# All $K$-rational points of $V$ form a $K$-variety

Here is a problem from Ernst Kunz's Introduction to Commutative Algebra and Algebraic Geometry:

Let $$L / K$$ be an extension of fields, $$V \subset \mathbb{A}^n(L)$$ an $$L$$-variety. Then the set $$V_k := V \cap \mathbb{A}^n(K)$$ of all $$K$$-rational is a $$K$$-variety in $$\mathbb{A}^n(K)$$

Here we can just deal with the case that V is a irreducible hypersurface defined by a irreducible polynomial $$F \in L[X_1, ..., X_n]$$.

What I've tried:

1. The most luciest case that $$F \in K[X_1, ..., X_n]$$, then problem solved. If not, suppose F has a zero point in $$\mathbb{A}^n(K)$$, does there exist a polynomial in $$K[X_1, ..., X_n]$$ associated with F?($$n = 1$$ does)

2. Let $$\mathfrak{J}_L (V)$$ be the set of all polynomials in $$L[X_1, ..., X_n]$$ which vanish on V, then $$\mathfrak{J}_L (V) = (F)$$ is an ideal of $$L[X_1, ..., X_n]$$. Consider $$\mathfrak{J}_L (V) \cap K[X_1, ...X_n]$$, it may be a zero ideal or a nonzero ideal of $$K[X_1, ...X_n]$$. For the latter, I want to find the special $$h \in K[X_1, ...X_n]$$ which $$h = FG$$ and $$h$$ does not contribute other zero points.

In fact, I have made no progress in the problem. Anyone can help?

• What have you tried? – Wuestenfux Aug 12 '19 at 15:54
• @Wuestenfux Thanks. I've added what I've tried, which I also failed. – JR Chan Aug 13 '19 at 8:53
• @JRChan Very good! Voted to reopen. – TheSimpliFire Aug 13 '19 at 9:16

Let $$L/K$$ be a field extension. Fix a $$K$$-basis $$\{e_\lambda\}_{\lambda\in \Lambda}$$ of $$L$$. Let $$V \subseteq \Bbb A^n(L):= L^n$$ be an $$L$$-variety defined by polynomials $$F_1,\dotsc, F_r \in L[X_1,\dotsc,X_n]$$. We can then write $$F_i = \sum_{\lambda\in \Lambda} f_{i,\lambda}\cdot e_\lambda,$$ where $$f_{i,\lambda}\in K[X_1,\dotsc,X_n]$$ for all $$i,\lambda$$. Notice that $$f_{i,\lambda}\neq 0$$ only for finitely many pairs $$(i,\lambda)$$. In other words: there exists a finite subset $$J\subseteq \{1,\dotsc,r\}\times\Lambda$$ such that $$(i,\lambda)\notin J \implies f_{i,\lambda} =0$$.
Now, for $$x\in \Bbb A^n(K)$$ we have the following equivalences: \begin{align*} x\in V\cap \Bbb A^n(K) &\iff F_i(x) = 0 \quad\text{for all 1\le i\le r}\\ &\iff f_{i,\lambda}(x) = 0 \quad \text{for all (i,\lambda)\in J} \end{align*} Therefore, $$V\cap \Bbb A^n(K)$$ is the $$K$$-variety defined by the polynomials $$f_{i,\lambda} \in K[X_1,\dotsc,X_n]$$ for $$(i,\lambda)\in J$$.