23. Solving equations by radicals

Transcription

1 23. Solvng equatons by radcals 23.1 Galos crteron 23.2 Composton seres, Jordan-Hölder theorem 23.3 Solvng cubcs by radcals 23.4 Worked examples Around 1800, Ruffn sketched a proof, completed by Abel, that the general quntc equaton s not solvable n radcals, by contrast to cubcs and quartcs whose solutons by radcals were found n the Italan renassance, not to menton quadratc equatons, understood n antquty. Ruffn s proof requred classfyng the possble forms of radcals. By contrast, Galos systematc development of the dea of automorphsm group replaced the study of the expressons themselves wth the study of ther movements. Galos theory solves some classcal problems. Ruler-and-compass constructons, n coordnates, can only express quanttes n repeated quadratc extensons of the feld generated by gven ponts, but nothng else. Thus, trsecton of angles by ruler and compass s mpossble for general-poston angles, snce the general trsecton requres a cube root. The examples and exercses contnue wth other themes. 1. Galos crteron We wll not prove all the results n ths secton, for several reasons. Frst, soluton of equatons n radcals s no longer a crtcal or useful ssue, beng mostly of hstorcal nterest. Second, n general t s non-trval to verfy (or dsprove) Galos condton for solvablty n radcals. Fnally, to understand that Galos condton s ntrnsc requres the Jordan-Hölder theorem on composton seres of groups (stated below). Whle ts statement s clear, the proof of ths result s techncal, dffcult to understand, and not re-used elsewhere here. 327

2 328 Solvng equatons by radcals [1.0.1] Theorem: Let G be the Galos group of the splttng feld K of an rreducble polynomal f over k. If G has a sequence of subgroups {1} G 1 G 2... G m = G such that G s normal n G +1 and G +1 /G s cyclc for every ndex, then a root of f(x) = 0 can be expressed n terms of radcals. Conversely, f roots of f can be expressed n terms of radcals, then the Galos group G has such a chan of subgroups. Proof: (Sketch) On one hand, adjuncton of n roots s cyclc of degree n f the prmtve n th roots of unty are n the base feld. If the n th roots of unty are not n the base feld, we can adjon them by takng a feld extenson obtanable by successve root-takng of orders strctly less than n. Thus, root-takng amounts to successve cyclc extensons, whch altogether gves a solvable extenson. On the other hand, a solvable extenson s gven by successve cyclc extensons. After n th roots of unty are adjoned (whch requres successve cyclc extensons of degrees less than n), one can prove that any cyclc extenson s obtaned by adjonng roots of x n a for some a n the base. Ths fact s most usefully proven by lookng at Lagrange resolvents. [1.0.2] Theorem: The general n th degree polynomal equaton s not solvable n terms of radcals for n > 4. Proof: The meanng of general s that the Galos group s the largest possble, namely the symmetrc group S n on n thngs. Then we nvoke the theorem to see that we must prove that S n s not solvable for n > 4. In fact, the normal subgroup A n of S n s smple for n > 4 (see just below), n the sense that t has no proper normal subgroups (and s not cyclc). In partcular, A n has no chan of subgroups normal n each other wth cyclc quotents. Ths almost fnshes the proof. What s mssng s verfyng the plausble clam that the smplcty of A n means that no other possble chan of subgroups nsde S n can exst wth cyclc quotents. We address ths just below. A group s smple f t has not proper normal subgroups (and maybe s not a cyclc group of prme order, and s not the trval group). A group G wth a chan of subgroups G, each normal n the next, wth the quotents cyclc, s a solvable group, because of the concluson of ths theorem. [1.0.3] Proposton: For n 5 the alternatng group A n on n thngs s smple. Proof: (Sketch) The trck s that for n 5 the group A n s generated by 3-cycles. Keepng track of 3-cycles, one can prove that the commutator subgroup of A n, generated by expressons xyx 1 y 1, for x, y A n, s A n tself. Ths yelds the smplcty of A n. [1.0.4] Remark: A smlar dscusson addresses the queston of constructblty by ruler and compass. One can prove that a pont s constructble by ruler and compass f and only f ts coordnates le n a feld extenson of Q obtaned by successve quadratc feld extensons. Thus, for example, a regular n-gon can be constructed by ruler and compass exactly when (Z/n) s a two-group. Ths happens exactly when n s of the form n = 2 m p 1... p l where each p s a Fermat prme, that s, s a prme of the form p = 2 2t + 1. Gauss constructed a regular 17-gon. The next Fermat prme s 257. Sometme n the early 19 th century someone dd lterally construct a regular gon, too.

3 Garrett: Abstract Algebra Composton seres, Jordan-Hölder theorem Now we should check that the smplcty of A n really does prevent there beng any other chan of subgroups wth cyclc quotents that mght secretly permt a soluton n radcals. A composton seres for a fnte group G s a chan of subgroups {1} G 1... G m = G where each G s normal n G +1 and the quotent G +1 /G s ether cyclc of prme order or smple. [1] [2.0.1] Theorem: Let {1} = G 0 G 1... G m = G {1} = H 0 H 1... H n = G be two composton seres for G. Then m = n and the sets of quotents {G +1 /G } and {H j+1 /G j } (countng multplctes) are dentcal. Proof: (Comments) Ths theorem s qute non-trval, and we wll not prove t. The key ngredent s the Jordan-Zassenhaus butterfly lemma, whch tself s techncal and non-trval. The proof of the analogue for modules over a rng s more ntutve, and s a worthwhle result n tself, whch we leave to the reader. 3. Solvng cubcs by radcals We follow J.-L. Lagrange to recover the renassance Italan formulas of Cardan and Tartagla n terms of radcals for the zeros of the general cubc x 3 + ax 2 + bx + c wth a, b, c n a feld k of characterstc nether 3 nor 2. [2] resolvent, havng more accessble symmetres. [3] Lagrange s method creates an expresson, the Let ω be a prmtve cube root of unty. Let α, β, γ be the three zeros of the cubc above. The Lagrange resolvent s λ = α + ω β + ω 2 γ The pont s that any cyclc permutaton of the roots alters λ by a cube root of unty. Thus, λ 3 s nvarant under cyclc permutatons of the roots, so we antcpate that λ 3 les n a smaller feld than do the roots. Ths s ntended to reduce the problem to a smpler one. Compute λ 3 = ( α + ωβ + ω 2 γ ) 3 [1] Agan, t s often convenent that the noton of smple group makes an excepton for cyclc groups of prme order. [2] In characterstc 3, there are no prmtve cube roots of 1, and the whole setup fals. In characterstc 2, unless we are somehow assured that the dscrmnant s a square n the ground feld, the auxlary quadratc whch arses does not behave the way we want. [3] The complcaton that cube roots of unty are nvolved was dsturbng, hstorcally, snce complex number were vewed wth suspcon untl well nto the 19 th century.

4 330 Solvng equatons by radcals = α 3 + β 3 + γ 3 + 3ωα 2 β + 3ω 2 αβ 2 + 3ω 2 α 2 γ + 3ωαγ 2 + 3ωβ 2 γ + 3ω 2 βγ 2 + 6αβγ = α 3 + β 3 + γ 3 + 3ω(α 2 β + β 2 γ + γ 2 α) + 3ω 2 (αβ 2 + βγ 2 + α 2 γ) + 6αβγ Snce ω 2 = 1 ω ths s α 3 + β 3 + γ 3 + 6αβγ + 3ω(α 2 β + β 2 γ + γ 2 α) 3ω(αβ 2 + βγ 2 + α 2 γ) 3(αβ 2 + βγ 2 + α 2 γ) In terms of the elementary symmetrc polynomals s 1 = α + β γ s 2 = αβ + βγ + γα s 3 = αβγ we have Thus, α 3 + β 3 + γ 3 = s 3 1 3s 1 s 2 + 3s 3 λ 3 = s 3 1 3s 1 s 2 + 9s 3 + 3ω(α 2 β + β 2 γ + γ 2 α) 3ω(αβ 2 + βγ 2 + α 2 γ) 3(αβ 2 + βγ 2 + α 2 γ) Nether of the two trnomals A = α 2 β + β 2 γ + γ 2 α B = αβ 2 + βγ 2 + γα 2 s nvarant under all permutatons of α, β, γ, but only under the subgroup generated by 3-cycles, so we cannot use symmetrc polynomal algorthm to express these two trnomals polynomally n terms of elementary symmetrc polynomals. [4] But all s not lost, snce A + B and AB are nvarant under all permutatons of the roots, snce any 2-cycle permutes A and B. So both A + B and AB are expressble n terms of elementary symmetrc polynomals, and then the two trnomals are the roots of whch s solvable by radcals n characterstc not 2. x 2 (A + B)x + AB = 0 We obtan the expresson for A + B n terms of elementary symmetrc polynomals. Wthout even embarkng upon the algorthm, a reasonable guess fnshes the problem: s 1 s 2 3s 3 = (α + β + γ)(αβ + βγ + γα) 3αβγ = α 2 β + β 2 γ + γ 2 α + αβ 2 + βγ 2 + γα 2 = A + B Determnng the expresson for AB s more work, but not so bad. AB = (α 2 β + β 2 γ + γ 2 α) (αβ 2 + βγ 2 + α 2 γ) = α 3 β 3 + β 3 γ 3 + γ 3 α 3 + α 4 βγ + αβ 4 γ + αβγ 4 + 3s 2 3 We can observe that already (usng an earler calculaton) α 4 βγ + αβ 4 γ + αβγ 4 = s 3 (α 3 + β 3 + γ 3 ) = s 3 (s 3 1 3s 1 s 2 + 3s 3 ) For α 3 β 3 + β 3 γ 3 + γ 3 α 3 follow the algorthm: ts value at γ = 0 s α 3 β 3 = s 3 2 (wth the s 2 for α, β alone). Thus, we consder α 3 β 3 + β 3 γ 3 + γ 3 α 3 (αβ + βγ + γα) 3 = 6s ( α 2 β 3 γ + α 3 β 2 γ + αβ 3 γ 2 + αβ 2 γ 3 + α 2 βγ 3 + α 3 βγ 2) [4] In an earler computaton regardng the specal cubc x 3 + x 2 2x 1, we could make use of the connecton to the 7 th root of unty to obtan explct expressons for α 2 β + β 2 γ + γ 2 α and αβ 2 + βγ 2 + α 2 γ, but for the general cubc there are no such trcks avalable.

5 Garrett: Abstract Algebra 331 = 6s 2 3 3s 3 ( αβ 2 + α 2 β + β 2 γ + βγ 2 + αγ 2 + α 2 γ ) = 6s 2 3 3s 3 (s 1 s 2 3s 3 ) by our computaton of A + B. Together, the three parts of AB gve AB = s 3 (s 3 1 3s 1 s 2 + 3s 3 ) + ( s 3 2 6s 2 3 3s 3 (s 1 s 2 3s 3 ) ) + 3s 2 3 = s 3 1s 3 3s 1 s 2 s 3 + 3s s 3 2 6s 2 3 3s 1 s 2 s 3 + 9s s 2 3 = s 3 1s 3 6s 1 s 2 s 3 + 9s s 3 2 That s, A and B are the two zeros of the quadratc x 2 (s 1 s 2 3s 3 )x + (s 3 1s 3 6s 1 s 2 s 3 + 9s s 3 2) = x 2 ( ab + 3c)x + (a 3 c 6abc + 9c 2 + b 3 ) The dscrmnant of ths monc quadratc s [5] = (lnear coef) 2 4(constant coef) = ( ab + 3c) 2 4(a 3 c 6abc + 9c 2 + b 3 ) = a 2 b 2 6abc + 9c 2 4a 3 c + 24abc 36c 2 4b 3 = a 2 b 2 27c 2 4a 3 c + 18abc 4b 3 In partcular, the quadratc formula [6] gves A, B = (ab 3c) ± 2 Then λ 3 = s 3 1 3s 1 s 2 + 9s 3 + 3ω(α 2 β + β 2 γ + γ 2 α) 3ω(αβ 2 + βγ 2 + α 2 γ) 3(αβ 2 + βγ 2 + α 2 γ) = a 3 + 3bc 9c + 3(ω 1)A 3ωB = a 3 + 3bc 9c + 3(ω 1) (ab 3c) + 2 = a 3 + 3bc 9c 3 2 (ab 3c) + (3ω 1 2 ) 3ω (ab 3c) 2 That s, now we can solve for λ by takng a cube root of the mess on the rght-hand sde: λ = 3 (rght-hand sde) The same computaton works for the analogue λ of λ wth ω replaced by the other [7] of unty λ = α + ω 2 β + ω γ The analogous computaton s much easer when ω s replaced by 1, snce prmtve cube root α + 1 β γ = s 1 = a Thus, we have a lnear system [5] When the x 2 27c 2 4b 3. coeffcent a vanshes, we wll recover the better-known specal case that the dscrmnant s [6] Whch s an nstance of ths general approach, but for quadratcs rather than cubcs. [7] In fact, there s no way to dstngush the two prmtve cube roots of unty, so nether has prmacy over the other. And, stll, ether s the square of the other.

6 332 Solvng equatons by radcals The lnear system α + β + γ = a α + ωβ + ω 2 γ = λ α + ω 2 β + ωγ = λ has coeffcents that readly allow soluton, snce for a prmtve n th root of unty ζ the matrx ζ ζ 2... ζ n 1 1 ζ 2 (ζ 2 ) 2... (ζ 2 ) n 1 M = 1 ζ 3 (ζ 3 ) 2... (ζ 3 ) n 1. 1 ζ n 1 (ζ n 1 ) 2... (ζ n 1 ) n 1 has nverse ζ 1 (ζ 1 ) 2... (ζ 1 ) n 1 1 ζ 2 (ζ 2 ) 2... (ζ 2 ) n 1 M 1 = 1 ζ 3 (ζ 3 ) 2... (ζ 3 ) n 1. 1 ζ n+1 (ζ n+1 ) 2... (ζ n+1 ) n 1 In the present smple case ths gves the three roots [8] α = a+λ+λ 3 of the cubc as β = a+ω2 λ+ωλ 3 γ = a+ωλ+ω2 λ 3 4. Worked examples [23.1] Let k be a feld of characterstc 0. Let f be an rreducble polynomal n k[x]. Prove that f has no repeated factors, even over an algebrac closure of k. If f has a factor P 2 where P s rreducble n k[x], then P dvdes gcd(f, f ) k[x]. Snce f was monc, and snce the characterstc s 0, the dervatve of the hghest-degree term s of the form nx n 1, and the coeffcent s non-zero. Snce f s not 0, the degree of gcd(f, f ) s at most deg f, whch s strctly less than deg f. Snce f s rreducble, ths gcd n k[x] must be 1. Thus, there are polynomals a, b such that af + bf = 1. The latter dentty certanly perssts n K[x] for any feld extenson K of k. [23.2] Let K be a fnte extenson of a feld k of characterstc 0. Prove that K s separable over k. Snce K s fnte over k, there s a fnte lst of elements α 1,..., α n n K such that K = k(α 1,..., α n ). From the prevous example, the mnmal polynomal f of α 1 over k has no repeated roots n an algebrac closure k of k, so k(α 1 ) s separable over k. [8] Agan, the seemng asymmetry among the roots s llusory. For example, snce λ s a cube root of somethng, we really cannot dstngush among λ, ωλ, and ω 2 λ. And, agan, we cannot dstngush between ω and ω 2.

7 Garrett: Abstract Algebra 333 We recall [9] the fact that we can map k(α 1 ) k by sendng α 1 to any of the [k(α 1 ) : k] = deg f dstnct roots of f(x) = 0 n k. Thus, there are [k(α 1 ) : k] = deg f dstnct dstnct mbeddngs of k(α 1 ) nto k, so k(α 1 ) s separable over k. Next, observe that for any mbeddng σ : k(α 1 ) k of k(α 1 ) nto an algebrac closure k of k, by proven propertes of k we know that k s an algebrac closure of σ(k(α 1 )). Further, f g(x) k(α 1 )[x] s the mnmal polynomal of α 2 over k(α 1 ), then σ(g)(x) (applyng σ to the coeffcents) s the mnmal polynomal of α 2 over σ(k(α 1 )). Thus, by the same argument as n the prevous paragraph we have [k(α 1, α 2 ) : k(α 1 )] dstnct mbeddngs of k(α 1, α 2 ) nto k for a gven mbeddng of k(α 1 ). Then use nducton. [23.3] Let k be a feld of characterstc p > 0. Suppose that k s perfect, meanng that for any a k there exsts b k such that b p = a. Let f(x) = c x n k[x] be a polynomal such that ts (algebrac) dervatve f (x) = c x 1 s the zero polynomal. Show that there s a unque polynomal g k[x] such that f(x) = g(x) p. For the dervatve to be the 0 polynomal t must be that the characterstc p dvdes the exponent of every term (wth non-zero coeffcent). That s, we can rewrte f(x) = c p x p Let b k such that b p = c p, usng the perfectness. Snce p dvdes all the nner bnomal coeffcents p! /!(p )!, ( ) p b x = c p x p as desred. [23.4] Let k be a perfect feld of characterstc p > 0, and f an rreducble polynomal n k[x]. Show that f has no repeated factors (even over an algebrac closure of k). If f has a factor P 2 where P s rreducble n k[x], then P dvdes gcd(f, f ) k[x]. If deg gcd(f, f ) < deg f then the rreducblty of f n k[x] mples that the gcd s 1, so no such P exsts. If deg gcd(f, f ) = deg f, then f = 0, and (from above) there s a polynomal g(x) k[x] such that f(x) = g(x) p, contradctng the rreducblty n k[x]. [23.5] Show that all fnte felds F p n wth p prme and 1 n Z are perfect. Agan because the nner bnomal coeffcents p!/!(p )! are 0 n characterstc p, the (Frobenus) map α α p s not only (obvously) multplcatve, but also addtve, so s a rng homomorphsm of F p n to tself. Snce F p s cyclc (of order n pn ), for any α F p n (ncludng 0) α (pn) = α Thus, snce the map α α p has the (two-sded) nverse α α pn 1, t s a bjecton. That s, everythng has a p th root. [23.6] Let K be a fnte extenson of a fnte feld k. Prove that K s separable over k. [9] Recall the proof: Let β be a root of f(x) = 0 n k. Let ϕ : k[x] k[β] by x β. The kernel of ϕ s the prncpal deal generated by f(x) n k[x]. Thus, the map ϕ factors through k[x]/ f k[α 1 ].

8 334 Solvng equatons by radcals That s, we want to prove that the number of dstnct mbeddngs σ of K nto a fxed algebrac closure k s [K : k]. Let α K be a generator for the cyclc group K. Then K = k(α) = k[α], snce powers of α already gve every element but 0 n K. Thus, from basc feld theory, the degree of the mnmal polynomal f(x) of α over k s [K : k]. The prevous example shows that k s perfect, and the example before that showed that rreducble polynomals over a perfect feld have no repeated factors. Thus, f(x) has no repeated factors n any feld extenson of k. We have also already seen that for algebrac α over k, we can map k(α) to k to send α to any root β of f(x) = 0 n k. Snce f(x) has not repeated factors, there are [K : k] dstnct roots β, so [K : k] dstnct mbeddngs. [23.7] Fnd all felds ntermedate between Q and Q(ζ) where ζ s a prmtve 17 th root of unty. Snce 17 s prme, Gal(Q(ζ)/Q) (Z/17) s cyclc (of order 16), and we know that a cyclc group has a unque subgroup of each order dvdng the order of the whole. Thus, there are ntermedate felds correspondng to the proper dvsors 2, 4, 8 of 16. Let σ a be the automorphsm σ a ζ = ζ a. By a lttle tral and error, 3 s a generator for the cyclc group (Z/17), so σ 3 automorphsm group. Thus, one reasonably consders s a generator for the α 8 = ζ + ζ 32 + ζ 34 + ζ 36 + ζ 38 + ζ ζ ζ 314 α 4 = ζ + ζ 34 + ζ 38 + ζ 312 α 2 = ζ + ζ 38 = ζ + ζ 1 The α n s vsbly nvarant under the subgroup of (Z/17) of order n. The lnear ndependence of ζ, ζ 2, ζ 3,..., ζ 16 shows α n s not by accdent nvarant under any larger subgroup of the Galos group. Thus, Q(α n ) s (by Galos theory) the unque ntermedate feld of degree 16/n over Q. We can also gve other characterzatons of some of these ntermedate felds. Frst, we have already seen (n dscusson of Gauss sums) that ( a ζ 17) a = 17 where ( ) a 17 s the quadratc symbol. Thus, 2 a mod 17 α 8 σ 3 α 8 = 17 α 8 + σ 3 α 8 = 0 so α 8 and σ 3 α 8 are ± 17/2. Further computaton can lkewse express all the ntermedate felds as beng obtaned by adjonng square roots to the next smaller one. 2 [23.8] Let f, g be relatvely prme polynomals n n ndetermnates t 1,..., t n, wth g not 0. Suppose that the rato f(t 1,..., t n )/g(t 1,..., t n ) s nvarant under all permutatons of the t. Show that both f and g are polynomals n the elementary symmetrc functons n the t. Let s be the th elementary symmetrc functon n the t j s. Earler we showed that k(t 1,..., t n ) has Galos group S n (the symmetrc group on n letters) over k(s 1,..., s n ). Thus, the gven rato les n k(s 1,..., s n ). Thus, t s expressble as a rato f(t 1,..., t n ) g(t 1,..., t n ) = F (s 1,..., s n ) G(s 1,..., s n ) of polynomals F, G n the s. To prove the stronger result that the orgnal f and g were themselves lterally polynomals n the t, we seem to need the characterstc of k to be not 2, and we certanly must use the unque factorzaton n k[t 1,..., t n ].

9 Wrte Garrett: Abstract Algebra 335 f(t 1,..., t n ) = p e pem m where the e are postve ntegers and the p are rreducbles. Smlarly, wrte π g(t 1,..., t n ) = q f qfn m where the f are postve ntegers and the q are rreducbles. The relatve prmeness says that none of the q are assocate to any of the p. The nvarance gves, for any permutaton π of ( ) p e pem m Multplyng out, q f qfn m π(p e ) q f = = pe pem m q f qfn m p e π(q f ) By the relatve prme-ness, each p dvdes some one of the π(p j ). These rng automorphsms preserve rreducblty, and gcd(a, b) = 1 mples gcd(πa, πb) = 1, so, symmetrcally, the π(p j ) s dvde the p s. And smlarly for the q s. That s, permutng the t s must permute the rreducble factors of f (up to unts k n k[t 1,..., t n ]) among themselves, and lkewse for the rreducble factors of g. If all permutatons lterally permuted the rreducble factors of f (and of g), rather than merely up to unts, then f and g would be symmetrc. However, at ths pont we can only be confdent that they are permuted up to constants. What we have, then, s that for a permutaton π π(f) = α π f for some α k. For another permutaton τ, certanly τ(π(f)) = (τπ)f. And τ(α π f) = α π τ(f), snce permutatons of the ndetermnates have no effect on elements of k. Thus, we have α τπ = α τ α π That s, π α π s a group homomorphsm S n k. It s very useful to know that the alternatng group A n s the commutator subgroup of S n. Thus, f f s not actually nvarant under S n, n any case the group homomorphsm S n k factors through the quotent S n /A n, so s the sgn functon π σ(π) that s +1 for π A n and 1 otherwse. That s, f s equvarant under S n by the sgn functon, n the sense that πf = σ(π) f. Now we clam that f πf = σ(π) f then the square root δ = = <j (t t j ) of the dscrmnant dvdes f. To see ths, let s j be the 2-cycle whch nterchanges t and t j, for j. Then s j f = f Under any homomorphsm whch sends t t j to 0, snce the characterstc s not 2, f s sent to 0. That s, t t j dvdes f n k[t 1,..., t n ]. By unque factorzaton, snce no two of the monomals t t j are assocate (for dstnct pars < j), we see that the square root δ of the dscrmnant must dvde f. That s, for f wth πf = σ(π) f we know that δ f. For f/g to be nvarant under S n, t must be that also πg = σ(π) g. But then δ g also, contradctng the assumed relatve prmeness. Thus, n fact, t must have been that both f and g were nvarant under S n, not merely equvarant by the sgn functon.

10 336 Solvng equatons by radcals Exercses 23.[4.0.1] Let k be a feld. Let α 1,..., α n be dstnct elements of k. Suppose that c 1,..., c n n k are such that for all postve ntegers l c α l = 0 Show that all the c are [4.0.2] Solve the cubc x 3 + ax + b = 0 n terms of radcals. 23.[4.0.3] Express a prmtve 11 th root of unty n terms of radcals. 23.[4.0.4] Solve x 4 + ax + b = 0 n terms of radcals.