Pointwise definable elements of the complex field. This is really two questions. Consider the structure $(\mathbb{C};+,-,*,0,1)$. Are the pointwise definable elements in that structure precisely the rational numbers? The second question is about the structure $(\mathbb{C};+,-,*,0,1,i)$. I conjecture that the pointwise definable elements in that structure are the complex numbers which have both a rational real part and a rational imaginary part. Are either or both of these conjectures true?
 A: Your intuition is exactly correct; an element $\alpha\in\mathbb{C}$ is $\varnothing$-definable if and only if $\alpha\in\mathbb{Q}$, and it is $\{i\}$-definable if and only if $\alpha\in\mathbb{Q}(i)$. More generally, $\alpha$ is $\varnothing$-algebraic or $\{i\}$-algebraic if and only if it lies in $\overline{\mathbb{Q}}$, the algebraic closure of $\mathbb{Q}$ in $\mathbb{C}$. There are a number of ways to see these facts; as Noah Schweber says, the easiest way is probably using automorphisms. (Recall that, if an element of some first-order structure is definable over some subset $A$ of parameters, then it will be fixed by every automorphism of the structure that fixes $A$ pointwise.) With Zorn's lemma and a bit of field theory, one can show that, for any subfield $F\subseteq\mathbb{C}$, two elements $\alpha,\beta\in\mathbb{C}$ are conjugate by an automorphism of $\mathbb{C}$ fixing $F$ pointwise if and only if one of the following holds:

*

*$\alpha$ and $\beta$ are both transcendental over $F$

*$\alpha$ and $\beta$ are both algebraic over $F$, and have the same minimal polynomial over $F$
Now, I assume you can show that every element of $\mathbb{Q}$ is $\varnothing$-definable, and that every element of $\mathbb{Q}(i)$ is $\{i\}$-definable. So we only need to show the reverse inclusions. Let $\alpha\in\mathbb{C}$. If $\alpha$ is transcendental over $\mathbb{Q}$ (resp. $\mathbb{Q}(i)$), then $\alpha/2$ is also transcendental over $\mathbb{Q}$ (resp. $\mathbb{Q}(i)$), and thus by condition 1 above we may find an automorphism of $\mathbb{C}$ fixing $\mathbb{Q}$ (resp. $\mathbb{Q}(i)$) and taking $\alpha$ to $\alpha/2$. In particular, $\alpha$ is not $\varnothing$-definable (resp. $\{i\}$-definable).
If $\alpha$ is algebraic over $\mathbb{Q}$ (resp. $\mathbb{Q}(i)$), let $f(x)$ be its minimal polynomial. Irreducible polynomials over a subfield of $\mathbb{C}$ have distinct roots in $\mathbb{C}$, so, if $\deg f>1$, then $f$ has another root $\beta\neq\alpha$. Now by condition 2 we can find an automorphism fixing $\mathbb{Q}$ (resp. $\mathbb{Q}(i)$) and taking $\alpha$ to $\beta$. In particular, once again, $\alpha$ is not $\varnothing$-definable (resp. $\{i\}$-definable). Thus, for $\alpha$ to be $\varnothing$-definable (resp. $\{i\}$-definable), $f$ must have degree 1; say $f=\lambda x+\mu$. Then $\alpha=-\mu/\lambda$, and hence lies in $\mathbb{Q}$ (resp. $\mathbb{Q}(i)$), as desired.

Edit: In answer to Mitchell Spector's question below, it is worth pointing out that the proof above is, in a certain sense, very strong overkill. The existence of the desired automorphisms of $\mathbb{C}$ relies on the axiom of choice, but the desired result still holds in $\text{ZF}$. The main point is that quantifier elimination for $(\mathbb{C},+,\cdot)$ can be proven choicelessly. In particular, the type of an element $\alpha\in\mathbb{C}$ over $\varnothing$ (resp $\{i\}$) is uniquely determined by the set of polynomials over $\mathbb{Q}$ (resp $\mathbb{Q}(i)$), if any, that it satisfies. Since minimal polynomials of algebraic elements over a subfield of $\mathbb{C}$ exist (no need for choice here), we get that two elements of $\mathbb{C}$ have the same type over $F=\mathbb{Q}$ (resp $F=\mathbb{Q}(i)$) if and only if:

*

*They are both transcendental.

*They are both algebraic, and have the same minimal polynomial over $F$.

Thus we can use exactly the same argument as above; $\alpha$ will be definable over $\varnothing$ (resp $\{i\}$) if and only if it is algebraic over $\mathbb{Q}$ (resp $\mathbb{Q}(i)$) and the only root of its minimal polynomial, if and only if it lies in $\mathbb{Q}$ (resp $\mathbb{Q}(i)$), as desired.
For some related questions about how much of the model theory of $\text{ACF}$ (and also $\text{RCF}$) can be carried out without choice, see this recent question of Joel Hamkins on mathoverflow, and this earlier question on math.stackexchange.

Edit 2: As one final point, as Alex Kruckman notes in the comments below, it's worth pointing out that an analogous fact holds for any set of parameters; the definable closure of a subset $A\subseteq\mathbb{C}$ is precisely $\mathbb{Q}(A)$, the subfield of $\mathbb{C}$ generated by $A$. The argument for this is exactly the same as above! But, as Alex points out, this fact does not quite extend to the case of algebraically closed fields of characteristic $p>0$ – we use in a crucial way the fact that every irreducible polynomial over a field of characteristic $0$ is separable in the argument above, which is not in general the case for fields of characteristic $p$!
