14 Symmetric Polynomials in This Section, We Consider a Special Type of Polynomials in Several Variables, Called Symmetric Polynomials

Home , Symmetric polynomial

14 Symmetric polynomials In this section, we consider a special type of polynomials in several variables, called symmetric polynomials. This is a very classic algebraic subject, which we will encounter later in great generality in Galois theory. In 12.3, we saw that for an integral domainR, a non-zero polynomialf R[X] ∈ has no more thann = deg(f) zeros inR. ForR=Z, the fundamental theorem of algebra 26.3 tells us thatf has exactlyn = deg(f) zeros, provided that we count them with multiplicity and are willing to consider zeros in a suitable ring larger thanZ itself, for exampleC. This is also true for an arbitrary integral domain, and in§21, we will construct the necessary “extensionﬁelds.” Even if the zeros off R[X] can only be found in a larger ringR � R, it turns ∈ ⊃ out that “symmetric expressions” in the zeros off lie inR itself. We will see how to calculate them without ever venturing outside the ground ringR. The methods we give are used by all computer algebra packages.

� General polynomial of degreen

We define a “general” polynomial of degreen by working in the ringR=Z[T 1,T2,. . ., Tn] of polynomials in then variablesT 1,T2,...,Tn. We can also allow other rings than Z for the ring of coefficients, but for the sake of simplicity, we will not do so. The e1 e2 en (total) degree of the monomialT 1 T2 ...Tn is equal toe 1 +e 2 +e 3 +...+e n, and we define the degree deg(f) of a non-zero polynomialf R as the maximum of the ∈ degrees of the monomials inf. If all monomials inf are of the same degreed, thenf is called homogeneous of degreed. An arbitrary polynomialf R of degreed can be ∈ written asf=f 0 +f 1 +f 2 +...+f d withf k homogeneous of degreek by combining the monomials of equal degree.

The general polynomialF n of degreen is the monic polynomial inR[X] with zeros the variablesT 1,T2,...,Tn:

n n k n k F = (X T )(X T )...(X T ) =X + ( 1) s X − R[X]. n − 1 − 2 − n − k ∈ �k=1 The coeﬃcientss R are called the elementary symmetric polynomials in the zeros k ∈ Ti off, and the Frenchman Fran¸coisVi`ete(1540–1603) already knew that fork= 1,2, . . . n, the polynomials k is equal to

sk = Ti1 Ti2 ...Tik , 1 i1

s1 =T 1 +T 2 +...+T n,

s2 =T 1T2 +T 1T3 +...+T 1Tn +T 2T3 +T 2T4 +...+T n 1Tn, −

49 Algebra II–§14 and

sn =T 1T2T3 ...Tn .

Note that, by deﬁnition, the general polynomial of degreen has coeﬃcients in the integral domainR 0 =Z[s 1, s2, . . . , sn] and that itsn zeros are the variables of the extension ringR=Z[T ,T ,...,T ] R . 1 2 n ⊃ 0 � Symmetric polynomials

A polynomial inR=Z[T 1,T2,...,Tn] (or, more generally, inA[T 1,T2,...,Tn] for a commutative ringA) is called symmetric (in the variablesT i) if it is invariant under all permutation of the variablesT i. Somewhat more formally, we can consider the natural action of the symmetric groupS n onR given by

(σf)(T ,T ,...,T ) =f(T ,T ,...,T ) forf R,σ S . 1 2 n σ(1) σ(2) σ(n) ∈ ∈ n The symmetric polynomials inR are then the polynomials inR that are invariant under the action ofS . For everyσ S , the mapf σf is an automorphism ofR, so that n ∈ n �→ we have an inclusionS Aut(R). It follows easily that the symmetric polynomials n ⊂ form a subringR R. 0 ⊂ Exercise 1. Verify this.

Since thekth elementary symmetric polynomials R is symmetric, we have the k ∈ inclusionZ[s , s , . . . , s ] R . The fundamental theorem for symmetric polynomials 1 2 n ⊂ 0 says that this inclusion is an equality: every symmetric polynomial is a polynomial in the elementary symmetric polynomials. This theme is elaborated further in Galois theory: polynomials with “many” symmetries are contained in “small” subrings.

14.1. Fundamental theorem. LetP R=Z[T ,T ,...,T ] be a symmetric poly- ∈ 1 2 n nomial. Then there is a unique way to writeP as a polynomial in the elementary symmetric polynomialss k.

We give a constructive proof, that is, a proof that gives an algorithm to write aP as an element ofZ[s 1, s2, . . . , sn]. Proof. We order the monomials inP lexicographically, as in a dictionary. In other e1 e2 en words, monomialsT 1 T2 ...Tn with the highest exponente 1 are in front, if two monomials have the samee 1, their order is determined bye 2, and so on. Now, letc T e1 T e2 ...T en withc Z 0 be theﬁrst term inP , lexicographically, · 1 2 n ∈ \{ } and letd=e +e +e +...+e be its degree. Then we havee e e ... e . 1 2 3 n 1 ≥ 2 ≥ 3 ≥ ≥ n After all, if this is not the case, then through a suitable permutation of theT i, we can transform this term into one that comes earlier lexicographically and that must also be inP because of the symmetry ofP ; this gives a contradiction. Now, form the monomial e1 e2 e2 e3 e3 e4 en 1 en en Σ=s 1 − s2 − s3 − . . . sn −1 − sn R − ∈

50 Algebra II–§14

of degree

e1 e 2 + 2(e2 e 3) + 3(e3 e 4) +...+(n 1)(e n 1 e n) + nen − − − − − − =e +e +e +...+e =d deg(P), 1 2 3 n ≤ withfirst termT e1 T e2 ...T en (lexicographically), and considerP =P cΣ R. Be- 1 2 n 1 − ∈ cause deg(Σ) deg(P ), we have deg(P 1) deg(P ), and all monomials inP 1 come ≤ e1 e2 en≤ later, lexicographically, thanT 1 T2 ...Tn . Since onlyfinitely many different monomials are possible when the degree is bounded, we see that by repeatedly subtracting

an element ofZ[s 1, s2, . . . , sn], we can reduce the polynomialP to 0. In other words, P is itself contained inZ[s 1, s2, . . . , sn]. To prove that there are not two ways to write a polynomial inZ[T 1,...,Tn] as a polynomial inZ[s 1, s2, . . . , sn], we must show that there is no non-zero polynomial g Z[X ,X ,...,X ] withg(s , s , . . . , s ) = 0. To do this, suppose that such a ∈ 1 2 n 1 2 n polynomial does exist. Write every monomial ing in the form

e1 e2 e2 e3 e3 e4 en 1 en en cX1 − X2 − X3 − ...Xn −1 − Xn , −

and consider the monomialM ing for which the correspondingn-tuple (e 1, e2, . . . , en) comesﬁrst lexicographically. When expandingg(s 1, s2, . . . , sn) as a polynomial in e1 e2 en Z[T1,T2,...,Tn], we see thatM leads to a term cT 1 T2 ...Tn that does not disappear, so we have a contradiction.

It follows from the uniqueness of representations in terms of the elementary sym-

metric polynomialss k thatZ[s 1, s2, . . . , sn] can be viewed as a polynomial ring in the variabless k: the elementary symmetric polynomials are algebraically independent. a1 a2 an We say that a monomials 1 s2 . . . sn has weighta 1 + 2a2 + 3a3 +...+ na n; more generally, the weight ofg Z[s , s , . . . , s ] is the maximum of the weights of ∈ 1 2 n the monomials ing. If all monomials ing have the same weightd, theng is said to be isobaric of weightd. Note that the weight ofg Z[s , s , . . . , s ] is nothing but ∈ 1 2 n the degree ofg as an element ofR=Z[T 1,T2,...,Tn]. The proof of 14.1 shows the following.

14.2. Corollary. A homogeneous symmetric polynomial inZ[T 1,T2,...,Tn] of degreed can be written in a unique way as an isobaric polynomial of weightd in

Z[s1, s2, . . . , sn].

An arbitrary symmetric polynomialP can be written as a sum of homogeneous poly-

nomialsP k of degreek; the polynomialsP k are symmetric because the action ofS n on Z[T1,T2,...,Tn] leaves the degree invariant. By 14.2, the polynomialP k can be written as an isobaric polynomial of weightk in the elementary symmetric polynomialss k.

� Calculations with symmetric polynomials There is a shortened notation for symmetric polynomialsP , deﬁned as follows: for

everyS n-orbit inP , we write a single representative preceded by the symbol n to � 51 Algebra II–§14

indicate that we take the sum over the monomials in theS n-orbit of the representative. In this notation, thekth elementary symmetric polynomials R is equal to k ∈ sk = T1T2T3 ...Tk. n � More generally, f withf Z[T ,T ,...,T ] is the notation for the sum of the n ∈ 1 2 n polynomials in theS -orbit off. Examples: � n 2 2 2 2 2 2 2 T1 T2 =T 1 T2 +T 1 T3 +T 2 T3 +T 1T2 +T 1T3 +T 2T3 3 � T1T2T3 =T 1T2T3 +T 1T2T4 +T 1T3T4 +T 2T3T4 4 � T1T2 =T 1T2 +T 1T3 +T 1T4 +T 2T3 +T 2T4 +T 3T4 4 � If we want to use the method from the proof of 14.1 to write a given symmetric

polynomial as a polynomial in thes k, then the short notation is often useful. After all, e1 e2 en if a monomialrT 1 T2 ...Tn occurs in a symmetric polynomialf Z[T 1,T2,...,Tn], e1 e2 en ∈ then n rT1 T2 ...Tn also occurs in that polynomial. 14.3. Examples. 1. Taken 2 andP=T 2 +T 2 +...+T 2 = T 2. ThenT 2 is � ≥ 1 2 n n 1 1 lexicographically the highest term inP , so we form � 2 2 2 s1 = (T1 +T 2 +...+T n) = T1 + 2 T1T2 n n � � and calculateP =P s 2 = 2 T T = 2s . In this case, we are done after one 1 − 1 − n 1 2 − 2 step, and wefindP=s 2 2s . Note thatP is homogeneous of degree 2 ands 2 2s 1 − 2 � 1 − 2 is isobaric of weight 2. 2. Now, taken 3 andP=T 3 +T 3 +...+T 3 = T 3. ThenT 3 is lexico- ≥ 1 2 n n 1 1 graphically the highest term inP , so we form � 3 3 3 2 s1 = (T1 +T 2 +...+T n) = T1 + 3 T1 T2 + 6 T1T2T3. n n n � � � 3 The coefficients 3 and 6 indicate how often a term occurs when we expands 1. Wefind P =P s 3 = 3 T 2T 6 T T T . The lexicographically highest term inP 1 − 1 − n 1 2 − n 1 2 3 1 is 3T 2T , so we subtract 3 times − 1 2 � − � s s = T T T = T 2T + 3 T T T 1 2 1 · 1 2 1 2 1 2 3 n n n n � � � � and obtain P =P + 3s s =P s 3 + 3s s = 3 T T T = 3s . 2 1 1 2 − 1 1 2 1 2 3 3 n � Conclusion:P=T 3 +T 3 +...+T 3 =s 3 3s s + 3s . Again, note thatP is 1 2 n 1 − 1 2 3 homogeneous of degree 3 ands 3 3s s + 3s is isobaric of weight 3. 1 − 1 2 3 Exercise 2. What happens in the casesn< 2 (in Example 1) andn< 3 (in Example 2)?

k k k See Exercise 24 for the representation of the sum of powersσ k =T 1 +T 2 +...+T n in terms of the elementary symmetric polynomials using Newton’s identities.

52 Algebra II–§14

� Discriminant

A common symmetric polynomial inZ[T 1,T2,...,Tn] is the discriminant

(14.4) Δ = (T T )2 n i − j 1 i

of the general polynomialF n of degreen. The polynomialΔ n is homogeneous of degreen(n 1), so by 14.2, it is a universal isobaric expression of weightn(n 1) − − inZ[s 1, s2, . . . , sn]. In other words, the discriminant of the general polynomialF n of n k degreen is a polynomial in the coeﬃcients ( 1) − s ofF . − k n After the uninteresting caseΔ 1 = 1, we have

Δ = (T T )2 = (T +T )2 4T T =s 2 4s , 2 1 − 2 1 2 − 1 2 1 − 2 a result that is also written asΔ(X 2 +AX+B)=A 2 4B. Using the method of 14.1, − we can expressΔ n in the elementary symmetric polynomials, and with some eﬀort, we obtain Δ =s 2s2 4s 3 4s 3s 27s2 + 18s s s 3 1 2 − 2 − 1 3 − 3 1 2 3 1 Δ = 4(s2 3s s + 12s )3 (2s3 72s s + 27s2s 9s s s + 27s2)2 , 4 27 2 − 1 3 4 − 2 − 2 4 1 4 − 1 2 3 3 � � formulas that are too unpleasant to remember. For large values ofn, the formulas for

theΔ n are, moreover, too unpleasant to write down.

Exercise 3. Verify thatΔ 4 is inZ[s 1, s2, s3, s4]. Next, letA be an arbitrary integral domain andf A[X] a monic polynomial of ∈ degreen. Then by 12.3, the polynomialf has at mostn zeros inA; in 21.13, we will prove that the number of zeros off in a suﬃciently large integral domainA � A is ⊃ exactlyn is the sense that n f= (X α ) − i i=1 � holds withα A �. IfA is a subring ofC such asZ,Q, orR, then by the fundamental i ∈ theorem of algebra 26.3, we can always takeA � =C. The discriminant off is now deﬁned as

(14.5) Δ(f)= (α α )2. i − j 1 i

SinceA � is an integral domain, we haveΔ(f) = 0 if and only iff has a double zero inA �. The homomorphismZ[T ,T ,...,T ] A � that sendsT toα mapsΔ onto 1 2 n → i i n Δ(f). Since the images of the elementary symmetric polynomialss k, which are sent to (plus or minus) the coeﬃcients off, are contained inA, we see thatΔ(f) is also an element ofA. In other words, the discriminant of a polynomialf A[X] is an element ∈ ofA, and it is given by a universal polynomial in the coeﬃcients off.

53 Algebra II–§14

14.6. Example. The general formula for the discriminant of a cubic polynomial is diﬃcult to remember, but the well-known formula

(14.7)Δ(X 3 + pX+q)= 4p 3 27q2 − − that arises by substituting (s , s , s ) = (0, p, q) in the general expression forΔ is 1 2 3 − 3 both easy to remember and easy to deduce. After all, by takingA=Z[p, q], we know that the discriminant is a universal polynomial inp andq. Since there are only two monomials inZ[s , s ] of weight 3(3 1) = 6, namelys 3 ands 2, there in fact exist 2 3 − 2 3 constantsc , c Z such that we haveΔ(X 3 + pX+q)=c p3 +c q2. We can easily 1 2 ∈ 1 2 calculatec 1 andc 2 by choosing a few suitable values forp andq. Namely, we have c = Δ(X 3 X)= 4 andc =Δ(X 3 1) = 27. 1 − − − 2 − − Exercise 4. Verify this.

� Resultant Usually, discriminants of polynomials of higher degree inA[X] are not calculated using

the general formula forΔ n; instead, the resultant is used. To avoid restrictions on division, if necessary, we replace the integral domainA with itsﬁeld of fractions; from now on, we assume thatA=K is aﬁeld. For polynomials

n m f=a (X α ) andg=b (X β ) − i − j i=1 j=1 � � inK[X] of degreen andm, respectively, the resultantR(f, g) is deﬁned by

n m (14.8)R(f, g)=a mbn (α β ). i − j i=1 j=1 � � The following properties are immediate from this deﬁnition: (R1)R(f, g)=( 1) mnR(g, f). −m n (R2)R(f, g)=a i=1 g(αi). (R3) Ifg 1 K[X] is of degreem 1 withg g 1 mod (f), then we have ∈ m �m1 ≡ R(f, g)=a − R(f, g1). Using these properties and division with rest inK[X], we can calculate resultants

without ever needing any explicit knowledge of the zerosα i andβ j that occur in the deﬁnition. Using (R1), if necessary, we may assume that we have deg(g) deg(f). ≥ We then use (R3) to replaceg with the remainderg A[X] of the division ofg byf. 1 ∈ We can lower the degree of the polynomials by repeating these steps, and as soon asf has degree 0 or 1 and we know the zerosα A, property (R2) gives the value of the i ∈ resultant. For a monic polynomialf= n (X α ), the derivative in a zeroα is equal to i=1 − i i � f �(αi) = (αi α 1)(αi α 2)...(α i α i 1)(αi α i+1)...(α i α n). − − − − − −

54 Algebra II–§14

If we take the product of these expressions fori=1,2, . . . , n, then after counting the number of factors 1, we see that − n(n 1)/2 (14.9)Δ(f)=( 1) − R(f, f �). − This is how we can use the resultant to calculate discriminants of polynomials inK[X].

14.10. Example. 1. TakeK=Q(p, q) andf=X 3 + pX+q. Then we have

Δ(X3 + pX+q)= R(X 3 + pX+q,3X 2 +p). − If we apply (R1) and then (R3) forf=3X 2+p,g=X 3+pX+q, andg =g (X/3)f= 1 − (2p/3)X+q, then this becomes 3 2 R(3X 2 + p, (2p/3)X+q). If we apply (R1) once − · more and recall thatg has a single zeroα= 3q/(2p), then property (R2) gives 1 − 2p 2 3q 2 Δ(X3 + pX+q)= 3 2 3 − +p = 4p 3 27q2, − · 3 2p − − � � � � � � in accordance with (14.7). 2. TakeK=Q andf=X 5 +X + 1. Then we have

Δ(X5 +X + 1) =R(X 5 +X+1,5X 4 + 1) =R(5X 4 + 1,X5 +X + 1) 4 = 54 R(5X 4 + 1, X + 1) · 5 4 4 5 4 = 54 5 − + 1 = 55 + 44 = 3381. · 5 · · 4 � � � � � � Sincef is irreducible inZ[X], the complex zeros off do not lie inZ orQ, and it is therefore not easy to give them “explicitly.” However, this is not at all necessary to calculate the discriminant.

55 Algebra II–§14

Exercises.

5. Suppose thatf Z[T ,T ,...,T ] is invariant under every exchangeT T of two ∈ 1 2 n i ↔ j variables. Isf necessarily symmetric? 6. Letα ,α , andα be the zeros ofX 3 X 1 inC, and setp =α k +α k +α k for all 1 2 3 − − k 1 2 3 k Z. Prove: the sequence p k k Z consists of integers that satisfyp 1 = 1,p 0 = 3, ∈ { } ∈ − − p1 = 0, and the recurrence relationp k =p k 2 +p k 3 fork Z. − − ∈ 7. Show: forf Z[T ,T ,...,T ] homogeneous of degreed andr Z, we have ∈ 1 2 n ∈ d f(rT 1, rT2, . . . , rTn) =r f(T 1,T2,...,Tn).

Conversely, isf homogeneous of degreed if this identity holds for allr Z? ∈ 8. Give an example of a ringA and a non-homogeneous polynomialf A[T ,T ,...,T ] ∈ 1 2 n that satisﬁesf(rT , rT , . . . , rT ) =r df(T ,T ,...,T ) for allr A for somed 1. 1 2 n 1 2 n ∈ ≥ n+d 1 e1 e2 en 9. Show that there are exactly n −1 distinct monomialsT 1 T2 ...Tn of degreed. − 2 3 10. Express the symmetric polynomials� � n T1 T2 and n T1 T2 in the elementary symmetric polynomials. � � 11. Show that the orbit of the polynomialB=T T +T T Z[T ,T ,T ,T ] under 1 2 3 4 ∈ 1 2 3 4 the action ofS 4 consists of three elementsB,B �, andB ��, and determine the cubic polynomial inZ[s 1, s2, s3, s4][X] with these three zeros. 12. Show that the discriminant of the cubic polynomial from the previous exercise is equal to the discriminantΔ 4 of the general polynomial of degree 4, and use this to determine Δ Z[s , s , s , s ]. 4 ∈ 1 2 3 4 n i m j 13. Show that the resultant of the polynomialsf= i=0 aiX andg= j=0 bjX of degreen andm inK[X] is equal to the (m+n) (m+n) determinant × � �

an an 1 a 0 − ··· an an 1 a 0  � − � m � ···. �  � .. �  � �  � � � an an 1 a 0�  � − ··· �  �bm bm 1 b 0 �  � − ··· � � bm bm 1 b 0 � n  � − ··· � � .. �  � . �  � � � bm bm 1 b 0 � � − ··· �  � �  � � � m+n �

in terms of the coeﬃcients off and� g. (Take all�� coeﬃcients left blank� equal to 0.) [Hint: show that this determinant also has properties (R1) and (R3).] n 1 n(n 1) n n 1 14. Prove: forn Z , we haveΔ(X +a) = ( 1) 2 − n a . ∈ >0 − − 15. Calculate the discriminant of the polynomialX 4 + pX+q Q(p, q)[X]. ∈ 16. For everyn> 1, determine an expression for the discriminant of the polynomialX n + pX+q Q(p, q)[X]. ∈ 56 Algebra II–§14

17. Letf Z[X] be a monic polynomial. Prove that the following are equivalent: ∈ a.Δ(f)= 0. � b. The polynomialf has no double zeros inC. c. The decomposition off inQ[X] has no multiple prime factors. d. The polynomialf and its derivativef � are relatively prime inQ[X]. e. The polynomialsf modp andf � modp are relatively prime inF p[X] for almost all prime numbersp. 18. Determine the “exceptional primes” in part (e) of the previous exercise forf=X 3 + X + 1 and for the polynomialX 7 + 7X + 1 from Exercise 12.23. 19. Letf Q[X] be a monic polynomial withn = deg(f) distinct complex zeros. Prove: ∈ the sign ofΔ(f) is equal to ( 1) s, where 2s is the number of non-real zeros off. − 20. Prove:X 3 + pX+q R[X] has three (counted with multiplicity) real zeros ∈ ⇐⇒ 4p3 + 27q2 0. ≤ 4 21. Express the sum of powers n T1 in the elementary symmetric polynomials. Does the value ofn matter? � 22. A rational functionf Q(T ,T ,...,T ) is called symmetric if it is invariant under ∈ 1 2 n all permutations of the variablesT i. Prove that every symmetric rational function is a rational function in the elementary symmetric functions. 1 2 23. Write n T1− and n T1− as rational functions inQ(s 1, s2, . . . , sn). k 24. (Newton’s� identities)� Show that the sums of powersσ k = n T1 satisfy

k 1 k� σk s 1σk 1 +s 2σk 2 ...+( 1) − sk 1σ1 + ( 1) ksk = 0 for 1 k n − − − − − − − ≤ ≤ and that we can use this to, inductively, write the sums of powersσ for 1 k n as k ≤ ≤ polynomials inZ[s 1, s2, . . . , sn]. Also show that fork>n, the sum of powersσ k can be written as a polynomial inZ[s 1, s2, . . . , sn] from the relation

n σk s 1σk 1 +s 2σk 2 ...+( 1) snσk n = 0. − − − − − − [Hint: determine the logarithmic derivativef /f R[[X]] off= n (1 T X) � ∈ i=1 − i ∈ R[X].] � 25. Can the method in the previous exercise also be used to write the sums of powersσ k fork< 0 as elements ofQ(s 1, s2, . . . , sn)?

26. Show that in terms of the sums of powersσ k, the discriminantΔ n can be written as

n Δn = det σi+j 2 . − i,j=1 � � Use this relation to calculateΔ 3 Z[s 1, s2, s3]. ∈ j 1 n [Hint: start with the Vandermonde-determinant det(Ti − )i,j=1.] 27. Show that the discriminantΔ Z[s , s , . . . , s ] of the general polynomial of degreen n ∈ 1 2 n is an irreducible polynomial inZ[s 1, s2, . . . , sn]. IsΔ n also irreducible in the ring C[s1, s2, . . . , sn]?