## 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 $Ssubseteq{mathbb F}.$ 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. $-ain S$ whenever $ain S.$
4. $a^{-1}in S$ whenever $ain S$ and $ane0.$
5. $S$ has at least two elements.

## 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.