Lemma 33.22.2. Let $k$ be an infinite field. Let $X$ be a scheme of finite type over $k$. Let $\mathcal{L}$ be a very ample invertible sheaf on $X$. Let $n \geq 0$ and $x, x_1, \ldots , x_ n \in X$ be points with $x$ a $k$-rational point, i.e., $\kappa (x) = k$, and $x \not= x_ i$ for $i = 1, \ldots , n$. Then there exists an $s \in H^0(X, \mathcal{L})$ which vanishes at $x$ but not at $x_ i$.

Proof. If $n = 0$ the result is trivial, hence we assume $n > 0$. By definition of a very ample invertible sheaf, the lemma immediately reduces to the case where $X = \mathbf{P}^ r_ k$ for some $r > 0$ and $\mathcal{L} = \mathcal{O}_ X(1)$. Write $\mathbf{P}^ r_ k = \text{Proj}(k[T_0, \ldots , T_ r])$. Set $V = H^0(X, \mathcal{L}) = kT_0 \oplus \ldots \oplus kT_ r$. Since $x$ is a $k$-rational point, we see that the set $s \in V$ which vanish at $x$ is a codimension $1$ subspace $W \subset V$ and that $W$ generates the homogeneous prime ideal corresponding to $x$. Since $x_ i \not= x$ the corresponding homogeneous prime $\mathfrak p_ i \subset k[T_0, \ldots , T_ r]$ does not contain $W$. Since $k$ is infinite, we then see that $W \not= \bigcup W \cap \mathfrak q_ i$ and the proof is complete. $\square$

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