Vanishing Ideal of a Linear Subspace Let $F$ be an infinite field. Let $V$ be a subspace of $F^n$. Let $V^{\perp}$ be the set of all linear functionals $F^n \rightarrow F$ that vanish on $V$. Let $I(V)$ be the vanishing ideal of $V$, i.e. the set of all polynomials of $F[x_1,\cdots,x_n]$ that vanish on $V$. Is it true that $I(V)$ is generated by $V^{\perp}$? If yes, how can we see that?
 A: Yes, $I(V)$ is generated by $V^\bot$. To prove this, we may assume after a change of coordinates that $V$ is spanned by the vectors $e_1,\ldots, e_k$ where $e_i = (0,\ldots,1,\ldots,0)$ is the $i$-th basis vector of $F^n$. Now let $f \in I(V)$, i.e. 
$$f(x_1,\ldots,x_k,0,\ldots,0) = 0 \quad \text{for all } x_1,\ldots,x_k \in F.$$
We can write $f$ in the form
$$f(X_1,\ldots,X_n) = g(X_1,\ldots,X_k) + \sum_{i=k+1}^n X_i h_i(X_1,\ldots,X_n)$$
where $g$ contains all the monomials of $f$ that do not involve any of $X_{k+1},\ldots,X_n$. Now we have
$$g(x_{1},\ldots,x_{k}) = 0\quad \text{for all } x_1,\ldots,x_k \in F.$$
The result then follows from the following general 

Lemma: Let $F$ be an infinite field and suppose $g \in F[X_1,\ldots,X_n]$ such that $g(x_1,\ldots,x_n) = 0$ for all $x_1,\ldots,x_n \in F$. Then $g = 0$.

Given the lemma, we get $g = 0$, so $f = \sum_{i=k+1}^n X_i h_i$ is contained in the ideal generated by the linear forms $X_i$.
The lemma itself is clear for $n= 1$ and follows for $n \geq 2$ by an induction argument.
