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So, Let K be the non-commutative field or division ring of the Quaternions. We will consider a sub-field of it L=Q[i,j] in other words, L=Q+Qi+Qj+Qk. Now let us consider the ring of skew polynomials of one variable over L, where multiplication is defined in the usual way and in the usual Quaternions. Let $f[x]=\sum_{i=0}^n a_ix^i$ that is an element of the polynomial ring where $a_i \in L$. in two of his Paper Ivan Niven, showed how to get the set of all the solutions( which is usually infinite) of such equations over K. Now the question is can we solve the same equations if the coefficients are elements of L. Ivan's method clearly does not work because he used the fact that $\mathbb{C} \subset K$ and $\mathbb{C}$ is algebraicly closed which is not our case here.

In conclusion there are three questions.

1) how to solve the polynomials over L.

2) if there are polynomials that does not have solutions over L can we get the solutions by finite extension.

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Can you give a link or a precise reference to Niven's paper? – Filippo Alberto Edoardo Apr 12 '12 at 9:30
yes sure, here you are… – Motaz Hammouda Apr 13 '12 at 23:31

Before an attempt to answer I'd like to add a few comments, some because they are simultaneously very interesting and hard to find in literature, some of them because they relate to the question. You decide which is which. :)

A. Niven's solution applies to polynomials $f(x)=\Sigma_{n=0}^N a_nx^n$, as you pointed out. Not all polynomial expressions have this form; for example, $xi+ix-1$ has no roots, where $i$ is a square root of $-1$.

B. Niven's proof that $f(x)$ has roots can be extended to $f(x)=g_N(x)+\Sigma_{n=0}^{N-1} h_n(x)$, where $g_N(x)$ is a monomial of degree $N$ and $h_n$ are polynomials of degree $n$. One just has to rephrase Gauss's proof for complex numbers, that relies on degrees of maps $S^1\to S^1$ (contraction of non-zero complex numbers to $U^1$) to degrees of maps $S^3\to S^3$ (contraction of non-zero quaternions to $Sp^1$). The picture that arises their is very interesting IMO.

C. Here's an interesting trick. Suppose that $q=xi+yj+zk+t$, where $i,j,k$ are the usual unit quaternions. Then you can recover $x,y,z,t$ as linear expressions of $q$:

$t = (q-iqi-jqj-kqk)/4$

$x = (q-iqi+jqj+kqk)/4$

$y = (q+iqi-jqj+kqk)/4$

$z = (q+iqi+jqj-kqk)/4$

This shows that any affine real algebraic variety in $R^4$ can be written up as the zero set of several quaternionic polynomials: just plug the above expressions over $q$ into the algebraic expression over the four real variables.

D. Notice that the above expressions for $x,y,z,t$ are rational. Therefore you can plug them into the equation for a rational variety: if $V\subset Q^4$ is the zero set of several equations in four variables $x,y,z,t\in Q$ then you just plug the above expressions and you get the same variety as the zero set of the corresponding equations in a single quaternionic variable $q\in L$. Here we identify $L$ with $Q^4$ in the obvious way.

E. Now, let's approach your question, let's restrict ourselves to $f(x)=\Sigma_{n=0}^N a_nx^n$, where $a_n\in L$. One of the things Niven showed that the zero sets of such polynomials consist of isolated points and 2-D spheres. The typical example for a sphere would be $f(x)=(x-a)^2+b$, where $a\in H$ is the center and $b\in R_+$ is the square of the radius.

F. So, in order to answer your question it's sufficient to ask ourselves: is it possible to embed $S^2\to R^3$ of imaginary quaternions so that (1) the center and the squared radius would be rational and (2) the intersection with $Q^3$ would be empty?

G. There is a result that the integers of the form $4n(8m+7)$ cannot be expressed as sums of three squares. So let's place the center of $S^2$ in the origin and pick radius 28. Then we have $f(x)=x^2+28$ that has no roots in $L$.

H. The above answers part (1) of the question, but obviously not part (2). I don't know off-hand how to extend this train of thought to finite extensions of $Q$.

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This is a very nice collection of information! – Jeff Strom Oct 4 '13 at 0:01

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