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?
Best Answer
A general element of $F[X]/(a_nX^n + \dotsb + a_0X^0)$ is of the form $c_{n-1}X^{n-1} + \dotsb + c_0$ where $X^n = \frac{a_{n-1}X^{n-1} + \dotsb + a_0X^0}{a_n}$. Multiplication is performed by multiplying two polynomials in this "general form" and then simplifying the result according to the rule $X^n = \frac{a_{n-1}X^{n-1} + \dotsb + a_0X^0}{a_n}$, which reduces it to the general form again.
An example of this is $\mathbb{R}[X]/(X^2 + 1)$, where the general form of an element is $c_0 + c_1X$, and $X^2 = -1$. This gives the complex numbers. Another example is $\mathbb{R}[X]/(X^2)$, where every element is of the form $c_0 + c_1X$ where $X^2 = 0$; this is the dual numbers. The general form of an element of $\mathbb{R}[X]/(X)$ is the general form of an element of $\mathbb{R}$ because $X = 0$ in this ring. I recommend studying the examples I just gave: the complex numbers, the dual numbers, the bog-standard real numbers when expressed as $\mathbb R[x]/(X)$. Also look at the split-complex numbers, which I haven't shown.
A general element of $\mathbb{Q}[X]/(X^2 - 2)$ is of the form $c_0 + c_1X$ where $X^2 = 2$.