If $p$ contains $a$, then $p/a$ should be prime in $A/a$, since $(A/a)/(p/a)\cong A/p$, and so both are integral domains.
If $p$ does not contain $a$, then this need not hold: for example, let $A=\mathbb Z$, let $a=(2)$, $p=(3)$. Then $A/a=\mathbb Z/2\mathbb Z$, and the image of $(3)$ is the entirety of the quotient ring, which is not prime.
This counterexample is not ideal (no pun intended) since only a triviality prevents the ideal from being prime. If someone can think of a counterexample where the image is a nonprime proper ideal, that would be best.
First, let us recall the following:
Proposition. Let $f\colon R\twoheadrightarrow R'$ be a surjective ring homomorphism and let $P$ be a prime ideal of the ring $R$, then $f(P)$ is a prime ideal of $R'$.
Proof. First, $f(P)$ is a subgroup of $R'$, since $f$ is a group homomorphism. Let $y\in f(P)$, there exists $x\in P$ such that $f(x)=y$ and let $r'\in R'$, since the map $f$ is surjective, there exists $r\in R$ such that $f(r)=r'$. Therefore, one gets:
$$r'y=f(rx)\in f(P),$$
since $f$ is a ring homomorphism and $rx\in P$, using that $P$ is an ideal. Finally, one has to establish that $f(P)$ is prime in $R'$. Finally, it is left to show that $f(P)$ is prime, to do so let $(y_1,y_2)\in{R'}^2$ such that:
$$y_1y_2\in f(P).$$
Since $f$ is surjective, there exists $r_i\in R$ such that $f(r_i)=y_i$, therefore using that $f$ is a ring homomorphism, one has:
$$f(r_1r_2)\in f(P).$$
Hence, since $P$ is prime, $r_1\in P$ or $r_2\in P$, namely $y_1\in f(P)$ or $y_2\in f(P)$. Whence the result. $\Box$
Remark. The image of an ideal by a ring homomorphism is not an ideal, for example the image of $\mathbb{Z}$ by the inclusion $\mathbb{Z}\hookrightarrow\mathbb{Q}$.
Now, let $\pi\colon R\twoheadrightarrow R/A$ be the canonical surjection, then $P\mapsto \pi(P)$ is a bijection between the set prime ideals of $R$ containing $A$ and the set of prime ideals of $R/A$. The inverse is given by: $P\mapsto \pi^{-1}(P)$, since $\pi$ is surjective.
Best Answer
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:
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:
and we complete the proof, since the set of all ideals will all be:
$$ \{0, \mathbb R\} \times \{0, \mathbb R\} \times \{0, \mathbb R\} $$