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}

\]