20190404, 22:10  #1 
Mar 2016
101100100_{2} Posts 
Quaternionenalgebra
A peaceful night for all persons,
You can split a quadratic irreducible polynomial in two complex lines. For example f(n)=n^2+1=( n+I ) ( nI ) Is it possible in a quaternion algebra to split ( n+I ) in two symmetrical (conjugated) m+ai+bj+ck ? Is there a geometrical explication what a line in the complex plane is in a quaternion field ? I am sure that someone has an answer to this question and thanks in advance for his efforts. Greetings from Dirichlet Bernhard 
20190405, 07:33  #2  
Dec 2012
The Netherlands
17×103 Posts 
Quote:
so there are no zero divisors. For any nonzero polynomials f,g over the quaternions, it follows that deg(fg)=deg(f)+deg(g), just as for complex numbers. Quote:
In the same way, quaternions can be thought of as linear maps from \(\mathbb{C}^2\) to \(\mathbb{C}^2\). If you are interested in this topic, I would recommend the article "Quaternions, Octonions and Normed Division Algebras" in the Princeton Companion to Mathematics https://press.princeton.edu/titles/8350.html 

20190407, 14:40  #3  
Feb 2017
Nowhere
2^{2}·3^{2}·139 Posts 
Quote:
Second, the ring of polynomials over any field is Euclidean (under polynomial division with quotient and remainder, the remainder being of lower degree than the divisor), so enjoys unique factorization whether the field is algebraically closed or not. Alas, with quaternion coefficients, unique polynomial factorization goes out the window. For example, in the quaternions, n^2 + 1 = (n + i)*(n  i) = (n + j)*(n  j) = (n + k)*(n  k). Also, with the quaternions, the "conjugate" of m+ai+bj+ck usually means maibjck, and the product of these is m^2 + a^2 + b^2 + c^2. 

20190408, 15:07  #4 
Romulan Interpreter
"name field"
Jun 2011
Thailand
9,787 Posts 
There are very nice and easy to grasp videos on youtube.
