305 -Fields (5)

February 27, 2009

At the end of last lecture we stated a theorem giving an easy characterization of subfields of a given field {\mathbb F}. We begin by proving this result.

Theorem 18. Suppose {\mathbb F} is a field and S\subseteq{\mathbb F}. If S satisfies the following 5 conditions, then S s a subfield of {\mathbb F}:

  1. S is closed under addition.
  2. S is closed under multiplication.
  3. -a\in S whenever a\in S.
  4. a^{-1}\in S whenever a\in S and a\ne0.
  5. S has at least two elements.

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275 -Positive polynomials

November 11, 2008

When studying local extreme points of functions of several (real) variables, the textbook asks in one of the exercises to consider the polynomial


Here we have P_x=2x+3y-6 and P_y=3x+6y+3, so the only critical point of P is (15,-8). Since P_{xx}=2 and the Hessian of P is 2\times 6-3^2=3>0, it follows that (15,-8) is a local minimum of P and, since it is the only critical point, it is in fact an absolute minimum with P(15,-8)=-63.

P being a polynomial, it is reasonable to expect that there is an algebraic explanation as for why -63 is its minimum, and why it lies at (15,-8). After all, this is what happens in one variable: If p(x)=ax^2+bx+c and a\ne0, then

\displaystyle p(x)=a\left(x+\frac b{2a}\right)^2+\frac{4ac-b^2}{4a},

and obviously p has a minimum at x=-b/2a, and this minimum is (4ac-b^2)/4a. 

The polynomial P of the example above can be analyzed this way as well. A bit of algebra shows that we can write

\displaystyle P(x,y)=\left(x-3+\frac32 y\right)^2+3\left(\frac y2+4\right)^2-63,

and it follows immediately that P(x,y) has a minimum value of -63, achieved precisely when both x-3+3y/2=0 and 4+y/2=0, i.e, at (15,-8).

(One can go further, and explain how to go in a systematic way about the `bit of algebra’ that led to the representation of P as above, but I will leave that for a future occasion.)

What we did with P is not a mere coincidence.  Hilbert’s 17th of the 23 problems of his famous address to the Second International Congress of Mathematicians in Paris, 1900, asks whether every polynomial P(x_1,\dots,x_n) with real coefficients which is non-negative for all  (real) choices of x_1,\dots,x_n is actually a sum of squares of rational functions. (A rational function is a quotient of polynomials.) A nonnegative polynomial is usually called positive definite, but I won’t use this notation here.

If Hilbert’s problem had an affirmative solution, this would provide a clear explanation as for why P is non-negative.

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