Let's suppose that $A$ is a finite-dimensional real algebra which also has the structure of an ordered ring. This means that for all $a, b, c$, we have

$$ a\leq b \implies a+c \leq b+c\\ (a \geq 0) \wedge (b \geq 0) \implies ab \geq 0 $$

Is it possible to have such an ordered ring with infinitely large elements, e.g. some element $\omega$ which is greater than all $r \in \Bbb R$?

Clearly this is possible for infinite-dimensional real algebras, such as the polynomial ring $\Bbb R[\omega]$, but what if it's finite?

I'm mostly interested in commutative rings but also interested in non-commutative ones.


1 Answer 1


No, infinite elements of an ordered $\mathbb{R}$-algebra must be transcendental (edit: over $\mathbb{R}$). If $\omega$ is algebraic over $\mathbb{R}$ then using that by definition $\omega > c$ for all $c \in \mathbb{R}$ we deduce that, for any $n$, we have $\frac{\omega^n}{n} > c_i \omega^i$ for arbitrary $c_i \in \mathbb{R}$ and $0 \le i \le n-1$, and adding these up gives

$$\omega^n > \sum_{i=0}^{n-1} c_i \omega^i.$$

Since this is true for arbitrary $n$ and $c_i$ it follows that $\omega$ is not the root of any real polynomial, and more generally is not the root of any polynomial with finite coefficients.

  • $\begingroup$ Thanks, but I don't see -- why does its transcendentalness require that no such $\omega$ can exist in a finite-dimensional real algebra? There are plenty of transcendental elements in finite dimensional real algebras... $\endgroup$ Oct 10, 2022 at 5:29
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    $\begingroup$ @Mike: every element of a finite-dimensional real algebra $A$ is algebraic over $\mathbb{R}$; if $\dim A = n$ then for any $a \in A$ the elements $\{ 1, a, \dots a^n \}$ must be linearly dependent, so $a$ satisfies a polynomial of degree $\le n$. $\endgroup$ Oct 10, 2022 at 6:21
  • $\begingroup$ Ah, algebraic over $\Bbb R$, ok. $\endgroup$ Oct 10, 2022 at 17:10

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