# Can any finite-dimensional real algebra be an ordered ring with infinitely large elements?

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.

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.
• 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... Oct 10, 2022 at 5:29
• @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 $\le n$. Oct 10, 2022 at 6:21
• Ah, algebraic over $\Bbb R$, ok. Oct 10, 2022 at 17:10