I'm reading through Marker's Model Theory: An Introduction, and I'm confused by the proof of Proposition 2.2.2. The result is as follows:
Let $T$ be an $\mathcal{L}$-theory with infinite models. If $\kappa$ is an infinite cardinal with $\kappa \geq |\mathcal{L}|$, then there is a model of $T$ of cardinality $\kappa$.
The proof goes as follows. Let $\mathcal{L}^*$ be the language $\mathcal{L} \cup \{c_\alpha : \alpha \in \kappa \}$, where the $c_\alpha$ are constant symbols. Let $T^*$ be the $\mathcal{L}^*$-theory $T^* = T \cup \{ c_\alpha \neq c_\beta : \alpha, \beta \in \kappa, \alpha \neq \beta \}$. Any model of $T^*$ will have at least $\kappa$ many elements. To prove that $T^*$ is satisfiable, we use the Compactness Theorem by showing that $T^*$ is finitely satisfiable. If $\Delta \subseteq T^*$ is a finite subset of $T^*$, then $\Delta$ is contained in $T \cup \{ c_\alpha \neq c_\beta : \alpha, \beta \in I \}$ for some finite subset $I$ of $\kappa$. Any infinite model of $T$ would model $\Delta$, so we know that $T^*$ is finitely satisfiable, and therefore satisfiable.
I have two questions about this. Firstly, it's not obvious to me that a model of $T^*$ would be of cardinality exactly $\kappa$, only at least $\kappa$. The $c_\alpha \neq c_\beta$ sentences ensure that $\kappa$ maps injectively into a model $\mathcal{M}$ by $\alpha \mapsto c_\alpha^\mathcal{M}$, but I don't see why that precludes the possibility that $M$ is of cardinality strictly greater than $\kappa$. At first, I took this to mean that there existed a model of cardinality at least $\kappa$ and it was just worded in a way that I misunderstood, but Marker uses this result to prove Vaught's test, where he invokes the existence of a model for $T$ of cardinality exactly $\kappa$.
Secondly, I don't see where the assumption that $\kappa \geq |\mathcal{L}|$ is used.
Any help clearing up either of these confusions would be greatly appreciated.
Your two concerns are related. We use $\kappa \geq |\mathcal{L}|$ in order to show that there is a model of size exactly $\kappa$.
The idea is to pick your favorite set of size $\kappa$ and then close it under all the operations of $\mathcal{L}$.
This is entirely analogous to finding the smallest subgroup containing a subset of a group $G$. We take $S \subseteq G$ and close it under the group operations.
Showing that taking this closure doesn't add too many elements (that is, we end with $\kappa$ many elements) uses the fact that we don't need to do this closure operation too many times -- after all, there are at most $\kappa$ many operations in $\mathcal{L}$ to worry about.
The actual process of taking this closure is a bit tedious, but isn't hard. You can find more information by googling around for the "Downward Lowenheim-Skolem Theorem"
I hope this helps ^_^