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Here is the relevant passage: enter image description here

Now, I am unsure what they mean by "passing to $R/I, \ldots$". They don´t seem to use that fact later? I mean, $R/I = R$ given that $I = 0$. So this seems entirely vacuous to me. Am I missing something?

Here is proposition 11: enter image description here

Ben123
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1 Answers1

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Statement 1: The nilradical of a commutative ring $R$ is the intersection of all the prime ideals of $R$.

Statement 2: The radical of an ideal $I$ of a commutative ring $R$ is the intersection of all prime ideals of $R$ that contain $I$.

Statement 1 is a special case of Statement 2, where $I = 0$.

  • The second paragraph in the proof of Proposition 12 is a proof of Statement 1.
  • The first paragraph in the proof of Proposition 12 explains that Statement 2 is implied by Statement 1 and Proposition 11.

Indeed, suppose that $I$ is an ideal in $R$. Then $(\text{rad }I) / I$ is the nilradical of $R / I$, by Proposition 11. Applying Statement 2 to the ring $R / I$, we see that $(\text{rad }I) / I$ is the intersection of all prime ideals of $R / I$.

But there is a bijective correspondence between prime ideals of $R / I$ and prime ideals of $R$ that contain $I$. The set of prime ideals of $R / I$ is precisely $$ \{ \mathfrak{p} / I : \mathfrak{p} \text{ prime ideal of } R \text{ such that } \mathfrak{p} \supset I \}.$$

Thus $$ (\text{rad } I) / I = \bigcap_{\mathfrak{p} \text{ prime in } R,\\\mathfrak{p} \supset I} (\mathfrak{p} / I) = \left( \bigcap_{\mathfrak{p} \text{ prime in } R,\\\mathfrak{p} \supset I} \mathfrak{p} \right) / I. $$

And hence, $$ \text{rad } I = \bigcap_{\mathfrak{p} \text{ prime in } R,\\\mathfrak{p} \supset I} \mathfrak{p} . $$

Kenny Wong
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