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Let $R$ be a commutative noetherian ring with no nonzero nilpotents. Let $p$ be a minimal prime of $R$. Could you help me to prove that $R_p$ is a field?

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Try proving the contrapositive. (If $R_p$ is not a field, it contains an element which is not invertible, so...?) –  Qiaochu Yuan May 28 '12 at 20:03
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3 Answers

up vote 4 down vote accepted

Hint: A ring is a field if and only if its only prime ideal is $(0)$. The prime ideals of $R_p$ are in one-to-one correspondence with the prime ideals of $R$ which are disjoint from $R\setminus p$, that is the primes contained in $p$. What can you conclude?

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If $qR_p$ is a prime ideal of $R_p$ then $q\subset p$ and so $p=q$, so the only prime ideal of $R_p$ is $pR_p$, am I right? How can I go on? I think I have to use the fact that the nilradical is the intersection of all the minimal primes and that there are only finitely many minimal primes because $R$ is noetherian, but I don't know how to put all this together, any other hints? –  Alex M May 28 '12 at 20:23
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@AlexM The nilradical of $R_p$ is the intersection of all minimal primes of $R_p$. But the only minimal prime of $R_p$ is $pR_p$, therefore $pR_p$ is the nilradical. But what is the nilradical of $R_p$? –  Alex Becker May 28 '12 at 20:30
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got it thank you, $\mathrm{nilrad}\;R_p=(\mathrm{nilrad}\;R)_p=0R_p$ and so $pR_p=0R_p$ –  Alex M May 28 '12 at 20:38
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It seems to me that the only thing that requires $R$ notherian is that the set on minimal primes is finite, but we didn't use it. The correspondence between prime ideals of $R_p$ and prime ideals of $R$ contained in $p$, the fact that the nilradical is the intersaction of the minimal primes and the fact that $\mathrm{nilrad}\;R_p=(\mathrm{nilrad}\;R)_p$ don't seem to require $R$ noetherian, what am I missing? –  Alex M May 28 '12 at 20:48
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A reduced local ring of dimension zero is a field. That the local ring is a localization is a red herring.As both Alexes and @QiL already said, noetherianness is irrelevant too. (+1 for the answer). –  Georges Elencwajg May 28 '12 at 21:05
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Here is a proof that uses nowhere the results of Noetherian rings. All the results I used below are from chapters 2 - 3 of Atiyah - Macdonald where Noetherian-ness has not come in yet! To prove your problem we use the following result from Atiyah - Macdonald:

A local ring that is absolutely flat is a field.

We already know that given any prime ideal $\mathfrak{p}$ that $R_\mathfrak{p}$ is a local ring and hence it suffices to prove that $R_\mathfrak{p}$ for a minimal prime ideal $\mathfrak{p}$ is absolutely flat. You have already observed above that $\textrm{nilrad} R_\mathfrak{p} = 0$, so if you can prove that every prime ideal in $R_\mathfrak{p}$ is maximal then this would imply (see at the bottom for a proof) that $$R_p/\textrm{nilrad}R_\mathfrak{p} \cong R_\mathfrak{p}$$ is absolutely flat, finishing the problem. Take a prime ideal $S^{-1}\mathfrak{q}$ of $R_\mathfrak{p}$. Then you have already observed that $\mathfrak{q}$ is a prime ideal of $R$ that must be disjoint from $R - \mathfrak{p}$, viz. it is completely contained in $\mathfrak{p}$. By minimality of $\mathfrak{p}$ we are forced to conclude that $\mathfrak{p} = \mathfrak{q}$ and hence $S^{-1}\mathfrak{q} = S^{-1}\mathfrak{p}$ is a maximal ideal. This concludes the proof.

$\hspace{6in} \square$


We prove this exercise first, and then use it to prove the main result I claimed above.

Exercise 3.10(ii) Atiyah - Macdonald: $A_\mathfrak{m}$ being a field for each maximal ideal $\mathfrak{m} \implies $ $A$ is absolutely flat.

Proof: Suppose you given an $A$ - module $M$. Then I claim given any maximal ideal $\mathfrak{m}$ one has that $M_\mathfrak{m}$ is a flat $A_\mathfrak{m}$ - module. Indeed this is clear because $M_\mathfrak{m}$ is just a vector space over $A_\mathfrak{m}$ and hence is isomorphic to $A_\mathfrak{m}^n$ for some $n$. Since $A_\frak{m}$ being a field is absolutely flat, in particular $A_\frak{m}$ as a module over itself is flat. Therefore by exercise 2.4 of Atiyah Macdonald

$$\bigoplus_{i=1}^n (A_\mathfrak{m})_i = A_\mathfrak{m}^n \cong M_\mathfrak{m}$$

is flat as an $A_\mathfrak{m}$ - module for any maximal ideal $\mathfrak{m}$. By proposition 3.10 of Atiyah - Macdonald we conclude that $M$ is flat as an $A$ - module and since $M$ was arbitrary this proves that $A$ is absolutely flat.

$\hspace{6in} \square$

Exercise 3.11 Atiyah - Macdonald: If every prime ideal of a ring $A$ is maximal then $A/\textrm{nilrad} (A)$ is absolutely flat.

By exercise 3.10(ii) Atiyah - Macdonald it suffices to prove that given any maximal $\mathfrak{m}$ of $A/\textrm{nilrad} A$ that $(A/\textrm{nilrad} A)_\mathfrak{m}$ is a field. Now a basic isomorphism in localisation tells us that

$$(A/\textrm{nilrad} A)_\mathfrak{m} \cong A_\mathfrak{m}/ \big(\textrm{nilrad}(A)\big)_\mathfrak{m}$$

so it suffices to prove that $\textrm{nilrad}(A)_\mathfrak{m}$ is a maximal ideal of $A_\mathfrak{m}$. Given what you have done above you should be able to complete this by the following steps:

  1. $\big(\textrm{nilrad}(A)\big)_\mathfrak{m} = \textrm{nilrad}({A_\mathfrak{m}})$
  2. There is a unique maximal ideal $Q$ of $A_\mathfrak{m}$ that contains $\textrm{nilrad}(A_\mathfrak{m})$.
  3. Recall the ideal correspondence between prime ideals of $A_\mathfrak{m}$ and prime ideals of $A$ that don't meet $A - \mathfrak{m}$.
  4. Use the fact that every prime ideal of $A$ is maximal and $A_\mathfrak{m}$ is a local ring to conclude your problem.
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@navigetor23 Hahahaha perhaps I used a nuclear weapon instead of a flyswatter......that's the way I thought about the problem though :D –  fpqc Jun 3 '12 at 22:14
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Maybe a little bit more clear than the first answer:

$R_p$ itself is reduced and local. Moreover $\dim R_p=0$, so we can ask the following: is it true that a local reduced ring with dimension 0 is a field? Answer: yes! Reduced means that the nilradical of $R$ is $(0)$, but in this case the nilradical is the only prime ideal of $R$, so the only prime ideal of $R$ is (0) which means that $R$ is a field.

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