In page 20 of Marker's Model Theory: An Introduction it's explained how $\mathbb{C}$ is a definable subset of the structure $\mathscr{M}= (\mathbb{C}(X), +, -, \cdot, 0,1)$ over the usual language of rings via the formula $\exists x \exists y (y^2 = v \wedge x^3 +1= v)$; the argument made there was using some theory of elliptic curves.
Is there any other way in which one can define $\mathbb{C}$ in $\mathscr{M}$ without making use of results about elliptic curves? I suspect that the answer tends more towards to "no" rather than to "yes"; in a previous example in the same page it's shown how to construct a $\emptyset$-definable ordering in $(\mathbb{Z}, +, -, \cdot, 0,1)$ over the language of rings using Lagrange's Theorem, so it seems that in general, if you can use some known machinery/results of your structure, then use it in your advantage. I'm just curious if anyone knows an alternative way of defining $\mathbb{C}$ in $\mathscr{M}$.
Here's a more elementary way of showing that $\mathbb{C}$ is definable in $\mathbb{C}(X)$. I found this solution in the book Model Theoretic Algebra by Jensen and Lenzing, which is a great reference for questions about definability in rings, fields, and modules. It's Proposition 3.3 on p. 34 of that book, and the method applies more generally to define $K$ in $K(X)$ whenever $K$ is a Pythagorean field (a field in which any sum of squares is a square) of characteristic $\neq 2$.
Consider the formula $\varphi(x)$: $$\exists y\, (1 + x^4 = y^2).$$ If $a\in \mathbb{C}$, then $1 + a^4\in \mathbb{C}$ is a square in $\mathbb{C}$, and hence also in $\mathbb{C}(X)$, so $\mathbb{C}(X)\models \varphi(a)$.
Conversely, suppose $a\in \mathbb{C}(X)$ and $\mathbb{C}(X)\models \varphi(a)$. Then there is some $b\in \mathbb{C}(X)$ such that $1 + a^4 = b^2$. Writing $a = p/q$ and $b = r/s$ in lowest terms, with $p,q,r,s\in \mathbb{C}[X]$, we have $1 + p^4/q^4 = r^2 / s^2$. Clearing denominators, $(q^4 + p^4)s^2 = r^2q^4$. Since $b = r/s$ is written in lowest terms, $s$ and $r$ are relatively prime. Thus $s^2 | q^4$, and hence $s | q^2$. Writing $q^2 = st$ for some $t\in \mathbb{C}[X]$, we have $q^4 = s^2t^2$. So $(q^4 + p^4)s^2 = r^2q^4 = r^2s^2t^2$, and $q^4 + p^4 = u^2$, where $u = rt$. Further, $p$, $q$, and $u$ are pairwise relatively prime, since a common irreducible factor of any two of the three would also divide the third, contradicting the assumption that $a = p/q$ is written in lowest terms.
Now it suffices to prove the following claim, since then $a = p/q\in \mathbb{C}$.
Claim: Suppose $p,q,u\in \mathbb{C}[X]$ satisfy $p^4 + q^4 = u^2$ and are pairwise relatively prime. Then $p,q,u\in \mathbb{C}$.
Proof: By induction on $\max(\deg(p),\deg(q))$. If $\max(\deg(p),\deg(q))\leq 0$, then $p,q\in \mathbb{C}$, so $u\in \mathbb{C}$ as well.
Now assume $\max(\deg(p),\deg(q))> 0$. By symmetry, we may assume $\deg(p) \leq \deg(q)$. Note that $2\deg(u) = \deg(u^2) \leq \max(\deg(p^4),\deg(q^4)) = 4\deg(q)$, so $\deg(u) \leq 2\deg(q) = \deg(q^2)$.
Rewriting $u^2 - q^4 = p^4$, we have $(u+q^2)(u-q^2) = p^4$. Now $(u+q^2)$ and $(u-q^2)$ are relatively prime, since a common irreducible factor would divide both $(u+q^2) + (u-q^2) = 2u$ and $(u+q^2) - (u-q^2) = 2q^2$, and hence would divide both $u$ and $q$. So each irreducible factor of $p$ divides exactly one of $(u+q^2)$ or $(u-q^2)$.
Since also any unit is a $4^\text{th}$ power in $\mathbb{C}$, it follows that $p$ factors as $p = \hat{p}\hat{q}$, where $(u+q^2) = \hat{p}^4$ and $(u - q^2) = \hat{q}^4$, and $\hat{p}$ and $\hat{q}$ are relatively prime. Then $2q^2 = (u + q^2)-(u-q^2) = \hat{p}^4 - \hat{q}^4$, so $\hat{u}^2 = \hat{p}^4 + (\zeta \hat{q})^4$, where $\hat{u} = \sqrt{2}q$ and $\zeta$ is a primitive $8^{\text{th}}$ root of unity.
We have $4\deg(\hat{p}) = \deg(\hat{p}^4) = \deg(u+q^2) \leq \deg(q^2) = 2\deg(q)$, where the inequality follows from the observation $\deg(u) \leq \deg(q^2)$ above. So $\deg(\hat{p}) \leq \deg(q)/2$. Similarly, $\deg(\hat{q}) \leq \deg(q)/2$. So $\max(\deg(\hat{p}),\deg(\zeta\hat{q})) < \deg(q) = \max(\deg(p),\deg(q))$.
Also, $\hat{p}$ and $\zeta\hat{q}$ are relatively prime, and hence $\hat{p}$, $\zeta\hat{q}$, and $\hat{u}$ are pairwise relatively prime, since a common irreducible factor of any two of the three would also divide the third.
By induction, $\hat{p}$, $\zeta\hat{q}$, and $\hat{u}$ are in $\mathbb{C}$. But then $\sqrt{2}q = \hat{u}$ and $p = \hat{p}\hat{q}$ implies $p,q\in \mathbb{C}$, and thus also $u\in \mathbb{C}$.