When $R$ is a subring of $R[\alpha]$ I am reading Artin's Algebra, Section 5 of Chapter 11.  The chapter begins by saying that we want to find a suitable ring extension $R[\alpha]$ that contains $R$ as a subring as well as $\alpha$.  He then proceeds with an example and a generalization of this to create a general ring extension $R[\alpha]$ in which $\alpha$ satisfies a polynomial relation $f(\alpha)=0$ by defining $R[\alpha]$ to be the quotient ring $R[x]/(f)$, where $(f)$ is the principal ideal generated by $f$.  $\alpha$ is identified by the image of $x$ under the quotient transformation.  
However, at the of the section it is brought up that we haven't actually shown that $R$ is a subring of $R[\alpha]=R[x]/(f)$, and proceeds to show that in general this isn't true.  The criterion mentioned for $R$ to be a subring of $R[\alpha]=R[x]/f$ is for the restriction of the quotient transformation to constant polynomials, say $\psi: R\rightarrow R[\alpha]$, is an injection. More importantly, that if $\psi$ is not an injection, then there exists no injective ring homomorphism of $R$ into $R[\alpha]$. 
Why is this enough?  
 A: The idea is that $R\to R[\alpha]$ is supposed to be an inclusion, so $R$ is a subring of $R[\alpha]$. This can fail if setting the polynomial $f$ equal to $0$ ends up "collapsing" $R$ itself a bit. However, just because the induced map $R\to R[x]/(f)$ isn't injective doesn't mean there isn't any injective map ...
Let $R=k[x_1,x_2,\cdots]$ be a polynomial ring in infinitely many variables. Let $f(T)=x_1$ be a constant polynomial in $T$, so $R[T]/(f)=R[x_2,x_3,\cdots,T]$ and the induced map $R\to R[T]/(f)$ is not injective (as $x_1\mapsto 0$), however there is clearly an injective map $R\to R[T]/(f)$ where $x_1\mapsto T$ and $x_i\mapsto x_i$ for $i>1$.
A: This is basically a statement of one way to define a subring.  More precisely, the first definition one learns is that if $R$ is a ring, then $S$ is called a subring of $R$ if $S \subseteq R$ and $S$ is a ring in its own right.  (If your definition of rings requires them to have $1$ then an additional requirement for a subring is often that $1_S = 1_R$, but this is irrelevant to the conversation).
Another way to define a subring of $R$ is to say that $S$ is a subring of $R$ if $S$ is a ring and there exists an injective homomorphism $\psi: S \to R$.  (If your definition of rings requires them to have $1$, then it is often required that $\psi(1_S) = 1_R$.)
That is, $S$ is a subring of $R$ according to the second definition if and only if $S$ is isomorphic to some $S'$, which is a subring of $R$ according to the first definition.
The second definition is the one operating here, since as formal objects $R$ is not a subset of $R[x] / (f)$.  But if $\psi: R \to R[x] / (f)$ given by $\psi(r) = r + f$ is injective, then $R$ is a subring of $R[x] / (f)$ according to this second definition.
