Is $\mathbb C$ the only extension of $\mathbb R$ with degree 2?

Question

Is $\mathbb C$ the only extension of $\mathbb R$ with degree 2?

$\mathbb C$ is the complex numbers, $\mathbb R$ is the real numbers.

I am pretty sure the answer is no: $\mathbb R(\sqrt{i+2})$ has degree 2 over $\mathbb R$, and $i$ is not in that extension and it is in $\mathbb C$.

Answer 1

Any algebraic extension of $\mathbb{R}$ is contained in $\mathbb{C}$, as $\mathbb{C}$ is the algebraic closure of $\mathbb{R}$. Therefore, $\mathbb{C}$ is the only extension of $\mathbb{R}$ of degree 2, up to $\mathbb{R}$-isomorphisms. The element $i$ is contained in your extension, since the extension contains $\sqrt{i+2}^2 = i+2$ and it contains $2$, so also $i+2-2 = i$.

Answer 2

An extension of $\mathbb{R}$ of degree $2$ is of the form $\mathbb{R}(\alpha)$, where $\alpha$ has minimal polynomial of degree $2$ over $\mathbb{R}$. It follows that $\alpha = x + iy \in \mathbb{C}$, where $y \neq 0$. Since $\mathbb{R}(\alpha)$ contains $\mathbb{R}$, we have $i \in \mathbb{R}(\alpha)$. Thus $\mathbb{R}(\alpha) = \mathbb{C}$.

More generally if any field extension of $\mathbb{R}$ contains a complex number that is not real, then it must contain $\mathbb{C}$. This shows that in your example, we actually have $\mathbb{R}(\sqrt{i+2}) = \mathbb{C}$. Furthermore, $\mathbb{C}$ is the only field extension of $\mathbb{R}$ that has finite degree (besides $\mathbb{R}$ itself).