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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 :-)2012-08-15
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    Where did you get the idea to call this "conjugation"?2012-08-15
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    @rschwieb: You could call it whatever you like, it just looks like a sort of conjugation roughly speaking.2012-08-15
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    @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 :)2012-08-15
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    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!2012-08-15
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    I know more about $f$ for the particular problem where this factorization question was born. In that context the question has the structure that $f$ is given, but $g$ is not known. However, there is some ambiguity in the choice of $f$ in my construction. I'm trying to sort out how to choose $f$ such that $g$ exists (using the notation in the question). In any event, perhaps the question is clear now?2012-08-15
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    An important property of complex conjugation is that $\overline{a} \overline{b} = \overline{ab}$. Do you require this property as well (or something similar, like $\overline{a} \overline{b} = \overline{ba}$)? Otherwise, I wouldn't quite call it conjugation, but rather some kind of inverse.2012-08-15
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    @Lieven, the only property I intend is the one I gave in the second formulation of the question. Notice that conjugation of $f(x,y) = x+iy$ gives $x-iy$ which is precisely the $g$ for that problem. The same pattern held for the other two-dimensional algebras I list. For that reason I used the term: conjugation. Also, multiplying the polynomials such that the coefficients of the product work out to real numbers is like $z \in \mathbb{C}$ having $z \bar{z} \in \mathbb{R}$, this is another way to see the $g$ as generalized conjugate. If this is troubling, let's say psuedo-conjugate, or just $g$.2012-08-16
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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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