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Let p be a prime. Prove that every commutative ring with the identity and p elements is a domain.

Any help or hints on how to get started would be great thanks

Thanks for any help

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The commutativity assumption is not needed. –  Pierre-Yves Gaillard Feb 16 '12 at 18:12

4 Answers 4

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By Lagrange, the order of the subgroup$\rm\: G = \{r:rs=0\}\:$ divides $\rm\:|(A,+)| = p\:$ prime.

Thus $\rm\:|G| = p\ (\Rightarrow 1\cdot s = 0)\:$ or $\rm\:|G| = 1\ (\Rightarrow\ rs\ne 0\ \ if\ \ r,s\ne0)$.

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HINT If the ring has $p$ elements, then its additive structure must be cyclic of order $p$; hence, since $1_R\neq 0$, $1$ will generate the ring additively. So every element of the ring is of the form $n\cdot 1_R$, with $0\leq n\lt p$. Show this determines the multiplication and that your ring must in fact be $\mathbb{F}_p$, the field with $p$ elements.

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For the multiplication, are you trying to show that every element has a multiplicative inverse? –  Idonknow Mar 24 '13 at 18:04

The surjective ringhomomorphism $\mathbb Z \rightarrow R$, $1\mapsto 1_R$ must have non-trivial Kernel $n\mathbb Z $, such that $\mathbb Z/n\mathbb Z = R$ has $p$ elements. Therefore $n=p$ and $R=\mathbb Z/p\mathbb Z=\mathbb F_p$

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The OP might have trouble showing that this homomorphism is surjective. –  Dylan Moreland Feb 16 '12 at 17:57

Consider $a\in R$ and the map $\mu : x \mapsto ax$. This is an endomorphism of the additive group of $R$. If $a\ne0$, the image of $\mu$ is not $0$ because $\mu(1)=a$. Since $R$ has $p$ elements, the image must be the whole of $R$. In particular, there is $b$ such that $ab=1$ and $R$ is a field. Alternatively, the kernel must be $0$ and $R$ is a domain.

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