Can two fields that have monomorphism to each other not be isomorphic? More precisely, let $F_1, F_2$ be fields, and we have none zero homomorphisms $f: F_1 \to F_2$ and $g: F_2 \to F_1$ (actually both are monomorphism since $F_1, F_2$ are fields), is there a possibility that $F_1$ is not isomorphic to $F_2$?
 A: To elaborate on the example given by perpetuallyconfused, indeed $\mathbb C$ and $\mathbb C(x)$ provide a counterexample.
Some additional elements.
$\mathbb C$ is algebraically closed: this is well known. $\mathbb C(x)$ is not. In particular the polynomial $p(t) = t^2-x \in \mathbb C(x)[t]$ can't have a root $\frac{r(x)}{s(x)} \in \mathbb C(x)$. If it was the case, you would have $r^2(x)=x s^2(x)$ with the contradiction that the left polynomial of the equality is having an even degree and the right one an odd one. Therefore $\mathbb C$ and $\mathbb C(x)$ are not isomomorphic.
Also, the identity is an obvious embedding $\mathbb C \hookrightarrow \mathbb C(x)$.
Regarding an embedding $\mathbb C(x) \hookrightarrow \mathbb C$, you have to know that Two algebraically closed fields are isomorphic if and only if they have the same transcendence degree over their prime fields (proof provided in the link). And also that the cardinality of the algebraic closure of an infinite field $F$ has the cardinality of $F$. As the cardinality of $\mathbb C(x)$ is the one of $\mathbb C$, the algebraic closure $\overline{\mathbb C(x)}$ of $\mathbb C(x)$ is isomorphic to $\mathbb C$ and therefore you can embed $\mathbb C(x)$ into $\mathbb C$.
A: So apparently, it is possible for this to fail, though I don't profess to understand the counterexamples given in the comments. You are asking if fields satisfy the Cantor–Schröder–Bernstein, and it appears that they do not.
Edit: I take that back, I think I understand one example, if not its construction. There is a natural inclusion $\mathbb{C} \hookrightarrow \mathbb{C}(x)$ and a (complicated) inclusion $\mathbb{C}(x) \hookrightarrow F$ where $F \cong \mathbb{C}$, where $\mathbb{C}(x)$ is the field of rational functions over $\mathbb{C}$. However, the complex field is algebraically closed whereas $\mathbb{C}(x)$ is (apparently) not, so they cannot be isomorphic.
