Let $\mathbb{K}$ be a field. Let $S\subseteq \mathbb{K}[x_1,\dots,x_n]$ be a set of polynomials. The variety defined by $S$ is the set, $V(S)=\{a\in \mathbb{K}^n:f(a)=0\:\forall f\in S\}$

For an algebraically closed field Nullstellensatz relates the radical of an ideal to its variety. If the field is not algebraically closed, can there be two radical ideals with the same zero set? I was trying to construct examples but I couldn't do it.

  • $\begingroup$ This depends on your definition of variety, which you should add to the question. $\endgroup$ – KReiser Oct 21 '19 at 7:17
  • $\begingroup$ @KReiser I have done that now. $\endgroup$ – cookiemonster Oct 21 '19 at 7:19
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    $\begingroup$ Hint: think about working over $\Bbb R$ and a sum of squares. $\endgroup$ – KReiser Oct 21 '19 at 7:29

You can look at $V(x^2+1)$ in $\mathbb{R}$, ie the empty set. $x^2+1$ is a radical ideal, since $\mathbb{R} [x]/ (x^2+1)$ is isomorphic to $\mathbb{C}$, a reduced ring. You can also look at $V(\mathbb{R}[x])$.


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