Basis of the free $R$-module $R[x]/(f)$

Question

If $f\in R[x]$ is any monic polynomial of degree $n$, then $R[x]/(f)$ has basis $\{1,x,…,x^{n−1}\}$.

Can someone explain this?

Answer 1

Recall that a module $M$ is said to be free with basis $B$ if:

1. The set $B$ is a generating set for $M$. That is, every element of $M$ is a "linear combination" of elements in $B$.
2. The set $B$ is "linearly independent" (virtually the same definition as in linear algebra).

It's not so hard to see that condition 1 is satisfied. In the quotient ring, any power of $x$ that is $x^n$ of higher is equivalent, after using the division algorithm with $f$, to some linear combination of the elements in the set you give in your question. Further, if they were linearly dependent, then we would have $$\sum_{i=0}^{n-1} c_i x^i =gf,$$ for some polynomial $g$, with $c_i\in R$. That is, we would have a linear combination that is zero in the quotient ring. But this is impossible, because the linear combination is a polynomial of degree $n-1$, while the right hand side has larger degree.

If $f$ is not monic, it might happen that $gf$ has degree lower than $n-1$. (If the leading coefficient is a zero-divisor, for example.)

Answer 2

the basis in question actually consists of the powers of $x$, i.e. $1,x,x^2,\dots,x^{n-1}$.

you may view this as a description of the process of division by $f$. any $g \in R[x]$ may be written as $fq+r$ where the degree of $r$ is $\lt n$. the elements of the quotient ring are represented by all such "remainders", with multiplication computed mod $f$