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I'm reading "Naive Lie Algebra" by John Stillwell, during introducing quaternion he mentioned:

The quaternion sum operation has the same basic properties as addition for numbers, namely $$q_1 + q_2 = q_2 + q_1 ,\text{(commutative law)}$$ $$q_1 + (q_2 + q_3 ) = (q_1 + q_2 ) + q_3 ,\text{(associative law)}$$ $$q + (−q) = \mathbf{0} \text{, where } \mathbf{0} \text{ is the zero matrix, (inverse law)}$$ $$q + \mathbf{0} = q.\text{(identity law)}$$ The quaternion product operation does not have all the properties of multiplication of numbers — in general, the commutative property $q_1 q_2 = q_2 q_1$ fails — but well-known properties of the matrix product imply the fol- lowing properties of the quaternion product: $$q_1 (q_2 q_3 ) = (q_1 q_2 )q_3 ,\text{(associative law)}$$ $$qq^{-1} = \mathbf{1}, \text{for } q \ne \mathbf{0} \text{, (inverse law)}$$ $$q\mathbf{1} = q, \text{(identity law)}$$ $$q_1 (q_2 + q_3 ) = q_1 q_2 + q_1 q_3 . \text{(left distributive law)}$$ Here $\mathbf{0}$ and $\mathbf{1}$ denote the $2 × 2$ zero and identity matrices, which are also quaternions.

So far so good.. but then he mentioned:

The right distributive law $$(q_2 + q_3 )q_1 = q_2 q_1 + q_3 q_1$$ of course holds too, and is distinct from the left distributive law because of the non-commutative product.

I'm lost here -- why the right distributive law of course holds? Is it that in ring's definition, if the addition is commutative, then we only need the left distribution law on product operation?

Arturo Magidin
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athos
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    It "of course holds" because it is true. The point Stillwell is making is that you cannot deduce right distributivity from left distributivity because multiplication is not commutative; so the two distributivity laws have to be stated separately and checked separately. It "of course holds" because it holds in matrices, which is how he has defined the quaternions. – Arturo Magidin Sep 30 '22 at 18:18
  • @ArturoMagidin I see, so the fact it holds comes from the definition via matrix. But, in general, would it be possible for a ring with commutative addition, and product operation that left distribution law holds, does not hold right distribution law? – athos Sep 30 '22 at 19:14
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    Yes, of course it is possible, but it wouldn't be a ring, now would it? If you don't require addition to be commutative, they are called "near rings". An example of a structure with two operations, a commutative "addition" and a product that left distributes but does not right distribute is given by ordinal arithmetic; you may object that the additive structure is only a monoid and not a group, though. An near ring with commutative addition is given here. – Arturo Magidin Sep 30 '22 at 19:22

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