I'm trying to understand the (concise) solution of the following exercise, given in my mathematical logic lecture. However, there are still some steps that I am struggling with.
Suppose $\mathcal{L}$ is a countable first order language and $\mathfrak{A}:=<A ; \cdots>$ an $\mathcal{L}-$ structure. Let $\mathcal{L}_{A}$ be the language $\mathcal{L} \cup C_{A}$, where $C_{A}:=\left\{c_{a}: a \in A\right\}$ is a new set of constants. Let $\hat{\mathfrak{A}}$ be the $\mathcal{L}_{A}$ -structure $<\mathfrak{A} ; a: a \in A>$ where $a=c_{a}^{\hat{\mathfrak{A}}}$ for all $c_{a} \in C_{A}$. The atomic diagram of $\mathfrak{A}$ denoted by $\operatorname{Diag}(\mathfrak{A})$ is defined to be the set of all $\mathcal{L}_{A}$ -sentences $\theta$ where $\hat{\mathfrak{A}} \models \theta$ and $\theta$ is either an atomic $\mathcal{L}_{A}$ -sentence or the negation of an atomic $\mathcal{L}_A$ -sentence. The elementary diagram of $\mathfrak{A}$ denoted by $\operatorname{Diag}^{+}(\mathfrak{A})$ is the set of all $\mathcal{L}_{A}$ -sentences $\theta$ where $\hat{\mathfrak{A}} \models \theta .$ Now, suppose $\mathfrak{A}$ is finite and $\mathcal{L}$ contains only finite many relation symbols (no function symbols and no constant symbols). Show $\operatorname{Diag}(\mathfrak{A})$ is a finite set. Conclude that there is an $\mathcal{L}$ -sentence $\theta$ such that if $\mathfrak{B} \models \theta$ then $\hat{\mathfrak{B}} \models \operatorname{Diag}^{+}(\mathfrak{A})$.
Proof: Clearly, $\mathcal{L}_{A}$ is a finite extension of $\mathcal{L}$ because $\mathfrak{A}$ is finite. Atomic $\mathcal{L}_{A}$ formulas (question 0: shouldn't this be sentences?) are either of the form of $c_{1} \doteq c_{2}$ or $\mathcal{R}\left(c_{1}, \ldots, c_{n_{R}}\right)$ for relation $\mathcal{R} \in \mathcal{L}$ and constants $c_{i} \in \mathcal{L}_{A}$. From this, it is easy to argue that there are only a finite set of $\mathcal{L}_{A}$ atomic or negation of atomic sentences. Thus $\operatorname{Diag}(\mathfrak{A})$ is finite, so consider $\mathcal{L}_{A}$ sentence $\psi:=\bigwedge_{\phi \in \operatorname{Diag}(\mathfrak{A})} \phi$. Clearly, $\hat{\mathfrak{A}} \models \psi$. Now consider a set of variables $\left\{v_{1}, v_{2}, \ldots, v_{n}\right\}$ where $n=|A|$ ($A$ is the domain of $\mathfrak{A}$). Then build a $\mathcal{L}$ -formula $\psi^{*}\left(v_{1}, \ldots, v_{n}\right)$ by substituting $v_{i}$ with constants $c_{a_{i}}$ in $\psi$ for $1 \leq i \leq n$.
Define $\theta:=\exists v_{1} \exists v_{2} \ldots \exists v_{n}\left(\psi^{*}\left(v_{1}, \ldots, v_{n}\right) \wedge\left(\forall v\left(\bigvee_{1<i<n} v \doteq v_{i}\right)\right)\right.$. Clearly, $\theta$ is an $\mathcal{L}$ -sentence and one can argue $\mathfrak{A} \models \theta$. Now assume $\mathfrak{B} \models \theta$. Clearly, $\mathfrak{B}$ has exactly $n$-many elements. One can prove by induction that every $\mathcal{L}_{A}$-sentence is equivalent an $\mathcal{L}_{A}$ sentence where no quantifier appears in the formula. From this, we conclude that the set of $\operatorname{Diag}^{+}(\mathfrak{A})$ is also finite modulo logically equivalent equivalence relation. Finally, we choose $b_{1}, \ldots, b_{n}$ an enumeration of elements of the domain of $\mathfrak{B}$ such that $\mathfrak{B} \models \psi^{*}\left[b_{1}, \ldots, b_{n}\right]$. Then for each $c_{a_{i}}$, choose the interpretation $c_{a_{i}}^{\hat{\mathfrak{B}}}=b_{i}=c_{b_{i}}^{\hat{\mathfrak{B}}}$.
Question 1: How can one argue that $\mathfrak{A} \models \theta$? Is this because $\hat{\mathfrak{A}} \models \theta$, and because $\theta$ is an $\mathcal{L}$-sentence, somehow we conclude $\mathfrak{A} \models \theta$?
Question 2: In the part "One can prove by induction that", isn't it enough to remark that an $\mathcal{L}_{A}$-sentence of the form $\forall v \phi(v)$ is equivalent to $\bigwedge_{1 \leq i \leq n} \phi\left(c_{a_{i}}\right)$ (using that $A$ is a finite set) and $\exists v \phi(v)$ is equivalent to $\bigvee_{1 \leq i \leq n} \phi\left(c_{a_{i}}\right)$? So there is no need for induction, as we just need to get rid of the quantifiers.
Question 3: Why does the above show that if $\mathfrak{B} \models \theta$, then $\hat{\mathfrak{B}} \models \operatorname{Diag}^{+}(\mathfrak{A})$?
Any partial answers (answering only one of the questions, for example) are more than welcome! Thanks!
Your ideas for questions 1 and 2 are correct:
Some comments on Q1:
The sentence $\hat{\mathfrak{U}} \models (\psi^*(c_1,...,c_n) \wedge(\forall v(\bigvee_{1 \leq i \leq n}v \dot{=} c_i)))$ is a statement about the interpretations of $c_i$ as elements of $\mathfrak{U}$. It implies more generally that this formula is true for some elements v_1,...,v_n i.e. there exist v_1,...,v_n such that the formula is true of them: $\hat{\mathfrak{U}} \models \exists v_1,...,\exists v_n (\psi^*(v_1,...,v_n) \wedge(\forall v(\bigvee_{1 \leq i \leq n}v \dot{=} v_i))).$ This is an $L$ sentence and so holds in the $L$-reduct of $\hat{\mathfrak{U}}$, $\mathfrak{U}$.
There is a clearer way to see that $\theta$ holds however. The $L_A$ sentence $(\psi^*(c_1,...,c_n) \wedge(\forall v(\bigvee_{1 \leq i \leq n}v \dot{=} c_i)))$ is true in $\hat{\mathfrak{U}}$ if and only if the $L$ formula $(\psi^*(a_1,...,a_n) \wedge(\forall v(\bigvee_{1 \leq i \leq n}v \dot{=} a_i)))$ is true of $a_1,...,a_n$ in $\mathfrak{U}$ where $a_i$ is the interpretation of $c_i$ in $\mathfrak{U}$. I.e. $\hat{\mathfrak{U}} \models (\psi^*(c_1,...,c_n) \wedge(\forall v(\bigvee_{1 \leq i \leq n}v \dot{=} c_i)))$ iff $\mathfrak{U} \models (\psi^*(a_1,...,a_n) \wedge(\forall v(\bigvee_{1 \leq i \leq n}v \dot{=} a_i)))$. Then note that $\mathfrak{U} \models (\psi^*(a_1,...,a_n) \wedge(\forall v(\bigvee_{1 \leq i \leq n}v \dot{=} a_i)))$ ensures that $\mathfrak{U} \models \theta$.
Some comments on Q2:
What is meant by "induction" here is really "induction on the structure of L_A sentences." More specifically one has to induct on the number of quantifiers present in the $L_A$ sentence. What your remark is doing is precisely the inductive step of this argument showing that if the sentence has $n+1$ quantifiers, it is "equivalent" to a sentence with only $n$. This is pedantic though.
Some comments on Q3:
Note that quantifier free sentences here are just finite boolean combinations of atomic formulas. Since the atomic diagram is finite what can you say about the cardinality of the quantifier free sentences?
Informally, theta holds all of the information about the atomic diagram. Since the quantifier free sentences are assembled from atomic sentences, and by the "inductive argument" more general sentences are equivalent to quantifier free sentences, theta holds all of the information about the elementary diagram.
Some comments about the "meaning of the problem":
In general, elementary equivalence is a strictly weaker notion than isomorphism. What this problem shows is that for finite structures elementary equivalence is enough to guarantee the existence of a bijective elementary embedding, which must be an isomorphism. This is a nice problem because it shows that elementary equivalence and isomorphism coincide for finite structures.