Constructing a Division Ring of Quaternions over K: Conditions and Solutions

[Ragna]

I don´t know how to solve it. Thanks beforehand

What condition has to fulfill a field $$K$$ so that we pruned to construct a ring of division of quaternions over $$K$$?

 

[mathwonk]

probably -1 should not be a square in K.

 

[matt grime]

The quartions are a degree 2 (non abelian) extension over the complex numbers:

H=C[j] with the rule ij=-ji

So, whether -1 is or is not a square is not important.

I can't say I understand the question at all. The only thing you need, surely, is for k to be a subfield of the quaternions. Is the extension supposed to be finite? What does pruned mean?

 

[mathwonk]

well they are also a 4 dimensional extension of the reals generated by 1 and 3 square roots of -1, so i thought maybe he was trying to describe which property of the reals was needed?

so if we just follow hamilton, we start from a field F and introduce 3 new elements i,j,k such that i^2 = j^2 = k^2 = -1, and set ij=k, jk=i. ki =j. ji=-k, kj=-i, ik=-j.

then extend this linearly as a definition of multiplication on the vector space F^4 with vector basis 1,i,j,k. the question becomes i guess whether this is division ring or not? so i would look at the formula for the inverse of an element and see if it makes sense, or needs some assumptions on F?

e.g. when defining the inverse of a complex number you need to divide by the sum of the squares of the coefficients, i.e. the ionverse of a+bi has denominator a^2 + b^2. so with a quaternion you seem to need that if 4 elements of your field are not all zero, then the sum of their squares is not zero. this requires e.g. that the field not admit a square root of -1.