A formula for the radical of $\mathbb{Z}/n\mathbb{Z}$.

Let I be an ideal in A. Then r(I)= { $x\in A | x^n \in I$ for some $n\in \mathbb{N}$}. $\phi : A \to A/I$ is the natural map. Then prove that $r(I) = \phi^{-1} (N_{A/I})$. N is nilradical there.

I am reading from notes in commutative algebra. $N_{A/I}$ means set of all x in A/I such that $x^n\subseteq I$ for some $n\in \mathbb{N}$ but how it's inverse image will be equal to r(I)?

Can you please tell?

Edit : Assuming the above result to be true I was asked to Find the r(I) in following cases.

(i) $\mathbb{Z}$ and $I= 9\mathbb{Z}$

(ii) $\mathbb{Z}$ and $I= 12\mathbb{Z}$.

Nil Radical in (i) is {0,3,6} and Nil Radical in (ii) is {0,6}.

but I am unable to understand how to find inverse image of such a set ie how to find inverse image of $\phi^{-1} $({0,3,6}) and $\phi^{-1}$( {0,6})?

Please help.

Solution 1:

Question: "Let $I$ be an ideal in $A$. Then $r(I)= { x∈A|x^n∈I \text{ for some }n∈N}$. Let $ϕ:A→A/I$ is the natural map. Then prove that $r(I)=ϕ^{−1}(N_{A/I})$. $N$ is nilradical there."

Answer: Let $p: A\rightarrow A/I:=B$ and let $N:=nil(B)$ be the nilradical of $B$. Let $J:=p^{-1}(N)$ be the inverse image ideal of $N$.

Claim: there is an equality of ideals $J =r(I)$.

Assume $x\in J$. It follows $p(x)\in N$ hence $p(x)^n=p(x^n)=0$ in $B$. This implies $x^n\in I$ and hence $x\in r(I)$ and $J \subseteq r(I)$. Assume $x\in r(I)$. It follows $x^n\in I$ and $p(x)^n=p(x^n)=0$ hence $p(x)\in N$ and $x\in J$.

If $I=(3^2) \subseteq R:=\mathbb{Z}$ it follows $(3)\subseteq R/(3^2)$ is the only prime ideal hence $rad(R/(3^2))=(3)$. If $J:=(12)=(3)(2^2)$ it follows from the chinese remainder lemma that

$$R/J \cong R/(3)\oplus R/(2^2)$$

and the prime ideals are $J_1:=(3)\oplus R/(2^2)$ and $J_2:=R/(3)\oplus (2)$ hence the radical is

$$rad(R/J)=J_1 \cap J_2=J_1J_2=(6)$$

since the intersection of coprime ideals equals the product. In general if $n:=\prod_{i=1}^n p_i^{r_i} \in R$ is any postitive integer with $p_i$ distinct primes, it follows the radical satisfies

$$rad(R/(n))=(p_1\cdots p_n).$$