Not valid projective algebraic set I need to prove that given a point $[a_0,..., a_n] \in \mathbb{P} ^n$, expressions of the form
$$V=V(x_0-a_0,...,x_n-a_n) \subseteq \mathbb{P} ^n$$
are not well defined algebraic sets. It's clear that the polynomials involved in the expression are not homogeneous, but I don't see how this allows me to conclude $V$ cannot be a well defined algebraic set. I've tried to reach a contradiction assuming that the ideal generated by $ x_0-a_0,...,x_n-a_n $ is homogeneous (i. e. it's generated by homogeneous polynomials), but I'm stucked. Any suggestion? 
 A: Think about it this way: For $V(f_1,..., f_k) \subseteq \mathbb P^n$ to be well defined, the corresponding vanishing locus in $\mathbb A^{n+1}$ has to be a cone. 
A: Though it hasn't been specified, I'll assume that we are working over a field $k$ which is not $\mathbb{F}_2$.
Remember that strictly speaking, $[a_0:\cdots:a_n]\in \mathbb{P}^n$ is the equivalence class of $(a_0,\ldots, a_n)\in \mathbb{A}^{n+1}$ under the quotient relation used to define projective space. So, to say a function $f$ vanishes on the whole equivalence class $[a_0:\cdots:a_n]$, it must be true that $f(a_0,\ldots, a_n)=0$ and moreover that for any $\lambda \in k^\times$ $f(\lambda a_0,\ldots, \lambda a_n)=0$ also. So, in this case just note that indeed $(a_0,\ldots, a_n)$ is a solution to your system of equations $x_i-a_i=0$ for $i=0,\ldots, n$ in $\mathbb{A}^{n+1}$ but this does not descend to the quotient, precisely because the solution set in $\mathbb{A}^{n+1}$ is a point - not a cone over the origin.
We can see this algebraically, too. Note that for this to make any sense at all, not all of the $a_i$ can be zero. Say that $a_0\ne 0$ with no loss of generality. The representative $(a_0,\ldots, a_n)$ of $[a_0:\cdots:a_n]$ satisfies the $x_i-a_i$, but taking any scalar $\lambda \ne 0,1$ we get that the representative $(\lambda a_0,\ldots, \lambda a_n)$ of $[a_0:\cdots:a_n]$ has 
$$\lambda a_0-a_0=(\lambda-1)a_0=0.$$
This is a contradiction, since $\lambda \ne 1$ and $a_0\ne 0$. Hence, it is not well-defined to say that $[a_0:\cdots:a_n]$ is in the vanishing set of $\{x_i-a_i\}_{i=0}^n$.
