## Wednesday, May 15, 2013

### Roots of Noncommutative Polynomial

When we study Riemann geometry, there is one
interesting object there. Its name is geodesic line.
It has two different definitions which ultimately lead
to the same object. Geodesic is line of extreme length.
Geodesic is such line that tangent vector
parallel transfers along it.

The reason for this identity is relation
between metric tensor and connection. Namely,
covariant derivative of metric is 0. As soon
this relation is broken in metric affine manifold,
we have two different lines.

I expect similar phenomenon in algebra of
polynomials. When we consider polynomial over commutative
ring (usualy this is a field) then we have two definition
of root. $$x=x_0$$ is root of polynomial $$p(x)$$
if $$p(x_0)=0$$. $$x=x_0$$ is root of
polynomial $$p(x)$$ if $$x-x_0$$ is divisor of
polynomial $$p(x)$$. However do these definition
lead to the same set when I consider polynomial
over noncommutative algebra?

The case with polynomial over quaternion algebra
$p(x)=(x-i)(x-j)+(x-j) (x-k)$
is relatively simple. Clearly, that $$p(j)=0$$.
This polynomial can be presented in the form
$p(x)=((x-i)\otimes 1+1\otimes (x-k))\circ(x-j)$
However it does not mean that polynomial $$x-j$$
is divisor of polynomial $$p(x)$$.

By the definition, polynomial $$x-j$$ is divisor
of polynomial $$p(x)$$ if we can write
$p(x)=q(x)(x-j)r(x)=(q(x)\otimes r(x))\circ (x-j)$
Here either $$q(x)$$ has power $$1$$ and $$r(x)$$
is scalar, either $$q(x)$$ is scalar and $$r(x)$$
has power $$1$$. But I do not think that there exist
polynomials $$q(x)$$ and $$r(x)$$ such that
$(x-i)\otimes 1+1\otimes (x-k)=q(x)\otimes r(x)$
From this example I see that if polynomial $$x-j$$
is divisor of polynomial $$p(x)$$, then $$x=j$$ is
root of polynomial $$p(x)$$. However oposite statement
can be wrong.

Now I want to consider the polynomial
$\begin{split} p(x)&=(x-k)(x-i) (x-j) +(x-i) (x-j) (x-k) \\ &+(x-k) (x-j)+(x-j) (x-k) \end{split}$
I see that $$p(j)=0$$ as well $$p(k)=0$$.
However I see now the problem. Namely.
$\begin{split} p(x)&=((x-k)(x-i)\otimes 1+ (x-i) \otimes (x-k) \\ &+(x-k)\otimes 1+1\otimes (x-k))\circ(x-j) \end{split}$
$\begin{split} p(x)=&(1\otimes(x-i) (x-j) +(x-i) (x-j)\otimes 1 \\ &+1\otimes (x-j)+(x-j)\otimes 1)\circ (x-k) \end{split}$
I can do something more with these equations. For instance
$\begin{split} p(x)&=(1\otimes(((x-i)\otimes 1)\circ (x-j)) +(x-i) (x-j)\otimes 1 \\ &+1\otimes (x-j)+(x-j)\otimes 1)\circ (x-k) \end{split}$