If $\sqrt{(f)}$=$\sqrt{(g)}$, then $f+\mathrm{Nil}(R)=g+\mathrm{Nil}(R)$? (The converse holds)

Question

We can define an equivalence relation on a commutative unital ring $R$ as follows $$f\sim g\iff\sqrt{(f)}=\sqrt{(g)}.$$

Now in $R/\mathrm{Nil}(R)$, for each element of the class $f+\mathrm{Nil}(R)$ (elements of the form $f+p$, where $p\in \mathrm{Nil}(R))$ it holds that $\sqrt{(f+p)}$=$\sqrt{(f)}$.

Indeed, if $P$ is a prime ideal containing $f+p$, then $P$ contains $f$ because $p$ $\in$ $\mathrm{Nil}(R)\subseteq P$. Conversely, if $P$ contains $f$ then of course contains $f+p$, because $p \in \mathrm{Nil}(R)\subseteq P.$ Their radicals, as the intersections of all prime ideals that contains them must be equal. So, $f+\mathrm{Nil}(R)\subseteq Γf$ , (if $Γf$ is the equivalence class of $f$ of the equivalence relation defined at first).

The problem that troubles me is the converse, if $Γf \subseteq f+\mathrm{Nil}(R)$. Equivalently, I'm asking if these two partitions of the ring $R$ (the one defined with radicals and the one from the quotient ring $R/\mathrm{Nil}(R)$) are the same.

Answer 1

The statement $$\sqrt{(f)}=\sqrt{(g)}\implies f+\operatorname{Nil}(R)=g+\operatorname{Nil}(R)$$ is, in general, false. Take the ring $\Bbb Z$ and $f=4$, $g=2$. Then clearly, $\sqrt{(4)}=\sqrt{(2)}=(2)$, but $4-2=2\notin (0)=\sqrt{(0)}=\operatorname{Nil}(\Bbb Z)$.

Answer 2

No, this is false: if $R$ is an integral domain (or even if $R$ is reduced) it would mean that if $\sqrt{(f)}=\sqrt{(g)}$, then $f=g$.

So, for instance, if $p$ is a prime, one has $\sqrt{\strut p\mathbf Z}=\sqrt{\strut p^2\mathbf Z}=p\mathbf Z$, yet, it doesn't imply that $p=p^2$ (which would mean $p$ is an idempotent in $\mathbf Z$ – which has not that many idempotents…).

Answer 3

Since $f\sim -f,$ this would mean that $2f\in \mathrm{Nil}(R)$ for all $f.$ So this is not going to be true in general.

For example, when $1+1\notin \mathrm{Nil}(R)$, we have $1\sim -1$ but $1+\mathrm{Nil}(R)\neq -1+\mathrm{Nil}(R)$.

More generally, for your converse to be true, $1+u\in \mathrm{Nil}(R)$ for any unit $u\in R$, since $f\sim -uf.$

There are additional cases where, for all units $u\in R$, $1+u\in \mathrm{Nil}(R).$ For example, $\mathbb Z_2[x]$ has only $1$ as a unit, and $1+1=0\in(0)=\mathrm{Nil}(R)$, but $\sqrt{(x)}=\sqrt{x^2}$.

A ring where your statement is true is $R=\mathbb Z/2^n\mathbb Z$. Here, $f+2^n\mathbb Z\sim g+2^n\mathbb Z$ if and only if $f-g$ is even, and $\mathrm{Nil}(R)=(2+2^n\mathbb Z)$.