In this answer to a question in Milnor, is says that if $e:C^\infty(M,\mathbb{R})\rightarrow \mathbb{R}$ is any ring homomorphism, then it is in fact an $\mathbb{R}$-algebra homomorphism, because $e(c)=c$ for any constant function $c$. He says "Indeed this must be true if $c\in \mathbb{Q}$".
Perhaps I am missing something obvious, but I do not see how it is automatically true that $e(c)=c$ for any $c\in \mathbb{Q}$. I understand the rest of the proof, but this little detail is driving me nuts!
First we can show $e(c)=c, \forall c\in\mathbb N$. $c(1)=1$ holds by the definition of ring homomorphism (which preserves identity, otherwise we may have $e(f)\equiv0, \forall f$, for which the current statement is false.)
And $e(n+1)=e(n)+e(1)=n+1$ by the induction hypothesis.
Now $e(-n)=-e(n)$ hence $e$ is identity on $\mathbb Z$, and finally $e(ab^{-1})=e(a)e(b)^{-1}=ab^{-1}$ for any $a, b\in\mathbb Z, a\not=0$.
This is a general phenomena that if $\rho:R\rightarrow S$ is a ring homomorphism where $R$ is an algebraic over a prime field $F$ (i.e. $F=\mathbb Q$ or $\mathbb F_p$) and $S$ is not the zero ring (i.e. $0_S\not=1_S$), then $S$ is also a $F$-algebra and $\rho$ is a $F$-algebra map, because the image of $F$ in $S$ must be isomorphic to $S$ since $S$ is a field which has no nontrivial quotient. In other words, there is a unique copy of $F$ in $R$, and since the automorphism group of $F$ is trivial, we don't have any room to move the elements around.
