## 9.23 The complex numbers

The fundamental theorem of algebra states that the field of complex numbers is an algebraically closed field. In this section we discuss this briefly.

The first remark we'd like to make is that you need to use a little bit of input from calculus in order to prove this. We will use the intuitively clear fact that every odd degree polynomial over the reals has a real root. Namely, let $P(x) = a_{2k + 1} x^{2k + 1} + \ldots + a_0 \in \mathbf{R}[x]$ for some $k \geq 0$ and $a_{2k + 1} \not= 0$. We may and do assume $a_{2k + 1} > 0$. Then for $x \in \mathbf{R}$ very large (positive) we see that $P(x) > 0$ as the term $a_{2k + 1} x^{2k + 1}$ dominates all the other terms. Similarly, if $x \ll 0$, then $P(x) < 0$ by the same reason (and this is where we use that the degree is odd). Hence by the intermediate value theorem there is an $x \in \mathbf{R}$ with $P(x) = 0$.

A conclusion we can draw from the above is that $\mathbf{R}$ has no nontrivial odd degree field extensions, as elements of such extensions would have odd degree minimal polynomials.

Next, let $K/\mathbf{R}$ be a finite Galois extension with Galois group $G$. Let $P \subset G$ be a $2$-sylow subgroup. Then $K^ P/\mathbf{R}$ is an odd degree extension, hence by the above $K^ P = \mathbf{R}$, which in turn implies $G = P$. (All of these arguments rely on Galois theory of course.) Thus $G$ is a $2$-group. If $G$ is nontrivial, then we see that $\mathbf{C} \subset K$ as $\mathbf{C}$ is (up to isomorphism) the only degree $2$ extension of $\mathbf{R}$. If $G$ has more than $2$ elements we would obtain a quadratic extension of $\mathbf{C}$. This is absurd as every complex number has a square root.

The conclusion: $\mathbf{C}$ is algebraically closed. Namely, if not then we'd get a nontrivial finite extension $K/\mathbf{C}$ which we could assume normal (hence Galois) over $\mathbf{R}$ by Lemma 9.16.3. But we've seen above that then $K = \mathbf{C}$.

Proof. See discussion above. $\square$

Comment #1097 by Matthieu Romagny on

A conclusion we can draw from the above is that...

Comment #1098 by Antoine Chambert-Loir on

Grammar (first line) : the field of complex numbers is algebraically closed.

Comment #3583 by Souparna Purohit on

A minor fix: Third line of the fourth paragraph: "... hence by the above $K^P = K$" should read "$K^P = \mathbf{R}$".

Comment #5018 by Anonymous on

There are actually two inputs from calculus: the intermediate value theorem has to be used again to show that every positive real number (and hence every complex number) has a square root.

Comment #5259 by on

@#:5018 Yes, this is true. But I don't think the text is misleading as it doesn't claim the intermediate value theorem is the only thing from calculus that is used (also I think you can use the intermediate value theorem to prove the existence of square roots, no?).

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