Why do the ideals have to be irreducible for finite fields?

Question

I was reading some examples of finite fields from my class notes:

$\Bbb Z_2[x]/\langle x^2+x+1\rangle$ (Where $x^2+x+1$ is an irreducible polynomial over $\Bbb Z_2$)

$\Bbb R[x]/I$ where $I$ is an irreducible polynomial over $\Bbb R$

etc.

Here, I don't understand why the ideals have to be necessarily irreducible over the corresponding fields ($\Bbb R,\Bbb Z_2$).

As far as I know, any polynomial divided by a $2$ degree polynomial leaves behind a $1$ degree polynomial as remainder (which would form equivalence classes of remainders). A $3$ degree polynomial leaves behind a $2$ degree polynomial as remainder (which again form equivalence classes of remainders).

Could someone please explain why the ideal has to be irreducible?

Answer 1

The polynomial $p(x)$ does not need to be irreducible in order for $\Bbb R[x]/\def\ang<#1>{\langle#1\rangle}\ang<p(x)>$ to be well defined. However, if $p(x)$ is reducible, then the ring $\Bbb R[x]/\ang<p(x)>$ will not have nice properties. For example, if $p(x)$ factors as $f(x)g(x)$, then the quotient ring will have zero divisors, as $f(x)g(x)=p(x)\equiv 0$. On the other hand, when $p(x)$ is irreducible, then $\Bbb R[x]/\ang<p(x)>$ is a field, which is very well behaved and well understood.

The same analysis applies when $\Bbb R$ is replaced with $\Bbb Z_2$ or any other field.