Potenzieren und Wurzelziehen in rationalen Quaternionenalgebren

Abstract

LetF be a field not of characteristic 2 andQ =F +F i +F j +F k the quaternion algebra overF whereij = -ji =k andi ² = α andj ² = β with 0 ≠ α, β ∈F fixed. (IfF = ℝ and α = β = - 1 thenQ is the division algebra of the Hamilton quaternions.) IfF = ℚ and Q is a division algebra then by embedding certain quadratic number fields inQ we derive an efficient formula to compute the powers of any quaternion. This formula is even true in general and reads as follows. If a, a₁, a₂, a₃ ∈F andn ∈ ℕ then$(a + a_1 i + a_2 j + a_3 k)^n = \frac{{(a + \omega )^n + (a - \omega )^n }}{2} + A_\omega \cdot (a_1 i - a_2 j + a_3 k)$ where ω ig a square root of αa₁ ² + βa ₂ ² - αβa ₃ ² in or overF and$A_\omega = \frac{{(a + \omega )^n - (a - \omega )^n }}{{2\omega }}(\omega \ne 0)$ andA ₀ =na ^n-1.

With the help of this formula and related ones we are able to solve the equationX ⁿ=q for arbitrary quaternionsq and positive integers n in case ofF = ℝ and hence in case ofF ⊂ ℝ as well. IfF = ℝ then the total number of all solutions equals 0, 1, 2, 4,n or ∞. (4 is possible even whenn < 4.) In case ofF = ℚ, which we are primarily interested in, there are always either at most six or infinitely many solutions. Further, for everyq ≠ 0 there is at most one solution provided thatn is odd and not divisible by 3. The questions when there are infinitely many solutions and when there are none can always be decided by checking simple conditions on the radicandq ifF = ℝ. ForF = ℚ the two questions are comprehensively investigatet in a natural connection with ternary and quaternary quadratic rational forms. Finally, by applying some of our theorems on powers and roots of quate-rions we also obtain several nice results in matrix theory. For example, for every k ∈ ℤ the mappingA ↦A ^k on the group of all nonsingular 2-by-2 matrices over ℚ is injective if and only ifk is odd and not divisible by 3.