TITLE: Give an intuitive explanation for polynomial quotient ring, or polynomial ring mod kernel
QUESTION [7 upvotes]: I learned how to see quotient groups intuitively when I learned of a group mod its commutator subgroup.  If we take a group and mod out all the elements that do not commute, we get a quotient group which is abelian. Simple, nothing technical.
Now I am having trouble seeing a polynomial ring mod some ideal.
Example $1$: Evaluation hom. at $0$: $\ \ \ \varphi: \mathbb Z[x] \to \mathbb Z$ where $\varphi(f(x))=f(0)$.  This hom. is surjective and its kernel is the principal ideal generated by $x$, that is $(x)$. So, $\mathbb Z[x]/(x) \cong \mathbb Z$.
Example $2$: Evaluation hom. at $\sqrt 2$: $\ \ \ \psi: \mathbb Q[x] \to \mathbb Q[\sqrt 2]$. We know $\mathbb Q[x]/(x^2-2) \cong \mathbb Q[\sqrt 2]$.
What is $\mathbb Z[x]/(x)$? What do the cosets look like?
What is $\mathbb Q[x]/(x^2-2)$? What do the cosets look like?
Intuitively, what is this telling me? Any other concrete examples for intuition of a quotient ring?

REPLY [2 votes]: In the polynomial rings you are exploring, there is a division algorithm. Suppose $p(x)$ is a monic polynomial and $q(x)$ is any polynomial, then over $\mathbb Z$ or $\mathbb Q$ we have $$q(x)=p(x)s(x)+r(x)$$ for some polynomial $s(x)$ is less than where the degree of $r(x)$ is less than the degree of $p(x)$. The elements of the quotient ring of $\mathbb Q[x]$ or $\mathbb Z[x]$ by the ideal generated by $p(x)$ can be thought of as the remainders $r(x)$. The construction is equivalent to setting all the multiples of $p(x)$ equal to zero.
If two remainders $r(x)$ and $r'(x)$ are in the same coset of the ideal then $r(x)-r'(x)$ is divisible by $p(x)$ - but $p(x)$ has a higher degree, so the difference must be zero and $r(x)=r'(x)$. The remainders can be added (no problem) and multiplied if the result is reduced to the remainder on division by $p(x)$ - we often say reduced mod $p(x)$.
This answer is just a little introduction to a big and important subject, and can't cover everything. You might like to think for yourself what happens over $\mathbb Z$ when $p(x)$ is not monic (say $p(x)=2x+1$). This doesn't make a difference over $\mathbb Q$ because $\frac 12\in \mathbb Q$. But $2$ is not invertible over $\mathbb Z$.

In your second example with $x^2-2$, the remainders are all linear of the form $ax+b$. Since $x^2-2$ has zero remainder, we can say $x^2=2$ and identify $x$ with one of the square roots of $2$ (note there are two possibilities which are algebraically equivalent - this is a significant fact - we cannot distinguish them algebraically). The elements can be identified with $a\sqrt 2+b$.