Can $V$ only have well-orderings definable with respect to a parameter?

Question

In this answer, Professor Hamkins gives a proof that for models $M$ of ZF, $M$ being a model of $\text{ZFC} + V = \text{HOD}$ is equivalent to there being a definable well-ordering of the universe:

https://mathoverflow.net/a/180734

His argument easily extends to an equivalence of these properties to $M$ having a well-ordering of the universe definable with respect to an ordinal parameter. So, if there is well-ordering of the universe with respect to some parameter $p,$ but there is not a well-ordering of $V$ definable without parameters, then necessarily $p \not \in \text{OD}$. Is this situation possible? My intuition is that it shouldn't be possible, since I don't think a non-ordinal parameter should be able to define something so fundamental when an ordinal cannot do the same.

Answer 1

Of course.

If you start with a model where there is a definable well-ordering, say $V=L$, and you add a single Cohen real $r$ you have that:

1. $V=L[r]$, so there is a definable well-ordering with a parameter $r$ (e.g. given two sets in $L[r]$ ask which one has a name appearing first in the order of the ground model, here $L$, that when interpreted with $r$ as the generic give you the two sets).
2. Since the Cohen forcing is homogeneous, $L=\mathsf{HOD}^L=\mathsf{HOD}^{L[r]}\neq L[r]$.

So while there is a well-ordering definable from a parameter, which in this case is a real number (read: a subset of $\omega$), there is no such well-ordering which is definable without parameters or from ordinals.

(This argument shows that any set forcing over a model with global well ordering from parameters will also have a global well ordering definable from a parameter. You can violate that with a class forcing, though.)

Answer 2

I just noticed this question, which I find quite interesting.

I thought I'd mention the following related result, which some readers may find interesting:

Theorem. The following are equivalent.

1. The universe is HOD of a set: $\exists b\ (V=\text{HOD}(b))$.
2. The axiom V=HOD is forceable.
3. Somewhere in the generic multiverse, the universe is HOD of a set.
4. Somewhere in the generic multiverse, the axiom V=HOD holds.

The proof is contained in my blog post, Being HOD-of-a-set is invariant throughout the generic multiverse.

In particular, it follows that the axiom V=HOD is a switch, in models for which $V=\text{HOD}(b)$, since it can be forced on and then off again as much as you like. If $V=\text{HOD}$ holds, then you can do the forcing in Asaf's answer, adding a Cohen real, and $V\neq\text{HOD}$ in the extension $V[c]$, but then you can force $V=\text{HOD}$ again in a further forcing extension. And furthermore, whenever $V=\text{HOD}(b)$, then you can force $V=\text{HOD}$, and the assertion $\exists b\ V=\text{HOD}(b)$ is invariant by forcing.