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If we define:

1) $R$ is a ring, $I$ is an ideal. We say that $I$ is prime if $I$ is not equal to $R$ and whenever $ab$ belongs to $I$ then either $a$ is in $I$ or $b$ is in $I$.

2) Let $R$ be a commutative ring and let a be an element of $R$. The set $\langle a\rangle=\{ar: r \in R\}$ is an ideal and any ideal of this form is called principal.

3) Let $R$ be an integral domain and let a be a non-zero element of $R$. We say that $a$ is prime, if $\langle a\rangle$ is a prime ideal.

But suppose $R$ is a finite integral domain. For any non zero $p$, $pa = pb$ iff $a = b$. Can we conclude that $\langle p\rangle = R$ (thus $p$ can't be prime)?

George Law
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user417839
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  • To answer your last paragraph: yes, and I'm pretty sure that the linked question and all of its duplicates all mention that "$r\mapsto pr$ is a bijection on $R$ so that $pr=1$ for some $r$", and after that $pR=R$ of course. – rschwieb Feb 19 '17 at 08:33

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You should prove that finite integral domains are fields. In this case, the only ideals are $(0)$ and the whole field, in which case all ideals are trivially prime and principal.

If the integral domain is not finite, there are principal ideals which do not generate the whole ring. An example is $k[x]$ for any field $k$, and the ideal generated by $(x)$. On the other hand, a principle ideal is not necessarily prime either. An example is again taking $k[x]$, but now with ideal $(x^2)$. Here we have $x^2 = x \cdot x$. $x \not \in (x^2)$, so this shows the ideal is not prime.

A. Thomas Yerger
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