Let $V[G]$ be a generic extension of $V$ by adding a new Cohen real (or generally a generic extension which adds new reals and do not blow up the power of the continuum). Working in $V[G],$ we can consider the following structures:

1) $(\mathbb{C}^{V[G]}, +, ., 0, 1)$; the field of complex numbers as computed in $V[G].$

2) $(\mathbb{C}^{V}, +, ., 0, 1)$; the field of complex numbers as computed in $V.$

Both of these structures are models of algebraically closed fields of characteristic zero and are of the size of the continuum, so it follows that

$(\mathbb{C}^{V[G]}, +, ., 0, 1) \cong (\mathbb{C}^{V}, +, ., 0, 1)$.

Question. Is it possible to define an explicit isomorphism between the above structures? or is it possible to show that there is no definable isomorphism?

  • $\begingroup$ Single-element transcendental extensions of isomorphic fields are isomorphic. That should be enough to inductively construct the required isomorphism, since at limits, simply taking the unions of the currently constructed objects (i.e. the partial isomorphism and sub-fields) will produce objects of the same type. Since you are allowing parameters from the extension, I feel like this should be enough to satisfy any definition of explicit; unless I'm missing something. $\endgroup$ – Not Mike Jan 6 '18 at 19:30

(1) If explicit = definable with parameters from $V$, then the answer is no after adding one Cohen real. For suppose, $\phi(x, y, w)$ ($w \in V$) defines a bijection $x \mapsto y$ from $V \cap \mathbb{C}$ to $V[c] \cap \mathbb{C}$. Then for some $a \in V$ and a Cohen condition $p$, $p \Vdash \phi(a, c, w)$. Let $n$ be outside the support of $p$ and $\pi$ an automorphism of Cohen forcing that acts by flipping the $n$th bit of $c$. Then $\pi(p) = p$ forces $\phi(a, c_n, w)$ where $c_n$ is obtained by flipping the $n$th bit of $c$. Hence $p \Vdash \phi(a, b, w)$ holds for more than one $b$ which is impossible.

(2) If explicit = definable with parameters from $V$ and a real parameter from $V[G]$, then adding $\aleph_1$ Cohen reals will give a negative answer.

  • $\begingroup$ Thanks for the answer, but I would like to allow parameters from the generic extension. $\endgroup$ – Mohammad Golshani Jan 5 '18 at 10:32

In this comment you specified that parameters from the generic extension are fair-game. In this case, the answer is yes, provided the continuum and it's successor have not changed size.

The crux of the idea is that, you can rather naively construct a $\mathbb{P}$-name for an isomorphism using a "respectable" $\mathbb{P}$-name of a well-ordering of $\mathbb{C}^{V[G]}$ with order-type $\vert \mathbb{C} \vert^{V}$.

The naive approach to the construction makes use of the following for the successor step (taking appropriate unions at limit stages.)

Lemma: Given sub-fields $F_0, G_0 \subset \mathbb{C}$, an isomorphism $\varphi_0:F_0 \rightarrow G_0$ and $\xi, \nu \in \mathbb{C}$. If $\xi$ is transcendental over $F_0$ and $\nu$ is transcendental over $G_0$, then there is an isomorphism $\varphi:F_0(\xi)\rightarrow G_0(\nu)$ such that $\varphi\vert_{F_0}= \varphi_0$ and $\varphi(\xi) = \nu$.

  • 3
    $\begingroup$ If we're allowing uncountable parameters, then why not just take the isomorphism itself as a parameter? $\endgroup$ – Joel David Hamkins Jan 7 '18 at 15:04
  • $\begingroup$ @JoelDavidHamkins At the time that idea seemed like cheating. However, right now I can't think of a solid reason to not just take the isomorphism as a parameter. $\endgroup$ – Not Mike Jan 7 '18 at 15:29
  • $\begingroup$ But it would seem to trivialize the "definability" of the isomorphism in this case. For this reason, it seems natural to allow only real parameters. $\endgroup$ – Joel David Hamkins Jan 7 '18 at 15:42
  • $\begingroup$ @JoelDavidHamkins thankfully I was vauge enough (i.e. "respectable $\mathbb{P}$-name" above) to allow for such a restriction. $\endgroup$ – Not Mike Jan 7 '18 at 16:01
  • $\begingroup$ Of course I mean real parameters as otherwise as it is stated by Hamkins, the answer is trivial. $\endgroup$ – Mohammad Golshani Jan 8 '18 at 8:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.