4

Hilbert's Nullstellensatz asserts, that if $k$ is an algebraically closed field and $I \subset k[x_1,\dots,x_n]$ is an ideal, then $I(V(I)) = \sqrt{I}$. So every polynomial that vanishes on the variety defined by $I$ is actually in the radical of $I$.

Obviously we always have the inclusion $\sqrt{I} \subset I(V(I))$.

Of course this is not true if $k$ is not algebraically closed, the standard example is $f = x^2 +1 \in \mathbb{R}$. Then $V(f) = \emptyset$, i.e. $I(V(f)) = k[x]$.

My question is: Are there any other examples where the Nullstellensatz does not hold, but $I$ actually has zeroes? Or is this the only class of counterexamples?

Jose Brox
  • 5,216
red_trumpet
  • 11,169
  • 7
    What about $f = x(x^2+1)$ so $V(f) = { 0 }$ and $I(V(f)) = \langle x \rangle$ while $\sqrt{\langle f \rangle} = \langle f \rangle$? – Daniel Schepler Aug 23 '18 at 17:40

1 Answers1

1

A different kind of example comes from finite fields. If $k=F_q$ then $a^q=a$ for any $a\in k$; hence the polynomials $x_i^q-x_i\in k[x_1,\ldots, x_n]$ vanish on all of $k^n$. So putting $I_q = \langle x_1^q-x_1,\ldots,x_n^q-x_n \rangle$, we have $$\sqrt 0 = 0 \varsubsetneq I_q \subseteq I(k^n) = I(V(0)).$$

Remark: We can actually show $I(k^n) = I_q$: Every $f\in k[x_1,\ldots,x_n]$ can be uniquely written as $f=g+r$, where $g\in I_q$ and $r$ has $\deg_{x_i} r\leq q-1$ for all $1\leq i\leq n$. By induction on the number of variables and counting the number of roots we can show that $r\in I(k^n)$ implies $r=0$; so if $f\in I(k^n)$ then $f=g\in I_q$ (details can be found for example in A note on Nullstellensatz over finite fields, sections 2.2 and 2.3).

Jose Brox
  • 5,216