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If I understand correctly, any quaternion algebra over the rationals is a noncommutative associative division algebra. I am currently working with implementations of quaternion algebras in MAGMA and have encountered the following situation:

If I generate a quaternion algebra D(i,j,k) := QuaternionAlgebra( Rationals()| 1, -2) and try to find an inverse to the element j+k, (j+k)$^{-1}$, MAGMA returns that it cannot take negative powers of non-units. In fact, we have that (j+k)$^2$ = 0 which should not be possible in an associative division algebra. Can anyone tell me what my mistake is?

Shouldn't all quaternion algebras generated by non-zero rational numbers a, b such that i$^2$ = a, j$^2$ = b be skew fields?

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    $\begingroup$ In a word NO. Whether you get a division algebra depends on the choice of $a$ and $b$. You always get a central simple algebra, but it could, for example, be isomorphic to the algebra of 2x2 matrices over $\Bbb{Q}$. $\endgroup$ – Jyrki Lahtonen Jun 12 '18 at 17:39
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    $\begingroup$ IIRC the elements of that quaternion algebra have reduced norms $$nr(a_1+a_2i+a_3j+a_4k)=a_1^2-aa_2^2-ba_3^2-ab a_4^2.$$ You should ascertain that this is never zero (non-trivially). One way of doing that is to select both $a$ and $b$ to be negative. With $a=1, b=-2$ you get $-ab=2$. Therefore $$nr(j+k)=2-2=0$$ and you have a non-unit. $\endgroup$ – Jyrki Lahtonen Jun 12 '18 at 17:45
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    $\begingroup$ Great script about quaternion algebras: math.uconn.edu/~kconrad/blurbs/ringtheory/quaternionalg.pdf For the case at hand, see in particular Theorems 4.14/4.16/4.21 (and example 4.19 for more non-skew field quaternion algebras over $\Bbb Q$). $\endgroup$ – Torsten Schoeneberg Jun 12 '18 at 20:54
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    $\begingroup$ Quite generally, a quaternion algebra with $a=1$ or $b=1$ is never a skew field, but always isomorphic to $M_2(F)$ over whatever field $F$ you work. See discussion at the beginning of section 4 of the above script. $\endgroup$ – Torsten Schoeneberg Jun 12 '18 at 20:56
  • $\begingroup$ @TorstenSchoeneberg the script indeed seems to be very helpful, thank you. $\endgroup$ – Jan Gerrit Jun 13 '18 at 9:45

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