Does there exist a field $K$ such that $\mathbb R \subsetneq K \subsetneq \mathbb C$? I'm thinking of unions of $\mathbb R$ with some subset of $\mathbb C$ but am not sure how to approach this without ending up with all of $\mathbb C$. Doe anyone have any suggestions?
 A: Any intermediate field between $\Bbb R$ and $\Bbb C$ is in particular an $\Bbb R$-vector subspace of $\Bbb C$. Since $\dim_{\Bbb R}\Bbb C=2$, it is either equal to $\Bbb R$ or to $\Bbb C$.
A: and just very simply if $z \in \mathbb{K} \subset \mathbb{C}$ but $z \notin \mathbb{R}$ then it must be of the form $a + ib$ with $a,b \in \mathbb{R}$ and $b \ne 0$. hence $i \in \mathbb{K}$ so $\mathbb{C} \subset \mathbb{R}(i) \subset \mathbb{K}$
A: No, since if $\mathbb R\subset K$, then $K$ is a vector space over $\mathbb R$ and in the same way $K\subset\mathbb C$ means that $\mathbb C$ is a vector space over $K$. Finally $\mathbb C$ is a vector space over $\mathbb R$ of dimension 2, and $K$ is a subspace. Therefore $K$ has dimension either 1 or 2. In the first case it is isomorphic to $\mathbb R$ and contains $\mathbb R$, so it is $\mathbb R$ and in the second - all of $\mathbb C$.
A: No: the dimension of $\mathbb{C}$ as a vector space over $\mathbb{R}$ is $2$. 
If $K$ is an extension field of $F$, denote by $[K:F]$ the dimension of $K$ as a vector space over $F$. Then, if $F_1\subseteq F_2\subseteq F_3$ are fields, we have
$$
[F_3:F_1]=[F_3:F_2][F_2:F_1]
$$
(assuming $[F_3:F_1]$ is finite then also $[F_3:F_2]$ and $[F_2:F_1]$ is finite and conversely).

Of course there's an elementary proof. Suppose $a+bi\in K$, $a+bi\notin\mathbb{R}$. Then $b\ne0$, so
$$
i=((a+bi)-a)b^{-1}\in K
$$
and therefore $K=\mathbb{C}$.
A: No. Look at the tower lemma. Otherwise $[\mathbb{C}:\mathbb{R}]=[\mathbb{C}:\mathbb{K}][\mathbb{K}:\mathbb{R}]>2$ which is a contradiction . By the way, take a look at the Artin Schreier theorem. It is magnificent.
A: suppose a field $K\subseteq \mathbb C $ contains $\mathbb R$ properly. Then it contains a non-real element $x+iy$, consequently it contains all elements of the form $z+iy$ with $z\in \mathbb R$.
We conclude it contains $z^2-y^2 + 2zyi$ for all $z\in \mathbb R$. And therefore it contains $2zyi$ for all $z\in \mathbb R$.
We conclude it contains $zi$ for every $z\in \mathbb R$, this shows $K=\mathbb C$.
