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Let $R$ be a ring with nilradical $N(R)$. How can one show that $R$ has just one prime ideal if and only if $R/N(R)$ is a field ?

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Can you prove one of the directions? – Tobias Kildetoft Sep 11 '13 at 19:16
This question is extremely on-topic and interesting, the topic being algebraic geometry. I consider the statement that it is homework insulting, since no solid proof for that is given. I vote to reopen. – Georges Elencwajg Sep 11 '13 at 20:43
@GeorgesElencwajg It has clear precedent here to use the term homework for all questions that have been given to the one asking them for them to solve. Whether it has been given by a book, by a teacher or something/someone else is not important for the idea that the best answers for questions of this type are the ones that provide the minimal amount of hint in order for the OP to solve it themselves (since presumably the OP is trying to learn). I agree that the phrasing of the closure reason is poor given that this is not something one can expect the OP to know. – Tobias Kildetoft Sep 12 '13 at 6:36
Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. Questions that do not provide this information are often closed even if they possess mathematical content. – Carl Mummert Sep 12 '13 at 12:17
Dear @Cantlog, this question on meta is probably the reason.The question was closed then reopened, then reclosed,... I don't know how many cycles there have been (and maybe will be!) – Georges Elencwajg Sep 12 '13 at 22:38
up vote 13 down vote accepted

Since you have tagged it "commutative algebra" I assume that your ring is commutative and has an identity element. First suppose $R$ has only one prime ideal. Then $R$ has only one maximal ideal and $N(R) = J(R)$ which mean that $R/N(R)$ is a field.

Conversely, suppose $R/N(R)$ is a field. This means that $N(R)$ is a maximal. On the other hand $N(R)$ is equal to intersection of prime ideal. SO it would be maximal only if it is the intersection of only one prime ideal which means that $R$ has only one prime ideal.

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@YACP I have edited the answer. How about know? – kian Sep 11 '13 at 19:38

Let us call a commutative ring $R$ a quasi-field if it satisfies the equivalent properties:

$\bullet$ $R$ has only one prime ideal.
$\bullet$ The nilradical $\operatorname {Nil}(R)$ is maximal.
$\bullet$ $R$ is local of dimension zero.

Noetherian quasi-fields are a dime a dozen and the ring $k[T_1,\cdots,T_n,\cdots]/(T_1^2,\cdots,T_n^2,\cdots)$ (where $k$ denotes a field) is a non-noetherian quasi-field.

An interesting appearance of quasi-fields is in the following result:

A point $\xi$ of a scheme $X$ is the generic point of an irreducible component of $X$ iff its local ring $\mathcal O_{X,\xi}$ is a quasi-field.

NB: This is not an answer to the question (which has already been answered satisfactorily) but a remark on the interest of this notion and propaganda for the terminology.

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Fantastic answer. Of the 15 users who went out of their way to prevent answers on this question, 10 did so after this answer was posted, in addition to the 5 who would have precluded it altogether. On MSE, it is more important to uphold a (far from unanimous) stylistic preference about how questions are posted, than to allow questions that lead to answers of significant long-term "interest to the community". – zyx Sep 12 '13 at 22:17
Thank you, @zyx. – Georges Elencwajg Sep 12 '13 at 22:26
Somebody downvoted this, but not any of the other answers. No good deed goes unpunished... (anyway, thanks again for the illuminating answer). – zyx Sep 12 '13 at 22:51
Dear @zyx, I really like your misanthropic but profound aphorism "no good deed goes unpunished". It reminds me of an exclamation I once read: "Why does he hate me? I didn't even help him..." – Georges Elencwajg Sep 12 '13 at 23:16

I assume your ring is commutative.

$R/N(R)$ is a field $\Leftrightarrow\,N(R)$ is a maximal ideal. But $$N(R)=\bigcap_{P\subset R}P,$$ the intersection being over all prime ideals in $R$. Such an intersection is a maximal ideal if and only if we are intersecting a single ideal.

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