Math 503: Abstract Algebra

Homework 3
Lawrence Tyler Rush
<me@tylerlogic.com>
February 15, 2014
http://coursework.tylerlogic.com/courses/upenn/math503/homework03

1


Define D(F) [x1,,xn] by
D (F) =   ∑   (xi - xj)2
        1≤i<j≤n
(1.1)

(a)


Existence Given equation 1.1, D(F) can alternatively be specified as

            ∑             2
D(F ) =            (xi - xj)
       i,j∈{1,...,n}, i<j

Because any σ Sn is bijective and because (xi - xj)2 = ((-1)(xj - xi))2 = (-1)2(xj - xi)2 = (xj - xi)2, then

     ∑                  2       ∑             2
            (xσ(i) - xσ(j)) =            (xi - xj) = D(F )
i,j∈{1,...,n}, i<j              i,j∈{1,...,n}, i<j

In other words, D(F) is symmetric. Thus there indeed exists a polynomial d [x1,,xn] such that d(s1,,sn) = D(F) since every symmetric polynomial in [x1,,xn] is contained in [s1,,sn].

Uniqueness There cannot exist two distinct d1,d2 [x1,,xn] such that d1(s1,,sn) = d2(s1,,sn) = D(F) because their existence would contradict the algebraic independence of s1,,sn [Lan02, p. 192] since it would imply d = d1 - d2 has d(s1,,sn) = 0.

(b)


Degree n = 2 To find an explicit formula for disc(f) where f(t) is a monic polynomial of degree two, we need to find d(z1,z2) [z1,z2] such that d(s1,s2) = (x1 - x2)2. Because of the existence/uniqueness proven above, we can find such a polynomial and there will be only one. So because s1 = x1 + x2, s2 = x1x2 for n = 2 and because

       2    2          2     2          2                  2
(x1 - x2) = x1 - 2x1x2 + x2 = (x1 + 2x1x2 + x2)- 4x1x2 = (x1 + x2) - 4x1x2

then we can deduce that d(z1,z2) = z12 - 4z2. Hence for any f(t) = t2 + at + b

disc(f) = d(- a,b) = a2 - 4b

Degree n = 3 To find an explicit formula for disc(f) where f(t) is a monic polynomial of degree three, we need to find d(z1,z2,z3) [z1,z2,z3] such that d(s1,s2,s3) = (x1 - x2)2(x1 - x3)2(x2 - x3)2. Again, because of the existence/uniqueness proven above, we can find such a polynomial and there will be only one.

We first notice that d(s1,s2,s3) will be a homogeneous polynomial of degree six in x1,x2,x3. Therefore, since s1 = x1 + x2 + x3, s2 = x1x2 + x1x3 + x2x3, and s3 = x1x2x3,

d(s1,s2,s3) = α1s2+ α2s3s2s1 +α3s3s3+ α4s3 + α5s2s2 +α6s2s4+ α7s6
               3                1     2     2 1       1     1

for some integers α1234567. We will find these values by analyzing the following expansion

(x1 - x2)2(x1 - x3)2(x2 - x3)2 = x41x22 - 2x31x32 + x21x42 - 2x41x2x3 + 2x31x22x3
                          + 2x21x32x3 - 2x1x42x3 +x41x23 + 2x31x2x23 - 6x21x22x23
                                3 2   4 2     3 3    2   3      2 3
                          + 2x1x2x3 + x2x3 - 2x1x 3 + 2x1x2x3 + 2x1x2x3
                          - 2x32x33 +x21x43 - 2x1x2x43 + x22x43
(1.2)

From this equation, it is immediately apparent that α6 = α7 = 0 since s2s14 and s16 would produce monomials containing x1 raised to a degree higher than 4, but there is no such monomial in the polynomial of equation 1.2. Now, given the expansions

   s23  =  x21x22x23
s3s2s1  =  x31x22x3 + x21x32x3 + x31x2x23 + 3x21x22x23 + x1x32x23 + x21x2x33 + x1x22x33
 s s3  =  x4x x + 3x3x2x  +3x2x3x  + x x4x  + 3x3x x2+ 6x2x2x2+ 3x x3x2+ 3x2x x3
  3 1      1 2 32 3 1 2 3 4   12 3    1 2 3    1 2 3    1 2 3    1 2 3    1 2 3
          +3x1x 2x3 + x1x2x3
   s32  =  x31x32 +3x31x22x3 + 3x21x32x3 + 3x31x2x23 + 6x21x22x23 + 3x1x32x23 + x31x33 + 3x21x2x33
          +3x1x22x33 + x32x33
 s2s2  =  x4x2 +2x3x3 + x2x4+ 2x4x x + 8x3x2x + 8x2x3x + 2x x4x + x4x2 +8x3x x2
  2 1      1 2 2 2122    1322   14 22 3  3 13 2 32   13 2 3  2 13 2 33 31 3 2 41 2 3
          +15x 1x2x 3 + 8x1x2x3 + x2x3 + 2x1x3 + 8x1x2x3 + 8x1x2x3 +2x2x3 + x1x3
          +2x1x2x43 + x22x43
we will strategically evaluate the coefficients of certain monomials in order to give us five linear equations that will allow us to solve for the α variables. So for example, the coefficient in equation 1.2 of x12x22x32 is -6, and in the expansions of s32, s3s2s1, s3s13, s23, and s22s12 the coefficients are 1, 3, 6, 6, and 15, respectively. Hence we have the equation
α1 + 3α2 +6α3 + 6α4 + 15α5 = - 6

In the same vein, we evaluate the coefficients for the monomials x13x22x3 to get

α2 + 3α3 + 3α4 + 8α5 = 2

x14x2x3 to get

α  + 2α = - 2
 3     5

x13x33 to get

α4 + 2α5 = - 2

and finally x14x32 to get

α5 = 1

Solving the above five equations we obtain α1 = -27, α2 = 18, α3 = -4, α4 = -4, and α5 = 1, which in turn informs us that

                2                3    3   2 2
d(z1,z2,z3) = - 27z3 + 18z3z2z1 - 4z3z1 - 4z2 + z1z2

Hence for any f(t) = t3 + at2 + bt + c

disc(f) = d(- a,b,- c) = - 27c2 + 18abc- 4a3c- 4b3 + a2b2

(c) Extra Credit


2 Newton Polynomials


(a) Find formulas for Newton polynomials p2, p3, and p4 in terms of s1 through s4


Since p2, p3, and p4 are symmetric polynomials over , they can be uniquely represented as a polynomial in terms of the elementary symmetric polynomials over . Furthermore, they are homogeneous polynomials of degrees 2, 3, and 4, respectively, and must therefore have the forms
p  =   a s2+ a s
 2      131   22
p3 =   b1s1 + b2s1s2 + b3s3
p4 =   c1s41 + c2s21s2 + c3s22 + c4s4
for integers ai, bi, and ci.

It’s clear that in order to obtain the appropriate values of p2, p3, and p4, a1 = b1 = c1 = 1 since none of the other clauses in the above equations will be able to account for the xi2, xi3, and xi4 in each of p1, p2, and p3. Hence, in determining the formula for pi, our approach will be to start with s1i ordered lexicographically and find the coefficient of the leftmost clause that isn’t in pi and then subtract the product of that coefficient with the appropriate clause on the right hand side of the equations above. We will know which are appropriate by looking at the lexicographical first term of each of the clauses:


ClauseFirst Term, Lexicographically


s12 x12
s2 x1x2
s13 x13
s1s2 x12x2
s3 x1x2x3
s14 x14
s12s2 x13x2
s22 x12x22
s4 x1x2x3x4

Table 2.1: First terms of elementary symmetric polynomials that generate Newton polynomials.

where these terms have been generated for n = 4.

So for the case of p2 we start with

 2    2          2                  2                         2
s1 = x1 + 2x1x2 + x2 +2x1x3 + 2x2x3 + x3 + 2x1x4 + 2x2x4 + 2x3x4 + x4

in which the first term x12 should remain since it is in p2, but we need to get rid of the second term 2x1x2. Through use of Table 2.1, this implies that a2 = -2 so that we are left with s12 - 2s2 = x12 + x22 + x32 + x42, and thus the formula for p2.

For the case of p3 we start with s13 = x14 + 3x12x2 + ⋅⋅⋅, leaving off the, currently unimportant, terms after the first two. Again through use of Table 2.1, the coefficient of 3 in the second term informs us that b2 = -3 so that we are left with s13 - 3s2s1 = x13 + x23 - 3x1x2x3 + ⋅⋅⋅, which finally informs us that b3 = 3, leaving us with p3’s formula: s13 - 3s2s1 + 3s2.

Finally, by following the same procedure regarding p4, we have

            s41  =  x41 + 4x31x2 + ⋅⋅⋅
     s4- 4s2s   =  x4- 2x2x2+ ⋅⋅⋅
 4    12    1 22      14   41 2  2
s1 - 4s1s2 + 2s2 = x1 + x2 - 4x 1x2x3 + ⋅⋅⋅
which therefore implies s14 - 4s12s2 + 2s22 + 4s4 = x14 + x24 + x34 + x44 and yields the formula for p4.

(b) Extra Credit


(c) Extra Credit


3


Let R be a commutative ring and M be an R-module.

(a) Show the existence of an R-linear “permutation action” of Sn on RnM


For each σ Sn, define fσ : Mn Mn by (x1,,xn)↦→(xσ(1),,xσ(n)) for x1,,xn M. With this definition
fσ((x1,...,xn )+ (y1,...,yn))  =  fσ(x1 + y1,...,xn + yn)
                           =  fσ(z1,...,zn)
                           =  (z   ,...,z   )
                              ( σ(1)     σ(n)             )
                           =  (xσ(1) +yσ(1),...),xσ((n) + yσ(n)  )
                           =   xσ(1),...,xσ(n) +  yσ(1),...,yσ(n)
                           =  fσ(x1,...,xn) + fσ (y1,...,yn)
for x1,,xn,y1,,yn M letting zi = xi + yi for each such i. Furthermore,
fσ(r(x1,...,xn))  =  fσ(rx1,...,rxn)

                =  f(σ(y1,...,yn) )
                =  (yσ(1),...,yσ(n)  )
                =   rxσ(1),...,rxσ(n)
                =  r(x   ,...,x    )
                      σ(1)     σ(n)
                =  rfσ(x1,...,xn)
for r R and x1,,xn M letting yi = rxi for each such i. Therefore each fσ is an R-multilinear map yielding, via the universal property of tensors, the existence of a unique R-linear homomorphism fσ : RnM Mn through which fσ factors. Hence fσ(x1 ⋅⋅⋅xn) = (             )
 xσ(1),...,xσ(n) for each σ Sn and therefore by defining the action of Sn on RnM by having σ act on x ∈⊗RnM by (ifσ)(x) where i : Mn →⊗RnM is the normal R-multilinear inclusion map, we obtain the desired R-linear permutation action, since i fσ satisfies the axioms of a group action, being that i and fσ are each homomorphisms. Note that we omit the use of σ x or x σ notation in light of the next part of this problem.

(b) Is the action previously defined a left or right action?


Set φσ = i fσ for each σ Sn so that φσ(x1 ⊗ ⋅⋅⋅⊗ xn ) = ifσ(x1 ⊗ ⋅⋅⋅⊗ xn) = (               )
 xσ(1) ⊗ ⋅⋅⋅⊗ xσ(n) for each x1,,xn M. For repetitious use later, we point out that φτσ = φτ φσ for all τ,σ Sn according to
            (      )
              ∑
φ τσ(x ) = φτσ    xi
         ∑     i
       =    φτσ(xi)
         ∑i
       =    xiτσ(1) ⊗ ⋅⋅⋅⊗ xiτσ(n)
          i
       = ∑  x      ⊗ ⋅⋅⋅⊗ x
          i  iτ(σ(1))       iτ(σ(n))
         ∑    (                 )
       =    φτ xiσ(1) ⊗ ⋅⋅⋅⊗ xiσ(n)
         ∑i
       =    φτ ∘ φσ(xi)
          i    ∑
       = φτ ∘ φσ  xi
                i
       = φτ ∘ φσ(x)
(3.3)

for each x = ixi = ixi1 ⋅⋅⋅xin ∈⊗RnM. With equation 3.3 in hand we see that if we were to attempt to make this action a left action then

τσ ⋅x = φ  x = φ φ (x) = φ (φ  (x )) = τ ⋅(φ (x)) = τ ⋅(σ ⋅x)
         τσ     τ σ      τ   σ          σ

and if we were to do so as a right action then

x ⋅τσ = φτσx = φ τφσ(x) = φτ (φ σ(x )) = (φ σ(x))⋅τ = (x⋅σ) ⋅τ

for each x ∈⊗RnM, implying not only that this action is a left action, but that it is not are right action.

(c)


(d) Show R[y1,,yn]~
=SR(M) for free R-module M with rank(M) = n


Since both R[y1,,yn] is a graded ring where each homogenous component of degree k is the ring of homogenous polynomials of degree k (denote it Rk[y1,,yn]) and SR(M) is a graded ring where the homogeneous components of degree k is Sk(M), it suffices to prove that Rk[y1,,yn] is isomorphic to Sk(M) for arbitrary k.

Let B = {v1,,vn} be a set of free generators on M and Y = {y1,,yn}. We will use to denote the tensor operation in Sk(M). Define α : Mk Rk[y1,,yn] and β : Y Sk(M) by

                    (            )
                  ∏   ∑n
α (m1, ...,mk ) =     (    δj(mi)xj)
                   i  j=1
        β(xi) =   vi ∙1∙ ⋅⋅⋅∙ 1
where δi HomR-mod(M,R) is the linear operator which takes vi B to 1 and all other elements of B to zero. With the product in its definition and because each factor in that product is the sum of linear maps, α is a symmetric k-multilinear map. Therefore, we have that two results:
  1. The existence of an R-module homomorphism α : Sk(M) Rk[y1,,yn] through which α factors. This comes by way of the universal property of symmetric multilinear maps.
  2. The existence of an R-algebra homomorphism β : Rk[y1,,yn] Sk(M) through which β factors. This is given by the universal property of polynomial algebras.

Pictorially we have the following commutative diagram.

   M k                     Sk (M )





ααββ-                         Rk [y ,⋅⋅⋅,y ]            Y
                               1     n

Now, let’s denote x-∙⋅⋅⋅∙x
◟   ◝◜  ◞j times Sk(M) by xj so that xj = xj 1∙-⋅⋅⋅∙-1
◟  ◝◜  ◞j-1 times since each element of Sk(M) is symmetric. Thus for the generators of Sk(M) and Rk[y1,⋅⋅⋅,yn]

--(-- a1   an )  --(--  ∙a1  --   ∙an)   -- ∙a1   ∙an    a1    an
α  β(y1 ⋅⋅⋅yn ) = α  β(y1)   ⋅⋅⋅β (yn)   ) = α(v1  ⋅⋅⋅vn  ) = y1 ⋅⋅⋅yn

and

β-(α-(v ∙ ⋅⋅⋅∙ v )) = β-(α(v ,...,v )) = β(x  ⋅⋅⋅x  ) = v ∙⋅⋅⋅∙v
      i1       ik          i1     ik       i1   ik    i1      ik

which implies that α and β are inverses of each other. Hence Sk(M) is isomorphic to Rk[y1,⋅⋅⋅,yn].

(e) Extra Credit


4


(a) Compute the character of ρk


Denote the symmetric and rotational generators of D2n by s and r, respectively. Since ρ is the homomorphism of the action of D2n acting on V 1 = 2, then for any arbitrary element sirj D2n, i ∈{0,1}, j ∈{0,,n - 1}
                  (      )i (    (2πj)      (2πj) )
ρ(sirj) = ρ(s)ρ(rj) =     1     cos(-n-)  - sin(-n)
                     1        sin 2πnj   cos 2πnj

and therefore

            ((      )i (    (2πj)      (2πj) ))   {
χρ(sirj) = Tr      1      cos(-n-) - sin( -n)     =   0  (2πj)  i = 1
                1        sin  2πnj   cos 2πnj          cos -n-   otherwise

Hence we can compute the character of ρk of any sirj D2n

    i j        (   i j)     (  i j)
χρk(s r)  =   Tr(ρk(s r )⋅⋅⋅χ ρ(s r))
         =   Tr ρ(sirj)⊗ ⋅⋅⋅⊗ ρ(sirj)
               (  i j)     (   ij )
         =   T{r ρ(s r ) ⋅⋅⋅Tr ρ(sr )
         =     0   (  )  i = 1
               cosk  2πjn-  otherwise

(b)


(c)


(d) Extra Credit


References


[Lan02]   S. Lang. Algebra. Graduate Texts in Mathematics. Springer New York, 2002.