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Let $A$ be a commutative ring with unity that is not an integral domain and $\mathcal{P}$ be any prime ideal of $A$. Then I know that $A_{\mathcal{P}}$ is not an integral domain using the correspondence between prime ideals of $A$ that does not meet $A \backslash \mathcal{P}$ and prime ideals of $A_{\mathcal{P}}$.

Just from curiosity, I wanted to prove it by a different approach, namely starting from $f, g \in A$ both non-zero and $fg=0$, I wanted to construct two non zero elements of $A_{\mathcal{P}}$ that multiply to $0$. Could someone possibly give me a hand on how one could do this? Thanks!

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    $\begingroup$ Are you trying to show that $A_P$ is never a domain? I don't think that's true. Look at something like $k \times k$. $\endgroup$ – Hoot Jan 12 '15 at 3:44
  • $\begingroup$ @Hoot I thought it was always a domain... Could you possibly tell me where the mistake is in the following argument? Suppose $A_{p}$ is a domain, then $(0)A_p$ is prime. Thus, $(0)A_p \cap A = (0)$ must be prime in $A$ by the correspondence. That means $A$ is a domain. Contradiction. $A_p$ is not a domain... $\endgroup$ – user192077 Jan 12 '15 at 3:49
  • $\begingroup$ I wonder what I am missing... $\endgroup$ – user192077 Jan 12 '15 at 3:51
  • $\begingroup$ I think you have to look closely at the correct definition of localization. For rings with zero-divisors, there may be some pitfalls. $\endgroup$ – Lubin Jan 12 '15 at 4:10
  • $\begingroup$ The argument you wrote in comment, there the problem lies. $(0)A_p \cap A \neq (0)$ in general. because the natural map $A \to A_p$ is not injective in general, if the ring is not a domain. $\endgroup$ – Krish Jan 12 '15 at 4:19
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Your claim is not true. Take $A := \prod_{i \geq 1} A_i,$ where each $A_i = \mathbb Z/2\mathbb Z.$ Then localization of $A$ at every prime ideal is $\mathbb Z/2\mathbb Z,$ which is a field. (Being integral domain is not a local property.)

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