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Given an ideal $I = langle x^3 - xrangle subseteq BbbR[x]$, determine the ideals in the quotient ring $BbbR[x]/I$.

I understand that the quotient ring is of the form $k[x_1...x_n]/I$ where I is an ideal in $k[x_1...x_n]$. I also understand that the ideals of the quotient ring are in one-to-one correspondence with the ideals of $k[x_1...x_n]$ containing $I$.

However, I'm really confused on how to actually go about finding the ideals of the quotient ring given a specific ideal. Any hints or suggestions would be appreciated, thank you.

Because $BbbR$ is a field, the polynomial ring $BbbR[x]$ in one indeterminate is a principal ideal domain. That means every ideal is of the form $langle frangle$ for some $finBbbR[x]$. Now use the fact that
$$langle x^3-xranglesubsetlangle frangle
qquadiffqquad
x^3-xinlangle frangle
qquadiffqquad
ftext divides x^3-x.$$

Note that we can factor $x^3 - x = x(x^2 - 1) = x(x + 1)(x - 1)$. Now, by the Chinese Reimander theorem:

$$mathbb R[X]/langle x^3 - xrangle = mathbb R[X] /langle x(x + 1)(x-1) rangle = mathbb R[X]/langle x rangle timesmathbb R[X]/langle x+1 rangle timesmathbb R[X]/langle x-1 rangle $$

We know that:

1. 
   $mathbb R[X]/langle x rangle simeq mathbb R$, since quotienting by $x$ kills all polynomials of degree 1 or higher. What's left are the reals


2. Similarly, $mathbb R[X]/langle x+1 rangle simeq mathbb R [X]/ langle x-1 rangle simeq mathbb R$.

So, the ring that we have is actually $mathbb R times mathbb R times mathbb R$.

Since $mathbb R$ is a field, it has only two idels: the zero ideal and the ring itself.

Now, note that:

1. the product of ideals is an ideal of the product ring


2. Every ideal of the product ring is a product of ideals

and we complete the proof, since the set of all ideals will all be:

$$
0, mathbb R times 0, mathbb R times 0, mathbb R
$$