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Suppose $x$, $y$, $z$ are three variables satisfying $yz=zy$, $zx=xz$, $xy=yzx$.

  1. Could anyone give me two (non-commutative) polynomials $f$ and $g$ in the above three variables such that the following equality holds: $$f(x,y,z)\cdot(3z^2+zy^2x+x^2)=g(x,y,z)\cdot(3+y^2x+x^2) $$

Especially, I expect $f$, $g$ to be some product of modified polynomials of the given ones, see remark 1.

  1. What about a more general one: solve $$ f(x,y,z)(B_0(y,z)z^{2n}+B_1(y,z)z^nx+x^2)=g(x,y,z)(B_0(y,z)+B_1(y,z)x+x^2) $$ for any given (non-commutative) polynomial $B_i(y,z)$ and some $n$.

Thanks in advance!


(Added)

I want to emphasize that as observed by eithil below, the nonzero solution $f, g$ always exist, for example, for the question 1, you can simply write $$f(x,y,z)=B_0(y,z)+B_1(y,z)x+B_2(y,z)x^2$$ $$g(x,y,z)=A_0(y,z)+A_1(y,z)x+A_2(y,z)x^2$$ and use the equality to solve $A_i, B_i, i=0,1,2$, but in general, the expression for these $A_i, B_i$'s look very messy, so what I really care is whether we can find $f, g$ to be in a "neat" form in the spirit of the example shown in remark 1.


Remarks:

1, I have tried to ask this question in MSE, but no further answer appeared besides the following example that I got by direct computation:

$$(3z^4+yz^4x+x^2)(3+yzx+x^2)(3z^2+yzx+x^2)=(3z^4+yz^3x+x^2)(3z^2+yz^3x+x^2)(3+yx+x^2)$$

Motivated by this example, I guess we can always find the solution $f$, $g$ to be products of some modified form of the given two (non-commutative) polynomials, but I have no idea whether there are some explanation for this (combinatorial) phenomenon.

2, If you like, you can think that we are working within some skew polynomial rings, just thinking $y$, $z$ lies in the base ring, and $x$ to be the monomorphism of the base ring. Or think we are working in the integral group ring of the discrete Heisenberg group.

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2 Answers 2

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Essentially, we're working in a version of the quantum plane, with the scalar $q$ replaced by the variable $z$. This is a Noetherian domain, so we can invert the set of non-zero elements on either the left or the right to obtain a division ring, and these two rings (left/right fractions) are equal.

Rephrased in these terms, you're asking if and how the right fraction \[\Big(B_0(y,z) + B_1(y,z)x + x^2\Big)\Big(B_0(y,z)z^{2n} + B_1(y,z)z^nx + x^2\Big)^{-1}\] can be rewritten as a left fraction $g(x,y,z)^{-1}f(x,y,z)$.

This is definitely always possible, for any $n$, since the ring of left fractions and the ring of right fractions are equal.

To find an expression for $f$ and $g$, you can use linear recurrence relations. Let $k(y,z)$ be the fraction field of the commutative polynomial ring $k[y,z]$, and $\alpha: k(y,z) \rightarrow k(y,z): y \mapsto zy, z \mapsto z$.

Define the Laurent power series ring $k(y,z)((x;\alpha))$ by \[k(y,z)((x;\alpha)) = \{ \sum_{i \geq n} a_i x^i : a_i \in k(y,z), n \in \mathbb{Z}, x a_i = \alpha(a_i)x\} \] Our ring of fractions $k(y,z)(x;\alpha)$ embeds into this larger division ring, and now you can easily generalise to this case the commutative result which says that a power series represents a fraction if and only if it satisfies a linear recurrence relation. I used this set of notes, which I found quite clear and easy to follow.

Keeping careful track of what side your fractions are on, you should be able to obtain results on recurrence relations for both left and right fractions. This should give you the computational tools to translate the right fraction above into a left one, for any polynomials $B_0$, $B_1$, $B_2$ and any $n$.

Note: technically the $f,g$ you obtain will be elements of $k(y,z)[x;\alpha]$. However, it is now easy to multiply through by the common denominator of the coefficients (which are elements of $k(y,z)$) to obtain $f,g \in k[y,z][x;\alpha]$.


Editing to add more detail to address comment below.

Based on the notes I linked to above, we find that a power series $\sum_{i \geq 0} a_i x^i$ represents a left fraction $g^{-1}f$ if and only if there exists some integer $n$ and fixed coefficients $c_1, \dots, c_n \in k(y,z)$ such that \[a_{i+n} = c_1 \alpha(a_{i+n-1}) + c_2 \alpha^2(a_{i+n-2}) + \dots + c_n \alpha^n(a_i)\] for all $i \geq 0$. In this case, $g$ has the form $1-\sum_{i=1}^n c_i x^i$ and $f$ has degree $\leq n-1$. (Note that some of the $c_i$ are allowed to be zero.)

The element $(1-xy)^{-1}$ mentioned below has the form \[1 + zyx + z^3y^2x^2 + z^6y^3x^3 + \dots = \sum_{i \geq 0} z^{\frac{i(i+1)}{2}}y^ix^i\] as a power series so its coefficients satisfy a linear recurrence relation of the form $a_{i+1} = zy\alpha(a_i)$. (Compare this to the statement above about the relation between the denominator of a fraction and its recurrence relation constants.)

Meanwhile, for a right fraction, it turns out to be easier to write the coefficients of the power series on the right. Now a power series $\sum_{i \geq 0}x^ib_i$ represents a right fraction $fg^{-1}$ if and only if there exists an integer $m$ and constants $d_1, \dots, d_m$ such that for all $i\geq 0$, the coefficients $b_i$ satisfy \[b_{i+m} = \alpha^{-1}(b_{i+m-1})d_1 + \dots + \alpha^{-n}(b_i)d_m.\]

I'm afraid I don't know if you'll end up with a modified version of the fraction you suggest: I don't have time to work through the details right now. Based on the techniques in the notes linked above, you should be able to write your right fraction quite easily as a power series with coefficients on the right (hint: there will only be three recurrence relation constants, and the last one will be zero), then switch coefficients over to the left and look for a recurrence relation on the left - I suspect this will be the tricky part. On the plus side, if it works you'll have a nice explanation as to why it works!

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  • $\begingroup$ thanks for such a detailed answer, do you think we can find the solution $f, g$ to be the product of the modified polynomials of the given ones as I expected? $\endgroup$
    – Jiang
    Commented Oct 28, 2013 at 12:02
  • $\begingroup$ Also, I doubt nice linear recurrence relation may not exists in this noncommutative setting, even for the simpliest rational, say, $(1-xy)^{-1}$. $\endgroup$
    – Jiang
    Commented Oct 28, 2013 at 13:04
  • $\begingroup$ You actually get some surprisingly nice relations, especially for fractions which have trivial numerator like that one: in that case you can just read off the recurrence relation directly. This comment box isn't going to let me format maths nicely so I'll edit an answer to your comment into my answer above; I'll also dig out my old notes on this stuff and see if I can get you some specific formulas. $\endgroup$
    – eithil
    Commented Oct 28, 2013 at 13:43
  • $\begingroup$ Ok, I expect to see your further answer, thanks again! $\endgroup$
    – Jiang
    Commented Oct 28, 2013 at 14:15
  • $\begingroup$ for the example $f(x,y,z)(3z^2+zy^2x+x^2)=g(x,y,z)(3+y^2x+x^2)$, using the above method, as you mentioned, we can only find the recurrence relation involving three constants, so the degree of $g$ is no bigger than 3, and we can not get the solution with expected forms. Anyway, thanks for your time and interest in this problem! $\endgroup$
    – Jiang
    Commented Oct 28, 2013 at 21:26
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This is not the answer that I expected, but at least it gives an algorithm to find some $ f, g$ such that $ f(x,y,z) a(x,y,z)= g(x,y,z) b(x,y,z)$ for given $a, b$.

See theorem 8 in the paper by Ore " Theory of non- commutative polynomials", Annals of Mathematics, vol.34, 1933, 480-508.

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