep 1 mod 2 k+2. ≥ ≡ 42. LetFbeafinitefield. Prove that the following two statements are equivalent: a. For any two subgroupsA andB ofF , we haveA B orB A. ∗ ⊂ ⊂ b. Either #F is equal to 2, 9, or a Fermat prime, or #F 1 is a Mersenne prime. − (k) 43. For a groupG andk Z 0, we define the subgroupG ofG recursively by ∈ ≥ G(0) =G andG (k+1) = [G(k),G(k)].

78 Algebra III– 25 § We callG solvable if there is afinite chain of subgroupsG=H H ... H = e 0 ⊃ 1 ⊃ ⊃ k { } ofG such that for everyi> 0, the groupH i is normal inH i 1 withH i 1/Hi abelian. − − a. Prove that forfiniteG, the definition given above is equivalent to Definition 10.14 in the syllabus Algebra 1. b. LetG be a group andN a normal subgroup ofG. Prove:G is solvable every ⇔ subgroup ofG is solvable N and G/N are both solvable there exists a (k) ⇔ ⇔ k Z 0 withG = e there exists afinite chain of subgroupsG=H 0 ∈ ≥ { }⇔ ⊃ H1 ... H k = e ofG that are all normal inG withH i 1/Hi abelian for all ⊃ ⊃ { } − i > 0.

44. LetI be a set,G i a solvable group for everyi I, andG= i I Gi. ∈ ∈ a. Prove: ifI isfinite, thenG is solvable. � b. IsG solvable in general? Give a proof or a counterexample. 45. LetK Lbea cyclic Galois extension with group σ , and letα L . Prove: ⊂ � � ∈ ∗ N (α)=1 there exists aβ L withα=σ(β)/β. L/K ⇐⇒ ∈ ∗ [Hint for = : imitate the construction of the Lagrange resolvent.] ⇒ The theorem in the previous exercise is called Hilbert’s Theorem 90, after Satz 90 from the Zahlbericht (1897) of David Hilbert (1862–1943). The more general theorem in the next exercise is also called Hilbert’s Theorem 90.

46. LetK Lbeafinite Galois extension with groupG, and letc:G L be a map. ⊂ → a. Prove: for allσ,τ G, we havec(στ)=c(σ) σ(c(τ)) if and only if there exists ∈ · aβ L such that for everyσ G, we havec(σ)=σ(β)/β. ∈ ∗ ∈ b. Show how the previous exercise follows from part (a).

79 26 Applications of Galois theory Galois theory is a useful tool that can be utilized in many situations. Nowadays, ideas about the invariance of “symmetric expressions” are common in mathematics, and there also exists, for example, a Galois theory of differential equations.19 This section gives several unrelated examples and also shows how certain expressions that are rational according to Galois theory can be calculated explicitly.

� Fundamental theorem of algebra As afirst application, we prove the fundamental theorem of algebra mentioned in 21.11, which says that thefieldC of complex numbers is algebraically closed. There are many proofs, which all use certain “topological arguments.” This is not surprising because the construction ofR fromQ using Dedekind cuts or Cauchy sequences is more a topological construction than an algebraic one, and unlikeC=R(i), thefieldQ(i) is not algebraically closed. As a topological argument, we use the intermediate value theorem for polynomials inR[X]: a polynomialf R[X] that takes on both a positive ∈ and a negative value has a real zero.

26.1. Lemma. Every polynomialf R[X] of odd degree has a real zero. For every ∈ field extensionR E of odd degree, we haveE=R. ⊂ Proof. Forf R[X] of odd degree, the valuesf(x) andf( x) have opposite signs ∈ − for sufficiently largex; by the intermediate value theorem,f then has a real zero. ForR E of odd degree andα E, the degree [R(α):R], as a divisor of [E:R], ⊂ α ∈ is also odd. Now,f R is an irreducible polynomial of odd degree. Because it has a zero inR, the polynomialf α is of degree 1. Wefindα R andE=R. R ∈ 26.2. Lemma. There is nofield extensionC E with[E:C] = 2. ⊂ Proof. To show thatC does not have any quadratic extensions, it suffices to show that every elementx C has a square root inC. If we writex=s+ it with s, t R, ∈ ∈ then solving the equationx=s+ it=(c+ di) 2 amounts tofinding c, d R with ∈ c2 d 2 =s − 2cd= t.

Fort = 0, the valuex=s is real and wefind cd = 0. Forx=s 0, we then take ≥ d = 0 andc= √x the real square root ofx; forx=s< 0, we takec = 0 andd= √ x − the real square root of x. − Fort= 0, we substituted= t/(2c) in thefirst equation, which gives 4c 4 4sc 2 � − − t2 = 0. Because the polynomial 4X4 4sX 2 t 2 is negative forX = 0 and positive for − − largeX, the intermediate value theorem implies that there indeed exists a real zeroc of this polynomial. This leads to the desired square root.

26.3. Fundamental theorem of algebra. ThefieldC is algebraically closed.

80 Algebra III– 26 § Proof. We must prove thatC has no non-trivial algebraic extensions. LetC L ⊂ befinite algebraic, and letM be the normal closure ofL overR. ThenR M is a ⊂ finite Galois extension, say with groupG. IfH is a 2-Sylow subgroup ofG in the sense of 10.7, then thefield of invariantsE=M H ofH is an extension ofR whose degree [E:R] = [G:H] is odd. By 26.1, we getE=R andG=H, soG is a 2-group. In particular, the subgroup Gal(M/C) G is a 2-group. By 10.17, the group Gal(M/C) ⊂ is therefore solvable. As in 10.14, this means that there exists a chain

Gal(M/C) =H H H ... H = 1 0 ⊃ 1 ⊃ 2 ⊃ ⊃ k

in which eachH i+1 is a subgroup of index 2 inH i. Under the Galois correspondence, this gives a chainC=E E E ... E =M of quadraticfield extensions. 0 ⊂ 1 ⊂ 2 ⊂ ⊂ k By 26.2, this chain has lengthk = 0, soM=L=C.

� Quadratic reciprocity

The calculation of the quadratic subfield ofQ(ζ p) for primep carried out in 24.11 allows us to explain a surprising symmetry in the “quadratic character” of primes modulo one another. We already came across a consequence of this phenomenon in Exercises 7.19 and 12.22: if 5 is a primitive root modulop, then we havep 2 mod 5. ≡± Ifp is an odd prime, thenF ∗ is a cyclic group of even orderp 1. The unique p − subgroupS F ∗ of index 2 consists of the remainders of squares modulop. It is the p ⊂ p kernel of the composed homomorphism

∼ F∗ 1 modp 1 p −→ �− � −→ {± } (p 1)/2 x x modp x − modp . �−→ �−→ p � �

x The symbol p , which lives in characteristic 0, is the Legendre symbol ofx modulop: it is 1 ifx is a� square� inF p∗ and 1 ifx is not a square inF p∗. There does not seem to − x be any symmetry inx andp in the definition of p ; nevertheless, around 1744, using numerical examples, Euler discovered a variant of� the� following result. 26.4. Quadratic reciprocity law. Letp andq be distinct odd primes. Then we have

p 1 q 1 p q − − = ( 1) 2 · 2 . q p − � � � �

p In words: ifp andq are not both 3 mod 4, then the symbol q can beflipped “upside down.” Forp q 3 mod 4, the sign changes. � � ≡ ≡ Euler was not able to prove this result, and his French colleague Legendre (1752– 1833), whose name is attached to the quadratic symbols, mistakenly denied that his own proof published in 1785 was incomplete. Gauss found thefirst correct proof of 26.4 in 1796 and later gave several “different” proofs of his theorema aureum.

81 Algebra III– 26 §

Proof. From 24.9 and 24.11, we know thatQ(ζ p) is a Galois extension ofQ with

group (Z/pZ)∗ =F p∗ and that the subgroupS p F p∗ of squares corresponds with the (p 1)⊂/2 intermediatefieldQ( √p ), wherep ∗ = ( 1) − p. This gives ∗ − q = 1 (q modp) S σ (√p ) = √p . p ⇐⇒ ∈ p ⇐⇒ q ∗ ∗ � � The automorphismσ q is a type of “lift to characteristic 0” of the Frobenius automor- phismF q on Fq. More precisely, ifζ is a primitivepth root of unity in an algebraic closure F ofF , then the reduction mapZ F has an extension q q → q r:Z[ζ ] F p −→ q a ζi a ζi i i p �−→ i i � � that satisfiesr σ q =F q r. The homomorphismr sends the quadratic Gauss sum ◦ ◦ 2 2 τ = √p Z[ζ ] to a zerow=r( √p ) ofX p ∗ F [X]. The zeros ofX p ∗ in F p ∗ ∈ p ∗ − ∈ q − q are distinct, soσ q leaves √p∗ invariant if and only ifF q leaves the elementw invariant. We now have

q 1 (q 1)/2 p∗ F (w)=w w − = 1 (p ∗ modq) − = 1 = 1, q ⇐⇒ ⇐⇒ ⇐⇒ q � � and wefind p 1 q 1 q p∗ − − p = = ( 1) 2 · 2 . p q − q � � � � � � Exercise 1. Prove: if 5 is a primitive root modulo a primep= 2, then we havep 2 mod 5. � ≡± � Symmetric polynomials

In the main theorem 14.1 for symmetric polynomials, we saw that forn Z 1, the ∈ ≥ “symmetric expressions” in the zeros of the general polynomial

n n k n k F = (X T )(X T )...(X T ) =X + ( 1) s X − n − 1 − 2 − n − k �k=1 of degreen can be written as polynomial expressions in the elementary symmetric polynomials s = T T ...T Z[T ,T ,...,T ] k i1 i2 ik ∈ 1 2 n 1 i1

natural action of the symmetric groupS n on the ringZ[T 1,T2,...,Tn] of polynomials overZ in then variablesT 1,T2,...,Tn given by

(σf)(T ,T ,...,T ) =f(T ,T ,...,T ) forf R andσ S , 1 2 n σ(1) σ(2) σ(n) ∈ ∈ n

the ring of invariants is equal toZ[s 1, s2, . . . , sn]. There is a Galois-theoretic formulation of this result, in terms of thefields of fractions of both rings, and also of the definition of the sign map given in 2.9.

82 Algebra III– 26 §

26.5. Theorem. For everyn Z 1, thefieldQ(T 1,...,Tn) is the splittingfield of the ∈ ≥ general polynomialF n of degreen overQ(s 1, . . . , sn). The extension

K=Q(s , . . . , s ) L=Q(T ,...,T ) 1 n ⊂ 1 n is Galois with groupS =S( T ,T ,...,T ). OverK, the polynomial n { 1 2 n} δ = (T T ) L n i − j ∈ 1 i

In 14.4, we came across the square of the polynomialδ n, which is contained inK because it is a symmetric function, as the discriminant

δ2 =Δ = (T T )2 n n i − j 1 i

δ = Δ = (T T )(T T )(T T ) 3 3 1 − 2 1 − 3 2 − 3 = (T 2T +T T 2 +T 2T ) (T 2T +T T 2 +T T 2). � 1 2 1 3 2 3 − 1 3 1 2 2 3

To obtain a radical expression forT 1 overK(δ 3), we adjoin a primitive third root of unityζ = 1 + 1 √ 3 toK(δ ), and as in 25.16, we form both Lagrange resolvents 3 − 2 2 − 3 U, V L(ζ 3) fromT 1: ∈ 2 U=T 1 +ζ 3T2 +ζ 3 T3 2 V=T 1 +ζ 3 T2 +ζ 3T3.

83 Algebra III– 26 § In terms of these resolvents ands =T +T +T K, we now have the expressions 1 1 2 3 ∈ 1 T = (s +U+V) 1 3 1 1 (26.6) T = (s +ζ 2U+ζ V) 2 3 1 3 3 1 T = (s +ζ U+ζ 2V) 3 3 1 3 3

2 because the three third roots of unity add up to 1 +ζ 3 +ζ 3 = 0. Exercise 2. Why do we haveUV K? ExpressUV ins , s , s . ∈ 1 2 3 The elementsU 3 andV 3 are inK(δ ,ζ ) =K( √Δ , √ 3), and even inK( √ 3Δ ) 3 3 3 − − 3 because the Lagrange resolventsU andV are invariant under theK-automorphism of

L(ζ3) of order 2 that both interchangesT 2 andT 3 and squaresζ 3. A short calculation now gives 9 27 3 U 3,V 3 = s3 s s + s 3Δ . 1 − 2 1 2 2 3 ± 2 − 3 � � If we explicitly substitute the third roots of unity forU �andV in 26.6, we obtain a

radical formula for theT i in terms of thes i. Exercise 3. How does the radical formula (25.19) follow from the obtained formula?

In degree 4, we can also express the zeros ofF 4 using radicals.

Exercises. See the next version of this syllabus.

84