Let $\mathcal{L}$ be the first-order language of theory of rings. Let A,B be two infinite fields. Suppose that A is existentially closed in B. We know that

A is existentially closed in B iff B can be embedded in a $\mathcal{L}$-structure B* which is elementary extension of A

Since A is existentially closed in B, A is also algebraically closed in B. The above proposition also tell us that A is algebraically closed in B* since B*/A is a regular extension. Here is my question :

Is B necessary algebraically closed in B* ?

I guess $\mathbb{C}$ and the rational function field $\mathbb{C}(T)$ may be a counterexample.


1 Answer 1


Your counterexample works. $\mathbb{C}$, being an existentially closed field, is existentially closed in $\mathbb{C}(T)$. So we can find an extension field $K$ with $\mathbb{C}(T)\subseteq K$ and $\mathbb{C}\preceq K$ (for example, $\overline{\mathbb{C}(T)}$ works). Then $K$ is algebraically closed, so $\mathbb{C}(T)$ is not algebraically closed in $K$ (for example, $K$ contains $\sqrt{T}$).


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