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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
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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
    
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
    
Err, yeah. My train of thoughts was disrupted by a phone and it seems that I continued my comment in an incompatible way to its beginning. I was trying to suggest that when reducing to the symmetric extension we exclude many real numbers from the universe. So perhaps the remainder of is small enough to be algebraically generated by the "canonical generics" (or some derivation of them), which would result in a transcendence degree which is either countable, or contains an infinite Dedekind-finite set (but not continuum itself). Your update seems to suggest otherwise. Which is good, thanks! –  Asaf Karagila Apr 2 '13 at 12:38
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