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I'm having a lot of difficulty understanding just what the Nullstellensatz is saying, and how it can be applied to a specific example that I cooked up.

Let $k$ be a field. Suppose that a polynomial $f(x_1, x_2, \ldots, x_n)$ vanishes on the subvariety $V(x_1 - x_2)$ of $\mathbb{A}^n$, by which I mean $f(a_1, a_2, \ldots, a_n) = 0$ for all $a_1, a_2, \ldots, a_n$ in $k$ with $a_1 = a_2$. By the Nullstellensatz, $f \in \sqrt{(x_1 - x_2)} = (x_1 - x_2)$ because the ideal is prime. That means $f$ is divisible by $x_1 - x_2$. So far, so good. (Right?)

Now let $k = \mathbb{F}_p$ and choose $f(x, y) = x^p - y$. By Fermat's Little Theorem, $f(a, a) = 0$ for all $a$ in $\mathbb{F}_p$, which means $f$ vanishes on the diagonal of $\mathbb{A}^2$. The above argument should imply that $x^p - y$ is divisible by $x - y$, but it isn't. What's going on?

Unit
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    You need an algebraically closed field for Hilbert‘s Nullstellensatz. – Qi Zhu Sep 22 '20 at 17:06
  • Are you sure? https://en.wikipedia.org/wiki/Hilbert%27s_Nullstellensatz#Formulation isn't clear on that requirement. – Unit Sep 22 '20 at 17:08
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    You found your own counterexample (a finite field is not algebraically closed). Moreover, in the link you provide, it is explicitly used that $K$ is algebraically closed: the correspondence is established over $K$ (not over $k$). I cite:Let k be a field (such as the rational numbers) and K be an algebraically closed field extension " " "In this way, we obtain an order-reversing bijective correspondence between the algebraic sets in $ K^n$ and the radical ideals of ${\displaystyle K[X_{1},\ldots ,X_{n}].}$" – GreginGre Sep 22 '20 at 17:12
  • I see that. But what about this line? "Hilbert's Nullstellensatz states that if $p$ is some polynomial in $k[X_1, …, X_n]$ that vanishes on the algebraic set $V(I)$, i.e. $p(x) = 0$ for all $x$ in $V(I)$, then there exists a natural number $r$ such that $p^r$ is in $I$." In my case, that means $f^r \in (x_1 - x_2)$, so $(x_1 - x_2)$ divides $f^r$, hence divides $f$. Is there a mistake somewhere? – Unit Sep 22 '20 at 17:18
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    Oh, $V(I) \subseteq K^n$. – Unit Sep 22 '20 at 17:27
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    One way to view the weak Nullstellensatz would be as a natural generalization of the Fundamental Theorem of Algebra - for which you of course need an algebraically closed field. – Qi Zhu Sep 22 '20 at 17:47

1 Answers1

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As was immediately pointed out in the comments, the Nullstellensatz only applies when $k$ is algebraically closed. Nevertheless, I'm still interested in the extent to which it's true that $x_i - x_j$ divides $f$ whenever $f(x_1, \ldots, x_n)$ vanishes on $V(x_i - x_j)$.

It turns out all we need is that $k$ be an infinite domain.

Vanishing Lemma. Let $k$ be an integral domain with $1$, and suppose $k$ is infinite. Let $f \in k[x_1, \ldots, x_n]$. If $f(a_1, \ldots, a_n) = 0$ for all $(a_1, \ldots, a_n)$ in $k^n$, then $f = 0$.

Proof. By induction on $n$ as described here, noting that the inequality $\# V(f) \le \deg f$ holds for any nonzero $f$ as soon as $k$ has no zero divisors. ▮

Proof of claim. Now suppose $f(x_1, \ldots, x_n)$ vanishes on $V(x_i - x_j)$. Without loss of generality, $i = 1$ and $j = 2$. The polynomial $$g(x_2, \ldots, x_n) = f(x_2, x_2, x_3, \ldots, x_n)$$ vanishes on $k^{n-1}$, so $g = 0$. But that means $f$ lies in the kernel of the evaluation homomorphism induced by $x_1 \mapsto x_2$, which is the ideal $(x_1 - x_2) \subset k[x_1, \ldots, x_n]$ (by the "factor theorem"). ▮

Unit
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