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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?

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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!)

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  • $\begingroup$ That helps. I now know that there are 6 ideals of $\mathbb{Z}_{63}$. However I only know what these are in the quotient form e.g. $\langle 7 \rangle / \langle 63 \rangle$ $\endgroup$ Commented May 26, 2020 at 23:25
  • $\begingroup$ @Group23: They are the same thing. $\langle 7\rangle/\langle 63\rangle$ is the ideal generated by $7$ in $\mathbb{Z}_{63}$. $\endgroup$ Commented May 26, 2020 at 23:43
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    $\begingroup$ “to contain is to divide” $\endgroup$ Commented May 26, 2020 at 23:45
  • $\begingroup$ What I meant is I don’t know to write these in non-quotient form. $\endgroup$ Commented May 26, 2020 at 23:50
  • $\begingroup$ @Group23: And as I indicated, they are just the ideal generated by the corresponding element. $\langle 7\rangle/\langle 63\rangle$ in $\mathbb{Z}/\langle 63\rangle$ is the same as $\langle 7\rangle$ in $\mathbb{Z}_{63}$. $\endgroup$ Commented May 27, 2020 at 0:30
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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 ;)

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  • $\begingroup$ Is $\mathbb{Z}_m$ definitely always a PID? $\endgroup$ Commented May 27, 2020 at 0:00
  • $\begingroup$ @Group23 Each ideal is always principal BUT the D in PID stands for Domain. So the answer is no. $\endgroup$ Commented May 27, 2020 at 1:47
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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 $7\mathbb{Z}_{63}$.

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  • $\begingroup$ As otherwise n is comprise to 63 and this n is a unit and so the ideal generated by n is the entire ring? $\endgroup$ Commented May 26, 2020 at 23:31

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