Let $L$ be a field and $K$ be a subfield of $L.$ Let $V$ be an affine variety in $\Bbb A^n (L).$ Then the coordinate ring $K[V]$ is defined by
$$K[V] = \frac {K[X_1,X_2, \dots ,X_n]} {\mathcal I(V)}$$
where $\mathcal I(V)$ is the vanishing ideal of $V.$ Now for any $(x_1,x_2,\dots,x_n) \in V$ set $p_x = \mathcal I_V \left ( \{(x_1,x_2, \dots,x_n) \} \right)$ the set of all functions $\varphi \in K[V]$ such that $\varphi (x_1,x_2, \dots, x_n) = 0.$ Then $p_x$ is a prime ideal $\ne K[V]$.
I don't understand the last sentence in bold letters mentioned in the lecture notes given by our instructor. Why is $p_x$ a prime ideal of $K[V]$? Let us take two functions $\varphi,\psi \in K[V]$ such that $\varphi \psi \in p_x.$ Then $\exists$ $f,g \in K[X_1,X_2, \cdots ,X_n]$ such that $\varphi = f + \mathcal {I} (V)$ and $\psi = g + \mathcal {I} (V).$ Since $\varphi \psi (x_1,x_2, \cdots, x_n) = 0$ so we have $(fg)(x_1,x_2, \cdots, x_n) \in \mathcal {I} (V).$ Therefore $f(x_1,x_2, \cdots, x_n) \cdot g(x_1,x_2, \cdots , x_n) \in \mathcal {I} (V).$ We have to show that either $f(x_1,x_2, \cdots , x_n) \in \mathcal {I} (V)$ or $g(x_1,x_2, \cdots , x_n) \in \mathcal {I} (V).$ Since $f(x_1,x_2,\cdots , x_n) \cdot g(x_1,x_2, \cdots , x_n)$ is a constant function belonging to the vanishing ideal $\mathcal I (V)$ so we must have $f(x_1,x_2,\cdots , x_n) \cdot g(x_1,x_2, \cdots , x_n) = 0.$ Since $f(x_1,x_2,\cdots , x_n),g(x_1,x_2, \cdots , x_n) \in K$ and $K$ is a field so we must have either $f(x_1,x_2,\cdots , x_n) = 0$ or $g(x_1,x_2,\cdots , x_n) = 0.$ So either $\varphi \in \mathcal {I} (V)$ or $\psi \in \mathcal {I} (V).$ So if we can show that $p_x \subsetneq K[V]$ we are through. But how can I show that?
Please help me in this regard. Thank you very much.