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Suppose that a set of sentences of a 1st order language has an infinite model $M$.

Under what conditions is there is a proper class-sized elementary extension of $M$?

How does the answer change if we begin with a proper class of sentences?

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The answer to your main question is that in ZFC there is always such a proper-class elementary-extension.

Theorem. In ZFC, every set-sized model in a set-sized first-order language has a proper-class elementary extension.

Proof. This is easiest to see in the case that the global axiom of choice holds, in other words if there is a class well-ordering of the universe. So let me first explain that case. Fix the global well-order and consider any fixed model $M_0$ in a set-sized first-order language. Using the upward Löwenheim-Skolem theorem, there is a proper elementary end-extension of $M_0$, and we may let $M_1$ be the least such model arising in the well-order. Continuing transfinitely, picking the least elementary-extension at each stage and unions at limit stages, we may build up an elementary chain $$M_0\prec M_1\prec\cdots\prec M_\alpha\prec\cdots$$ by a definable procedure whose union will be a proper-class elementary extension of each of them and in particular of $M_0$, as desired.

But my next observation is that in ZFC you don't actually need global choice. If we fix $M_0$, then by the axiom of choice, we may code $M_0$ by a set of ordinals $A$. Consider the inner model $L[A]$, which satisfies ZFC and global choice. Since $A$ codes $M_0$, we may undertake the argument of the previous paragraph inside $L[A]$ to get a proper-class elementary extension of $M_0$. In the original universe $V$, then, we get an $A$-definable proper class elementary-extension of $M_0$, as desired. QED

Your second question, however, can fail in some models. I claim that it is possible to have a definable proper-class-sized theory $T$ in a model of ZFC, such that every subset of $T$ has a model, and so in particular the theory is consistent, but there is no definable (allowing parameters) model of all of $T$. For example, assume we are working in a model of ZFC in which there is no definable linear order. (I explained how to construct such a model in my answer to Asaf Karagila's question, Does ZFC prove the universe is linearly orderable?.) Let $T$ be the theory of a linear order $<$, with a constant symbol $\hat a$ for every object $a$. Every restriction of $T$ to only set-many constants will have a model, since in ZFC every set is linearly orderable, but in this model there is no definable model of all of $T$, since there is no definable linear ordering of the universe.

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    $\begingroup$ Hrm... cough cough... any consistent theory with infinite models. :-) $\endgroup$
    – Asaf Karagila
    Jan 23 '16 at 16:03
  • $\begingroup$ If global choice fails completely (namely, there is no parameter from which can define a global choice function), can't you just look at the language with constants $\check a$ for every set $a$ and a function symbol $F$ such that $F(\check a)\in\check a$ for all $a\neq\varnothing$? This will be set-consistent, but not class consistent. $\endgroup$
    – Asaf Karagila
    Jan 23 '16 at 16:08
  • $\begingroup$ Asaf, that doesn't work, since in first-order logic, you can't guarantee that $F(\check a)$ is $\check b$ for some $b$. I could make a model by inventing a new object $c$ that is not a $\check b$ and setting $F(\check a)=c$ for every $a$. But your idea would work in $L_{\infty,\infty}$ logic. $\endgroup$ Jan 23 '16 at 16:40
  • $\begingroup$ Ah, yeah, you're right. $\endgroup$
    – Asaf Karagila
    Jan 23 '16 at 16:47
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    $\begingroup$ I realized that one may get rid of the global AC assumption by coding the model and working in L[A]. So the answer in ZFC is just plain yes, as far as getting an elementary extension of a fixed model. $\endgroup$ Jan 23 '16 at 17:47

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