Can we construct the complex numbers by using $i^2 = -1$ as an axiom? Informally we can construct the complex numbers by starting with the real numbers and adding a new element $i$ which satisfy $i^2 = -1$. If we want to formalize this construction we usually use quotient fields, which is analogous to the informal construction, but a bit different on a technical level. However the informal construction still works and is enough for us to start doing computations and algebra with complex numbers. I was wondering if it was possible to make it formal, e.g like this:
We define an algebraic structure $ℂ$ such that:

*

*$ℂ$ is a field.


*$ℝ$ is a subfield of $ℂ$.


*There exist an element $i$ in $ℂ$ such that $i*i=-1$
Does this work as a rigorous construction? To me this doesn't look any less rigorous then e.g the peano axioms, and it captures everything from the informal construction.
 A: The problem with your definition is that every single field containing the complex numbers as a subfield  also satisfies your three conditions, so  while the complex numbers satisfy your properties there are also many other non-isomorphic fields that that satisfy your property.  I suggest adding a fourth axiom that if $i^2=-1$, then the function $\mathbb{R}^2\to \mathbb{C}$ which maps each pair $(x,y)$ to $x+iy$ is bijective. Any field satisfying these four axioms would be isomorphic to the complex numbers. Alternatively as MJD points out you could instead add a fourth axiom that $\mathbb{C}$ is a subset of any field satisfying the three axioms. Then also any field satisfying the four axioms would be isomorphic to the complex numbers.
A: I think the other answers miss an important part of the issue.  You asked:

Does this work as a rigorous construction?

The comments hint at the problem: You cannot just make up some axioms and then assume that there is necessarily a family of objects satisfying them.  Because sometimes there isn't.
An example that comes up on this site with great frequency: people ask if you can take the real numbers and add a new element $\infty$ which satisfies $\infty\cdot 0 = 1$.  In form, this looks almost exactly like what you want to do: “start with the real numbers and add a new element $i$ which satisfies $i^2=−1$.”  But the two cases are quite different: $i$ works, and $\infty$ doesn't.  There is no reasonable extension of the real numbers that has such an element.  (Of course you can insert one, but you can't also preserve the associative, distributive, or additive inverse laws, so what you get no longer looks much like an extension of the real numbers.)
People similarly ask on this site if they can add a number $\epsilon$ to the real numbers with the property that $0 < \epsilon \le r$ for all positive real numbers $r$. And the answer is no, they cannot, it does not work. (Consider $\epsilon\cdot \frac12$.)
So the strategy of "real numbers, and add a new number such that (whatever)” can't be pursued without some care.
Some of the comments to your question point this out: you can make up whatever rules you want, but then you have to prove that there is some system that actually follows those rules, and that still contains something recognizable as the real numbers.  The usual constructions of the complex numbers do this.  We take some concrete set of objects (ordered pairs of reals, $2×2$ matrices, sets of polynomials, etc.) and show that they satisfy the properties that you wanted:

*

*We can define addition and multiplication operations on these objects; for example using the ordered pairs we define $(a,b)\cdot(c, d) = (ac-bd, ad+bc)$.

*We can identify a subset of the objects that behave just like the real numbers as regards addition and multiplication: in this case the real number $r$ corresponds to $(r, 0)$.

*There is some object $i$ that has the property $i\cdot i=-1$. In this case $i=(0, 1)$ works.  ($i=(0, -1)$ also works.)

The concrete objects are essential here, because without them, you don't know that your definition isn't inconsistent or even incoherent.
A: Define the complex numbers as $\Bbb C =\{a+bi\mid a,b\in\Bbb R\}$, where the sums are just formal sums (without evaluation).
Define addition componentwise as
$$(a+bi)+(c+di) = (a+c)+(b+d)i$$
and multiplication (multiply out)
$$(a+bi)\cdot(c+di) = ac+adi+bci+bdi^2$$
which becomes with $i^2=-1$
$$(ac-bd)+(ad+bc)i$$
One can show that $\Bbb C$ is a field.
