Today we prove the very basic theorem
Theorem(FTA). $\mathbb{C}$ is algebraically closed.
We fix a $p\in\mathbb{C}[X]$ such that $\deg p\geq 1$, and suppose that $p$ has no roots. The simplest proof utilizes the idea of the maximum modulus principle.
Proof 1. There exists some $R>0$ satisfying $\vert z\vert>R\Longrightarrow\vert p(z)\vert>\vert p(0)\vert$. Let $f=1/p$. Then $\vert f\vert$ must attain a maximum on the disk $\overline{B(0,R)}$, but it’s not on the boundary, a contradicting with the maximum modulus principle.
One of the most beloved (well, I’m not actually sure) and intuitive proof is the following.
Proof 2. First, WLOG, write $p(z)=z^n+r(z)=z^n+O(z^{n-1})$.
Identify $S^1$ with $\{\vert z\vert=R\}\subset\mathbb{C}$, and take $R$ sufficiently large so that $z\in S^1\Longrightarrow\vert r(z)\vert<\vert z^n\vert$. Therefore the line segment $[z^n,p(z)]$ does not contain $0$ for all $z\in S^1$.
This means there’s a homotopy of $z^n\simeq p(z)$ of maps $S^1\to\mathbb{C}\setminus\{0\}$. Clearly, being restricted from a contractible space, $p(z)$ is null-homotopic. But $z^n$ is not; it has $\deg=n$ if we identify $\mathbb{C}\setminus\{0\}\simeq S^1$.
We also have the algebra-flavored proof, generalized to the notion of real closed fields. The following definitions sets the stage for us.
Definition. An ordered field is a field $F$ with a subset $P$ called the positive cone, satisfying
It follows that
Define $x\leq y\Longleftrightarrow y-x\in P\cup\{0\}$. Then $(F,\leq)$ is a totally ordered set.
$1\in P$, so $\operatorname{char}(F)=0$; more generally,
Any sum of squares is $\geq 0$.
In particular, $-1<0$ and is thus not a square, so
Corollary. $\mathbb{C}$ is not an ordered field.
Definition. An ordered field $F$ is called a real closed field, if
Lemma 1. Every element of $F(\sqrt{-1})$ is a square.
Proof. Observe that $a$, $b\in F\Longrightarrow a+b\sqrt{-1}=(x+y\sqrt{-1})^2$, where
\[x=\sqrt{\frac{a+\sqrt{a^2+b^2}}{2}},\] \[y=\frac{b}{2x}.\]
We need the following lemma for the final proof.
Lemma 2. Any finite $p$-group $G$ has subgroups of all possible orders.
Proof. Use induction on the order $\vert G\vert$. When $\vert G\vert=p$ there’s nothing to prove. In general, since the center $Z$ of $G$ is nontrivial, we have either $Z=G$, i.e. $G$ is abelian; or $Z$ has a smaller order than $G$. In the first case, the result follows from that theorem blah blah blah on a PID, and the other case we can pass to $G/Z$ and $Z$ and use the induction hypothesis.
Theorem’. If $F$ is a real closed field then $F(\sqrt{-1})$ is algebraically closed.
Proof. Fix a nonconstant polynomial $p\in F(\sqrt{-1})[X]$. Consider the splitting field $E$ of $p\cdot(X^2+1)$ over $F$, forming the tower of extensions $E\vert F(\sqrt{-1})\vert F$. From the discussion above we know
- The are Galois extensions. Let $G=\operatorname{Gal}(E\vert F)$.
Let $H\leq G$ be the Sylow $2$-subgroup of $G$. By the Galois correspondence, $E^H\vert F$ is a subextension of odd degree, hence $E^H=F$. Therefore $G$ is a $2$-group.
It has a subgroup $\operatorname{Gal}(E\vert F(\sqrt{-1}))$. Suppose for the contrary that it is nontrivial. By Lemma 2, there is a subextension $L\vert F(\sqrt{-1})$ of degree $2$. However, in view of Lemma 1, this is not possible!
Therefore $E=F(\sqrt{-1})$, i.e., $p$ splits in $F(\sqrt{-1})$.
Post date: 2025/02/22
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