How should I understand $R[x]/(f)$ for a ring $R$? The following is a proposition in Artin's Algebra: 

Proposition 11.5.5 Let $R$ be a ring, and let $f(x)$ be a monic polynomial of positive degree $n$ with coeeficients in $R$. Let $R[\alpha]$ denote the ring $R[x]/(f)$ obtained by adjoining an element satisfying the relation $f(\alpha)=0$. Then 
the set $(1,\alpha,\cdots, \alpha^{n-1})$ is a basis of $R[\alpha]$ over $R$: every element of $R[\alpha]$ can be written uniquely as a linear combination of this basis, with coefficients in $R$. 


I don't understand the polynomial $f(x)$ in the theorem here. Consider $R={\Bbb Z}$. Let $f(x)=x^2+1$ and $g(x)=(x^2+1)^2$. Then we have
$$
f(i)=g(i)=0.
$$
Then by the theorem above, ${\Bbb Z}[i]$ is not only ${\Bbb Z}[x]/(f)$ with basis $(1,i)$, but also ${\Bbb Z}[x]/(g)$, with basis $(1,i,i^2,i^3)$. What is wrong here?

When $R$ is a field $F$, and if $\alpha\in E$ is algebraic over $F$ where $E$ is an extension field of $F$, then we do have
$$
F[\alpha]\cong F[x]/(f)
$$ 
where $f$ is the minimal polynomial of $\alpha$. But in Proposition 11.5.5, there seems to be no requirement for the degree of $f$. (I thought this might be implicitly in the phrase "the relation $f(\alpha)=0$". But I'm not sure.)

[Added:] Does one have any other references about this proposition? (It seems that some assumption in the proposition is missing according to the comments.)
 A: Let $I = (f)$. Then "let $R[\alpha]$ denote the ring $R[x]/I$" means that $\alpha := x+I\,$ and $\,R[\alpha]\,$ denotes the subring of $R[x]/I$ generated by $R$ and $\alpha$, i.e. the smallest subring containing both. Clearly this is the whole ring $R[x]/I$ since   $\,g(x)+I = g(x+I) = g(\alpha)\in R[\alpha].\ $ Furthermore,  notice that $\,0 = g(\alpha) = g(x+I) = g(x)+I$ $\iff$ $g\in I = (f)$ $\iff$ $\,f\mid g\,$ in $R[x].\,$  Thus $\,\alpha\,$ serves as a  "generic" root of $\,f\,$ over $R$ since it satisfies $f$ but no smaller degree polynomials. We can view the ring $R[x]/(f)$ as the most general (universal) way of "adjoining" a root of $f$ to $R$. Here "adjoining" has a technical meaning, which I elaborate on below (from a prior answer).
More generally, if $\rm\,R \subset S\,$ are rings and $\rm\,s\in S\,$ then $\rm\,R[s]\,$ denotes the ring-adjunction of $\rm\,s\,$ to $\rm\,R\,,\,$ i.e. the smallest subring of $\rm\,S\,$ containing both $\rm\,R\,$ and $\rm\,s\,.\,$ Equivalently $\,\rm R[s]$ is the image of $\rm\,R[x]\,$ under the evaluation map $\rm\,x\mapsto s.\,$ It is the set of all elements that can be written as polynomials in $\rm\,s\,$ with coefficents in $\rm\,R.\,$ The notation for the polynomial ring $\rm\,R[x]\,$ is  the special case where $\rm\,x\,$ is transcendental over $\rm\,R\ $ (an "indeterminate" in old-fashioned language),$\ $ i.e. $\rm\, x\,$ isn't a root of any polynomial with coefficients in $\rm\,R\,$. One may view $\rm\,R[x]\,$ as the adjunction of a universal (or generic) element $\rm\,x\,$, in the sense that any other adjunction $\rm\,R[s]\,$ is a ring-image of $\rm\,R[x]\,$ under the evaluation homomomorphism $\rm\, x\to s\,.\ $ For example, if $\rm\,R \subset S\,$  are fields then $\rm\,R[s]\cong R[x]/(f(x))\,$ where $\rm\,f(x)\,$ is the minimal polynomial of $\rm\,s\,$ over $\rm\,R\,.\,$ Essentially this serves to faithfully ring-theoretically model $\rm\,s\,$ as a "generic" root $\rm\,x\,$ of the minimal polynomial $\rm\,f(x)\,$ for $\rm\,s\,.\,$ Polynomial rings may be characterized by the existence and uniqueness of such evaluation maps ("universal mapping property"), e.g. see any textbook on Universal Algebra, e.g. Bergman.
