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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?

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I was reluctant to tag this under some algebraic tags, but feel free to add them. –  Asaf Karagila Apr 2 '13 at 8:36

2 Answers 2

up vote 9 down vote accepted

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.

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That's great! Now I wonder what happens when we reduce ourselves to Cohen's first model's analogue (without requiring $V=L$ that is). I suspect that the transcendence degree there won't even be will either be $\omega$ or infinite Dedekind-finite cardinal $+\omega$. –  Asaf Karagila Apr 2 '13 at 10:43
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Could you explain your comment? Without AC, does the concept of transcendence degree make sense? After all, I think it is possible that there is no transcendence base. If you mean a model something like $V(\mathbb{R}^{V[c]})$, then the argument of my update paragraph shows that there are sets of size continuum of mutually transcendental reals in the extension, which seems related. –  Joel David Hamkins Apr 2 '13 at 11:49
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Will there be a party when Joel hits 100k points? –  Andrej Bauer May 6 at 9:39
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@AndrejBauer I certainly hope so! But we're all far flung. On a separate note, I wonder if someday we might organize some kind of MO conference or perhaps an MO special session at one of the big AMS meetings. –  Joel David Hamkins May 6 at 10:50
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You should make sure to hit 100k during the Vienna summer logic meetings. We'll buy you a Sachertorte. –  Andrej Bauer May 6 at 12:55

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.

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Thanks Mohammad. It seems to me that in its essence this is a similar argument to the one Joel gives at the end of his answer, since you essentially generate an almost disjoint family and takes a trace, so to speak. –  Asaf Karagila May 6 at 7:35

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