# In the Proj construction, are the homogeneuos prime ideals determined up to an element in the base ring?

Consider the point $[a_0,...,a_n]\sim [a_0\lambda,...,a_n\lambda] \in \Bbb P^k_n$. How do you write the corresponding homogeneous prime ideal in the graded ring $S:=k[x_0,...,x_n]$?

Well, the straightforward homogenization in any variable seems to work because the $a_i$ are not all zero ($\Rightarrow$ so we do not end up with the irrelevant ideal!). Thus, we can homogenize, for example in $x_0$, the corresponding prime ideals in $\operatorname {Spec} S$ to get the homogeneous prime ideal in $\operatorname{Proj} S$.

$$(x_0-a_0x_0,...,x_n-a_nx_0) \sim (x_0-a_0\lambda x_0,...,x_n-a_n\lambda x_0) \in \operatorname {Proj} S.$$

But how do you show that these two prime ideals are indeed the same? My intuition is to let $x_0 \mapsto \lambda x_0$, but I don't know how make this argument make sense and how to deal with the first coordinate transforming like $x_0-a_0x_0\mapsto \lambda x_0-a_0\lambda x_0\neq x_0-a_0\lambda x_0$.

The ideal attached to this point is, by definition, the ideal generated by all homogeneous polynomials that vanish at this point. You can check that this ideal is generated by the elements $a_i x_j - a_j x_i$ for $0 \leq i,j\leq n$.

We know that at least one $a_i \neq 0$; for ease of notation, suppose it is $a_0$.

Then this ideal is in fact generated by $x_i - (a_i/a_0) x_0$, as you might expect.