## Math 503: Abstract Algebra

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

### 1

Define D(F) [x1,,xn] by
 (1.1)

#### (a)

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

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

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

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

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,

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

 (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

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

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

x14x2x3 to get

x13x33 to get

and finally x14x32 to get

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

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

### 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
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:

 Clause First 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

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

which therefore implies s14 - 4s12s2 + 2s22 + 4s4 = x14 + x24 + x34 + x44 and yields the formula for p4.

### 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
for x1,,xn,y1,,yn M letting zi = xi + yi for each such i. Furthermore,
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) = 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 = ifσ = for each x1,,xn M. For repetitious use later, we point out that φτσ = φτ φσ for all τ,σ Sn according to
 (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

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

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

#### (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 = {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

where δi HomR-mod(M,R) is the linear operator which takes vi to 1 and all other elements of 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.

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

and

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

### 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}

and therefore

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

### References

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