TITLE: The transcendence degree of $\mathbb R$ after adding a Cohen
QUESTION [11 upvotes]: 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?

REPLY [2 votes]: 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.