Homogeneous models in a strongly minimal theory I am trying to prove that every infinite model of a complete strongly minimal theory T is homogeneous.
Clearly, if $k:M \to M $ is a partial elementary map with $|k|<|M|$ and $M$ is a model of $T$, it is sufficient to find for all $a\in M$ an partial elementary map extending $k$, defined on $a$. In this way, using a back-and-fourth type argument, we are able to construct the required automorphism.   
If $a$ is algebraic over the domain of $k$, this is trivial.
If not, clearly it is sufficient to prove that there exists a $b\in M$ which is not algebraic over the image of $k$. If such $b$ does not exist, then $M=\text{acl}(\text{im}(k))$ and $|M|=|\text{acl}(\text{im}(k))|=|\text{acl}(\text{dom}(k))| \le |L|+|\text{dom}(k)|+\omega < |k|$ if $|k|> |L|+\omega $. However, if $|k|\le |L|+\omega$, I can't find a contradiction. Any suggestions? Is it better to attack the problem in a different way? 
Also I'm thinking of another possibility to do the exercise, distinguishing between cases $|\text{acl}(\text{dom}(k))|$ finite and infinite. In the first case it is simple to find $b$. In the second, let $A$ be a basis of $M$ over $\text{dom}(k)$. I claim that there exists a basis $B$ of $M$ over $\text{im}(k)$ of the same cardinality as $A$ (in this case, it is simple to extended our $k$ to an automorphism of $M$.)  If not, for example $|A|<|B|$ then let $h:A\to B$ be any injection of $A$ into $B$. The $h$ is elementary, and we can extended $h$ to an isomorphism $g$ from $M=\text{acl}(A\cup \text{dom}(k))$ to $\text{acl}(h[A]\cup \text{im}(k))$. Then $M$ has two bases $B$ and $h[A]$ over $\text{im}(k)$ of different cardinality, contradiction. However I do not understand where I use $|k|<|M|$...
Thanks in advance.
 A: First of all, the statement is not true in general if $L$ is uncountable. For a counterexample in an uncountable language, look at the theory of $\mathbb{R}$-vector spaces in the language $L = \{+,-,0\}\cup\{a\mid a\in\mathbb{R}\}$, where the $a$ are unary function symbols for scalar multiplication.
Let $M$ be an $\mathbb{R}$-vector space of countable dimension (so $M\cong \oplus_{i\in\omega}\mathbb{R}$, and $|M| = 2^{\aleph_0}$), and let $B = \{b_i\mid i\in\omega\}$ be a basis. Let $k$ be the partial elementary map with domain $B$ that sends $b_i$ to $b_{i+1}$ for all $i$. We cannot extend $k$ to an automorphism of $M$, since every element of $M$ is in the span of $\text{dom}(k)$, but $b_0$ is not in the span of $\text{im}(k)$. But $|B|<|M|$, so $M$ is not homogeneous.

OK, so let's agree to assume $L$ is countable.
The argument you propose isn't quite right. The issue appears when you conclude for contradiction that $B$ and $h[A]$ are both bases of $M$ of different cardinality over $\text{im}(k)$. The injection $h:A\to B$ can be extended to a partial elementary map $g$ from $M$ to itself, but there's no reason why this map should be surjective. That is, $h[A]$ will be algebraically independent over $\text{im}(k)$, and it will be a basis for $g[M]$, but $g[M]$ might be a proper elementary submodel of $M$.

However, your idea of splitting into the finite and infinite cases is a good one. I'll sketch a correct argument now:
Pick a basis $B$ for $\text{dom}(k)$ (note $|B|\leq |\text{dom}(k)| < |M|$), and use the fact that $k$ is partial elementary to show that $k[B]$ is a basis for $\text{im}(k)$. As you noted, the case when the new element $a$ is algebraic over $B$ is easy, so let's pick $a\in M$ not algebraic over $B$. We need to find a target $a'\in M$ not algebraic over $k[B]$.
Case 1: $B$ is infinite. Then $|\text{acl}(k[B])| \leq |k[B]| + |L| = |B| < |M|$ (this is where we use countable language), so there is an element of $M$ which is not algebraic over $k[B]$, and we're done.
Case 2: $B$ is finite. Suppose for contradiction that there is no $a'\in M$ not algebraic over $k[B]$, i.e. $M = \text{acl}(k[B])$. Then $k[B]$ is a basis for $M$. But on the other hand, $\text{acl}(B) \neq M$, since $a$ is not algebraic over $B$. So we can extend $B$ to a strictly larger basis for $M$ (i.e. by throwing in $a$ and possibly other stuff). Hence $M$ has two bases of different size, contradiction.
