# Construction of $\mathfrak{J}$

I have what could be taken as a quite naive way of understanding the complex field. I simply define the ring isomorphism: $$$$\phi:\mathbb{R}^2\longmapsto\mathfrak{J}$$$$ Where $$\phi$$ is, of course, a bijective map, and $$\mathfrak{J}$$ is the algebraic structure $$(\mathbb{C},\oplus,\odot)$$, following the axioms, given $$\mathfrak{n}_1,\mathfrak{n}_2\in\mathbb{R^2}$$, where $$\mathfrak{n}_j=(x_j,y_j):$$ \begin{align} &(i)\ \phi(\mathfrak{n}_1+\mathfrak{n}_2)=\phi(\mathfrak{n}_1)\oplus\phi(\mathfrak{n}_2)=(x_1+x_2,\ y_1+y_2)\\ &(ii)\ \phi(\mathfrak{n}_1\cdot\mathfrak{n}_2)=\phi(\mathfrak{n}_1)\odot\phi(\mathfrak{n}_2)=(x_1x_2-y_1y_2,\ x_2y_1+x_1y_2) \end{align} From which known identities such as $$i^2=1$$ follow naturally from simply defining $$\mathfrak{n}=i=(0,1)$$. Thing is, I see nobody treating it like that, which leads me to suspect that's something poor about it.

I'm bringing this up because it's often for me to see people treating complex numbers identities like something really impressive, or jaw-dropping, and for, me, understanding it as I just defined, there's no secret, or something obscure about it.

It's more often, however, to formally define it as the quotient ring between the real polynomials $$\mathbb{R}[x]$$ and the polynomial $$x^2+1$$. Even though it's motivation looks quite clear, I don't see why it should be more appreciated, or wanted, then the one I proposed.

So, is the definition I showed valid? If so, why isn't it used, or taught? I ask because, it feels like it would clear the myth that the average joe has on the matter, and, well, is literally the way I've found to understand and convince myself about the complex field, and, yet, I might be terribly wrong. Any help to that discussion will be appreciated. Thank you, in advance.

• I'm not sure why you're defining a ring isomorphism between two rings if your goal is to define a single ring $\mathbb C$. It is relatively common to define $\mathbb C$ as the structure $(\mathbb R^2,\oplus,\otimes)$ with the operations defined as you gave (but without $\phi$), so I worry I'm missing a subtlety of your question. Feb 25 at 0:34
• Well, I did that in order to explicitly construct $\mathbb{C}$ as an isomorphic structure to $\mathbb{R}^2$. I used $\phi$ just to explicitly define the isomorphism. Feb 25 at 0:37
• Related, not an answer: math.stackexchange.com/questions/3244132/… Feb 25 at 1:53
• @JohannWagner Please see my answer. As rings, those two are NOT isomorphic. Feb 25 at 3:58

I have no idea whether this is valid or not, but will assume that it is, and will address (part of) your next question, "Why isn't it taught this way?"

tl;dr: "mathematics" is not the same thing as "mathematics education".

In many parts of the world complex numbers are part of the secondary school syllabus. There are various reasons for doing complex numbers, and doing it the way you suggest would delay this until such time as students are capable of understanding a much more abstract approach. Moreover, it is usually very difficult to understand an abstraction unless you have concrete examples of it. So the sensible educational path is to learn complex numbers (whether rigorously or not) and then use them to understand ring isomorphisms (and field axioms, and many other things), not the other way round.

Among other things, how is anyone who has not seen complex numbers expected to make sense of your apparently arbitrary definition of multiplication? (Emphasis: apparently arbitrary: I know it is not really arbitrary.) Why is there a $$+$$ in one component and a $$-$$ in the other? Why are the $$x$$s and $$y$$s placed exactly where they are? (Why have pairs and not triples?)

This is far from an isolated example. Surely one is not going to expect 6-year-olds to learn fractions as equivalence classes of pairs of integers; or later years to learn real numbers as Dedekind cuts, or equivalence classes of sequences of rationals (that is, equivalence classes of sequences of equivalence classes of pairs of integers).

The best (educational) way in to any mathematical topic is through the parts which are at a medium level. Then one proceeds forwards to more difficult (and, hopefully, important) sections; and backwards to the more fundamental concepts, providing a rigorous and formal basis for the subject. Of course not all students will do all of this.

In terms of a normal educational path, I think it is entirely appropriate for "the average joe" to see complex numbers as something jaw-dropping: we have known and understood for years that it is not possible to take the square root of a negative number, and now we find that it's possible after all!

Of course none of this has any bearing on the mathematical validity of your approach, and I'm very happy to accept that this is a correct formal way to do it. But as far as "why is it not taught this way?", there are clearly many good reasons.

• Yes. Very good, balanced explanation. :) Feb 25 at 1:44

So, is the definition I showed valid?

No. You never defined $$\mathbb{C}$$, you assumed it was defined and defined a map from $$\mathbb{R}^2$$ to it. In fact, $$\mathbb{R}^2$$ (a ring with pointwise addition and multiplication) is not even isomorphic as a ring to $$\mathbb{C}$$ (whose addition is pointwise, but whose multiplication is not). To properly define $$\mathbb{C}$$, you must take a set (use $$\mathbb{R}^2$$), define two operations on it ($$+,\cdot$$ as you have defined them), and show that the set with these operations satisfy the properties of a ring.

If so, why isn't it used, or taught?

It is taught, e.g. in my first complex analysis course. See for instance An Introduction to Complex Function Theory by Bruce Palka; this definition of $$\mathbb{C}$$ is the first thing discussed.

Let's suppose there is a preference for the algebraic definition. Now let's speculate as to why it is preferred. Perhaps it is because it is immediately clear what is going on: we are creating a 'formal' square root of $$-1$$ and considering the numbers created by evaluating all real polynomials at this imaginary number. With only a modest background in algebra, we know instantly that these form a field containing $$\mathbb{R}$$. To me, the alternative definition does not immediately belie a square root of $$-1$$. Or that it is a field.

There's probably a more elaborate pedagogical answer, but I'm not qualified to give it.