Determining the ideals of a quotient ring

Question

Given an ideal $I = \langle x^3 - x\rangle \subseteq \Bbb{R}[x]$, determine the ideals in the quotient ring $\Bbb{R}[x]/I$.

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.

Answer 1

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

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

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

To expand on why $\mathbb{R}[X]/\langle x \rangle$ contains only real numbers, notice that for any $p(x) \in \mathbb R[X]$, we can find polynomials $q(x), r(x)$ such that $p(x) = q(x) \cdot x + r(x)$ (by division). Now, we also know that $degree(r(x)) < degree(x)$. If This were not the case, then I could have my quotient be some polynomial, and thereby "remove" higher degree terms.

But now, in the quotient ring $\mathbb R[X]/ \langle x \rangle$, we know that $x \simeq 0$, and hence $p(x) = q(x) \cdot x + r(x) \simeq r(x)$. Since $r(x)$ has degree 0, it's "just a real number".

Now see that the exact same argument will hold for $\langle x + 1\rangle$ and $\langle x - 1 \rangle$, since all we depended on was the degree of $x$.

In general, a ring $\mathbb R [X] / \langle p(x) \rangle$ can only have polynomials of degree less than the degree of $p(x)$, since any polynomial of equal or higher degree can be factorized into some multiple of $p(x)$ plus a remainder. In the quotient ring, $p(x)$ goes to zero, so the multiple of $p(x)$ goes to zero, so all that's left is the remainder.

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:

$$ \{0, \mathbb R\} \times \{0, \mathbb R\} \times \{0, \mathbb R\} $$

Answer 2

Because $\Bbb{R}$ is a field, the polynomial ring $\Bbb{R}[x]$ in one indeterminate is a principal ideal domain. That means every ideal is of the form $\langle f\rangle$ for some $f\in\Bbb{R}[x]$. Now use the fact that $$\langle x^3-x\rangle\subset\langle f\rangle \qquad\iff\qquad x^3-x\in\langle f\rangle \qquad\iff\qquad f\text{ divides }x^3-x.$$