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Old 2019-04-04, 22:10   #1
bhelmes
 
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Default 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
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Old 2019-04-05, 07:33   #2
Nick
 
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Quote:
Originally Posted by bhelmes View Post
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 View Post
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
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Old 2019-04-07, 14:40   #3
Dr Sardonicus
 
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Quote:
Originally Posted by bhelmes View Post
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.
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Old 2019-04-08, 15:07   #4
LaurV
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There are very nice and easy to grasp videos on youtube.
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