I am working through a Calculus book and I found an exercise, which I am not able to solve:

Let $K$ be a field of rational functions over $\mathbb{R}$ and let $a,b$ be arbitrary in $\mathbb{Q}$ but not all $b$ are $0$. And let $m,n$ be arbitrary in $\mathbb{N}$. We define $$P:=\left.\left\{\frac{\sum_{i=0}^n a_ix^i}{\sum_{i=0}^m b_ix^i}\;\right|\;a_nb_m\lt 0\right\}.$$

Is $P$ a prepositive cone?

Does anybody have a hint. My idea was to express $-1$ as a sum of squares but that didn't really work for me.

  • 2
    $\begingroup$ I am really curious as to what "calculus book" this is taken from. $\endgroup$ – Pete L. Clark May 10 '11 at 20:04
  • $\begingroup$ It's a german calculus book "Analysis Band 1, Erhardt Behrends", first chapter :S $\endgroup$ – monoid May 10 '11 at 22:20

The inequality $a_nb_m<0$ goes the wrong way. Every square in the field should be in the cone, since we are trying to produce an order on the field compatible with the field structure. But it is obvious that for example $1$, also $x^2$, are not expressible as a quotient of the type you mention.

Are you sure that the inequality is not $a_nb_m>0$? That will in fact produce a field ordering.

What's Hilbert's 17th problem doing in a calculus book?

  • $\begingroup$ Yes I am sure, that the inequality goes the right way. For example, how do you show, that $1$ is not expressible in $P$? $\endgroup$ – monoid May 10 '11 at 22:22
  • $\begingroup$ @monoid: Suppose to the contrary that there exist polynomials $A(x)$, $B(x)$ such that their lead coefficients are of opposite signs (that's your $a_nb_m<0$) and such that $A(x)/B(x)$ (as a polynomial) is the same as the polynomial $1$. Then $A(x)-B(x)$ is the $0$ polynomial. This cannot be the case if the lead coefficients are of opposite sign. $\endgroup$ – André Nicolas May 10 '11 at 22:34
  • $\begingroup$ I'm not an expert on prepositive cones, but isn't $-1$ in $P$? $a_0 = -1$, $b_0 = 1$, and all other $a, b = 0$. $a_0b_0 < 0$. $\endgroup$ – Adam Saltz May 10 '11 at 22:56
  • $\begingroup$ But if all other $a$ and $b$ are equal $0$, then $a_nb_m = 0$ but it should be $a_nb_m < 0$. And I tried to express $-1$ as a sum of squares. For example $x^2 = -1$. If that's possible then there is no order on that specific field and therefore no prepositive cone exists. $\endgroup$ – monoid May 10 '11 at 23:42
  • $\begingroup$ @user6312 : So your approach is $\frac{\sum_{i=0}^n a_ix^i}{\sum_{i=0}^m b_ix^i} = 1 \Leftrightarrow \sum_{i=0}^n a_ix^i = \sum_{i=0}^m b_ix^i \Leftrightarrow \sum_{i=0}^n a_ix^i - \sum_{i=0}^m b_ix^i = 0$ $\endgroup$ – monoid May 10 '11 at 23:55

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