Does there exist a subfield $K\subsetneq \mathbb{C}$ such that $x$ is transcendental over $K$ for all $x\notin K$?

Question

My conjecture is no.

Proof: Suppose, to the contrary, that such a field $K\subsetneq \mathbb{C}$ exists. Then for all $x\notin K$, $p(x)\neq 0$ for all $p(t)\in K[t]\backslash\{0\}$. Screeching halt.

Any hint would be appreciated.

Answer 1

Hint:

If $K \subsetneq \mathbb{C}$ is algebraically closed, then every $x \in \mathbb{C}$ algebraic over $K$ actually lies in $K$ (do you see why?). So that means every $x \in \mathbb{C} \setminus K$ must be transcendental over $K$.

Now: do you know of any algebraically closed proper subfields of $\mathbb{C}$?

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I hope this helps ^_^