Elements of $\mathbb{R}[x]/f$ for polynomial $f$ of degree $n$

Question

I was thinking about cosets of this ring and my intuition is this ring is made of all possible polynomials of degree $n-1$. Is that right? Would I need to assume something more about $f$ or ring itself for it to be true? (Asking for polynomial with real coefficients, but can it be generalized?). I think I was able to prove that any polynomial of degree bigger than degree of $f$ can be written in the form $hf + g$ where deg $g<n$

Answer 1

I think your intuition is correct. If $F$ is a field, then we have the following fact:

Theorem: For every $g,f\in F[x]$, with $f\ne0$, there exist unique $q,r\in F[x]$, such that $g=qf+r$ and either $r=0$ or $\deg(r)<\deg(f)$.

The $q$ in this theorem is the called the "quotient", and the $r$ is called the "remainder". The algorithm for finding $q$ and $r$ is often called "long-division". So the theorem is often referred to as "the division algorithm".

Let $F$ be a field and let $f$ be a non-zero polynomial in $F[x]$. Then for every $g\in F[x]$, there is a unique $q,r\in F[x]$ with $g=qf+r$ and either $r=0$ or $\deg(r)<\deg(f)$. Note that $g=qf+r$, implies that $g\equiv r$ (mod $f$). So every $g\in F[x]/(f)$ is congruent to a polynomial of degree at most $n-1$.

Also note that for polynomials $p,q\in F[x]$, if the degree of $p$, $q$ is at most $n-1$, then the degree of $p-q$ will also be at most $n-1$ (or $p-q=0$). Hence for $p,q\in F[x]$ with degree at most $n-1$, the following are equivalent:

$$p\equiv q\,(\text{mod }f)$$ $$f\text{ divides }p-q$$ $$p-q=0$$ $$p=q$$

What this shows is that if $p,q$ are distinct polynomials of degree at most $n-1$, then they are not congruent modulo $f$.

It follows that the elements of $F[x]/(f)$ correspond to polynomials of degree at most $n-1$.

Note: in fact $F[x]/(f)$ is a ring, but if we want to know the exact structure of this ring, then we will need to know more about $F$ or more about $f$.