Symmetric Polynomials in Multiple Variables

20 Apr

In the previous blog post on the quadratic formula, I discussed how the coefficients of a polynomial in one variable are symmetric expressions in the roots, and that solving for the roots in terms of the coefficients requires that one break the symmetry. A question that arises is: what kind of symmetric expressions involving the roots can you form with the coefficients of a polynomial? After all, you need to know which symmetric expressions involving the roots you’re allowed to form and try to render asymmetric.

Consider the case of a cubic polynomial

\large P(x) = x^3 + bx^2 + cx + d

Letting the roots be ‘r’, ‘s’ and ‘t’, we can write the polynomial as

\large (x - r)(x - s)(x - t)

Multiplying out, we get

\large x^3 - (r + s + t)x^2 + (rs + rt + st)x - rst

 So that

\large -b = r + s + t

 \large c = rs + st + rt

 \large -d = rst

 The theorem is that if

\large Q(r, s, t)

is invariant under permutations of the roots, then it can be written as a polynomial in the coefficients ‘b’, ‘c’ and ‘d’ of P(x). For example, consider the case

\large Q(r, s, t) = r^3 + s^3 + t^3

We can express this in terms of ‘b’, ‘c’ and ‘d’ as follows. In order to involve the cubes of ‘r’, ‘s’ and ‘t’, we have to cube -b:

\large (-b)^3 = r^3 + s^3 + t^3 + \cdots

Here the ellipses denote the sum of

\large 3 \left[r^2 s + r^2 t + s^2 r + s^2 t + t^2 r + t^2 s \right]

and

\large 6rst

The last of these is nothing but -6d. In order to involve the second to last expression, we need to take the product of ‘3’, ‘-b’, and ‘c’. Doing so gives the second to last expression, in addition to

\large 3rst

which is nothing but -3d. Putting this all together, we get

 \large Q(r, s, t) =-b^3 -3bc - 9d

Still more generally, if P(x) is a degree n polynomial with leading coefficient 1, then any symmetric polynomial in the roots of P(x)  can be written as a polynomial in the coefficients of P(x). This theorem was essentially known to Isaac Newton, who seems to have been the first to discover it. The proof is left to the reader (-:

Advertisements

2 Responses to “Symmetric Polynomials in Multiple Variables”

Trackbacks/Pingbacks

  1. The Fundamental Theorem of Galois Theory | Math is Beauty - April 21, 2013

    […] and we discussed how to break this symmetry in the special case of a quadratic polynomial. In a subsequent post, we characterized the polynomial expressions of the coefficients as precisely the polynomial […]

  2. Moving Up the Ladder of Subfields of a Splitting Field | Math is Beauty - April 25, 2013

    […] the elements of H, so the coefficients of P(x) are symmetric with respect to H, so by the Fundamental Theorem of Galois Theory the coefficients are in […]

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: