Difficulty understanding how an element of a quotient ring/field can be represented a certain way...

Question

This is the proposition I'm given, which I don't really understand:

Let $p(x)=p_0 + p_1x + ... + p_nx^n$ be an irreducible polynomial over a field $F$, so that $ E = {f[x]}/{\lt p(x)\gt} $ is a field. Let $\theta : F[x] \to E$ be the natural homomorphism and let $\alpha = \theta(x) = x+ \lt p(x) \gt$. Then any element of $E$ can be represented uniquely in the form $a_0 + a_1\alpha + ... + a_n \alpha^{n-1}$, where $a_0,a_1, ..., a_{n-1} \in F$

So I understand that $p(x)$ is irreducible over a field $F$ therefore $\lt p(x) \gt$ is a maximal ideal. $F[x]$ must also be a field and therefore $E$ is also a field.

I see that $\alpha$ is the image some $x$ by the natural homomorphism from $\theta:F[x] \to E$, but I'm struggling to understand what this actually means when looking at the following example:

Show ${\mathbb{Z}_{13}[x]}/{\lt x^2 - 2 \gt} $ and ${\mathbb{Z}_{13}[x]}/{\lt x^2 - 5 \gt} $ are isomorphic.

In the solution it says that all elements of ${\mathbb{Z}_{13}[x]}/{\lt x^2 - 2 \gt} $ can be written as $a_0 + a_1\alpha$, where $a_0,a_1 \in \mathbb{z}_{13}$, which i understand from the proposition. But it says $\alpha^2=2$ and that's the part I'm struggling to understand. Furthermore ff we let $p(x)=x^2 -2 $, then $\alpha$ in this case is a root of $p(x)$ obviously. But $\alpha = x +\lt p(x) \gt$ so how can this squared equal $2$? $2$ is not an ideal, so how can squaring an ideal be $2$? In fact what exactly does $x + \lt p(x) \gt$ mean? I'm assuming its not $ \lt p(x) +x \gt$, so how can we add $x$ to a set?

Answer 1

More properly, $\alpha ^2 +<p(x)>=2+<p(x)>.$. This can be seen by $\alpha^2-2=0$ So, identifying $\alpha=x$, we have $\alpha^2=2$ MOD $p(x)$ means $x^2-2\in<(p(x))>$ which is trivially the 0.

$x+<p(x)>$ means the coset of the ideal $<(p(x))>$ formed by taking any polynimal $q(x)$ and then polynomials of the form $x+p(x)q(x)$. Which you can see when we mod out by $p(x)$, we get $x$.

Answer 2

Maybe you’ve forgotten the definition of addition and multiplication in the factor ring (quotient ring) $k[x]/(p)$. The elements are cosets $q+(p)$, where $q$ may be any element of $k[x]$, i.e. any polynomial. You may interpret $q+(p)$ as the set of all things of the form $q$ plus an element of the ideal.

Now, you must remember that a coset does not have a unique representation in the form $q+(p)$ unless you make some further restrictions on $q$. The typical restriction in this situation is to demand that your $q$ be of degree less than that of $p$. You don’t need to do that, though: you may be happy with the ambiguity. So, for instance, in your $\mathbb F_{13}[x]/(x^2-2)$, you may write the square of $x+1+(p)$ as $x^2+2x+1+(p)$, or you may perform Euclidean division on $x^2+2x+1$ and get $x^2+2x+1=1\cdot(x^2-2)+2x+3$. The quotient in your division is $1$ and the remainder is $2x+3$. So you can also write the square of $x+1+(p)$ as $2x+3+(p)$. The usual, and much more economical, way of writing this, is $(x+1)^2\equiv2x+3\pmod{x^2-2}$.

Now, to add two cosets, you may choose an element of each, add these two, and take the coset that contains that sum. In our situation, if the two summands started out of degree less than $\deg(p)$, the sum will be also, so you can just leave things without rewriting. Add $3x+2+(p)$ to $5x-4+(p)$ to get $8x-2+(p)$.

Multiplying two cosets is similar: choose an element in each coset, multiply these two, and take the coset that contains the product. As I said before, you can write the product of $3x+2+(p)$ and $5x-4+(p)$ as$15x^2-2x-8+(p)$, and if we’re still working in $\Bbb F_{13}[x]$, that will be $2x^2+11x+5+(p)$. But if your ideal is $(x^2-2)$ and you want to write your coset with the representative of degree less than $2$, you’ll have to divide again and get the remainder $11x+9$.

Now that you know how to add and multiply cosets, it should be clear why the square of the coset $x+(x^2-2)$ in $\Bbb F_{13}[x]/(x^2-2)$ is also $2+(x^2-2)$. Hope this helps.