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Let $f_1, \cdots, f_k$ be homogeneous polynomials over $\mathbb{Q}[x_0, \cdots, x_n]$. They define an projective variety $X$ over $\mathbb{P}^n(\mathbb{C})$, namely their set of zeros $$X = Z(f_1, \cdots, f_k) = \{(x_i) \in \mathbb{P}^n(\mathbb{C})| f_k(x_i) = 0\text{ for all }k\}$$ Many times we are interested in finding zeros $(x_i)$ in $\mathbb{P}^n(\mathbb{Z})$ or $\mathbb{P}^n(\mathbb{Q})$.

But a necessary condition for those zeros to exist is for them to exist on a finite extension $K$ of $\mathbb{Q}$. Thus, I may wonder what are the $K$-points of $X$. Has is been known/studied techniques for finding those points? Supposing $K$ is Galois over $\mathbb{Q}$, do the Galois group $G$ can provide any insight on this?

The Weil conjectures related the homology $H_i(X) = H_i(X, \mathbb{Q})$ of $X$ with coefficients of the zeta function $Z_p(\bar{X}, t) = \exp\left(\sum_{n\geq 1} N_{p^n}(X)\frac{t^n}{n}\right)$. Thus, topological, homological and arithmetical aspects unite. Maybe, if $X$ is an abelian variety (like an elliptic curve), we can see the $K$-points of $X$ as $G$-module? Has anyone ever made sense of something like $H^q(G, X)?$

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    $\begingroup$ First, $\mathbb P^n(\mathbb Z)$ is equal to $\mathbb P^n(\mathbb Q)$, but it is often interesting to study integer solutions to non-homogeneous equations. Second, it is quite standard to first study $X(K)$, or more generally $X(\overline{\mathbb Q})$, and then use Galois theory to isolate the points in $X(\mathbb Q)$. For example, if $E/\mathbb Q$ is an elliptic curve, in order to compute the rank of $E(\mathbb Q)$ in practice, one generally works in a field $K$ where $E$ has a cyclic subgroup defined over $K$. $\endgroup$
    – Joe Silverman
    Commented Feb 2, 2018 at 15:09
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    $\begingroup$ When $X$ is an abelian variety, then $H^1(\text{Gal}(\overline{\mathbb Q}/Q),A(\overline{\mathbb Q}))$ is called the Weil-Chatalet group of $A$. It has a conjecturally finite subgroup, called the Tate-Shafarevich group, that is used in a fundamental way to prove that $A(\mathbb Q)/mA(\mathbb Q)$ is finite (the weak Mordell-Weil theorem). $\endgroup$
    – Joe Silverman
    Commented Feb 2, 2018 at 15:12

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