Why should I consider the components $j^2$ and $k^2$ to be $=-1$ in the search for quaternions?

I'm reading a paper about Hamilton's discovery of quaternions and it explains why he failed in his 'theory of triplets' where he tried to make a vector with $3$ dimensions, as an analogy to the complex field, where we can see a number as a $2$ dimensional vector.

In this paper, he explains why is impossible to create a field with $3$ components, that is an extension of the complex field (in other words, it respects addition, and multiplication in the same way...). Here it is:

![enter image description here][1]

As you can see, it goes through all the possibilities and proves that it is impossible.

**The paper, however, does not explain why $j^2=-1$. It could be anything! Why $-1$?**

The article itself is pretty intuitive, but this aspect kills me.

Later, in the article, it says that we should instead consider a 4th component called $k$, such that $k^2=-1$ (also, $i$ and $j$ too).

[Here is the paper][2]

**EDIT**: [this][3] paper by Rupert Shuttleworth turned out to be extremely helpful (mirrored [here][4] on archive.org)

It seems to me that we can achieve a contradiction more quickly without assuming this anyway: Since $(1, i, j)$ is a basis for the field, $ij = a + bi + cj$ for some unique $a, b, c \in \mathbb{R}$. Then, on the one hand $i^2 j = -j$, and on the other it is

$i(ij) = i(a + bi + cj) = -b + ai + c(ij) = -b + ai + c(a + bi + cj) = (-b + ac) + (a + bc)i + c^2 j .$

Comparing coefficients of the $j$ term gives that $c^2 = -1$, which is not true for any $c \in \mathbb{R}$.

Anyway, the question is certainly not an idle one: Indeed, it leads to something interesting when carrying out this sort of analysis for four-dimensional algebras over $\mathbb{R}$, which famously yields the quaternion algebra $\mathbb{H}$ (NB that since $\mathbb{H}$ is not commutative, it is not a field but a division algebra).

Interestingly, one can also try to construct such an algebra taking $i^2 = -1$ but $j^2 = k^2 = 1$ and find there's a coherent and interesting way to define an associative product, giving an algebra $\widetilde{\mathbb{H}}$ sometimes called the split quaternions. This is perhaps nonobviously isomorphic (as an $\mathbb{R}$-algebra) to the ring $M(2, \mathbb{R})$ of $2 \times 2$ real matrices. Unlike $\mathbb{H}$ this is not a division ring (there are nonzero matrices that square to the zero matrix), but like $\mathbb{H}$ it has a nondegenerate quadratic form $Q$ that satisfies $Q(xy) = Q(x) Q(y)$, namely the determinant.

I don't know if this is what Hamilton had in mind, but: suppose you've decided that what you want to do is to write down a division algebra $D$ over $\mathbb{R}$ bigger than the complex numbers $\mathbb{C}$. Any element $d \in D$ of $D$ generates a subalgebra $\mathbb{R}[d]$ of $D$ which is a commutative division algebra (in other words, a field) over $\mathbb{R}$ and hence (because $\mathbb{C}$ is algebraically closed) must be isomorphic to either $\mathbb{R}$ or $\mathbb{C}$.

If $d \not \in \mathbb{R}$, then this subalgebra $\mathbb{R}[d]$ must be isomorphic to $\mathbb{C}$. Any such isomorphism realizes $d$ as a complex number $a + bi, b \neq 0$, and hence up to a real change of coordinates we might as well have chosen $d$ to correspond to $i$ under this isomorphism; that is, to satisfy $d^2 = -1$. With a bit more work this argument shows that $D$ is generated by elements that square to $-1$.