Does the rational power series ring $\mathbb{Q}[[X]]$ embed as a ring into the field of real numbers? The title says it all. I'm wondering if the power series ring $\mathbb{Q}[[X]]$ (with rational coefficients) embeds as a ring into the field of real numbers. There are various topologies one might consider here, but I'm curious if there is an algebraic embedding.
 A: $\def\QQ{\mathbb{Q}}\def\RR{\mathbb{R}}$The answer is no!
Lemma Let $f(x) \in \QQ[[x]]$ with $f(0) =c^2$ for some nonzero rational $c$. Then $f(x)$ is a square in $\QQ[[x]]$.
Proof Use the Taylor series for $\sqrt{c^2+u}$ about $u=0$. $\square$
Therefore, if $\phi : \QQ[[x]] \to \RR$ is a ring homomorphism, then $\phi(1/n^2 + x) = 1/n^2 + \phi(x)$ must be a square for every positive integer $n$, and so $1/n^2 + \phi(x) \geq 0$ for every positive integer $n$.
Similarly, $\phi(1/n^2 - x) = 1/n^2 - \phi(x) \geq 0$ for every positive integer $n$.
So $-1/n^2 \leq \phi(x) \leq 1/n^2$ and we conclude that $\phi(x)=0$, so $\phi$ cannot be an embedding.
A: Some commentary on David's very nice answer that might provide some useful context. If $D$ is a subring of $\mathbb{R}$ then it must be an integral domain and its field of fractions $\operatorname{Frac}(D)$ must also embed into $\mathbb{R}$ so must be an ordered field. As an ordered field it must be Archimedean in the sense that it has no infinitesimals (elements $\varepsilon$ satisfying $0 < \varepsilon < \frac{1}{n}$ for all $n \in \mathbb{N}$) since $\mathbb{R}$ is Archimedean. So we can rule out the possibility of an integral domain $D$ admitting an embedding into $\mathbb{R}$ if we can show that any ordering on $D$ contains an infinitesimal. (Of course many integral domains will not admit any ordering at all, which would be an easier argument, but this one does. The relevant keyword here is "formally real".)
This is what David's argument does: the key is that squares are always non-negative, and from there David shows that this implies that $x \in \mathbb{Q}[[x]]$ is always infinitesimal, which is nicely intuitive. An example of an ordering is the lex ordering where a positive element is an element whose first nonzero coefficient is positive; for convergent power series this corresponds to the ordering where a positive element is an element which is positive on a sufficiently small neighborhood of the origin, which is an ordering that makes sense more generally for germs of real-valued continuous functions.
The corresponding question for embeddings into $\mathbb{C}$ is much easier: here we can appeal to the fact that uncountable algebraically closed fields of characteristic $0$ are classified up to isomorphism by their cardinality. From here it suffices to check that $\mathbb{Q}[[x]]$ is small enough that the algebraic closure of its field of fractions is at most as large as $\mathbb{C}$, which is true, and then the conclusion follows. If we tried to imitate this argument for $\mathbb{R}$ we'd be led to constructing the real closure of the field of fractions of $\mathbb{Q}[[x]]$ (which I believe is Puiseux series with real algebraic coefficients) but unfortunately real closed fields don't enjoy the same property of being classified by their cardinality, and the presence or absence of infinitesimals (and towers of infinitesimals, e.g. we can have elements which are infinitesimally small with respect to other infinitesimals) is one way to see this. This has been discussed on MO previously, at Are there as many real-closed fields of a given cardinality as I think there are?.
This suggests an interesting follow-up question:

The $\mathbb{C}$ argument shows that the subrings of $\mathbb{C}$ are exactly the integral domains of characteristic $0$ with cardinality at most the continuum. Is there a similar characterization of the subrings of $\mathbb{R}$? They must be integral domains admitting an Archimedean order with cardinality at most the continuum; is this enough?

