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I'm trying to find these ideals without success.

I have a hint: $\mathbb C[\mathbb Z_3] \simeq \frac{\mathbb C[X]}{(X^3 - 1)}$, considering $\mathbb C[X]$ is a PID.

Edit: How can I prove that ideals of $\frac{\mathbb C[X]}{(X^3 - 1)}$ are the ideals of $\mathbb C[X]$ containing $(X^3 - 1)$?

user26857
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Christian Coronel
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    what does $\Bbb C[\Bbb Z_3]$ stand for? – Sine of the Time Jul 31 '23 at 00:27
  • $\mathbb C[\mathbb Z_3] = \displaystyle \left{{\sum_{n\in{\mathbb Z_3}}{a\cdot n}: a\in\mathbb C}\right}$ – Christian Coronel Jul 31 '23 at 00:32
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    Use the correspondence theorem – Sine of the Time Jul 31 '23 at 02:06
  • Ok. Using this theorem, I obtained that the prime ideals of $\mathbb C[\mathbb Z_3]$ are $J = (1 \cdot \overline{1} - w \cdot \overline{0})$ and $w$ is a cube root of unity. Thanks – Christian Coronel Jul 31 '23 at 04:37
  • @rschwieb why not? I found the prime ideals of $\mathbb C[X]$ that contains $(x^3 - 1)$ and there's only three: $(x - w)$ with $w$ a cube root of unity. That means I have $I_1, I_2$ and $I_3$, $w_1, w_2$ and $w_3$ different cube roots of unity, then $I_1 = (x - w_1)$, $I_2 = (x - w_2)$ and $I_3 = (x - w_3)$. Then "I took" these ideals in $\frac{\mathbb C[X]}{(X^3 - 1)}$ using the correspondence theorem and described them as ideals of $\mathbb C[\mathbb Z_3]$ because there are isomorphic rings. – Christian Coronel Jul 31 '23 at 20:56
  • @ChristianCoronel Excuse me: I appear to have completely misread the field in question, possibly because I was jumping around between posts looking for a duplicate. I see what you mean now. – rschwieb Aug 01 '23 at 03:10

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In your edit you are just quoting the correspondence theorem which is available everywhere.

In a polynomial ring like $\mathbb C[x]$ that amounts to factoring $x^3-1$ over $\mathbb C$. The maximal ideals will correspond to irreducible factors. You should find exactly three.

Because this ring is Artinian (it's finite dimensional over a field after all) its prime ideals will be exactly its maximal ideals.

rschwieb
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