TITLE: Universe cardinals and models for ZFC
QUESTION [7 upvotes]: I'm reading through Joel David Hamkins' set theory lecture notes.  On page 14, on the subject of inaccessible cardinals and submodels of ZFC in $V$, he defines a universe cardinal to be a cardinal $\gamma$ such that $V_\gamma$ models ZFC.  For example, if an inaccessible cardinal $\kappa$ exists, then $\kappa$ is a universe cardinal.  But furthermore, by the Löwenheim-Skolem theorem, there are then lots of models of ZFC inside $V_\kappa$, and Hamkins says that this proves that $\kappa$ is the $\kappa$th universe cardinal.  Specifically, $\{\gamma < \kappa: V_\gamma \prec V_\kappa\}$ is evidently closed in $\kappa$, and is "unbounded by Löwenheim-Skolem".

This is confusing to me.  The Löwenheim-Skolem theorem says that $V_\kappa$ contains submodels of arbitrary cardinality.  But in order to have any universe cardinals at all, you need models of ZFC of the form $V_\gamma$, which don't seem to come from Löwenheim-Skolem.  Why should these exist?  Is there something you can do to a model to put it in the form $V_\gamma$?
Relatedly, it's also true (this is in Kanamori's Higher Infinite) that the only models for second-order ZFC are $V_\kappa$ for $\kappa$ inaccessible.  Can someone explain the difference between these two concepts?  I'm finding the distinction between first-order and second-order Replacement a little hair-splitting.

I hope I've phrased this the right way -- I'm a set theory rookie.  Thanks in advance!

REPLY [3 votes]: Miha gave a very good answer to the first question. Let me add a bit on the second question.
The difference between first-order and second-order replacement is subtle. First-order replacement is a schema, and it says that if $f$ is an internally definable function with a set domain, then its range is a set; whereas the second-order replacement axiom is a single axiom, and it states that whenever $f$ is a subset of $V_\kappa$ which is a function, and its domain is a set of $V_\kappa$, then its range is an element of $V_\kappa$ as well.
If $V_\kappa$ satisfies the second-order axiom, then it $\kappa$ to be regular. To see this note that:

If $A\subseteq\kappa$ is an unbounded set, then $A\notin V_\kappa$.
Every unbounded set of ordinals is the range of a function which enumerates it.

Now if $\kappa$ is singular, there is some $\gamma<\kappa$, and a function $f\colon\gamma\to\kappa$ which enumerates an unbounded set. This function $f$ is a subset of $V_\kappa$, and its domain is an element of $V_\kappa$, but its range is not.