Why is it that Complex Numbers are algebraically closed?

Question

I find it curious that Complex Numbers give enough flexibility to be algebraically closed, where the reals, rational numbers do not. For the reals it is easy to see that they cannot be used to solve equations like $x^2 + 1 =0$. Geometrically, one can look at the number line as see that any $x$ squared yields a positive number which when added to one cannot get you back to zero. In the complex case, however, we are working with the plane. In this case exponents stretch and rotate any given $x$. It is easy to therefore see in the particular circumstance that if $x=i$ that $x^2$ rotates it to $-1$ which when added to one yields the desired result (i.e. $0$). So because the Complex Numbers are algebraically closed, I conclude that any polynomial equation with complex coefficients my be solved by choosing one or more $x$'s in the plane and rotating them and stretching them such that they will combine using the given coefficients to produce the RHS.

Question: Why is it that we do not need a larger space than the plane to solve Complex polynomial equations?

I have tried to find a sufficient answer through Google, but was not able. I also searched M.SE and could not find a sufficient answer. I am not a mathematician, so I am looking for an intuitive answer if possible.

Answer 1

I think your question can be boiled down to understanding why the fundamental theorem of algebra is true. As has already been pointed out above there are many different ways to prove this and you should try to understand a few of them. However your question is not really about the mechanics of those different proofs but rather you are asking 'Why are the complex numbers enough..?'

It's a good question. The truth is that each of those different proofs is giving an argument about why they are enough -- from a slightly different perspective. Depending on your background you may find one more intuitive than another. I have two suggestions on how to get a better feel for the FTOA:

• Perhaps what you are looking for in an intuitive answer is something visual. The nice thing here is that you can make awesome colorized pics which reveal the structure of complex valued functions. Check out this unpublished paper by Daniel J. Velleman at Amherst. It's a great visual walk through of a few approaches to the FTOA. It's really a very nice read and the plots bring it all together in a way that many people feel is intuitive. If you're good with coding then you can leverage something like SAGE or mathematica to make some plots of your own and understand the reasoning of the FTOA with your own examples too!

• If that doesn't lock it in then take a stab at reading through Fine and Rosenberger's book. You can find it in most university libraries with a strong math department. They'll walk you through the FTOA from three different perspectives; algebra, complex analysis, and topology. It's a longer approach perhaps but I suspect it will bring a lot of mathematical loose ends together for you.

Best of luck and success in your studies! You've asked a great question and that's where it all starts.

Answer 2

Of course there are many proofs, and perhaps some others will post the most attractive proofs, but I think you are looking for an intuitive explanation that would somehow make the result seem less surprising.

One such explanation, I think, is the simple observation that reals already go a long way towards being algebraically closed---they are a real closed field---since every odd-degree polynomial over $\mathbb{R}$ has a root in $\mathbb{R}$. This follows immediately from the intermediate value theorem, since in the large scale every odd degree polynomial moves from $-\infty$ to $\infty$ or conversely and hence must cross the axis.

Answer 3

One definition of the complex numbers is that they are the algebraic closure of the reals.

In other words, start with the reals and write down some degree-$d$ polynomial that has fewer than $d$ roots (incl multiplicities). Call one of these roots $x$. Now extend the reals with $x$. After you continue this (infinite) process, you have $\mathbb{C}$.

If you're willing to buy this as the definition of $\mathbb{C}$, then it's trivial to see that it's algebraically closed.

Answer 4

Here are 3 facts that I think provide some kind of intuition :

• To show that $\mathbb{C}$ is algebraically closed, you only need to show that real polynomials have a root in $\mathbb{C}$ : because of Taylor's formula, a complex polynomial taking real numbers to real numbers is actually real. Now if $P \in \mathbb{C}[X]$ is a complex polynomial, then $P\overline{P}$ is a real polynomial and has the same roots as $P$.

• All odd degree real polynomials have a root in $\mathbb{R}$ (because they are continuous and the limits at $-\infty$ and $\infty$ have different signs).

• Because of the quadratic formula, solving degree 2 equations only requires taking square roots, which is always possible in $\mathbb{C}$ because of the geometric interpretation and because square roots of positive numbers exist in $\mathbb{R}$ (if you write $z = \rho e^{i \theta}$, then a square root of $z$ is given by $\sqrt{\rho} \ e^{i \theta /2}$).

From these 3 facts (notice that they use some analysis) and some clever algebraic manipulations (which you can find on Wikipedia in the section "Algebraic proofs"), you can deduce that $\mathbb{C}$ is algebraically closed. This is in my opinion the closest you can get to an intuition.

Answer 5

Let $p(x)$ be an irreducible polynomial in $\Bbb R[x]$. It may be factored as $p(x)=\prod (x-\alpha_i)$ in $\Bbb R^a$, and hence $p'(x)=\sum_j\prod_{i\ne j}(x-\alpha_i)$. For each linear factor of $p(x)$, there is exactly a term in the sum that is not divisible by it, and so $\gcd(p,p')=0$. Thus, every irreducible polynomial in $\Bbb R[x]$ is separable, and every finite extension of $\Bbb R$ is Galois.

Let $K/\Bbb R$ be a finite extension. It is Galois. Its Galois group $G=Gal(K/\Bbb R)$ has order $[K:\Bbb R]=[K:\Bbb R(i)][\Bbb R(i):\Bbb R]$, and $[\Bbb R(i):\Bbb R]=2$, so $2\mid [K:\Bbb R]$. Therefore, $G$ has a Sylow 2-subgroup $H$. Denote its fixed field by $K^H(=\{a\in K\mid \sigma a=a\;\forall \sigma\in H\})$, then $[G:H]=[K^H:\Bbb R]$ by the Galois correspondence. This quantity is an odd integer since $H$ is a 2-subgroup with maximal order.

By the primitive element theorem, $K^H=\Bbb R(\beta)$ for some $\beta\in K^H$, and hence $[K^H:\Bbb R]=[\Bbb R(\beta):\Bbb R]$, the degree of the minimal polynomial of $\beta$ over $\Bbb R$, is odd. But every odd degree polynomial in $\Bbb R[x]$ has a root in $\Bbb R$, so the minimal polynomial, being irreducible, must have degree $1$. In other words, $K^H=\Bbb R$ and $[K:\Bbb R]=|G|=|H|=[K:K^H]$, which is a power of 2. Consequently, $[K:\Bbb R(i)]$ is also a power of 2.

Suppose that $[K:\Bbb R(i)]\ne 1$ i.e. $Gal(K/\Bbb R(i))$ is not trivial. Then it has a subgroup of index 2, which corresponds to an extension $E$ of degree 2 over $\Bbb R(i)$. Then the minimal polynomial of an element $\gamma$ in $E\setminus \Bbb R(i)$ is of the form $x^2+bx+c$ where $b,c\in \Bbb R(i)$. By completing the square, we have $(x+\dfrac b2)^2=\dfrac{b^2}4-c$. The right hand side lies in $\Bbb R(i)$ and thus has a square root in $\Bbb R(i)$, so the original polynomial actually is reducible, yielding a contradiction. Therefore, $K=\Bbb R(i)$, which completes the proof.