Let $k$ be an algebraically closed field and fix $(a_1,\dots,a_n)\in k^n\setminus \{0\}$.
If $f \in(x_1-a_1,\dots,x_n-a_n)\subseteq k[x_1,\dots,x_n]$ is homogeneous we have $f(a_1,\dots,a_n)=0\implies f(\lambda a_1,\dots,\lambda a_n)=0$ so $$f(a_1:\dots :a_n)=0 \text{ for } (a_0:\dots :a_n)\in \mathbb{P}^n.$$ Then as $I_+(\{(a_1:\dots:a_n)\})=(\{x_ia_j-x_ja_i\}_{ij})$ we have $f\in (\{x_ia_j-x_ja_i\}_{ij})$.
I want to know if this is still true when $k$ is not algebraically closed. So my question is
Is it true that every $f\in (\{x_i-a_i\}_i)$ homogeneous belongs to the ideal $(\{x_ia_j-x_ja_i\}_{ij})$ without the assumption of $k$ alg. closed?
Yes. Without loss of generality let $a_1\neq 0$ and observe that $$f(x_1,\dots,x_n)-f\left(x_1,\tfrac{a_2}{a_1}{x_1},\dots,\tfrac{a_n}{a_1}{x_1}\right)\in I_+\subseteq I$$ By homogeneity $f\left(x_1,\tfrac{a_2}{a_1}{x_1},\dots,\tfrac{a_n}{a_1}{x_1}\right)=\lambda x^{\text{deg}(f)}$ for some constant $\lambda\in k$, and it follows that $$0=f\left(a_1,\tfrac{a_2}{a_1}{a_1},\dots,\tfrac{a_n}{a_1}{a_1}\right)=\lambda a_1^{\text{deg}(f)}\implies \lambda=0$$ Implying in turn that $f(x_1,\dots,x_n)\in I_+$ $\blacksquare$