TITLE: Nullstellensatz for non-algebraically closed fields
QUESTION [7 upvotes]: I'm trying to prove that the Nullstellensatz holds for non algebraically closed fields, when the variety is taken over the algebraic closure. Let $R=K[x_1,...,x_n]$ and $\overline{K}$ the algebraic closure of $K$. I was able to prove that $\sqrt{I}\subseteq \mathcal{I}_R(\mathcal{V}_{\overline{K}^n}(I))$ for any ideal $I$. I'm struggling a bit with the other direction. My attempt goes as follows:
It is clear that given any ideal $J$, $V_{K^n}(J)\subseteq  \mathcal{V}_{\overline{K}^n}(J)$. Applying $\mathcal{I}_{\overline{K}^n}$ reverses the order, so we have:
$$\mathcal{I}_{\overline{K}^n}(V_{K^n}(J))\supseteq  \mathcal{I}_{\overline{K}^n}(\mathcal{V}_{\overline{K}^n}(J))=\sqrt{I}$$
where the equality is just the normal Nullstellensatz. I don't know if this idea seems fruitful since I haven't been able to get the reverse inclusion. Any ideas on how to show this direction would he highly appreciated.

REPLY [5 votes]: This result appears in Section 11.2.1 of my commutative algebra notes under the name Semirational Nullstellensatz (which is not standard, but seems broadly reasonable).  The proof is indeed nontrivial.  (It uses something that I call Lang's Lemma.)  Perhaps there is a more straightforward approach -- indeed, the exercise at the end of the section seem to ask about this -- and if so I would be interested to know.