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, a typical textbook exercise is to consider the polynomial

P(x,y)=x^2+3xy+3y^2-6x+3y-6.

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 let’s leave that for now.)

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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