For Ring of Quaternions, does $a + bi +cj + dk = 0$ imply $a,b,c,d = 0$, where $a,b,c,d \in \mathbb R$?

Question

Of course, if $a,b,c,d = 0$, then $a + bi +cj + dk = 0$. But having trouble with other direction.

Answer 1

This is true but I don't agree with the comments that it's true "by definition"; it depends on how you define the quaternions. Here are $4$ different definitions:

1. The set of formal $\mathbb{R}$-linear combinations $a + bi + cj + dk$, with multiplication determined by $ij = k, jk = i, ki = j, i^2 = j^2 = k^2 = -1$, and the condition that $i, j, k$ anticommute with each other. With this approach $1, i, j, k$ are linearly independent by definition, but you need to prove separately that multiplication is associative.

2. The $\mathbb{R}$-algebra generated by three generators $i, j, k$ subject to the relations $i^2 = j^2 = k^2 = ijk = -1$ (you can also remove $k$ and use two generators). With this approach the quaternions are associative by definition but you need to prove that $1, i, j, k$ are linearly independent. You can do this by embedding the quaternions into a matrix algebra, which can also be used as a definition:

3. (and 4) As either a specific set of $2 \times 2$ complex matrices or a specific set of $4 \times 4$ real matrices; you can consult Wikipedia for details. With this approach the quaternions are associative by definition but you still need to prove that $1, i, j, k$ are linearly independent although it's a lot easier to check this with matrices.