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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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The answer to your 1st question is no: in general that is impossible. The answer to your 2nd question is yes: take $f$ to be a real number... You probably want to be more stringent about what you want :-) – Mariano Suárez-Alvarez Aug 15 '12 at 17:05
Where did you get the idea to call this "conjugation"? – rschwieb Aug 15 '12 at 17:09
@rschwieb: You could call it whatever you like, it just looks like a sort of conjugation roughly speaking. – James S. Cook Aug 15 '12 at 18:03
@Mariano Suarez-Alvarez: I realize it is generally impossible, I'd like to find if it is possible for anything nontrivial besides the examples I gave. The case of a constant real polynomial is not what I'm looking for, as you correctly suspect :) – James S. Cook Aug 15 '12 at 18:08
My point is that you should try to be very explicit about what you know and what you want to know. For example, it is clear that what you write as your first question is not really your question! – Mariano Suárez-Alvarez Aug 15 '12 at 18:25

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