Symmetric Polynomials Involving Discriminant Poly

[mathjam0990]

Question: Let τ = (i, j) ∈ Sn with 1 ≤ i < j ≤ n. Prove: δ(r_τ(1) , . . . ,r_τ(n) ) = −δ(r₁, . . . ,r_n)

Note: Discriminant Polynomial δ(r₁,r₂,...,r_n) = ∏ (r_i - r_j) for i<j

I am pretty confused on where to begin. Based on the note, does −δ(r₁, . . . ,r_n) then = (r₁-r₂)(r₁-r₂)·····(r₁-r_n)(r₂-r₃)(r₂-r₄)····(r₂-r_n)····(r_n-1-r_n) ?

Also, since τ = (i, j) is a transposition, does that suggest τ(1) = j (because i "goes to" j), then τ(2) = i (because i "goes to" j and j "goes to" i) ? Sorry I feel like I've got this all mixed up. Could anyone provide detailed insight please?

Thanks in advance!

 

[Deveno]

Question: Let τ = (i, j) ∈ Sn with 1 ≤ i < j ≤ n. Prove: δ(r_τ(1) , . . . ,r_τ(n) ) = −δ(r₁, . . . ,r_n)

Note: Discriminant Polynomial δ(r₁,r₂,...,r_n) = ∏ (r_i - r_j) for i<j

I am pretty confused on where to begin. Based on the note, does −δ(r₁, . . . ,r_n) then = (r₁-r₂)(r₁-r₂)·····(r₁-r_n)(r₂-r₃)(r₂-r₄)····(r₂-r_n)····(r_n-1-r_n) ?

Also, since τ = (i, j) is a transposition, does that suggest τ(1) = j (because i "goes to" j), then τ(2) = i (because i "goes to" j and j "goes to" i) ? Sorry I feel like I've got this all mixed up. Could anyone provide detailed insight please?

Thanks in advance!

Let's just look at an example: take $n = 3$.

Then $\delta(x_1,x_2,x_3) = (x_1-x_2)(x_1-x_3)(x_2-x_3)$.

Suppose we examine the result of $\sigma = (1\ 2)$ applied to $\delta$, that is, we look at:

$(x_{\sigma(1)} - x_{\sigma(2)})(x_{\sigma(1)} - x_{\sigma(3)})(x_{\sigma(2)} - x_{\sigma(3)})$

We have:

$\sigma(1) = 2$
$\sigma(2) = 1$
$\sigma(3) = 3$, so we get:

$(x_2 - x_1)(x_2 - x_3)(x_1 - x_3) = (x_2 - x_1)(x_1 - x_3)(x_2 - x_3) = -\delta(x_1,x_2,x_3)$.

Note the only factor that changed sign in this example is $(x_1 - x_2)$. In general (for higher $n$) what will happen is that $(x_i - x_j)$ will change sign, factors that involve neither $x_i$ nor $x_j$ will remain the same, factors where $i,j < k$ and $k < i,j$ will keep the same sign, so the only factors that will change sign (besides ($x_i - x_j)$) are when $i < k < j$ where the two factors $(x_i - x_k),(x_k - x_j)$ will change sign together.

 

[mathjam0990]

$(x_2 - x_1)(x_2 - x_3)(x_1 - x_3) = (x_2 - x_1)(x_1 - x_3)(x_2 - x_3) = -\delta(x_1,x_2,x_3)$.

Note the only factor that changed sign in this example is $(x_1 - x_2)$. In general (for higher $n$) what will happen is that $(x_i - x_j)$ will change sign, factors that involve neither $x_i$ nor $x_j$ will remain the same, factors where $i,j < k$ and $k < i,j$ will keep the same sign, so the only factors that will change sign (besides ($x_i - x_j)$) are when $i < k < j$ where the two factors $(x_i - x_k),(x_k - x_j)$ will change sign together.

This is the part I wasn't seeing! Thanks a lot for breaking that down!