I'm working on the following question:

Let $R = \mathbb{Z}_{27}$. Find $N(\langle 9\rangle)$ where $N$ is the radical of $\langle 9\rangle$.

I have so far:

$N(\langle 9\rangle) = \{x \in \mathbb{Z}_{27} | \ x^n \equiv \{0,9,18\} \mod 27\}$

Is there a quick way of checking which $x$ satisfy this? The only way I can think of is take each element of $\mathbb{Z}_{27}$ and check this condition by multiplying to itself until I see a pattern.


1 Answer 1


You could always try reasoning. What I mean is that in this case, it’s obvious $3^2\in (9)$, so $(3)$ is contained in the radical of $(9)$, and $(3)$ is a maximal ideal in $\mathbb Z_{27}$, so it is exactly the radical of $(9)$.

It should also be said that the nilradical refers to the radical of the zero ideal. Usually you call what you’re describing “the radical” of the ideal. Of course, that radical of $I$ corresponds to the nilradical of the quotient ring $R/I$.

  • $\begingroup$ why does it follow form $3^2 \in \langle9\rangle$ that the entire ideal $\langle 3 \rangle \subset N(\langle9\rangle)$? $\endgroup$
    – yoshi
    Commented Jun 5, 2018 at 11:13
  • 1
    $\begingroup$ @yoshi By definition, it means $3$ is in the radical of $(9)$, and therefore the entire ideal generated by $3$ is in $(9)$. $\endgroup$
    – rschwieb
    Commented Jun 5, 2018 at 11:23

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