Prove that $(0)$ is a radical ideal in $\mathbb{Z}/n\mathbb{Z}$ iff $n$ is square free

Question

Let $n>1$ be an integer. Prove that $0$ is a radical ideal in $\mathbb{Z}/n\mathbb{Z}$ if and only if $n$ is a product of distinct primes to the first power (i.e., $n$ is square free). Deduce that $(n)$ is a radical of $\mathbb{Z}$ if and only if $n$ is a product of distinct primes in $\mathbb{Z}$.

I don't really get it. For example, $8$ is not a square-free integer; so take $\mathbb{Z}/8\mathbb{Z}$. Why is $0$ not a radical ideal in this ring but is in $\mathbb{Z}/6\mathbb{Z}$?

Answer 1

First, let's have a look at your first example.

If $\{0\}$ is going to be a radical ideal in $\mathbb{Z}/8\mathbb{Z}$, then it has to be equal to its own radical. But it is not, because $2^3 = 0$ in $\mathbb{Z}/8\mathbb{Z}$, so at least we have that $2\in\sqrt{\{0\}}$.

Generally, what you need to consider are the nilpotent elements of the ring $\mathbb{Z}/n\mathbb{Z}$. If there are any non-zero nilpotent elements, then $\{0\}$ is not a radical ideal. In particular, suppose that $n=p_1^{k_1}\ldots p_s^{k_s}$, where $k_i>1$ for some $i$, consider the element $m:=\text{rad}(n) = p_1\ldots p_s$, and see if you can figure out a power of $m$ that gives you the zero element.

Also, if $n=\text{rad}(n)$, there will be no non-zero nilpotent elements in your ring. The key lies of course in that powers of a number $x<n$ has a hard time being a multiple of $n$, when $n$ consists only of distinct primes. This is perhaps easiest to obtain by a contradiction argument.

Let me know, if I should provide any more hints, or if it is clear by now.

Answer 2

Here's another approach (for one direction). Suppose $n$ is square free and $n=p_1\cdots p_s$ is the prime canonical factorization of $n$.

By the Chinese Remainder theorem for commutative rings, we have the isomorphism $\mathbb{Z}/n\mathbb{Z}\cong \mathbb{Z}/p_1\mathbb{Z}\times \cdots \times \mathbb{Z}/p_s\mathbb{Z}$.

Recall $\mathbb{Z}/p\mathbb{Z}$ is a finite field of $p$ elements. This shows that there are no nilpotent elements that are nonzero. But the radical of $0$ is the set of nilpotent elements so, the radical must be itself.

Answer 3

If $\phi: A \to B$ is a ring homomorphism and $J$ is an ideal of $B$, then $\phi^{-1}(rad(J))=rad(\phi^{-1}(J))$.

In particular, $\phi^{-1}(rad(0))=rad(\phi^{-1}(0))=rad(\ker\phi)$.

Now consider the canonical ring homorphism $\phi:\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}$.

Then, $\phi^{-1}(rad(0))=rad(\ker\phi)=rad(n\mathbb{Z})$ and so $rad(0)=0$ in $\mathbb{Z}/n\mathbb{Z}$ iff $rad(n\mathbb{Z})=n\mathbb{Z}$ iff $n$ is square free.