You actually already proved the difficult part.
If $J$ is maximal, then by the correspondence of ideals you mention $J'$ has to be maximal, too. If it weren't, there would be a proper ideal $T'\supset J'$, which would give a proper ideal $T\supset J$.
Now assume $J$ prime and let $a'b'\in J'$, where $a',b'$ are the classes in $R/I$ of some $a,b\in R$. This means that there are $i_a,i_b\in I$ such that $(a+i_a)(b+i_b)=ab+ai_b+bi_a+i_ai_b\in J+I=J$, as $I\subseteq J$. Thus $ab\in J$ and, say, $a\in J$. Hence $a'\in J'$, so $J'$ is prime.
Finally, suppose $J$ radical and let $(a')^r\in J'$ for some $a\in R$, $r\in\Bbb N$. Then there are some $i_a,i\in I$ such that, by binomial expansion and since $I$ is an ideal, $(a+i_a)^r=a^r+i\in J+I=J$. Hence $a^r\in J$, so $a\in J$. Therefore $a'\in J'$ and $J'$ is radical.
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:
- $\mathbb R[X]/\langle x \rangle \simeq \mathbb R$, since quotienting by $x$ kills all polynomials of degree 1 or higher. What's left are the reals
- 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:
- the product of ideals is an ideal of the product ring
- 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:
$$
\{0, \mathbb R\} \times \{0, \mathbb R\} \times \{0, \mathbb R\}
$$
Best Answer
Note that $\Bbb C[x,y]/(x^2-y^3,x+2y-3)\cong \mathbb C[y]/((3-2y)^2-y^3).$ Furthermore, since $(3-2y)^2-y^3$ has three distinct roots over $\Bbb C$, we have $\mathbb C[y]/((3-2y)^2-y^3)\cong \Bbb C^3,$ which has maximal ideals $(\Bbb C,\Bbb C,0),(\Bbb C,0,\Bbb C),$ and $(0,\Bbb C,\Bbb C).$ Now, find the images of these in $\Bbb C[x,y]/(x^2-y^3,x+2y-3).$