Is it true that $\mathbb{C}(x) \equiv \mathbb{C}(x, y)$? It is easily seen that any two consecutive entries in the tower of fields given below are not elementarily equivalent in the language of rings:
$$\mathbb{Q} \subseteq \mathbb{Q}(\sqrt{2}) \subseteq \mathbb{R} \subseteq \mathbb{C} \subseteq \mathbb{C}(t).$$
For example, $\mathbb{C} \not\equiv \mathbb{C}(t)$ since (for example) the sentence $$\forall x \ \exists y \ x = y^2$$ holds in $\mathbb{C}$ but not in $\mathbb{C}(t)$. So it seems natural to ask: are $\mathbb{C}(t)$ and $\mathbb{C}(t_1,t_2)$ elementarily equivalent? More generally, for what fields $\mathbb{F}$ is it true that $\mathbb{F}(t) \equiv \mathbb{F}(t_1,t_2)$?
There are many well-known results involving the elementary equivalence of fields of rational functions. For example, it is known that given a field $K$ which admits a unique ordering, $K(x) \equiv \mathbb{Q}(x)$ implies $K \cong \mathbb{Q}$.
I have considered trying to use the Keisler-Shelah isomorphism theorem to prove or disprove that $\mathbb{C}(t) \equiv \mathbb{C}(t_1,t_2)$, but it is not obvious as to whether or not ultrapowers corresponding to $\mathbb{C}(t)$ and $\mathbb{C}(t_1,t_2)$ are isomorphic.
 A: Theorem 2.40 in the book Model Theoretic Algebra by Jensen and Lenzing gives a negative answer. This book is a great reference for questions like this, you'll probably find it very interesting. For those without access to the book, I'll summarize the reason that $\mathbb{C}(x) \not\equiv \mathbb{C}(x,y)$.
A field $K$ is called a $C_i$-field if every homogeneous polynomial of over $K$ of degree $d$ in $n$ variables, such that $n>d^i$, has a nontrivial zero in $K^n\setminus \{(0,\dots,0)\}$. It's not too hard to show that a field is $C_0$ if and only if it is algebraically closed.  
Now it's a fact that $K$ is $C_i$ if and only if $K(x)$ is $C_{i+1}$. Jensen and Lenzing don't prove this, instead citing Chapter XI of Ribenboim's book L'arithmetique des corps, which might be difficult to track down... It would be nice if someone could post a more accessible reference. You can actually find the direction "if $K$ is $C_i$ then $K(x)$ is $C_{i+1}$" as Theorem 3.3.9 in this thing I wrote once, but the converse is crucial here.
Now $K_1 = \mathbb{C}(x)$ is $C_1$ but not $C_0$ (it's not algebraically closed), so $K_2 = \mathbb{C}(x,y)$ is $C_2$ but not $C_1$. Hence there is some $d^2\geq n > d^1$ and some homogeneous polynomial $p(\overline{x})$ in $n$ variables of degree $d$ over $K_2$ which has no nontrivial zero over $K_2$. Fixing $n$ and $d$ and quantifying over the coefficients, the existence of such a polynomial is described by a first-order sentence which is true in $K_2$ but not in $K_1$.
This argument generalizes to show that $\mathbb{C}(x_1,\dots,x_n)\not\equiv \mathbb{C}(x_1,\dots,x_m)$ for any $n\neq m$.
