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Let $K$ be an algebraically closed field, $S \subset R = K[X_1,X_2,...,X_n]$ and $J = (S)$ the ideal generated by $S$. Let $V$ and $I$ denote the usual correspondences between $R$ and $K^n$

The Nullstellensatz states (among other things) that for any $J \subset R$:

$$ I(V(J)) = \sqrt{J}$$

Since $V(S) = V(J),$ we have $\sqrt{J} = \sqrt{S}.$

This proof can't be correct. Take $S = \{1\}$ then $J = R$ but

$$ \sqrt R = R $$ and

$$ \sqrt{\{ 1\}} = \{g \vert \exists n : g^n = 1 \} \neq R.$$

Where did I go wrong ?

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  • $\begingroup$ The problem is the definition of the radical of $1$. You are mistakenly taking the radical of the singleton $\{1\}$, when you need to be taking the radical of the ideal generated by 1. So it should read $\{g \vert \exists n : g^n \in (1) = R \}$. Notice I took out your equal sign and replaced it with \in symbol. And of course this is the whole ring by taking $n = 1$ for all elements $g$. $\endgroup$ May 9, 2020 at 18:54
  • $\begingroup$ @NicholasRoberts Do you mean that $$ \sqrt{J} := \{ g \vert \exists n : g^n \in (J) \}$$ and not $$\sqrt{J} = \{ g \vert \exists n : g^n \in J \}$$ ? $\endgroup$
    – Digitallis
    May 9, 2020 at 19:03
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    $\begingroup$ It's the second one you wrote. You can only take radicals of ideals. The singleton set $\{1\}$ is not an ideal. $\endgroup$ May 9, 2020 at 19:05
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    $\begingroup$ @NicholasRoberts this looks like an answer to the question to me. Would you care to record this below? $\endgroup$
    – KReiser
    May 9, 2020 at 21:30
  • $\begingroup$ I have posted as an answer @KReiser Thank you. $\endgroup$ May 9, 2020 at 22:06

1 Answer 1

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As per request, I will write my comment as an answer. The problem here is that you have (mistakenly) taken the radical of the singleton set $\{1\}$. The singleton set $\{1\}$ is not an ideal. We would rather take radicals of ideals because then it can be shown quite easily that the radical of an ideal is also an ideal. Thus, we should take the radical of the ideal generated by $\textbf{1}$ which is denoted as $\sqrt{(1)}$.

So the correct formulation of your last line would be: $$\sqrt{(1)} = \{g \vert \exists n : g^n \in (1) = R \} = R.$$

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