What do elements of this set look like?

Question

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

Answer 1

Do you know what a quotient ring is? Let's see in particular your example.

$\mathbb{F}[x]/(f)$ consists of elements of the form $h+(f)$ where $h\in\mathbb{F}[x]$. That is not your standard addition. Moreover two elements $h+(f)$, $g+(f)$ are equal if and only if $f|h-g$.

Therefore for $f(x)=x^3+x+1$, you can represent all the elements of $\mathbb{F}[x]/(f)$ as elements of the form $h+(f)$ with polynomials h with degree $\leq 2$ since for every $h\in\mathbb{F}[x]$ with degree $\geq 3$ there is $g\in\mathbb{F}[x]$ and $r\in\mathbb{F}[x]$ such that $h(x)=g(x)f(x)+r(x)$ with $degr \leq 2$ and therefore $h+(f)=r+(f)$ in $\mathbb{F}[x]/(f)$.

Answer 2

An element of the quotient ring is an equivalence class of the original ring modulo elements of the ideal. So for instance, one of the elements of $\mathbb{F}[x]/(x^3+x+1)$ is the equivalence class $[2x+x^3+x+1]$. But since $(2x+x^3+x+1) - (2x) = x^3 + x+1$, that is, an element of the ideal $(x^3+x+1)$, the equivalence classes are the same: $[2x + x^3 + x + 1] = [2x]$.

By the division algorithm, every equivalence class has a unique representative with degree less than 3. So it's commonplace to write “$2x$” instead of “$[2x]$” and simply cancel $x^3+x+1$ whenever we see it.

Answer 3

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 $$(x^2+1)(x^2+x-1)=x^4+x^3-x^2+x^2+x-1\\ =x(x^3)+x^3+x^2+x-1\\ =x(-x-1)+(-x-1)+x^2+x-1\\ =-x^2-x-x-1+x^2+x-1\\=-x-2$$

Answer 4

You are correct that the elements of $\mathbb F[x]/\langle f\rangle $ are elements of the form $h(x) + \langle f\rangle$ - or more generally, for a quotient ring, that the elements of $R/I$ are of the form $r + I$ for $r\in R$. However, I think you might be confused about what this means. Note that, to avoid ambiguity, I'm using $\langle f \rangle$ to be the ideal generated by $f$ - that is, the set of values of the form $\langle f\rangle = \{p\cdot f : p\in F[x]\}$.

Let's take a concrete example: consider $\mathbb Z[x]/\langle x^2+1\rangle $. The elements of this ring are by definition sets of polynomials. This is to say that neither $x$ nor $x^3$ are elements of the ring - rather, they belong to sets like $$x+\langle x^2+1\rangle = \{x,x^2+x+1,2x^2+x+2,x^3+2x,\ldots\}$$ $$x^3+\langle x^2+1\rangle = \{x^3,x^3+x^2+1,2x^3+x,-x,\ldots\}$$ where each set contains every expression of the form $x + p(x)(x^2+1)$, for instance. Note that we can also have non-obvious relations - like, above, we can see that $x^3+\langle x^2+1\rangle = -x + \langle x^2+1\rangle$ since $x^3 = -x + x\cdot (x^2+1)$.

So, naturally, from this definition, we can't talk about degree because the elements of $\mathbb Z[x]/(x^2+1)$ aren't polynomials, but are rather sets of polynomials. This is the same way that $\mathbb Z/2\mathbb Z$ is not the set $\{0,1\}$, but is better thought of as the ring where we do arithmetic on $\{\text{even},\text{odd}\}$ where the labels each stand in for classes of integers by parity.

However, whenever we consider a quotient like $\mathbb F[x]/\langle f\rangle$ there is a canonical representative of each class $g + \langle f\rangle$: one may prove that each such class contains a unique polynomial of minimum degree within that class so long as either $\mathbb F$ is a field or $f$ is monic. For instance, in the example above the class $x^3+\langle x^2+1\rangle$ contains a single linear polynomial: $-x$. In fact, the division theorem says that if $g,f$ are polynomials over a field $F$ then there is some pair of polynomials $p,q$ such that $$g(x) = p(x)f(x) + q(x)$$ where the degree of $q$ is less than the degree of $g$ - the proof being by polynomial long division. In the set $g + \langle f\rangle$ one will find out that $q(x)$ is the polynomial of lowest degree.

It is common to equate the whole class of polynomials with this representative - hence why some people think of $\mathbb Z[x]/(x^2+1)$ to consist of linear polynomials with some weird multiplication rule, but this is not the definition and I think this is where your confusion arises from.