Exercises on symmetric polynomials

1. Prove that the ring 𝕂[x₁, x₂, ..., x_n] with 𝕂 a field, is an integral domain.

2. Express (if possible) the following polynomials in terms of symmetric elementary functions.

   1. x₁²x₂ + x₂²x₃ + x₃²x₁

   2. x₁³ + x₂² + x₂³ + x₃² + x₃³ + x₁²

   3. x₁² + x₂² + x₃²

   4. x₁²x₂² + x₁²x₃² + x₂²x₃²

3. Let f(x) the polynomial x³ + 3x² − 6x + 3. Given the roots α₁, α₂, α₃, determine the monic polynomial with roots

   1. 1/α₁², 1/α₂², 1/α₃²

   2. and the monic polynomial with roots α₁², α₂², α₃².

4. Prove that the set of symmetric polynomials in a field 𝕂 forms a subring of the polynomial ring 𝕂[x₁,..., x_n].

5. If possible express the following function in terms of elementary symmetric functions

   x/y + y/z + z/x

6. Prove that the subset S̃ of 𝕂(x₁,..., x_n) of rational symmetric functions is a field.

7. Prove that if f/g is symmetric, then any rational function equivalent to it is symmetric as well.

8. Prove in detail that 𝕂(x₁,..., x_n) is a field with respect to the operations previously defined, that is the field of quotients of 𝕂[x₁,..., x_n].

Solutions

1. We can represent an element of 𝕂[x₁, x₂, ..., x_n] as

   f(x₁, x₂, ..., x_n) = ∑ a_i1i2...in x₁^i₁ x₂^i₂ ... x_n^i_n

   with deg f = m. The zero element is the polynomial with all c_k = 0 and the multiplicative identity 1, has c₀ = 1, c_k= 0 for k > 0. Every nonzero polynomial has a degree.

   If we take a second polynomial g = ∑ b_i1i2...in x₁^i₁ x₂^i₂ ... x_n^i_n, deg g = n and consider the product f·g, then

   deg(f·g) = deg(f) + deg(g),   for f,g ≠ 0 in 𝕂[x₁, x₂, ..., x_n]

   follows because the leading nonzero term of f·g is a_m b_n x^m+n, with a_m,b_n ≠ 0. Obviously 𝕂[x₁, x₂, ..., x_n] has no zero divisors and is an integral domain.

2. 1. It is not symmetric (the combination of elementary symmetric function is a symmetric function).

   2. It's a symmetric polynomial with highest term x₁³, i.e. 300. φ₁ = σ₁³= (x₁ + x₂ + x₃)³, f₁ = f − φ₁.

      φ₁ = (x₁ + x₂ + x₃)³ = x₁³ + x₂³ + x₃³ + 3(x₁²x₂ + x₂²x₃ + x₃²x₁ + x₁x₂² + x₂x₃² + x₃x₁²) + 6x₁x₂x₃

      f₁ = f − φ₁ = −3(x₁²x₂ + x₂²x₃ + x₃²x₁ + x₁x₂² + x₂x₃² + x₃x₁²) − 6x₁x₂x₃ = −3∑_{i ≠ j} x_i²x_j − 6x₁x₂x₃

      whose highest term is −3x₁²x₂ hence 210.

      φ₂ = −3σ₁σ₂ = −3(x₁ + x₂ + x₃)(x₁ x₂ + x₁x₃ + x₂x₃) = −3∑_{i ≠ j} x_i²x_j − 9x₁ x₂ x₃.

      f₂ = −3∑_{i ≠ j} x_i²x_j − 6x₁x₂x₃ + 3∑_{i ≠ j} x_i²x_j + 9x₁ x₂ x₃ = 3x₁ x₂ x₃.

      φ₃ = 3σ₁¹⁻⁰ σ₂¹⁻¹ σ₃¹⁻⁰ = 3σ₃ = + 3x₁ x₂ x₃.

      f₃ = 3x₁ x₂ x₃ − 3x₁x₂x₃ = 0

      f = φ₁ + φ₂ + φ₃ = σ₁³ − 3σ₁σ₂ + 3σ₃.

   3. The leading term is x₁² hence 200.

      φ₁ = σ₁² = (x₁ + x₂ + x₃)² = x₁² + x₂² + x₃² + 2(x₁x₂ + x₂x₃ + x₁x₃)

      f₁ = f − σ₁² = −2(x₁x₂ + x₂x₃ + x₁x₃)

      with leading term −2x₁x₂ hence 110.

      φ₂ = −2σ₁^1-1 σ₂^1-0 = −2σ₂ = −2(x₁x₂ + x₂x₃ + x₁x₃)

      And f₃ = f₂ −φ₂ = 0

      Thus f = φ₁ + φ₂ = σ₁² − 2σ₂

   4. It's not a symmetric polynomial.

3. Using Viete's relations

   1. a₁ = 3 = −(α₁ + α₂ + α₃)

      a₂ = −6 = (α₁ α₂ + α₁α₃ + α₂α₃)

      a₃ = 3 = −α₁ α₂ α₃

      (x − 1/α₁²)(x − 1/α₂²)(x − 1/α₃²)

      expanding this polynomial we have

      x³ − x² (α₁² α₂² + α₁² α₃² + α₂² α₃²) /(α₁ + α₂ + α₃)² − 1/(α₁α₂ α₃)² + x[(α₁² α₂ α₃)² + (α₁ α₂² α₃)² + (α₁ α₂ α₃²)²] / [(α₁α₂)² (α₂α₃)² (α₁α₃)²]

      using Viète's relations above we notice that the constant term is −1/9. The coefficient of quadratic term has 9 as denominator; To obtain the numerator we notice it is a symmetric polynomial, call it g

      g = α₁² α₂² + α₁² α₃² + α₂² α₃²

      we express it in terms of elementary symmetric functions. The laeding term of g₁ is α₁² α₂²; we have

      φ₁ = σ₂² and g₁ = g − φ₁ = −2(α₁ α₂α₂α₃ + α₁ α₃ α₁ α₂ + α₁α₂α₂ α₃)

      The leading term of g₁ is −2α₁ α₂α₂α₃ which yields to

      φ₂ = −2 σ₁^2-1 σ₂^1-1σ₃ = −2σ₁σ₃

      Since g₂ = g₁ − φ₂ = 0, we can express g = σ₂² −2σ₁σ₃ = 36 − 18 = 18. By dividing by the denominator 9 we have the coefficient 18/9 = 2

      For the coefficient of the x term we notice that the denominator is (α₁ α₂α₃)⁴ = 81. The numerator can be simplified using the 3rd Viète's relation as k = 9 (α₁² + α₂² + α₃²) which is also a symmetric polynomias that can be expressed in terms of elementary symmetric polynomials; the leading term is 9α₁².

      φ₁ = 9σ₁² = 9[α₁² + α₂² + α₃² + 2 (α₁α₂ + α₁α₃ + α₂α₃)].

      k₁ = k − φ₁ = −18 (α₁α₂ + α₁α₃ + α₂α₃)

      The leading term of k₁ is −18α₁α₂, so

      φ₂ = −18 σ₂ = −18 (α₁α₂ + α₁α₃ + α₂α₃)

      k₃ = 0, so k = φ₁ + φ₂ = σ₁² −18 σ₂ = 9*(-3)² −18*(-6) = 189.

      So the coefficient of the x term is 189 / 9 = 7/3.

      x³ −2x² + 7/3x −1/9

   2. We express the polynomial as

      (x −α₁²)(x − α₂²)(x − α₃²)

      expliciting it we have

      x³ −x²(α₁² + α₂² + α₃²) + x (α₁²α₂² + α₃²α₁² + α₃²α₂²) + α₁² α₂²α₃²

   Using Viète's relations above we have for the constant term 9.

   The quadratic coefficient can be calculated using the method above used for k. Since we've already calculated k and they differ only for a 9 factor we just have to devide the result by 9: 189/9 = 21.

   The coefficient of the x term is the same as the g polynomial calculated above which is 18. So the monic polynomial is

   x³ −21x² + 18x −9

4. Since the property to be symmetric is preserved under addition and multiplication of polynomials, the symmetric polynomials form a subring of 𝕂[x₁,..., x_n]. Moreover 1, 0, -1 are symmetric polynomials and the axiomatic definitions of a ring are thus verified.

5. This function is not symmetric.