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Suppose $\mathcal{A}$ is an associative algebra over $\mathbb{R}$. Furthermore, let $f(x_1, \dots , x_n) \in \mathcal{A}[x_1, \dots , x_n]$.

Preliminary Question: Is it possible to find $g(x_1, \dots , x_n) \in \mathcal{A}[x_1, \dots , x_n]$ such that $f(x_1, \dots , x_n)g(x_1, \dots , x_n)\in \mathbb{R}[x_1, \dots , x_n]$?

Let me show a few examples where nice answers are known:

  • $\mathcal{A} = \mathbb{C}$ take $f(x,y) = x+iy$. Take $g(x,y)=x-iy$ then $f(x,y)g(x,y) = x^2+y^2 \in \mathbb{R}[x,y]$.

  • $\mathcal{A} = \mathbb{R}\oplus j \mathbb{R}$ with $j^2=1$. If $f(x,y) = x+jy$ then set $g(x,y)=x-jy$ then $f(x,y)g(x,y) = x^2-y^2 \in \mathbb{R}[x,y]$.

  • $\mathcal{A} = \mathbb{R}\oplus \eta \mathbb{R}$ with $\eta^2=0$. If $f(x,y) = x+\eta y$ then set $g(x,y)=x-\eta y$ then $f(x,y)g(x,y) = x^2 \in \mathbb{R}[x,y]$.

I'm interested in examples where $dim_{\mathbb{R}}(\mathcal{A}) \geq 3$. Certainly for some polynomials $f$ it is not possible to find a $g$ such that $fg$ has real coefficients. Consider $\mathcal{A} = \mathbb{R}\oplus j\mathbb{R} \oplus j^2\mathbb{R}$ where $j^3=1$. If $f(x,y,z) = x-\frac{1}{2}j z-\frac{1}{2}j^2y$ then there does not exist $A,B,C \in \mathcal{A}$ such that $g(x,y,z)=Ax+By+Cz$ gives $f(x,y,z)g(x,y,z) \in \mathbb{R}[x,y,z]$. Here's why: $ \begin{align}\biggl(x-\frac{j}{2} z-\frac{j^2}{2}y\biggr)(Ax+By+Cz) &= Ax^2+\bigg(B-A\frac{j^2}{2}\bigg)xy+\bigg(C-A\frac{j}{2}\bigg)xz \\ & \qquad+ \bigg(-C\frac{j^2}{2}-B\frac{j}{2}\bigg)yz-B\frac{j^2}{2}y^2-C\frac{j}{2}z^2 \end{align}$ It is not possible to simulaneously choose coefficients of $x^2,xy,xz,yz,y^2,z^2$ from $\mathbb{R}$. I begin to think the details of $f$ are not terribly relevant. Thus, I give the freedom to choose both $f$ and $g$ in the final form of my question:

Question: Given a particular associative algebra $\mathcal{A}$ with dimension $n \geq 3$ over $\mathbb{R}$ do there exist $f(x_1, \dots, x_n) \in \mathcal{A}[x_1,\dots, x_n]$ such that a conjugate $g(x_1, \dots , x_n) \in \mathcal{A}[x_1, \dots , x_n]$ exists? In other words, do there exist nonconstant pairs $f(x_1, \dots , x_n),g(x_1, \dots , x_n)\in \mathcal{A}[x_1, \dots , x_n]$ such that $f(x_1, \dots , x_n)g(x_1, \dots , x_n) \in \mathbb{R}[x_1, \dots , x_n]$?

Thanks in advance for your insights.

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    For what it's worth, in many formulations of supernumbers the conjugate reverses the order of the product; $(vw)^* = w^*v^*$. Ah, you would allow that. Inverse? Well, I'm not so sure it's going to be unique... let's just call it $g$.2012-08-16

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