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.

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    $\begingroup$ You have just one indeterminate. There's a very useful description of all ideals of $\mathbb{R}[x]$, because it is a principal ideal domain. $\endgroup$ – egreg Apr 24 '19 at 16:01

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:

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

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    $\begingroup$ This answer is not correct. $X^{2}-1$ factors as $(X-1)(X+1)$ over $\mathbb{R}$. The Chinese Remainder Theorem therefore shows that $\mathbb{R}[X]/\langle X^{3}-X \rangle \cong \mathbb{R} \oplus \mathbb{R} \oplus \mathbb{R}$. $\endgroup$ – Alex Wertheim Apr 24 '19 at 17:11
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    $\begingroup$ Looks mostly correct to me! (The factorization is $(X^{3}-X) = X(X^{2}-1)$, not $X(1-X^{2})$. But this is a minor point, which doesn't affect the computation of the quotient up to isomorphism. $\endgroup$ – Alex Wertheim Apr 24 '19 at 17:21
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    $\begingroup$ Indeed a product of ideals is an ideal of the product ring. But to be sure that you have all ideals, you need the converse, that every ideal of the product ring is a product of ideals. $\endgroup$ – Servaes Apr 24 '19 at 17:33
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    $\begingroup$ @SiddharthBhat is there a way of doing this without using the chinese remainder theorem? The reason is that we didn't cover this in class, so I was wondering if there was a different method $\endgroup$ – Masha Apr 25 '19 at 17:22
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    $\begingroup$ @SiddharthBhat can you please elaborate on why $\mathbb{R} / [x]]$ gets rid of everything but the reals? How does it get rid of polynomials with a degree higher than 1? And going off of that how does that work for x+1 and x-1 as well? Thank you! $\endgroup$ – Masha Apr 26 '19 at 1:45

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 $$\langle x^3-x\rangle\subset\langle f\rangle \qquad\iff\qquad x^3-x\in\langle f\rangle \qquad\iff\qquad f\text{ divides }x^3-x.$$

  • $\begingroup$ Can you please explain what you mean by an indeterminate and what a principal ideal domain is? $\endgroup$ – Masha Apr 25 '19 at 18:32
  • $\begingroup$ @Masha An indeterminate is often also called a variable; the symbols $x$, $y$ and $z$ in the polynomial ring $\Bbb{R}[x,y,z]$, for example. $\endgroup$ – Servaes Apr 25 '19 at 20:11
  • $\begingroup$ @Masha A principal ideal domain is a domain in which every ideal is principal; every ideal is generated by a single element. $\endgroup$ – Servaes Apr 25 '19 at 20:12
  • $\begingroup$ Ok, so if I understand correctly the polynomial ring in one variable is a domain in which every ideal is generated by a single element? So for example if $\mathbb{R}[x] = x^2$ then every ideal of that ring is just generated by a single element, x? $\endgroup$ – Masha Apr 25 '19 at 23:13

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