${\rm Spec}(A/\mathfrak{a})={\rm Spec}(A)$ if and only if $\mathfrak{a}$ is generated by nilpotent elements

Question

I have been introduced to the Zariski topology and I cant solve this problem:

Let $A$ be a commutative ring with unity and $\mathfrak{a}$ an ideal of $A$, we define ${\rm Spec}(A) $ as the set of prime ideals of $A$, and ${\rm Spec}(A/\mathfrak{a})$ as the set of primes ideals of $A/a$ (which is homeomorphic to the zeros of $\mathfrak{a}$). Knowing this prove:

${\rm Spec}(A/\mathfrak{a})={\rm Spec}(A)$ if and only if $a$ is generated by nilpotent elements.

Answer 1

Here are a few hints: the (prime) ideals of $A/\mathfrak{a}$ correspond to the (prime) ideals of $A$ containing $\mathfrak{a}$. You can use this idea to get a continuous injective map $\mathrm{Spec}(A/\mathfrak{a}) \to \mathrm{Spec}(A)$.

Now, this map is surjective (and in fact a homeomorphism) exactly when all of the prime ideals of $A$ contain $\mathfrak{a}$. This means $$\mathfrak{a}\subset \bigcap_{\mathfrak{p}\in \mathrm{Spec}(A)} \mathfrak{p}.$$ However, there is a characterization of $\mathfrak{N}_A$ (the nilradical of $A$) as $$ \mathfrak{N}_A = \bigcap_{\mathfrak{p}\in \mathrm{Spec}(A)}\mathfrak{p}. $$ Putting these ideas together solves the problem.