Module of $R$-valued functions on an infinite set is not countably generated Let $R$ be an integral domain and $X$ be an infinite set.  Let $R^X$ be the set of all functions $f: X \rightarrow R$, viewed as an $R$-module in the usual manner: for $\alpha \in R$, $\alpha f: x \in X \mapsto \alpha f(x)$.
For a short paper I'm writing, I would like to use the fact that $R^X$ is not a countably generated $R$-module.  In the case that $R$ is a field, this is a classic tricky linear algebra question which has been discussed on this site before.  Namely, for any field $X$ 
and any infinite set $X$ one has
$\operatorname{dim}_K K^X = (\# K)^{\# X} \geq 2^{\# X} > \aleph_0$.
An especially elegant proof of this result is given here.  This proof does not directly carry over to the case of an integral domain, but maybe another proof would?  Another plausible line of attack is to try to deduce the result over $R$ from the result over its fraction field $K$.  But I have not gotten that to work: the snag is that, since $X$ is infinite, not every function $f: X \rightarrow K$ can be scaled to land in $R$.
Added: Since $R^X$ is uncountable, the result holds when $R$ is countable for cardinality reasons.  
What I really want to show is that there is no surjective $R$-algebra map from a polynomial ring $R[t_1,\ldots,t_n]$ to $R^X$ (viewed as an $R$-algebra under pointwise multiplication, or in other words a big Cartesian product of copies of $R$)  when $X$ is infinite.  When $R$ is Noetherian, this follows easily from the Hilbert Basis Theorem.  My idea for showing it in the general case is to establish the above "fact" (which, I suppose, I am not completely sure is true, but if it's false that would be more interesting than what I am trying to prove!).  If someone sees a different way to show it, please do tell.
Finally, because this is being used to establish an auxiliary result in a short paper in "proper generality", I would prefer not to have to write out the argument if it is more than half a page or so.  So a reference to where this is established in the literature would be ideal.
 A: Take just a numerable set of entries, and forget the other for the moment. Now consider the following fact: there's a continuos chain in the part of $\mathbb{N}$(this is clear once you consider $(-\infty,x) \bigcap \mathbb{Q}, x \in \mathbb{R}$). So call F this family. To each A in F, you associates $1_A$(that is you make a bjection with the numerable entries and $\mathbb{N}$ and you consider that, and the others just put 0). Now it's clear they are independent on R, indeedi take a linear combination, and on the maximum element(i mean the $1_A$ associated to the biggest part A) there are some entries where appear just is coefficient, so is coefficent is 0 and so on.
Maybe it helps also to your aims. Let me know
A: Let me explain why Donald's answer really does solve my question.  Recall $R$ is a domain with fraction field $K$.
Step 1: In the power set $2^X$ there is a subset $\mathcal{C}$ which is totally ordered and of continuum cardinality.  (One can do this for $X = \mathbb{Q}$ by considering the collection of all rational numbers in the interval $(-\infty,x)$ for $x \in \mathbb{R}$.  Adjusting by a bijection, this works for any countably infinite set $Y$.  And for any infinite set $X$ there is an injection $\iota: Y \hookrightarrow X$ which induces an embedding of partially ordered sets $2^{\iota}: 2^Y \hookrightarrow 2^X$, so take $2^{\iota}(\mathcal{C})$.  
If $\varnothing \in \mathcal{C}$, we remove it and still get a chain of continuum cardinality.
Step 2: For each $A \in \mathcal{C}$, let $1_A$ be the characteristic function of $A$.  Then $\{1_A\}_{A \in \mathcal{C}}$ is an $R$-linearly independent set.
Proof: If not, there are $A_1,\ldots,A_n \in \mathcal{C}$ and $\alpha_1,\ldots,\alpha_n \in R$ such that $\alpha_1 1_{A_1} + \ldots + \alpha_n 1_{A_n} \equiv 0$.  We may order the $A_i$'s such that $A_1 \subset \ldots \subset A_n$ and thus there is $x \in A_n \setminus \bigcup_{i=1}^{n-1} A_{i}$.  Evaluating the linear dependence equation at $x$ shows that $\alpha_n = 0$.  Arguing in a similar manner -- or, by induction on $n$ -- we find that $\alpha_{n-1} = \ldots = \alpha_1 = 0$.
Step 3: Let $F =\bigoplus_{i \in \mathbb{Z}^+} R$ be a free $R$-module of rank $\aleph_0$.  If $R^X$ were countably generated as an $R$-module, there would be a surjective homomorphism of $R$-modules $\Phi: F \rightarrow R^X$.  Under any homomorphism of $R$-modules, the preimage of a linearly independent set is a linearly independent set, so by Step 2, $\Phi^{-1}(\{1_A\}_{A \in \mathcal{C}})$ is an uncountable $R$-linearly independent subset of $F$.  Tensoring from $F$ to $F \otimes_R K$ we get an uncountable $K$-linearly independent subset of $F \otimes_R K$, which is a $K$-vector space of countably infinite dimension: contradiction.
