The transcendence degree of $\mathbb R$ after adding a Cohen

Question

Let $V\models\sf ZFC$, and let $V[r]$ be a generic extension obtained by adding one Cohen real, or equivalently $\omega$ Cohen reals.

It is clear that $\Bbb R^{V[r]}$ and $\Bbb R^V$ have the same cardinality, and that $\Bbb R^{V[r]}\setminus\Bbb R^V$ also have that cardinality.

Furthermore any generic real must be transcendental (I think, because we don't change the rationals), so $\Bbb R^{V[r]}$ is a transcendental extension of $\Bbb R^V$. But what's the transcendence degree of this extension?

It clearly cannot be finite, because adding one Cohen real adds $\omega$ pairwise generic Cohen reals, and the above argument gives us that they cannot be algebraically dependent. But is it $\aleph_0$, or is it $2^{\aleph_0}$? Maybe something else?

Answer 1

The transcendence degree of $\mathbb{R}^{V[c]}$ over $\mathbb{R}$ is $2^{\aleph_0}$, the largest it could be. The reason is that adding a single Cohen real $c$ adds a family of continuum many Cohen reals, which are finitely-mutually generic. That is, in the extension, there is a family of continuum many reals, such that any finitely many of them are mutually $V$-generic over the ground model. It follows that they are also mutually transcendental.

To get the family, consider the following forcing: we force with finite $\{0,1\}$-labeled binary branching trees, ordered by extension. That is, the conditions are finite trees whose nodes are labeled with $0$ or $1$. The generic filter provides us with a full labeling of the tree $2^{\lt\omega}$. Let $B$ be the set of reals arising from the sequence of labels on paths through the tree. So $B$ has size continuum, but a density argument shows that every real in $B$ is a $V$-generic Cohen real. The reason is that if we are given any labeled tree and any dense set in Cohen forcing, then we may extend the tree in such a way that every branch up to the top level of the extended tree is in the dense set. Similarly, a slightly more complicated argument shows that any finitely many of the reals in $B$ are mutually $V$-generic: for any condition $t$, consider the reals arising from the top level of $t$ and any dense set in the forcing to add that many Cohen reals. We may extend $t$ to a condition $t'$ such that any collection of branches though $t'$ with one branch going through each top node of $t$ is in the dense set. So this forcing adds the family of continuum many finitely-mutually generic Cohen reals.

Finally, we complete the argument by realizing that the tree forcing is countable and hence just the same as the forcing to add a single Cohen real. So adding a single Cohen real adds a family of size continuum of finitely-mutually generic Cohen reals, and these are mutually transcendental over the ground model reals.

An alternative way to think about the forcing is to take the single Cohen real $c$ and use it to label the nodes of the full tree $2^{\lt\omega}$.

Note that the reals in the family of continuum many finitely-mutually generic reals cannot be fully mutually generic, since adding a single Cohen real does not add a generic for adding continuum many Cohen reals. These are a kind of fake mutually generic reals, which are only mutually generic if you consider them finitely at a time.

Update. Here is one more way to think about this forcing, which I believe may provide an answer to your comment. Let $\cal A$ be an almost-disjoint family of subsets of $\omega$, of size continuum. So these are infinite sets, but any two have finite intersection. Now, add a single Cohen real $c$, and for each $I\in\cal{A}$, let $c_I$ be the real arising by the pattern of the digits of $c$ on $I$. It follows easily that each $c_I$ is a $V$-generic Cohen real, and furthermore, for any finitely many $I_0,\ldots,I_n$, the corresponding reals $c_{I_k}$ are mutually $V$-generic. Thus, we have a continuum-sized family of mutually $V$-generic Cohen reals. And this way of thinking seems to work also in a ZF world.

Answer 2

I think the following argument might be easier than Hamkins answer. It also does not use AC. So suppose $g$ is a Cohen real. We want to show that there are continuum many mutually generic Cohen reals in $V[g]$.

In $V$, fix a canonical enumeration $F: 2^{<\omega} \to \omega$ such that if $|s|<|t|,$ then $F(s)<F(t).$ For any $t\in (2^\omega)^V$, define $g_t$ by $g_t(n)=g(F(t\restriction n)).$ Then $(g_t: t\in (2^\omega)^V)$ is as required.