 mersenneforum.org Quaternionenalgebra
 Register FAQ Search Today's Posts Mark Forums Read 2019-04-04, 22:10 #1 bhelmes   Mar 2016 1011001002 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 ) ( n-I ) 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   2019-04-05, 07:33   #2
Nick

Dec 2012
The Netherlands

17×103 Posts Quote:
 Originally Posted by bhelmes Is it possible in a quaternion algebra to split ( n+I ) in two symmetrical (conjugated) m+ai+bj+ck ?
No - the quaternions form a division ring (also known as a skew field)
so there are no zero divisors. For any non-zero polynomials f,g over the
quaternions, it follows that deg(fg)=deg(f)+deg(g), just as for complex numbers.

Quote:
 Originally Posted by bhelmes Is there a geometrical explication what a line in the complex plane is in a quaternion field ?
You can think of complex numbers as linear maps from $$\mathbb{R}^2$$ to $$\mathbb{R}^2$$ since multiplication by a complex number corresponds geometrically with rotating and scaling.
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   2019-04-07, 14:40   #3
Dr Sardonicus

Feb 2017
Nowhere

22·32·139 Posts Quote:
 Originally Posted by bhelmes 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 ) ( n-I ) Is it possible in a quaternion algebra to split ( n+I ) in two symmetrical (conjugated) m+ai+bj+ck ?
First, the complex field is "algebraically closed," so any nonconstant polynomial with complex coefficients can be split into linear factors.

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 m-ai-bj-ck, and the product of these is m^2 + a^2 + b^2 + c^2.   2019-04-08, 15:07 #4 LaurV Romulan Interpreter   "name field" Jun 2011 Thailand 9,787 Posts There are very nice and easy to grasp videos on youtube.  Thread Tools Show Printable Version Email this Page

All times are UTC. The time now is 16:05.

Mon Oct 25 16:05:19 UTC 2021 up 94 days, 10:34, 0 users, load averages: 1.32, 1.35, 1.24