February 15, 2014

http://coursework.tylerlogic.com/courses/upenn/math503/homework03
Define D(F) ∈ ℤ[x

| (1.1) |

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

Because any σ ∈ S_{n} is bijective and because (x_{i} - x_{j})^{2} = ((-1)(x_{j} - x_{i}))^{2} = (-1)^{2}(x_{j} - x_{i})^{2} = (x_{j} - x_{i})^{2},
then

In other words, D(F) is symmetric. Thus there indeed exists a polynomial d ∈ ℤ[x_{1},…,x_{n}] such that d(s_{1},…,s_{n}) = D(F)
since every symmetric polynomial in ℤ[x_{1},…,x_{n}] is contained in ℤ[s_{1},…,s_{n}].

Uniqueness
There cannot exist two distinct d_{1},d_{2} ∈ ℤ[x_{1},…,x_{n}] such that d_{1}(s_{1},…,s_{n}) = d_{2}(s_{1},…,s_{n}) = D(F) because their existence
would contradict the algebraic independence of s_{1},…,s_{n} [Lan02, p. 192] since it would imply d = d_{1} - d_{2} has
d(s_{1},…,s_{n}) = 0.

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(z_{1},z_{2}) ∈ ℤ[z_{1},z_{2}] such that d(s_{1},s_{2}) = (x_{1} - x_{2})^{2}. Because of the existence/uniqueness proven above, we
can find such a polynomial and there will be only one. So because s_{1} = x_{1} + x_{2}, s_{2} = x_{1}x_{2} for n = 2 and
because

then we can deduce that d(z_{1},z_{2}) = z_{1}^{2} - 4z_{2}. Hence for any f(t) = t^{2} + 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(z_{1},z_{2},z_{3}) ∈ ℤ[z_{1},z_{2},z_{3}] such that d(s_{1},s_{2},s_{3}) = (x_{1} - x_{2})^{2}(x_{1} - x_{3})^{2}(x_{2} - x_{3})^{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(s_{1},s_{2},s_{3}) will be a homogeneous polynomial of degree six in x_{1},x_{2},x_{3}. Therefore, since
s_{1} = x_{1} + x_{2} + x_{3}, s_{2} = x_{1}x_{2} + x_{1}x_{3} + x_{2}x_{3}, and s_{3} = x_{1}x_{2}x_{3},

for some integers α_{1},α_{2},α_{3},α_{4},α_{5},α_{6},α_{7}. We will find these values by analyzing the following expansion

| (1.2) |

From this equation, it is immediately apparent that α_{6} = α_{7} = 0 since s_{2}s_{1}^{4} and s_{1}^{6} would produce monomials containing
x_{1} raised to a degree higher than 4, but there is no such monomial in the polynomial of equation 1.2. Now, given the
expansions

In the same vein, we evaluate the coefficients for the monomials x_{1}^{3}x_{2}^{2}x_{3} to get

x_{1}^{4}x_{2}x_{3} to get

x_{1}^{3}x_{3}^{3} to get

and finally x_{1}^{4}x_{3}^{2} to get

Solving the above five equations we obtain α_{1} = -27, α_{2} = 18, α_{3} = -4, α_{4} = -4, and α_{5} = 1, which in turn informs us
that

Hence for any f(t) = t^{3} + at^{2} + bt + c

Since p

It’s clear that in order to obtain the appropriate values of p_{2}, p_{3}, and p_{4}, a_{1} = b_{1} = c_{1} = 1 since none of the other clauses
in the above equations will be able to account for the x_{i}^{2}, x_{i}^{3}, and x_{i}^{4} in each of p_{1}, p_{2}, and p_{3}. Hence, in determining the
formula for p_{i}, our approach will be to start with s_{1}^{i} ordered lexicographically and find the coefficient of the leftmost clause
that isn’t in p_{i} 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 |

s_{1}^{2} | x_{1}^{2} |

s_{2} | x_{1}x_{2} |

s_{1}^{3} | x_{1}^{3} |

s_{1}s_{2} | x_{1}^{2}x_{2} |

s_{3} | x_{1}x_{2}x_{3} |

s_{1}^{4} | x_{1}^{4} |

s_{1}^{2}s_{2} | x_{1}^{3}x_{2} |

s_{2}^{2} | x_{1}^{2}x_{2}^{2} |

s_{4} | x_{1}x_{2}x_{3}x_{4} |

where these terms have been generated for n = 4.

So for the case of p_{2} we start with

in which the first term x_{1}^{2} should remain since it is in p_{2}, but we need to get rid of the second term 2x_{1}x_{2}. Through use of
Table 2.1, this implies that a_{2} = -2 so that we are left with s_{1}^{2} - 2s_{2} = x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2}, and thus the formula for
p_{2}.

For the case of p_{3} we start with s_{1}^{3} = x_{1}^{4} + 3x_{1}^{2}x_{2} + , 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 b_{2} = -3 so that we are left with
s_{1}^{3} - 3s_{2}s_{1} = x_{1}^{3} + x_{2}^{3} - 3x_{1}x_{2}x_{3} + , which finally informs us that b_{3} = 3, leaving us with p_{3}’s formula:
s_{1}^{3} - 3s_{2}s_{1} + 3s_{2}.

Finally, by following the same procedure regarding p_{4}, we have

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

For each σ ∈ S

Set φ

| (3.3) |

for each x = ∑
_{i}x_{i} = ∑
_{i}x_{i1} ⊗⊗ x_{in} ∈⊗_{R}^{n}M. 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 ∈⊗_{R}^{n}M, implying not only that this action is a left action, but that it is not are right action.

Since both R[y

Let = {v_{1},…,v_{n}} be a set of free generators on M and Y = {y_{1},…,y_{n}}. We will use ∙ to denote the tensor operation in
S^{k}(M). Define α : M^{k} → R^{k}[y_{1},…,y_{n}] and β : Y → S^{k}(M) by

- The existence of an R-module homomorphism α : S
^{k}(M) → R^{k}[y_{1},…,y_{n}] through which α factors. This comes by way of the universal property of symmetric multilinear maps. - The existence of an R-algebra homomorphism β : R
^{k}[y_{1},…,y_{n}] → S^{k}(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} ∈ S^{k}(M) by x^{∙j} so that x^{∙j} = x^{j} ∙_{j-1 times} since each element of S^{k}(M)
is symmetric. Thus for the generators of S^{k}(M) and R^{k}[y_{1},,y_{n}]

and

which implies that α and β are inverses of each other. Hence S^{k}(M) is isomorphic to R^{k}[y_{1},,y_{n}].

Denote the symmetric and rotational generators of D

and therefore

Hence we can compute the character of ρ_{k} of any s^{i}r^{j} ∈ D_{2n}