Is there a ring $K$ such that $\mathbb R\subsetneqq K\subsetneqq \mathbb C$? Is there a ring $K$ such that $\mathbb R\subsetneqq K\subsetneqq \mathbb C$? and if $K$ is a local ring, is there such a ring? I need this result be negative to prove a question I'm working.
I need help
Thanks in advance
 A: No. for every element $\alpha\in\mathbb{C}-\mathbb{R}$ the ring $\mathbb{R}\left[\alpha\right]$,
i.e. the smallest subring of $\mathbb{C}$ that contains $\mathbb{R}$
and $\alpha$, can be shown to satisfy $\mathbb{R}\left[\alpha\right]=\mathbb{C}$.
It is enough allready to prove that it contains $i$, and $\alpha\in\mathbb{C}-\mathbb{R}$ tells us that $\text{Im}\alpha\neq0$.

edit:
To make things complete. If $\alpha\in\mathbb{C}-\mathbb{R}$ then
evidently $\left\{ r+s\alpha\mid r,s\in\mathbb{R}\right\} =\mathbb{C}$
and any subring $K$ of $\mathbb{C}$ that contains $\mathbb{R}$
and $\alpha$ will contain this set.
A: If $L/K$ is an algebraic extension, then every ring $R$ with $K \subseteq R \subseteq L$ is a field. In fact, $R$ is an integral domain, and $K \to R$ is integral, hence $R$ is a field by a general fact.
But by degree considerations, there is no non-trivial intermediate field between $\mathbb{R}$ and $\mathbb{C}$.
A: Assume $K$ is a subring of $\mathbb{C}$ over $\mathbb{R}$ and we
show that it must be one of them.
If all elements of $K$ are real then $K=\mathbb{R}$ and otherwise
there is $\alpha\in K\setminus\mathbb{R}$.
Consider the ring $K=K[\alpha]$ - clearly 
$$
\mathbb{R}[\alpha]\subseteq K[\alpha]\subseteq\mathbb{C}[\alpha]
$$
but since $\alpha\in\mathbb{C}\setminus \mathbb{R}$ we have that 
$$
\mathbb{R}[\alpha]=\mathbb{C}[\alpha]=\mathbb{C}
$$
and thus 
$$
K=K[\alpha]=\mathbb{C}
$$
A: Any ring containing a field is a vector space over this field.
