Is the space of real analytic functions a freely generated algebra? Let us consider the space $C^{\omega}(\mathbb{R})$ of all the functions $f \colon \mathbb{R} \to \mathbb{R}$ which are analytic on the whole real line. It is clear that $\mathcal{C}^\omega(\mathbb{R})$ is an algebra, because it is closed under addition, multiplication by scalars and inner multiplication. However, is it true that $\mathcal{C}^\omega(\mathbb{R})$ is freely $\mathfrak{c}$-generated? That is:

Is there a set $S \subseteq \mathcal{C}^\omega(\mathbb{R})$ of cardinality $\mathfrak{c}$ such that the algebra generated by $S$ is $\mathcal{C}^\omega(\mathbb{R})$ and whenever a polynomial $P \in \mathbb{R}[X_1, \dotsc, X_p]$ with $P(0, \dotsc, 0) = 0$ satisfies $P(s_1, \dotsc, s_p) = 0$ for some $s_1, \dotsc, s_p \in S,$ then $P = 0?$

If so, how can it be proven?
 A: This is certainly false but it seems surprisingly difficult to disprove. Golasinski and Henriksen's Residue Class Rings of Real-Analytic and Entire Functions proves that every residue field of $C^{\omega}(\mathbb{R})$ (quotient by a maximal ideal) is a real closed field. But the ring of polynomials in $\mathfrak{c}$ variables over $\mathbb{R}$ does not have this property: its residue fields consist of every field extension of $\mathbb{R}$ which is generated as an algebra by at most $\mathfrak{c}$ variables, and in particular contains $\mathbb{R}(x)$ (which is $\mathfrak{c}$-dimensional as a vector space), which is not real closed.
Edit: Never mind, here's a much easier argument. The ring of polynomials in any number of variables over $\mathbb{R}$ has the property that its group of units is exactly the invertible scalar multiples of the identity $\mathbb{R}^{\times}$ (because a nonconstant polynomial has positive degree and so its product with another nonzero polynomial also has positive degree). But there are many invertible real analytic functions which are not scalar multiples of the identity, e.g. $\exp(x)$.
