Four dimensional field over complex numbers

Question

A guy in Facebook claims he's come up with an algebraic field extension to the complex plane. He's defined the unit multiplications as $i^2=-1$, $j^2=i$ and $k^2=-i$. This implies that $ij=ji=k$, $ik=ki=-j$ and $jk=kj=-1$.

Furthermore complex numbers are seemingly a subset of this space, since each complex number $z = a + bi$ can be represented as $a + bi + 0j + 0k$.

So far I've checked that both addition and multiplication in this space are indeed associative and commutative, the addition and multiplication are distributive, and that $0$ and $1$ are the additive and multiplicative inverses respectively. I've yet to find an example of a non-inversible element.

Could someone help explain what's going on?

Answer 1

Your Facebook guy seems to have stumbled upon some interesting identities among $8$-th roots of unity. To wit, $$ j = e^{\pi i/4} = \tfrac{\sqrt{2}}{2} (1+i) \qquad \text{and} \qquad k = e^{3\pi i/4} = \tfrac{\sqrt{2}}{2} (-1+i) $$ satisfy all of your identities within $\mathbb{C}$. (Notice that $i = j^2$ and $k = j^3$.) This had to be the case since $\mathbb{C}$ is algebraically closed. This means that any $\mathbb{R}$-linear combination of $1, i, j, $ and $k$ can actually be expressed just in the span of $1, i$.

Note: the pair $-j, -k$ would also work.

Answer 2

If $j^2=i$, then $j$ is a squareroot of $i$. This implies that $j\in \mathbb{C}$ as $\mathbb{C}$ is algebraically closed. Furthermore, if $ij=k$, then $k\in \mathbb{C}$ as well. This is not an extension, there are no elements being added.

Answer 3

This is not an answer but an answer to some comments which is simply too long to be a comment itself:

Intuitively this construction feels like a four dimensional space, but it's not. Is there some graphic or layman's explanation that would make it more clear what is actually happening in this construction?

I do not have a graphical explanation. The crucial point indeed is that the space is not four dimensional. If you are familiar with group presentations, a very similar issue occurs. Suppose you have a group generated by a bunch of elements ($a,b,c$) and you are giving a bunch of relations in between those elements ($a^2b=cb, \dots$). It is possible that you have so many relations that one does not really need all generators (for example $a^2b=cb$ implies that $a^2=c$ and thus $c$ is redundant). In this you at first might be tempted to think that this group needed three generators, but actually less suffice.

Your facebook guy did something similar. At first glance it looks like you need 4 independent vectors te express everything, but these vectors are really not independent. That's the crux of the current story.

Or alternatively, what would be the implications if we were to suppose that $z^2=i$ had more than two solutions?

If $z^2-i=0$ has more than two solutions, you cannot be working in a field. Indeed, you can easily prove that any polynomial $P(X)\in K[X]$ of degree $n$ over a field $K$ cannot have more than $n$ roots. The proof is straightforward as soon as you have the division algorithm for polynomials over a field.

Edit: I'll include a proof of the latter statement. Let $P(X)\in K[X]$ be a polynomial. We can write $$P(X)=\sum_{i=0}^na_iX^i$$ with each $a_i\in K$ and $a_n\neq 0$. First note that we may assume that $a_n=1$ by multiplying the polynomial by $a_n^{-1}$ (here I use that $K$ is a field). Furthermore, doing so does not change the roots of $P(X)$ (that's a claim that you can prove as an exercise).

Now I claim the following: If $c\in K$ is a root of $P(X)$ (i.e. $P(c)=0$), then $$P(X)=(X-c)Q(X)$$ for some polynomial $Q(X)\in K[X]$ with $\deg(Q(X))=\deg(P(X))-1$. Indeed, by the division algorithm for polynomials, one can write $$P(X)=(X-c)Q(X)+R(X)$$ where $Q(X),R(X)\in K[X]$ and $\deg(Q(X))\leq \deg(P(X))$ and $\deg(R(X))<\deg(P(X))=1$. By assumption, $P(c)=(c-c)Q(c)+R(c)=R(c)=0$. On the other hand, $\deg(R(X))=0$ and thus $R(X)=0$ as it is a constant polynomial. Thus $P(X)=(X-c)Q(X)$ as required.

Hence we now showed that if $c$ is a root of $P(X)$, then we find that $P(X)=(X-c)Q(X)$ with $\deg(Q(X))=\deg(P(X))-1$. Now assume that $P(X)$ has more roots than its degree. Iterating the above procedure, you find that $P(X)=(X-c_1)(X-c_2)\dots (X-c_m)Q(X)$ with $m>\deg(P(X))$, but then $\deg(P(X))\geq m$, a contradiction!

Answer 4

You're looking at the extension $\Bbb{C}\subset\Bbb{C}[j,k]$ where $j$ and $k$ satisfy $j^2=i$ and $k^2=-i$. The key is that you are requiring this to be an extension of fields. Over a field, a polynomial of degree $d$ has at most $d$ roots. As $j$ and $k$ are zeros of $X^2\pm i\in\Bbb{C}[X]$, and we can find two roots of both quadratics in $\Bbb{C}$, it follows that $\Bbb{C}[j,k]$ is isomorphic to (a subfield of) $\Bbb{C}$.

More explicitly, we can find the polar forms of these roots from the fact that $\pm i=e^{\pm\tfrac{\pi}{2}i}$. It follows that $$j=\pm e^{\tfrac{\pi}{4}i}\qquad\text{ and }\qquad k=\pm e^{-\tfrac{\pi}{4}i},$$ because these four complex numbers are precisely the roots of $X^2\pm i$.

Similarly we could find the Cartesian forms of these roots by solving $(a+bi)^2=\pm i$. This is equivalent to the simultaneous equations $$a^2-b^2=0\qquad\text{ and }\qquad 2ab=\pm1.$$ The former implies that $b=\pm a$ and hence the latter reduces to $2a^2=\pm1$, or equivalently $a=\pm\sqrt{\pm\tfrac12}$, which we can express more concisely as $a=i^k\tfrac{1}{\sqrt{2}}$. Tracing our steps back then shows that $$j=\pm\frac{1}{\sqrt{2}}(1+i)\qquad\text{ and }\qquad k=\pm\frac{1}{\sqrt{2}}(1-i).$$ Either way we see that $\Bbb{C}[j,k]$ is isomorphic to a subfield of $\Bbb{C}$, and hence isomorphic to $\Bbb{C}$ itself.