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i) Why it's a unit can prove this proposition ii)see picture

  • 1
    $\begingroup$ (1) $S^{-1}{\frak q}$ is an ideal of $S^{-1}A$. If an ideal contains a unit, it is the whole ring. (Have you seen this fact? If not, try to prove it; it is straightforward.) $\endgroup$
    – anon
    Mar 26, 2014 at 22:30
  • $\begingroup$ Dumb question: what's the r() notation mean? I have a guess but I may as well ask. $\endgroup$
    – rschwieb
    Mar 26, 2014 at 22:42
  • $\begingroup$ @rschwieb r() mean radical $\endgroup$
    – amateur
    Mar 26, 2014 at 22:45

1 Answer 1


Some hints:

1) First note that in a Noetherian ring every ideal contains a power of its radical. Hence there exists some $n>0$ such that $p^n \subset q$. Now if $x \in S \cap p$ then $x^n \in S \cap q$ because $S$ is multiplicatively closed. This implies that $S^{-1}q = S^{-1} A$.

2) If $A$ is an integral domain, i.e. it has no zero divisors, then the canonical map $A \rightarrow S^{-1}A$ is injective and so we may say that $q \subset S^{-1} A$. But the correct relation in general is simply $S^{-1} q \subset S^{-1} A$.

3) In general, $S^{-1}q$ is not maximal, it is just primary. To prove this claim, we just check that the definition of primary is true. So let $(x_1/s_1)(x_2/s_2) \in S^{-1} q$ and suppose that $\frac{x_1}{s_1} \not\in S^{-1}q$. This latter condition implies that $s x_1 \not\in q$ for any $s \in S$. Then $x_1 x_2 / s_1 s_2 \in S^{-1} q$ and so $\frac{x_1 x_2}{s_1 s_2} = \frac{y}{t}$ where $y \in q, t \in S$. Then there exists some $s_3 \in S$ such that $s_3 t x_1 x_2 = s_1 s_2 s_3 y$. The right hand side is inside $q$ and so $s_3 t x_1 x_2 \in q$. Now $(s_3 t) x_1$ can not be inside $q$, because this would contradict the implication of our hypothesis that $s x_1 \not\in q, \forall s \in S$. Thus $x_2^n \in q$ for some $n$ and so $(x_2/s_2)^n \in S^{-1}q$.

  • $\begingroup$ I have changed the notation and checked the definition of primary.q/s=q1/s1*q2/s2,I still don't figure how to prove "q1/s1∈q/s or (q2/s2)^n ∈q/s for some n" $\endgroup$
    – amateur
    Mar 26, 2014 at 23:16
  • $\begingroup$ I figure out the secound question,the reason is A-S is an ideal.I still don't figure out 3rd question,more detail please.Thanks $\endgroup$
    – amateur
    Mar 26, 2014 at 23:28
  • $\begingroup$ Can you please number your questions 1) 2) 3) e.t.c. so that we can communicate more easily? $\endgroup$
    – Manos
    Mar 27, 2014 at 0:08
  • $\begingroup$ Sure,I have numbered them now.You said that it's not maximal but just primary.I still don't know how to prove it's primary. $\endgroup$
    – amateur
    Mar 27, 2014 at 21:21
  • $\begingroup$ @program666: I have added a detailed proof. $\endgroup$
    – Manos
    Mar 28, 2014 at 17:46

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