There are only two non isomorphic rings with $p$ elements

Question

Prove that for any prime $p$ there are only two non isomorphic rings with $p$ elements.

I have found out there are up to two rings of order p , they are $\mathbb Z_p$ and $\mathbb C_p$. Please help in doing the proof!!

see http://math.stackexchange.com/questions/1043373/any-ring-of-prime-order-commutative — Krish, Jan 04 '15 at 17:27
it was proved that there are exactly two distinct ring structure over the group $\mathbb{Z}/p\mathbb{Z}.$ — Krish, Jan 04 '15 at 17:31
@RossMillikan But how to show that $R\cong (\Bbb Z_p,+,×)$ if $R$ does not have a unit element, i.e $1\neq 0,$ where $×$ is such an operation so that $a.b=0,\forall a,b\in \Bbb Z_p$ ? — Thomas Finley, Dec 19 '23 at 13:11

score 4 · Answer 1 · answered Jan 04 '15 at 17:32

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By $\mathbb{C}_p$ you mean the zero ring, right?

In fact, if $R$ is a non-zero ring of order $p$, let $a \in R$ such that $a R \ne \{ 0 \}$.

The function $R \to R$ given by $x \mapsto a x$ is a group homomorphism. Looking at the kernel, one sees that it has to be injective, thus bijective. In particular there is an element $e$ such that $a e = a$. Since $a \ne 0$, we have $a a \ne a 0 = 0$, so also $R a \ne \{ 0 \}$, and the map $R \to R$ given by $x \mapsto x a$ is bijective. So there is $f$ such that $f a = a$. Now it is easy to see that $f$ is a left unit, and $e$ is a right unit. So $e = f e = e$ is the unit, and the rest is not difficult.

Please try and fill up the details.

PS The answer in the posted link is definitely sleeker.

answered Jan 04 '15 at 17:32

Andreas Caranti

68,873

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how can zero ring have p elements?? – Ankit Ghosh Jan 04 '15 at 17:34
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@AnkitGhosh: "zero ring" means $ab = 0, \forall a, b.$ it's different from the trivial ring ${0 }$ – Krish Jan 04 '15 at 17:37
ok I have understood it.... thank you – Ankit Ghosh Jan 04 '15 at 17:39

There are only two non isomorphic rings with $p$ elements

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