The Fundamental Theorem on Symmetric Polynomials: History's First Whiff of Galois Theory

0.11 MB·English

by  Ben Blum-Smith

THE FUNDAMENTAL THEOREM ON SYMMETRIC POLYNOMIALS: HISTORY’S
FIRST WHIFF OF GALOIS THEORY
BENBLUM-SMITHANDSAMUELCOSKEY
3
1
0
2 EvaristeGalois’(1811–1832)shortlifeisoneoftheclassicromantictragediesofmathemat-

n icalhistory. TheteenageGaloisdevelopedarevolutionarytheoryofequations,answering
a morefullythaneverbeforeacenturies-oldquestion: whycan’twefindaformulaforsolv-
J
ingquinticpolynomialsanalogoustothequadratic,cubicandquarticformulas? Thenhe
0
died in a duel, apparently over the honor of a woman [Stillwell,2010, p. 290] before his
3
twenty-first birthday. His discoveries lay in obscurity, ignored by the French academy,

] for 15 years, until Joseph Liouville encountered them, recognized their importance, and
O
madethemknown. Fromthatpointtheyquicklytranscendedtheproblemtheywerede-
H
signedtosolve,andreshapedthelandscapeofmodernmathematics, initiatingthestudy
.
h ofbothgroupsandfields. Thesetwoconceptsarenowutterlyfoundationalinalgebraas
at well as in many otherfields ofmath: number theory,topology,complex function theory,
m andgeometry,tonameafew.
[
Thisstoryistoldandretoldinpopularizationsofmathematics. Lessfrequentlydiscussed

1
is the actual content of Galois’ discoveries. These are usually reserved for a second-
v
semestercourseinadvancedundergraduateorgraduatealgebra. Thisarticleisintended
6
1 togivethereaderalittlewhiffoftheflavor ofGalois’workthroughatheoremthatplays
1 a unique role in it. This theorem appears to have been understood, or at least intuited
7
and used, by Newton, as early as 1665. By the turn of the nineteenth century it was re-
.
1
gardedaswellknown. ForGaloishimself,itwastheessentiallemmaonwhichhisentire
0
theoryrested. However,it was notproperlyprovenor even precisely stated until thenine-
3
1 teenthcentury. This theoremis now known as the FundamentalTheoremon Symmetric
: Polynomials(FTSP).
v
i
X In this article we wish to give a brief look into the beginnings of the theory of Galois by
r examiningthistheoremanditsproof. Weapproachthetheoremfromapedagogicalpoint
a
of view, and in doing so we hope to showcase the pleasure of mathematical discovery,
as well as provide a classroom module for other instructorsand students. Our narrative
arose out of an informal inquiry-based course in group theory and the historical foun-
dations of Galois theory.1 During one session, we posed the problem of trying to prove
thetheorem“fromscratch”beforelearningthestandardproof. Ourworkrevealedaway
1ThiscoursewasgivenbyBentoasmallgroupofteachersandmathematiciansincludingSamuel,Kayty
Himmelstein,JesseJohnson,JustinLanier,andAnnaWeltman.
1
SYMMETRICPOLYNOMIALS 2
of looking at the problem that was new to us, and we realized that it could be used to
constructadifferentproofthatgavenewinsightintothestandardone.
Galois theory has been substantially reformulated since Galois’ time to make its argu-
ments more elegant and more general. The modern body of results known as Galois
theory is thus very different from what Galois himself developed, and it is this modern
formulationthatisusuallytreatedinuniversityclasses. Forexample,Galois’ownreliance
ontheFTSPhasbeenreplacedwiththeelementarytheoryofvectorspacesoverafield,a
theoryunavailable inthe1820’s. Forourpurposes,whatmakesthetheoremimportantis
notonlythehistoricalfactofitscentralityinGalois’originalformulationofhistheory,but
also something else. The theory’s essential insight is that there is a connection between
symmetry and expressibility. (We will make our meaning clear in the next section.) The
FTSPistheearliestknowntheoremtohintatthisconnection. Byexplainingandproving
it,wehopetogivethereaderatasteofwhatGaloistheoryisallabout.
1. THE BACK STORY
The FTSP states that any polynomial in n variables x ,...,x that is invariant under all
1 n
permutationsofthe variables (i.e., symmetric) is representablein a unique way as a poly-
nomialinthenelementarysymmetricpolynomials,
σ = x + +x
1 1 n
···
σ = x x +x x + +x x = ∑x x
2 1 2 1 3 n 1 n i j
··· − i<j
σ = ∑ x x x
3 i j k
i<j<k
.
.
.
σ = x x ...x
n 1 2 n
Formally:
Theorem 1 (Fundamental Theorem on Symmetric Polynomials). Any symmetric polyno-
mial in n variables x ,...,x is representable in a uniqueway as apolynomial in the elementary
1 n
symmetricpolynomials σ ,...,σ .
1 n
For example, since the polynomial d = (x x )2 is unchanged by transposing the two
1 2
−
variables, the theorem guarantees an expressionfor d in terms of σ = x +x and σ =
1 1 2 2
x x . Inthiscasetheexpressioniseasytofind: d = (x +x )2 4x x = σ2 4σ .
1 2 1 2 − 1 2 1 − 2
The importance of the theorem to the theory of equations stems from the fact known
as Vieta’s theorem, that the coefficients of a single-variable polynomial are precisely the
elementarysymmetricpolynomialsinitsroots:
SYMMETRICPOLYNOMIALS 3
Theorem2(Vieta’sTheorem). Letp(z)beannthdegreemonicpolynomialwithrootsα ,α ,...,α .
1 2 n
Letσ ,...,σ bethenelementarysymmetricpolynomialsintheα . Then
1 n i
p(z) = zn σ zn 1+σ zn 2 +( 1)nσ
1 − 2 − n
− −··· −
The proof is a straightforward computation, but its significance belies its ease. With this
fact in hand, the FTSP becomes the fact that given any polynomial equation p(z) = 0,
any symmetric polynomial in its roots is actually a polynomial in its coefficients, and
consequently its value can be written down without (in fact, on the way to) solving the
equation. Continuing the example from above, if x and x are therootsof a monic qua-
1 2
draticpolynomial, thenthatpolynomialis p(z) = z2 σ z+σ and d isitsdiscriminant.
1 2
−
Thetheoremguaranteedthatthediscriminant(definedasthesquareofthedifferencebe-
tween the roots) would have an expression in terms of the coefficients. This of course is
key to the quadratic’s solution: √d is the difference between the roots and σ is the sum
1
oftheroots;andtherootsthemselvescan bededucedfromthesetwovalues. Since dcan
beexpressedintermsofthecoefficients,itfollowsthattherootscanbetoo.
Thisis theformin which thetheoremplayeditsseminal historicalrole. Itwas treatedas
well-known by the time of Galois, even though published proofs (or even precise state-
mentsofthetheoreminitsgeneralform)didnotappearuntilthemid-nineteenthcentury
[Edwards,1984, p. 8]. For a discussion ofsome of its historical applications, seeGreg St.
George’sdelightfulessay“SymmetricPolynomialsintheWorkofNewtonandLagrange”
in Mathematics Magazine [St.George,2003]. It continues to form the basis of modern re-
search,forinstanceintoquestionsofcomplexityregardingasymmetricpolynomial’srep-
resentationintermsofelementarysymmetricpolynomials[Gaudryetal.,2006].
Thewayoflookingat thistheoremthatwewish tohighlightis, as wementionedabove,
the way it intimates the central insight of Galois theory, the connection between symme-
try and rational expressibility. We have a polynomial p(z), whose coefficients we know.
Evenifwedon’tknowtheroots,theFTSPtellsusthatsymmetric expressionsin theroots
are rationally expressible in the coefficients. As a corollary, if the coefficients of p(z) are
rational numbers, then every symmetric expression in the roots (for instance the sum of
theirsquares)is rational as well. Symmetryguaranteesrational expressibility. In thelast
sectionwewillindicatehowthisfitsintothebiggerpictureofGaloistheory.
In our course on Galois theory, we did not approach the FTSP directly, but rather sidled
up to it by considering some problems of historical significance that implicitly depend
on it. The first was a problem of Newton: given two polynomials f,g, how can one
determinewhethertheyhavearootincommonwithoutfindingtheroots? (Thisproblem
is discussedat lengthin Greg St.George’sessay.) Thesecondwas posedby Gauss in his
Disquisitiones Arithmeticae: given a polynomial f, without finding its roots determine a
polynomial gwhoserootsarethesquares,orcubes,etc.,oftherootsof f.
Participants solvedboth oftheseproblems for polynomials of low degree. Thesolutions
were accomplished by writing desired expressions in the roots, which turned out to be
SYMMETRICPOLYNOMIALS 4
symmetric, and then expressing these in terms of the coefficients instead. For example,
they considered Gauss’ problem for a quadratic: if f(z) = z2 σ z+σ , write down g
1 2
−
whose roots are the squares of f’s. In this case, if α ,α are the roots of f, then α2,α2 are
1 2 1 2
therootsof g,sothat
g = (z α2)(z α2) = z2 (α2+α2)z+α2α2
− 1 − 2 − 1 2 1 2
Towritedownthispolynomialwithoutactually solving f,itwouldbenecessarytohave
expressions for the coefficients α2 +α2,α2α2 in terms of f’s coefficients σ ,σ . You may
1 2 1 2 1 2
enjoylookingforthemyourselfbeforereadingthenextline.
α2+α2 = σ2 2σ
1 2 1 − 2
α2α2 = σ2
1 2 2
These activities called attention to the fact that such expressions exist in every case we
considered. Thus the participants began to suspect that something like the FTSP would
be true. It was clear that any expression in the roots of a polynomial would have to be
symmetricto be expressiblein terms ofthe coefficients (since the coefficients are already
symmetric). Butitwasmuchlessclearthatanysymmetricexpressionintherootswould
beexpressibleinthecoefficients.
2. THE TWO AND THREE VARIABLE CASE
In this section we start to tackle the question of why any symmetric expression in the
roots is expressible in the coefficients from the na¨ıve point of view of the participants. It
is natural to begin with the special cases in which the polynomial has just two and then
three variables. The participants were able to cobble together proofs in these two cases
overthecourseoftwomeetings.
To start, let p(x,y) be a polynomial which is symmetric in x and y. We want to show
that it can be expressedas a polynomial in σ = x+y and σ = xy. Taking an arbitrary
1 2
monomial xmyn which appears in p(x,y), we will try to eliminate it by expressing it in
terms of σ and σ . Renaming the variables if necessary, we can suppose that m n. If
1 2
n > 0, then we can already write xmyn as σnxm n, so it suffices to deal with mon≥omials
2 −
oftheformxn. Forthis,notethatthesymmetryof p(x,y) impliesitsconjugatemonomial
yn is also a term of p(x,y), so we can deal with xn +yn together. Now, we immediately
recognize xn+yn asthefirstandlasttermsofσn = (x+y)n. Hence,wehavethat
1
n n
xn+yn = σn xyn 1 xn 1y
1 −(cid:18)1(cid:19) − −···−(cid:18)n 1(cid:19) −
−
= σn σ q(x,y) ,
1 − 2
where q(x,y) is a polynomial of degree n 2. In other words, we have shown that an
−
inductiononthedegreeof p(x,y) willsucceed.
SYMMETRICPOLYNOMIALS 5
Inthecaseofthreevariables,letp(x,y,z)beapolynomialwhichissymmetricinx,y,z. We
wishtoexpress p(x,y,z) asafunctionofσ = x+y+z,σ = xy+xz+yz,andσ = xyz.
1 2 3
Again consider an arbitrary monomial xmynzp in p(x,y,z), where for convenience we
assume that m n p. If p > 0 then we can write xmynzp as σpxm pyn p, leaving a
≥ ≥ 3 − −
monomial with just two variables to deal with. In other words, we only need to treat
monomials of the form xmyn. Now, all of the conjugate monomials xnzm, xmzn, xnzm,
ymzn and ynzm are also found in p(x,y,z). In analogy to the two variable case, we now
recognizethatthesearealltermsof
σm nσn = (x+y+z)m n(xy+xz+yz)n .
1 − 2 −
Thus,wecanwrite
xmyn+xnzm+xmzn+xnzm+ymzn+ynzm = σm nσn q(x,y,z) .
1 − 2 −
Butunlikethetwovariablecase,notalloftheleftoverterms(whichwedenotedq(x,y,z))
have a common factor. However, it is easy to see that any of the terms from q(x,y,z)
which does involve just two variables must be a conjugate of xkyl, where m > k l > n
≥
andk+l = m+n. Sowhilewehavenotreducedthedegreeineverycase,incaseswhere
wehavenotwehaveimprovedthesituationinonekeyway: wehavereducedthespread
between the exponents. In other words, this time we will succeed using an induction
which takes into account both the degree and the spread between the exponents in the
caseofmonomialswithjusttwovariables.
Itisnaturaltotrytogeneralizethismethodtofourandmorevariables,buttherearesome
difficulties. For starters, it is not clear what the “spread between the exponents” would
mean when there are more than two variables in play! While it would have been nice
to let the discussion unfold and try to turn this into a general proof, the instructor (Ben)
decided in the interest of time to wrap up the FTSP by presenting one of the standard
arguments.
3. A CLASSICAL PROOF
In this section we present the proof of the FTSP found in the beautiful presentation of
Sturmfels [Sturmfels,2008], who attributes the proof to Van der Waerden. In the next
section,wewillreturntotheparticipants’proofidea.
ProofoftheFTSP. Let f be the symmetric polynomial to be represented. The set of f’s
terms of a given degreeis itself a symmetric polynomial and if we can representeach of
theseasapolynomialin theσ, wecan represent f; thusnothingislostbyassumingthat
i
f ishomogeneous.
Now,orderthetermsof f lexicographically. Thatis,putthetermwiththehighestpower
of x first, and if there is a tie, decide in favor of the term with the most x , and so on.
1 2
Formally, define axi1xi2 ...xin > bxj1xj2 ...xjn if i > j , or if i = j and i > j , or if
1 2 n 1 2 n 1 1 1 1 2 2
SYMMETRICPOLYNOMIALS 6
i = j ,i = j andi > j ,etc.,andthenorderthetermsof f sothatthefirsttermis>the
1 1 2 2 3 3
>
secondwhichis thethird,andsoon.
Because f issymmetric,foreverytermcxi1xi2 ...xin init,italsocontainsallpossibleterms
1 2 n
that look like this one except with the exponentspermuted (its “conjugates”). It follows
thattheleadingtermof f,sayc xi1xi2...xin,hasi i i . Welet
1 1 2 n 1 ≥ 2 ≥ ··· ≥ n
g1 = c1σ1i1−i2σ2i2−i3···σnin−11−inσnin .
−
Then g is symmetric, and it is easy to see that it has the same leading term as f. Thus
1
f g is symmetric with a “lower” leading term, which we denote c xj1xj2 xjn. As
− 1 2 1 2 ··· n
before, it follows from the symmetry that j j i . Thus we can let g =
1 2 n 2
≥ ≥ ··· ≥
c2σ1j1−j2σ2j2−j3···σnjn, sothat g2 has thesame leadingtermas f −g1, and f −g1−g2 has a
leadingtermthatislowerstill.
Continue in like manner. The algorithm must eventually terminate with no terms re-
maining, becausethereare onlyfinitelymanypossiblemonomials xi1xi2 xin ofagiven
1 2 ··· n
degreeinthefirstplace. Thuswemustcometoapointwherewehave f g g
1 2
− − −···−
g = 0. Then f = g +g + +g is thedesiredrepresentationof f as apolynomial in
k 1 2 k
···
theσ.
i
To prove its uniqueness, it is sufficient to show that the zero polynomial in x ,...,x is
1 n
representable uniquely as the zero polynomial in σ ,...,σ . This is so because no two
1 n
distinctproductsofelementarypolynomials σk1 σkn havethesameleadingterm. (The
···
leading term of σ1k1 ···σnkn is x1k1+···+knx2k2+···+kn···xnkn, and the map (k1,...,kn) 7→ (k1 +
+k ,...,k ,k )isinjective.) Thustheleadingtermsinasumofdistinctproductsof
n n 1 n
··· −
elementarysymmetricpolynomialscannotcancel;sosuchasumcannotequalzerounless
(cid:3)
itisempty.
While this lexicographic-order argument is both elegant and simple, there is something
conceptually opaque about it (something that troubled the first author on the train ride
home from class.) It conjures in one’s mind an image of the terms of f neatly ordered
lexicographically and then picked off one-by-one, left to right, by our careful choice of
g ,...,g . However,since f andg ,...,g areallsymmetric,thetermsarenotreallybeing
1 k 1 k
pickedoffoneatatime. Whenweform f g notonlyistheleadingtermc xi1xi2 xin
− 1 1 1 2 ··· n
gettingcanceled,butsoareallofitsconjugates(forinstance,the“trailingterm”c xinxin 1 xi1).
1 1 2− ··· n
Somehow, the lexicographic ordering is obscuring the symmetry between all the conju-
gates by picking out one as the leading term, even while it exploits this symmetry to
make the proof work. It is this unsatisfying situation that led us to return to our idea of
“spreadbetweentheexponents”fromtheclass.
SYMMETRICPOLYNOMIALS 7
4. SPREADINESS PROOF
Wenowreturntotheideasofourproofinthetwoandthreevariable caseanddevelopit
into a complete argument. Recall that to generalize our ideas, we first needto overcome
thedifficultyofdecidingwhatthe“spreadbetweentheexponents”meanswhenthereare
alargernumberofvariables. Indeed,findingthisdefinitionisthelinchpinofourstrategy.
Once itis done,itallows ustoprovethetheorembybuilding an algorithmthat picksoff
the monomials with the most spread-out exponents first. The algorithm is identical in
spirit and similar in practice to the standard one, but uses spread-out-ness rather than
lexicographicordertodeterminewhichmonomialstocanceloutfirst. Sowhileweavoid
thearbitrarychoicesimposedbythelexicographicordering,ourproofstillhighlightsthe
truekeyfeatureofthestandardalgorithm.
Our first idea for defining “spread” was to use the highest exponent minus lowest. Un-
fortunately, a simple computation shows this will not work in general. To get a deeper
insight,noticethattheleadingtermsinthelexicographicorderingarealsotheoneswhere
theexponentsareinsomesensethemostspreadout. Ourdefinitionshouldthereforetake
thisintoaccount.
Definition1. Givenamonomial xi1 xin defineitsspreadinesstobethesumi2+ +i2.
1 ··· n 1 ··· n
Thisisequivalentto(inthesensethatfortermsofagivendegreeitisanincreasingfunc-
tion of) both the variance of the set of exponentsin a given monomial xi1...xin, and the
1 n
heightofthecenterofgravityofthemonomialwhenitispicturedasapileofbricks,with
a stack of i bricks corresponding to each x . (We will show this below.) Moreover, it
k k
has the advantage of being a nonnegativeinteger, allowing us to useit as the basis of an
induction.
Thekeyfacttoestablishisthatjustasc xi1xi2...xin,withi i i ,istheleading
1 1 2 n 1 ≥ 2 ≥ ··· ≥ n
term of c1σ1i1−i2σ2i2−i3···σnin when the terms are ordered lexicographically, it and all its
conjugates also have strictly greater spreadiness than the rest of the terms of this latter
product.
Theorem 3 (Spreadiness Lemma). Given i ,...,i with i i i , the terms of
1 n 1 2 n
≥ ≥ ··· ≥
σ1i1−i2σ2i2−i3···σnin withmaximumspreadiness areprecisely x1i1x2i2···xnin anditsconjugates.
Centerofgravityproof. In this argument we identify a monomial xj1xj2 ...xjn with a se-
1 2 n
quence of stacks of heights j ,...,j of identical bricks. We first compute that for terms
1 n
takenfromσ1i1−i2σ2i2−i3···σnin,thespreadinessisan increasinglinear functionoftheverti-
cal coordinate (y) of the center of gravity of its corresponding brick configuration. Sup-
posingthateachbrickhasunitmass,thentheverticalcoordinateofthecenterofgravityis
givenbythesumoverthebricksofeachbrick’sheight,dividedbythenumberofbricks.
If we suppose the first brick of each stack lies at a height of 1 and each brick has unit
SYMMETRICPOLYNOMIALS 8
height, then the stack of height j contributes 1+2+ +j = j1(j1+1) to the sum. The
1 ··· 1 2
fullverticalcoordinateyofthecenterofgravityisthengivenby
1 j (j +1) j (j +1)
y = 1 1 + + 1 n
d (cid:18) 2 ··· 2 (cid:19)
1
= j2+ +j2 +j + +j
2d 1 ··· n 1 ··· n
(cid:0) (cid:1)
1
= (s+d)
2d
wheredisthenumberofbricks(i.e.,thedegree),and sisthespreadiness. Sos = 2dy d
−
andsincedisfixed,sisanincreasinglinearfunctionofyasclaimed.
Next,weobservethatallofthetermsofσ1i1−i2σ2i2−i3···σnin canbeobtainedfromx1i1x2i2···xnin
bymoving somebricks horizontally (and droppingthemontothetopofthestackbelow
ifnecessary). Theconjugatesof xi1xi2 xin arepreciselythosetermsinwhicheachlayer
1 2 ··· n
of bricks rests completely on top of the layer below it before any dropping takes place.
Thus bricks will fall for precisely those terms that are not conjugates of xi1xi2 xin. See
1 2 ··· n
Figure1.
Finally,weappealtothesimplefactthatgivenanyphysicalconfigurationofbricks,mov-
(cid:3)
ingsomebrickstolowerpositionsdecreasesthecenterofgravity.
x x x x x x x x x x x x x x x
1 2 3 4 5 1 2 3 4 5 1 2 3 4 5
FIGURE 1. Left: the target term x15x22x32x4. Center: another generic term
fromtheproductσ3σ σ ;inthispicturethetermx3x x x2x3isrepresented.
1 3 4 1 2 3 4 5
Right: thesamegenerictermwiththebricks“fallen;” ithasalowercenter
ofmassthanthetargetterm.
Once this is established, the proofof the fundamental theoremfollows the outline of the
standardargumentgivenabove.
ProofoftheFTSPusingtheSpreadinessLemma. Let f bethesymmetricpolynomialtoberep-
resented. Asabove,welosenothingbyassuming f ishomogeneous.
Thealgorithmproceedsasinthestandardproofexceptwithspreadinessplayingtherole
of lexicographic order. Pick any term of f with maximum spreadiness s and consider it
1
and its conjugates. Form the product of elementary symmetric polynomials g that has
1
thesetermsasitstermsofmaximumspreadiness. (Ifthetermsof f havecoefficientc and
1
SYMMETRICPOLYNOMIALS 9
exponentsi1 ≥ i2 ≥ ··· ≥ in, then g1 = c1σ1i1−i2σ2i2−i3...σnin as always.) Then since these
termsare theonly termsof g with spreadinessas high as s by theSpreadinessLemma,
1 1
f g containsfewertermsofspreadinesss than f does,possiblyzero.
1 1
−
Continuinginlikemannerbeginningwith f g ,formingg andthen f g g ,etc.,we
1 2 1 2
− − −
getanalgorithmthatmustterminatebecauseateachstage,eitherthemaximumspreadi-
nessorthenumberoftermswiththisspreadinesshasbeendecreased.
Theuniquenessoftherepresentationfollowsexactlyasit didinthestandardproof. Dis-
tinctproductsofelementarysymmetricpolynomialswillhavedistincttermsofmaximum
spreadiness because of the injectivity of the map (k ,...,k ) (k + +k ,...,k ).
1 n 1 n n
7→ ···
Thereforecompletecancellationisimpossible: anynonzeropolynomialintheelementary
(cid:3)
symmetricpolynomialswillbenonzerowhenmultipliedout.
As an aside, we mentioned above that spreadiness is also equivalent to variance. To see
this, we compute that for terms taken from σ1i1−i2σ2i2−i3···σnin, the spreadiness s is an in-
creasinglinearfunctionofthevariance σ2 oftheset j ,j ,...,j . Indeed,
1 2 n
{ }
1
σ2 = j2+ +j2 µ2
n 1 ··· n −
(cid:0) (cid:1)
Here n is fixed and so is the mean µ, being a function of just n and the degree d. Thus,
s = nσ2+nµ2 isanincreasinglinearfunctionofσ2.
5. THE FTSP IN GALOIS’ WORK
We promised earlier to contextualize the FTSP in the bigger picture of Galois theory by
showinghowitisanexampleofalargerphenomenon. FTSPsaysthatexpressionsthatare
completelysymmetricarecompletelyrationallyexpressible. InhisseminalessayMe´moiresur
les conditions de re´solubilite´ des e´quations par radicaux, Galois proveda seriesofresultsthat
tiepartialtypesofsymmetrytopartiallyrationaltypesofexpressibilityaswell. First,we
justify our up-till-now flip use of phrases like “rationally expressible” (since the FTSP is
onlyastatementaboutpolynomials; nodivisionallowed)extendingtheFTSPtorational
functions:
Theorem4(FTSPforrationalfunctions). Anyrationalfunctioninx ,...,x thatissymmetric
1 n
in x ,...,x isarationalfunctionoftheelementarysymmetricpolynomials σ ,...,σ .
1 n 1 n
Proof. Let f besuchafunction. Itisaquotientofpolynomials f = P/Q. LetQ ,Q ,...,Q(n!)
′ ′′
betheresultofpermutingthevariablesin Qineverypossibleway. Then
PQ Q Q(n!)
′ ′′
f = ···
QQ Q Q(n!)
′ ′′···
Thedenominatorofthisexpressionisinvariantunderallpermutationsofthex’sbycon-
i
struction, and f is as well by assumption. It follows that the numerator is also invariant
SYMMETRICPOLYNOMIALS 10
(symmetric). Thus f is here expressed as a quotient between symmetric polynomials,
(cid:3)
whicharepolynomialsintheσ bytheFTSP.
i
WhatGaloisdidwastoproveaseriesofsimilarstatementsaboutrationalfunctionsofthe
x thatarepartiallyratherthanfullysymmetric. Westatethemwithoutproof:
i
Theorem 5. If f is a rational function of x ,...,x that is symmetric under all permutations of
1 n
the x’sthatfix x ,thenitisexpressible asarationalfunctionofσ ,...,σ and x .
i 1 1 n 1
Theorem 6. If V is a rational function of x ,...,x that is not left fixed by any nontrivial per-
1 n
mutationofthe x’s,theneveryrationalfunctionofthe x’sisexpressible asarationalfunctionof
i i
theσ’sandV.
i
We can summarize the connections between symmetry and rational expressibility in a
tableasfollows.
Ifyouareinvariantunder... thenyouarerationallyexpressiblein...
allpermutations σ ,...,σ
1 n
allpermutationsfixing x σ ,...,σ and x
1 1 n 1
anysubset,ornopermutationsatall σ ,...,σ andV
1 n
Toculminate, lettingthe x’s representelementsofagivenfield(Galois himselfwasgen-
i
erally thinking of subfields of C) rather than abstract symbols, and therefore letting the
σ’s representthe coefficients of the polynomial that has the x’s as roots, Galois asserted
i i
hisfamous
Theorem 7 (Galois’ Proposition I). Let f be a polynomial with coefficients σ ,...,σ . Let
1 n
x ,...,x be its roots. Let U,V,... be some numbers that are rational functions of the x’s that
1 n i
wesupposeareknowntous. Then,thereexistsagroup Gofpermutationsofthe x ’ssuchthatthe
i
rational functions of the x ’s that are fixed under all the permutations in G are exactly those that
i
arerationallyexpressible intermsofσ ,...,σ andU,V,....
1 n
If you have studied Galois theory, this formulation may feel unfamiliar to you. To see
that it is really the same thing you have seen before, consider that the set of quantities
thatarerationalfunctionsofσ ,...,σ formsafield(f’scoefficientfield);similarlyforthe
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set of rational functions of x ,...,x (f’s splitting field). The set of rational functions of
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σ ,...,σ ,U,V,... issomeextensionofthecoefficientfieldcontainedinthesplittingfield.
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So a modern way to state Galois’ Proposition I is that given a polynomial f and a given
extension K of its coefficient field contained in its splitting field, there is a permutation
group G acting on the roots of f such that the fixed field of G is K. This powerful and
famousresulttiesinaveryprecisewaythatwhichisrationallyexpressible(theelements
ofafield)toagiventypeofsymmetry(thegroup).
We hope to have shown you that the FTSP contains the first whisper of this connection.
If you are interested to learn more, Harold Edwards’ recent article in the Notices of the