Give an intuitive explanation for polynomial quotient ring, or polynomial ring mod kernel

Question

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?

Answer 1

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 $X^2 = 2$.

Answer 2

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 $\mathbb Z$.

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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 $a\sqrt 2+b$.