Finding all the maximal ideals of $\mathbb{Z}_{63}$

Question

How do I go about finding all the maximal ideals of this ring ?

I realise that all ideals are subgroups with respect to addition. Therefore, since $\mathbb{Z}_{63}$ is cyclic then every subgroup, and so every ideal, will be generated by a single element. I also realise that $\langle n \rangle \subseteq \langle m \rangle \iff m \vert n $.

I want to conclude then that all the ideals generated by prime numbers are maximal but this doesn’t seem right as $\langle 2 \rangle = \mathbb{Z}_{63}$

Is there a better method to find the maximal ideals?

Answer 1

Note that since $2$ is invertible in $\mathbb Z_{63}$ then the ideal generated by $2$ is everuthing.

Hint Show that an ideal $I$ is non-trivial if and only if $$ I = \langle d \rangle \mbox{ with } d|63 $$ And you are somehow right that the key is prime numbers, but not ALL primes ;)

Answer 2

Hint: $\mathbb{Z}_{63}$ is the quotient of $\mathbb{Z}$ by the ideal $(63)$. By the Lattice Isomorphism Theorems, there is a one-to-one, inclusion preserving correspondence between the ideals of $\mathbb{Z}_{63}$ and the ideals of $\mathbb{Z}$ that contain $(63)$.

In $\mathbb{Z}$, we have that $(a)\subseteq (b)$ if and only if $b|a$ (if you don’t know that yet, then prove it!)

Answer 3

For any positive integer $n$, the maximal ideals of $\mathbb{Z}_n$ correspond to the primes dividing $n$. In particular, the ring $\mathbb{Z}_{63}$ has exactly two maximal ideals, namely $3\mathbb{Z}_{63}$ and $7\mathbb{Z}_{63}$.