finding all rings of order $p^3$

Asked Oct 05 '14 at 13:54

Active Oct 05 '14 at 13:54

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Classify all commutative rings with $1$ of order $p^3$.

I only observed $1$ can have order $p,p^2,p^3$. If its $p^3$ then $\mathbb{Z}/(p^3)$. Otherwise will it be just be $\mathbb{Z}/(p^2)\times \mathbb{Z}/(p)$ and $\mathbb{Z}/(p) \times \mathbb{Z}/(p) \times \mathbb{Z}/(p)$ ?

asked Oct 05 '14 at 13:54

dragoboy

1,891

Certainly there are more examples, like $(\mathbb{Z}/p) [X]/(X^3)$. – Andrew Dudzik Oct 05 '14 at 13:59
Maybe the most interesting one is $(\mathbb{Z}/p^2)[pX]$. – Andrew Dudzik Oct 05 '14 at 14:01
Yup, but how to classify all ? – dragoboy Oct 05 '14 at 14:06
Such a ring must contain a prime subring, that is, Z/(p). Then you can investigate the homo. from Z/(p)[x] to the original ring. – Sky Oct 05 '14 at 14:16
This question really needs context. But I would say: most examples need only one generator over $\mathbb{Z}$, so are quotients of $\mathbb{Z}[X]$, and the remaining examples are vector spaces over $\mathbb{Z}/p$, so you can work with a basis. – Andrew Dudzik Oct 05 '14 at 14:19
@Sky Not quite, the subring generated by $1$ could be $\mathbb{Z}/p^2$ or $\mathbb{Z}/p^3$. – Andrew Dudzik Oct 05 '14 at 14:20
Sorry I have ignored. But I think this problem can be worked out this way. – Sky Oct 05 '14 at 14:25
Context: I saw for $p^2$ thats why was thinking how about the case $p^3$ ... and isn't there any concrete form ? – dragoboy Oct 05 '14 at 16:13

finding all rings of order $p^3$

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