<s>[INST] What do elements of this set look like? I was under the impression that elements of quotient rings such as $\mathbb F[x]/(f)$ were of the form $h(x)+(f)$ where $h(x)\in\mathbb F[x]$. Is this correct? If so could somebody explain why elements of $\mathbb F[x]/(x^3+x+1)$ have degree 2 or less? Why isn’t $2x+x^3+x+1$ an element? Is my understanding of what $\mathbb F[x]/(f)$ means wrong? I have looked at several answers and still don’t understand
 [/INST] In fact, in the quotient ring $R/I$, some elements are eaten by the ideal $I$.
For example, in the case of $\mathbb{F}[x]/(x^3+x+1)$, since
$$x^3+x+1\in(x^3+x+1)\text{ therefore }\\ x^3+(x^3+x+1)=-x-1+(x^3+x+1)$$
and thus
$$ax^3+(x^3+x+1)=-ax-a+(x^3+x+1)$$
also we have
$$x^4+(x^3+x+1)=x(-x-1)+(x^3+x+1)=-x^2-x+(x^3+x+1)$$
and $$x^5+(x^3+x+1)=x(-x^2-x)+(x^3+x+1)\\=-x^3-x+(x^3+x+1)\\=x+1-x+(x^3+x+1)=1+(x^3+x+1)$$
and so on.
As you see, any monomial of degree greater than or equal to $3$, reduces to a polynomial of degree at most $2$.
In fact the ring $\mathbb{F}[x]/(x^3+x+1)$ is
$$\{a+bx+cx^2:a,b,c\in\mathbb{F}, x^3=-x-1\}$$
For example, the product of $x+1$ and $x^2+x$ is
$$(x^2+1)(x^2+x-1)=x^4+x^3-x^2+x^2+x-1\\
=x(x^3)+x^3+x^2+x-1\\
=x(-x-1)+(-x-1)+x^2+x-1\\
=-x^2-x-x-1+x^2+x-1\\=-x-2$$
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