Archimedean Representation Theorem for modules over a commutative ring

Abstract

Pólya’s Positivstellensatz and Handelman’s Positivstellensatz are known to be concrete instances of the abstract Archimedean Representation Theorem for (commutative unital) rings. We generalise the Archimedean Representation Theorem to modules over rings. For example, consider the module of all symmetric matrices with entries in a polynomial ring, also known as matrix polynomials. Pólya’s Positivstellensatz and Handelman’s Positivstellensatz had been generalised by Scherer and Hol, and Lê  and Du’ respectively to matrix polynomials, using the method of effective estimates from analysis. We show that these two Positivstellensätze for matrix polynomials are concrete instances of our Archimedean Representation Theorem in the case of the module of symmetric matrix polynomials over the polynomial ring.

Appendix: Proof of Proposition 4

There is nothing essentially original due to the author below—the following proof of Proposition 4 is that of Burgdorf et al. (2012, p. 123), and reproduced verbatim below only for the convenience of the reader.

The author’s only contribution is the observation that G being contained in A (and hence an ideal) is never used in the proof. As mentioned in loc. cit., precedents of Proposition 4 can be found in the work of Bonsall, Lindenstrauss and Phelps (Bonsall et al. (1966), Theorem 10), Krivine (1964b, Theorem 15) and Handelman (1985, Proposition 1.2).

Let A, S, G, M, u be as in Proposition 4. Given a map $\Phi : G \rightarrow {\mathbb {R}}$, we associate to each $f \in A$ satisfying $\Phi (f \cdot u) \ne 0$ a map $\Phi _f: G \rightarrow {\mathbb {R}}$ given by

$$\begin{aligned} \Phi _f(s) := \frac{\Phi (f \cdot s)}{\Phi (f \cdot u)}{} & {} (\forall s\in G). \end{aligned}$$

(10)

The reader can verify that if $\Phi$ is a state of (G, M, u) and $p \in S$ satisfies $\Phi (p \cdot u) > 0$, then $\Phi _p$ is also a state of (G, M, u). Furthermore, if $\Phi$ is a state and $p_1, p_2 \in S$ satisfy $\Phi (p_1 \cdot u), \Phi (p_2 \cdot u) > 0$, so that $p_1 + p_2 \in S$ and $\Phi ((p_1 + p_2) \cdot u )> 0$, then $\Phi _{p_1 + p_2}$ is a proper convex combination of the states $\Phi _{p_1}$ and $\Phi _{p_2}$:

$$\begin{aligned} \Phi (p_1 \cdot u) \Phi _{p_1} + \Phi (p_2 \cdot u) \Phi _{p_2} = \Phi ((p_1 + p_2) \cdot u) \Phi _{p_1 + p_2}. \end{aligned}$$

(11)

Proof of Proposition 4

Since $S \subseteq A$ is archimedean and u is an order unit of (G, M), it suffices to show that (1) holds whenever $f \in S$ and $s \in M$.

Let $f \in S$ and $s \in M$ be given. Then $f \cdot u\in S \cdot M \subseteq M$. Hence $\Phi (f \cdot u) \ge 0$. There are two cases: either $\Phi (f \cdot u) = 0$ or $\Phi (f \cdot u) > 0$.

Case 1: $\Phi (f \cdot u) = 0$. Then u being an order unit of (G, M) gives a positive integer $n$ such that $0 \le _M s \le _M nu$. Since M is closed under the S-action, $0 \le _M f \cdot s \le _M nf \cdot u$. Now $\Phi$ is monotone with respect to $\le _M$, so

$$\begin{aligned} 0 \le \Phi (f \cdot s) \le n\Phi (f \cdot u) = 0, \end{aligned}$$

(12)

forcing $\Phi (f \cdot s) = 0$ so that both sides of (1) equals to zero in this case.

Case 2: $\Phi (f \cdot u) > 0$. Since $S \subseteq A$ is archimedean, there is some positive integer $n$ such that $n- f \in S$. But increasing $n$ if necessary, we may suppose that $n> \Phi (f \cdot u)$. Then

$$\begin{aligned} \Phi ((n- f) \cdot u) = n\Phi (u) - \Phi (f \cdot u) = n- \Phi (f \cdot u) > 0. \end{aligned}$$

(13)

Since $f, n- f\in S$ and $\Phi (f \cdot u), \Phi ((n- f) \cdot u ) >0$, we may apply (11) to conclude that $\Phi _n$ is a proper convex combination of $\Phi _f$ and $\Phi _{n- f}$. But $\Phi _n= \Phi$ (by direct calculation, using the fact that $n$ is a scalar), so the purity of $\Phi$ implies that $\Phi _f = \Phi$, which is just (1). $\square$