$\mathbb{C}$ is the smallest algebraically closed field containing $\mathbb{R}$

Question

How can I prove that $\mathbb{C}$ is the smallest algebraically closed field containing $\mathbb{R}?$

I know that the field $F$ is algebraically closed if every polynomial with complex coefficients have a root in F.

Answer 1

If $\mathbb{R} \subset F$ and $F$ is algebraically closed, then all roots of $z^2+1$ should belong to $F$.

Answer 2

An even stronger true statement is that there are no intermediary fields $F$ such that $\mathbb R \subsetneq F \subsetneq \mathbb C$.

Proof: Let $z \in F \setminus \mathbb R$. Then $z$ is of the form $x + iy$ for some $x, y \in \mathbb R$, and since $z \notin \mathbb R$, $y$ must be nonzero. Then since $F$ is a field containing $\mathbb R$, we may perform some operations on $z$ using real numbers and stay within $F$:

$$i = \frac{z - x}{y}$$

And thus for any $x' + i y' \in \mathbb C$,

$$x' + i y' = x' + (\frac{z - x}{y})y'$$

Hence $x' + i y' \in F$, and hence $F$ contains all of $\mathbb C$.

Answer 3

The fact that $\mathbb{C}$ is algebraically closed is the fundamental theorem of algebra. Any algebraically closed field $F$ containing $\mathbb{R}$ must contain $i$, because the polynomial $x^2 + 1 = 0$ must have a root. Then since $\mathbb{R} \subseteq F$ we must have $a + bi \in F$ for all $a, b \in \mathbb{R}$. Hence $\mathbb{C} \subseteq F$.

Answer 4

Any field is a linear space over any subfield, but we know that

$$\;\dim_{\Bbb R}\Bbb C=2\;\implies \;$$

there can't be any other field $\;K\;$ s.t. $\;\Bbb R\subsetneqq K\subsetneqq\Bbb C\;$ by plain linear algebra...and this is even stronger than what you need.