For a subfield $K$ of $\Bbb C$ with $K\nsubseteq \Bbb R$, show $K$ is dense in $\Bbb C$.

Question

Let $K $ be a subfield of $\mathbb C$ not contained in $\mathbb R$. Is $K$ dense in $\mathbb C$?

My problem is I have never used the concept of dense set in algebra and neither have any idea about using sequences here. Please give some hints

Answer 1

If it is a subfield it must contain the rationals, an if it contains a number $a+bi$ with $b\neq 0$ then multiply by a rational to get $c+(1+\epsilon) i$ where $\epsilon$ is small. Then subtract a rational to get $\delta +(1+\epsilon )i$. This gives that $i$ is in the closure. I think it is easy to complete the argument, either by a generalization of the above or by noting that $\overline{K}$ is a field.

Answer 2

1) Let $K \subset \mathbb{C}$ be a field. Then $K \supset \mathbb{Q}$, so $K$ is a $\mathbb{Q}$-vector space.

2) For any $\mathbb{Q}$-subspace $V$ of Euclidean space $\mathbb{R}^N$, the closure of $V$ in $\mathbb{R}^N$ (in the Euclidean topology!) is $V_{\mathbb{R}}$, the $\mathbb{R}$-subspace it generates. More concretely, there are $v_1,\ldots,v_k \in V$ such that $V = \{ a_1 v_1 + \ldots + a_k v_k \mid a_i \in \mathbb{Q}\}$ and then $V_{\mathbb{R}} = \{ a_1 v_1 + \ldots + a_k v_k \mid a_i \in \mathbb{R}\}$. Since $V_{\mathbb{R}}$ is a closed subset containing $V$, one containment is clear. Writing things in terms of bases as above makes the other containment easy to see.

3) So if $K \subset \mathbb{C}$ is a field, $\overline{K}$ is an $\mathbb{R}$-subspace containing $\overline{\mathbb{Q}} = \mathbb{R}$. Thus $\mathbb{R} \subset \overline{K} \subset \mathbb{C}$. By dimension considerations the only possibilities are $\overline{K} = \mathbb{R}$ and $\overline{K} = \mathbb{C}$. The former occurs if and only if $K \subset \mathbb{R}$.