[proofplan] We compare two nonzero coordinate representatives of the same projective point. The equality test for [homogeneous coordinates](/page/Homogeneous%20Coordinates) gives a nonzero scalar $\lambda \in k^\times$ relating the two representatives. Homogeneity of $F$ then shows that the two evaluated values differ by the nonzero factor $\lambda^d$, and multiplication by a nonzero field element preserves and reflects vanishing. [/proofplan]

[step:Relate the two coordinate representatives by a nonzero scalar]Let \begin{align*} a=(a_0,\ldots,a_n) \in k^{n+1}\setminus\{0\} \end{align*} and \begin{align*} b=(b_0,\ldots,b_n) \in k^{n+1}\setminus\{0\} \end{align*} be two representatives of the same point of $\mathbb{P}^n_k$, so \begin{align*} [a_0:\cdots:a_n]=[b_0:\cdots:b_n]. \end{align*} By [citetheorem:9486], there exists $\lambda \in k^\times$ such that \begin{align*} b_i=\lambda a_i \end{align*} for every index $i \in \{0,\ldots,n\}$.[/step]

[guided]We begin by translating equality of projective points into an algebraic statement about tuples. The objects \begin{align*} a=(a_0,\ldots,a_n) \end{align*} and \begin{align*} b=(b_0,\ldots,b_n) \end{align*} are nonzero elements of $k^{n+1}$, and the hypothesis says that they determine the same projective point: \begin{align*} [a_0:\cdots:a_n]=[b_0:\cdots:b_n]. \end{align*} The relevant fact is precisely the equality test for homogeneous coordinates. By [citetheorem:9486], equality of these two projective points is equivalent to the existence of a scalar $\lambda \in k^\times$ such that each coordinate of $b$ is obtained by multiplying the corresponding coordinate of $a$ by $\lambda$. Therefore there exists $\lambda \in k^\times$ with \begin{align*} b_i=\lambda a_i \end{align*} for every $i \in \{0,\ldots,n\}$. The condition $\lambda \in k^\times$ is essential for the final vanishing argument: it means that $\lambda$ is nonzero and hence invertible in the field $k$.[/guided]

[step:Use homogeneity to compare the two polynomial values] Since $F$ is homogeneous of degree $d$, every monomial appearing in $F$ has total degree $d$. Thus there is a finite index set \begin{align*} A \subset \{\alpha=(\alpha_0,\ldots,\alpha_n)\in \mathbb{Z}_{\ge 0}^{n+1}:\alpha_0+\cdots+\alpha_n=d\} \end{align*} and coefficients $c_\alpha \in k$ such that \begin{align*} F=\sum_{\alpha\in A} c_\alpha X_0^{\alpha_0}\cdots X_n^{\alpha_n}. \end{align*} Using $b_i=\lambda a_i$ for all $i$, we compute \begin{align*} F(b_0,\ldots,b_n)=\sum_{\alpha\in A} c_\alpha(\lambda a_0)^{\alpha_0}\cdots(\lambda a_n)^{\alpha_n}. \end{align*} For each $\alpha\in A$, the total exponent of $\lambda$ is $\alpha_0+\cdots+\alpha_n=d$, so \begin{align*} F(b_0,\ldots,b_n)=\lambda^d\sum_{\alpha\in A} c_\alpha a_0^{\alpha_0}\cdots a_n^{\alpha_n}. \end{align*} Therefore \begin{align*} F(b_0,\ldots,b_n)=\lambda^d F(a_0,\ldots,a_n). \end{align*} [/step]

[step:Conclude that vanishing is preserved and reflected] Because $\lambda \in k^\times$ and $k$ is a field, $\lambda^d \in k^\times$. Hence multiplication by $\lambda^d$ is injective on $k$. From \begin{align*} F(b_0,\ldots,b_n)=\lambda^d F(a_0,\ldots,a_n), \end{align*} we obtain \begin{align*} F(a_0,\ldots,a_n)=0 \implies F(b_0,\ldots,b_n)=0. \end{align*} Conversely, if $F(b_0,\ldots,b_n)=0$, then \begin{align*} \lambda^d F(a_0,\ldots,a_n)=0. \end{align*} Multiplying by $(\lambda^d)^{-1}\in k$ gives \begin{align*} F(a_0,\ldots,a_n)=0. \end{align*} Thus \begin{align*} F(a_0,\ldots,a_n)=0 \iff F(b_0,\ldots,b_n)=0. \end{align*} Since the representatives $a$ and $b$ were arbitrary homogeneous coordinate representatives of the same projective point, the condition $F(a_0,\ldots,a_n)=0$ is independent of the chosen homogeneous coordinates. [/step]