# The Fundamental Theorem of Algebra – Galois Theory-style

I’ve been having a new look into Galois theory during the past couple of months. Just to settle my curiousity, not because I really needed to. Actually, I learned about Galois theory about 8 years ago in an undergraduate lecture. I was way too undergraduate then, so I didn’t look into the deeper details then – all that I really memorized were the basic ideas of what the theory was supposed to be about. Then again, every book on mathematics will tell you that Galois theory is the most elegant branch of mathematics you can find. And it will allow to prove many of the neat theorems that stand in the center of attention of classical mathematics, such as “you can only find a general solution for equations of degree 4 or less” and “squaring the circle is impossible”. So now, I spent some time diving into the theory to really understand the solutions to these problems, and the background of why they are the way they are. When I stumbled over the field-theoretic proof of the Fundamental Theorem of Algebra, the proof seemed so nice that I felt the need to share this 🙂

I remember reading about this proof, when I first learned about Galois theory years ago. Then, it seemed very awkward and hard. But then again, I lacked the background that allowed me now to grasp all the details of the theory. All of a sudden, the proof doesn’t seem hard anymore, much the contrary. But of course, I have been messing around with the details so thoroughly, so I can’t blame you, if you don’t agree with me on my notion of elegance here. As always, either you don’t get the proof at all, or you consider it trivial. There is pretty much no room for anything in between.

The Fundamental Theorem of Algebra says that any polynomial $p(x):=x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$, where $a_i\in\mathbb{C}$ and $n>0$, has some root $\alpha\in\mathbb{C}$, which means $p(\alpha)=0$. One says: $\mathbb{C}$ is algebraically closed. The theorem is wrong for the real numbers, because the polynomial $x^2+1$ doesn’t have any zeroes in $\mathbb{R}$.

The shortest proof that I know of comes from complex analysis. Suppose the theorem is wrong, then $\frac{1}{p}$ is a holomorphic function (because $p$ is, and $p$ doesn’t vanish anywhere by assumption). Besides, $p$ is bounded, because it can’t approach $0$ too closely (because of continuity). Then Liouville’s theorem says $p$ is constant. Thus, $n=0$ – contradiction. q.e.d.

Now, we’ll have a look at the Galois theory proof. I can’t give an exhausting account on the theorems you’d need to prove first, I’ll just focus on the elegant part of the thing here. The idea isn’t mine anyway, we have to bow to Artin for getting these nice ideas. The presentation of the following proof is relying on chapter 8 of the book “Galois theory” by David Cox. I can’t recommend this book strong enough to any beginner in Galois theory.

It will be enough to prove that any polynomial $f\in\mathbb{R}[x]$ splits completely over $\mathbb{C}$ (because for given complex polynomials $p$, you can consider the real polynomial $f:=p\overline{p}$). There is a splitting field $L\supset\mathbb{R}$ of $f$, this field extension is separable because we’re dealing with characteristic $0$. Since any splitting field is a normal field extension, we have a Galois extension. This is about the nicest kind of field extension I have ever met – the entire theory builds on the nice properties of these extensions. In particular, the group of automorphisms of this field extension $G=\mathrm{Gal}(L/\mathbb{R})$ has as many elements as the vector-space-dimension of $L$ over $\mathbb{R}$. Now, let’s look at the subgroup $H\subset G$, which shall be $\{\mathrm{id}\}$ if $G$ is odd, and which shall be a Sylow-$2$-group if $G$ is even.

This used to be the tricky part for me. One shouldn’t be too afraid of the Sylow theorems – what we need of them here is, there is a subgroup whose order is the highest possible order of $2$ (which means $|H|=2^k$ if $|G|=2^km$ and $m$ is odd). There may be several of these groups, but there is at least one of them. This is a partial reverse of Lagrange’s theorem. One can prove this by using some group operations on $G$, which don’t matter too much here. The surprising thing for me is: the structure of groups is pretty strict. Once you impose a group structure on a set, you give up a surprisingly big lot of freedom for what you can do with the set (such as: making useful subsets – you can’t just take any subset you like).

Now comes the pretty part of the proof (at least if you can enjoy Galois theory – if not… well, sorry):

Any subset of $G$ has a fixed field in $L$, let’s call this one $L_H$. By the fundamental theorem of Galois theory, $[L_H :\mathbb{R}]=[G:H]=\frac{|G|}{|H|}$. By construction of $H$, this is odd. Now, any field extension of $\mathbb{R}$ is generated by some primitive element $\alpha\in L_H$ (this is the theorem of the primitive element). This $\alpha$ has a minimal polynomial with degree $[L_H :\mathbb{R}]$. But any polynomial of odd degree has a real root, because of the mean value theorem. So, $\alpha$ is real already. So, there’s no choice but $L_H=\mathbb{R}$. Then again, $H=G$ by the Galois correspondence, and then the order of $G$ is a power of $2$, $|G|=2^m$, say.

From here on, the proof is shuffling around the Galois correspondence. This looks straight-forward, but you mustn’t lose your nerves doing it.

Of course, if $m=0$, nothing is to be done, $G$ is trivial, so $L=\mathbb{R}$ and all roots of the polynomial are real.

More interestingly, if $m\geq1$, $G$ is a group of order of a prime power. Those groups are solvable, because by Lagrange’s theorem, all of its subgroups will have some lower power of the same prime as their order. Since these groups must by cyclic, we’ll arrive at a tower of groups $\{\mathrm{id}\}=G_n\subset G_{n-1}\subset\cdots\subset G_1\subset G_0=G$ and this corresponds to a tower of fields $\mathbb{R}=L_{G_0}\subset L_{G_1}\subset\cdots\subset L$. Each of these fields has dimension $2$ over the previous one by Galois correspondence. In particular, $\mathbb{R}=L_{G_1}$ has dimension $2$, so it is generated by some $\beta\in L_{G_1}$ whose minimal polynomial is quadratic with no real roots. But by the quadratic solution formula, this polynomial has two complex roots – there is no choice but $L_{G_1}\cong \mathbb{C}$. So far, so good.

Now suppose, $m\geq 2$. Then, $L_{G_1}\subset L_{G_2}$ is an extension of $L_{G_0}\cong \mathbb{C}$ of degree $2$. So, there should be some $\gamma$ with quadratic minimal polynomial over $\mathbb{C}$. But quadratic complex polynomials already have their roots in $\mathbb{C}$. So, $L_{G_1}\cong \mathbb{C}$. The same argument applies to the rest of the tower of fields, so we’ll arrive at $m=1$. In the end we find, that $m>0$ yields $m=1$, thus $|G|=2$ and $L=\mathbb{C}$. The polynomial $f\in\mathbb R[x]$ splits completely over $\mathbb{C}$. q.e.d.

The nice thing for me about this proof was, that you can boil down the possibly terrifyingly complicated polynomial $f$ to some Galois group of even order. Even more, there can’t be any non-trivial case, unless there is no odd prime involved at all. Note however, you can’t do the proof without calculus, you need the mean value theorem here (this is this proof’s way of mentioning the completeness of the reals).

I am still about to understand, in what way this proof of the Fundamental Theorem of Algebra is related to the inductive proof given in Chapter 3 of Cox’s book. The induction is remarkable for increasing the degree of the considered polynomial not by 1 in the inductive step, but by a power of $2$ each time. I guess that this corresponds to the powers of the Galois groups involved here, but I still need some time to figure this one out. Let’s see if this train of thought will “make my day” once I get it…