Non-saturated algebraically closed field I've always thought of saturated models as a generalization of algebraically closed fields, viewing types as generalized polynomials. 
I was wondering, in the particular case of (infinite) fields (thought of as models of the f.o. theory of fields), whether one can have an algebraically closed, non-saturated model? (The converse implication is obviously true.)
Looking at $\mathbb{C}$, $\text{Aut}(\mathbb{C})$ seems to be rich enough so as to guarantee saturation (although I'm not 100% sure of this), so that wouldn't work as an example. 
 A: An algebraically closed field $K$ is determined up to isomorphism by its characteristic and its transcendence degree over the prime field. And $K$ is saturated if and only if it has infinite transcendence degree. 
Fix a characteristic $p$ (prime or $0$) and let $F$ be the prime field ($\mathbb{F}_p$ if $p$ is prime, or $\mathbb{Q}$ if $p = 0$). For uncountable $\kappa$, there is exactly one algebraically closed field of characteristic $p$ and cardinality $\kappa$ (it must have transcendence degree $\kappa$ over $F$), and it is saturated. In particular, $\mathbb{C}$ is saturated. 
Among countable algebraically closed fields of characteristic $p$, the possible transcendence degrees are $0, 1, 2, \dots, \aleph_0$. The fields of finite transcendence degree are not saturated, but the one of infinite transcendence degree is saturated. 
Proof: Suppose $K$ has transcendence degree $n$ over $F$. Then $K\cong \overline{F(t_1,\dots,t_n)}$. There is a consistent type $p(x)$ with finitely many parameters $\{t_1,\dots,t_n\}$ expressing that $x$ does not satisfy any non-trivial polynomial in $F(t_1,\dots,t_n)$, and this type is not realized in $K$, since every element of $K$ is algebraic over $F(t_1,\dots,t_n)$, so $K$ is not saturated. For the converse, suppose $K$ has infinite transcendence degree over $F$. Let $p(x)$ be a type with parameters from $A$, with $|A|<|K|$. Then either $p(x)$ is algebraic (because it contains some non-trivial polynomial equation) in which case $p$ is realized in $K$, or $p(x)$ is the type of an element transcendental over $A$ (because it contains no non-trivial polynomial equations). In the latter case, since $|A| < |K|$, and $K$ has infinite transcendence degree over $F$, $K$ also has infinite transcendence degree over the relative algebraic closure of the field generated by $A$ in $K$, so we can realize the type in $K$. 
These facts hold more generally for all uncountably categorical theories, with the notion of transcendence degree replaced by dimension of a strongly minimal set over the prime model.

Extra comment: I think the proper model-theoretic generalization of algebraically closed fields is the notion of existentially closed models, rather than the notion of saturated models. Of course, these two notions are similar in a lot of ways, but the important difference is that existentially closed models have solutions to conditions described by quantifier-free formulas (like a polynomial equation $f(x) = 0$), whereas saturated models have solutions to conditions described by complete types (which are much more expressive, e.g. a type $p(x)$ can say "$x$ is transcendental", which can't be expressed by a single formula). Indeed, algebraically closed fields are exactly the existentially closed fields, but they are not necessarily saturated. 
