# How exactly is $i=\sqrt{-1}$ related to $\mathbb{C}$ being a closed algebraic field?

There are many known proofs of why $\mathbb{C}$ (field of complex numbers) is algebraically closed (for example Cauchy's proof )

However:

how does introducing the solution to the equation $x^2 + 1 = 0$ (imaginary number $i$) makes this possible or is intimatelly related to it?

Thanx

• It's the only thing $\Bbb R$ is missing to get algebraically closed. – blue May 26 '14 at 21:17
• In fact, for any polynomial with coefficients in $\Bbb{R}$ which doesn't have a root in $\Bbb{R}$, adjoining a 'formal root' of this polynomial to $\Bbb{R}$ yields a field isomorphic to $\Bbb{C}$. – Inactive - avoiding CoC May 26 '14 at 21:25
• No, the fields would be isomorphic, not the roots. You have the tag field-theory, try reading some of the basics, you might enjoy it. – blue May 26 '14 at 21:38
• Adjoining any root of a polynomial to $\mathbb R$ that isn't in $\mathbb R$ yields $\mathbb C$. $x^2 + 1$ happens to be one of the simplest polynomials with that property, so it's not surprising that it happens to be the first one that we stumbled upon. So the question is: why is it that $\mathbb R$ is only one root away from being algebraically closed? – Jack M May 26 '14 at 21:41
• @NikosM. Google "real closed fields." – Pedro Tamaroff May 26 '14 at 22:12

Adding a square root of any other negative number would work just as well. For $a>0$, if $x$ satisfies $x^2+a=0$, then $y=x/\sqrt a$ satisfies $y^2+1=0$. This is only trivially different.

You could also do things like add on a solution to $x^4+1=0$, and then it would square to a solution to $x^2+1=0$, etc. (Edit: To clarify, as Servaes said, any polynomial that doesn't have a root in $\mathbb R$ would work. )I don't know if you would consider this sort of thing "trivially different" or if it answers your question. The thing is that adding on a root to $x^4+1=0$ (call it $j$) basically gives you the same complex numbers (the algebraic closure is essentially unique). In this particular example, every number of the form $a+bj$ can essentially (well, you can choose some signs) be written as $\left(a+\frac{b}{\sqrt 2}\right)+\frac{b}{\sqrt 2}i$ and every number of the form $a+bi$ can essentially be written as $(a-b)+\left(\sqrt 2 b\right)i$.

I don't know whether this is what you're looking for or not, but the Artin-Schreier Theorem basically states that if you only need to add on finitely many things to make your field algebraically closed, then adding on a zero of $x^2+1$ will do the job. A nice write-up is here.

• yes i think it almost answers, it all boils down to even powers because then there is a problem with negative numbers, yet all these are essentially "trivial" alternatives to x^2+1 = 0. A question, if a > 0 how can we take sqrt(a) before defining imaginary number or is it non-issue? – Nikos M. May 26 '14 at 21:35
• @Nikos $\sqrt\pi$ is the number that you square to get $\pi$. Every positive number has a positive square root: If you believe it for rationals, just take the limit of a sequence of rational numbers whose square is not quite big enough, but whose squares tend to the number in question. – Mark S. May 26 '14 at 21:39
• oops yeap sorry misread :) – Nikos M. May 26 '14 at 21:41
• The Artin-Schreier Theorem gives a nice answer, will wait a little and if nothing else arrives will accept the answer, thank you very much! – Nikos M. May 26 '14 at 21:44

Other replies have answered the meat of this question. I just wanted to point out that if you start talking about matrix equations, instead of just polynomials, you end up with more fun things that extend the complex numbers.

http://en.wikipedia.org/wiki/Quaternion

http://www.math.mcgill.ca/bsmith/Revisedpaper.pdf

• great, thank you very much! – Nikos M. May 26 '14 at 21:45
• It's important to note (and the links clarify this) that the Quaternions are not what you get when you have the same goal (algebraic closure) but happen to be working with matrices, but rather something reminiscent of the complex numbers you can get when you change/weaken the goal. – Mark S. May 26 '14 at 21:54