Let $A$ be a ring, $\operatorname{Spec}(A) = \{ \mathfrak{p} \mid \mathfrak{p}\textrm{ is a prime ideal of }A\}.$ Given an ideal $I$ of $A$, define $V(I) = \{ \mathfrak{p} \in \operatorname{Spec}(A) \mid \mathfrak{p} \supset I \}.$ Recall that the radical of $I$ is defined as $\sqrt{I} = \{ a \in A \mid a^n \in I\textrm{ for some }n\in\Bbb N\} $ and the radical of $A$ is defined by $\sqrt{A} = \sqrt{\{0\}}.$ With these definitions I want to prove:
"If $\operatorname{Spec}(A) = V(\sqrt{A}),$ then $\operatorname{Spec}(A) =\{ \{ 0 \} \}$," or
"If $\operatorname{Spec}(A) = V(\sqrt{A}),$ then $\operatorname{Spec}(A) =V(\{ 0 \} ).$"
I don't know if these statements are true, however I tried to prove them using only the definitions above and was not able to succeed.
Is there some counterexample, and if not, can I get help for the proof?
thanks.
You should be able to find the following in more detail in any standard text on algebraic geometry.
It is always true that $\operatorname{Spec}(A) = V\left(\sqrt{(0)}\right),$ so the "if" part of each statement is always satisfied. This is because $\sqrt{(0)}$ consists of the nilpotent elements of $A,$ and every prime ideal must contain all nilpotents of the ring.
So, we have a counterexample to your first statement: take any ring which is not a field. This ring necessarily has a maximal ideal $\mathfrak{m}$ if it is not the zero ring, and as $A$ is not a field, $\mathfrak{m}\neq (0).$ If $A$ is the zero ring, it has no prime ideals at all, so that is also a counterexample.
Your second statement boils down to the claim $\operatorname{Spec}(A) = V\left(\sqrt{(0)}\right) = V((0)).$ This is true; note that $0$ is the only element of $(0).$ As every prime ideal of $A$ necessarily contains $0,$ we have the desired equality.
More generally, if $I$ is an ideal, then $I\subseteq\sqrt{I},$ so that $V(\sqrt{I})\subseteq V(I)$ (if $\mathfrak{p}\supseteq\sqrt{I},$ then $\mathfrak{p}\supseteq I$ as well). In fact, $V(\sqrt{I}) = V(I).$ Indeed, let $a\in \sqrt{I},$ and suppose $\mathfrak{p}\supseteq I.$ Then $a^n\in I$ for some $n,$ thus $a^n\in\mathfrak{p}.$ But this implies that $a\in\mathfrak{p}$!