The first paragraph of [this question](https://mathoverflow.net/questions/116433/2-questions-on-nagatas-counterexample-kf-1-f-rkg-1-g-s-vs-kf) shows the construction of the first counter example to Hilbert's 14th Problem. There, we start from a prime field $P$ of arbitrary characteristic, i.e., $P=\Bbb Q$ or $P=\Bbb F\_p$ for some prime $p$, and adjoin 48 algebraically independent elements $a\_{ij}$, $i=1,2,3$, $j=1,...,16$, of some field $k$ containing $P$.


Now for the proof of the example to work, Nagata uses the following geometric lemma:



> 
> Let $p\_1,...,p\_{16}$ be **independent generic points** of the projective plane $S$ **over the prime field** $P$. For any curve $C$ of degree $d$, the sum of the multiplicities of $p\_i$ on $C$ is less than $4d$.
> 
> 
> 


At first I had some problems with the word "generic" here, but it seems to mean "general" as in "in general position", since this paper is from the fifties and hence some uncommon and no-longer-used terminology occurs. I can imagine what this means in this case, but I'm not really sure about the exact definition; 4 points of a projective plane are in general position if no 3 of them lie on a line, this is what I know. I also found a definition for 6 points, then additionally no 6 of them shall lie on a conic. How do these generalize to even more points?


Then, I find "independent" confusing there. Is it maybe that "independent generic" = "in general position", i.e., the two words only occur together?


I'd also have thought that we're talking about the projective plane $\Bbb P\_P^2$ because of "over the prime field $P$"; then the points $p\_i$ would be of the form $(a\_0:a\_1:a\_2)$ with $a\_j\in P$. But in the actual proof, we come across the $k$-algebra $k[y\_1,y\_2,y\_3]$ (for the exact construction you could read the linked question, but I think it has no real impact on my actual problem here), and interpret it as a homogeneous coordinate ring of "the projective plane $S$", and consider the sixteen points $p\_j=(a\_{1j}:a\_{2j}:a\_{3j})$. I quote: *Then the* $p\_i$ *are independent generic points of* $S$ *over the prime field* $P$.


What does this mean? To me, $k[y\_1,y\_2,y\_3]$ can be considered as homogeneous coordinate ring of the projective plane $\Bbb P\_k^2$, and the $p\_i$ lie in there. What does the "over the prime field $P$" part do there? The $p\_i$ are no points of $\Bbb P\_P^2$ at least.


Any help in the fight against my confusion is greatly appreciated!



Independent generic/general points over some prime field
You should look in Weil's "Foundations of algebraic geometry"; it isn't as impenetrable as you may fear. He does give all necessary definitions, the key one for your occurring on page 3.


Weil's notion of "generic point" was one of his technical innovations, and it is very close to the scheme-theoretic notion, so much more powerful than the weaker concept of "point in general position". A defect is that his (positive-dimensional) varieties have lots of generic points, so in that sense scheme theory has a better notion, but Weil's work is where many important notions in scheme theory were first introduced (in an embryonic form that nonetheless often involved most of the key algebraic difficulties). 


In Weil's "Foundations", for each characteristic we fix a *universal domain*: an abstract field $\mathbf{K}$ of infinite transcendence degree over the prime field. A "field" by definition is a *subfield* $k$ of $\mathbf{K}$ over which $\mathbf{K}$ has infinite transcendence degree. For Weil, if $X$ is a geometrically integral (and separated) $k$-scheme of finite type then a *point* of $X$ is an element $x \in X(\mathbf{K})$ (a $k$-morphism $x:{\rm{Spec}}(\mathbf{K}) \rightarrow X$). Denoting by $[x]$ the image in $X$ of this $k$-morphism, Weil says $x$ is *generic* over $X$ if $[x]$ is its generic point. 


The field $k(x)$ is defined to be the residue field of $X$ at $[x]$ (so it is a finitely generated extension of $k$) and because $x$ is a $k$-morphism from ${\rm{Spec}}(\mathbf{K})$ we see that there is a canonically associated $k$-embedding $k(x) \hookrightarrow \mathbf{K}$ (so $k(x)$ is a "field"). Note that *distinct* points $x, y \in X(\mathbf{K})$ can satisfy $[x] = [y]$ in $X$ and even $k(x) = k(y)$ inside $\mathbf{K}$, namely when the canonical $k$-embeddings of $k(x)$ and $k(y)$ into $\mathbf{K}$ do not respect the equality of images inside $\mathbf{K}$. For example: taking $\mathbf{C}$ to be the universal domain in characteristic 0, generic points of the affine line over $k = \mathbf{Q}$ are identified with elements of $\mathbf{C}$ that are transcendental over $\mathbf{Q}$, so $\pi$ and $\pi+1$ are distinct generic points of the affine line over $\mathbf{Q}$ even though $\mathbf{Q}(\pi) = \mathbf{Q}(\pi+1)$ as subfields of $\mathbf{C}$ (such equality is stronger than merely saying that the fields $\mathbf{Q}(\pi)$ and $\mathbf{Q}(\pi+1)$ are $\mathbf{Q}$-isomorphic). 


Weil says that points $x, y \in X(\mathbf{K})$ are *independent over $k$* if the image domain of the canonical $k$-algebra map $k(x) \otimes\_k k(y) \rightarrow \mathbf{K}$ defined using multiplication via the canonically associated $k$-embeddings into $\mathbf{K}$ (!) has fraction field whose transcendence degree over $k$ is equal to the sum of those of $k(x)$ and $k(y)$ over $k$. (See page 3 of the Foundations, especially the third paragraph.) For example, if $k(x) \otimes\_k \overline{k}$ is a domain (equivalently, $[x]$ has Zariski closure in $X$ that is geometrically integral over $k$, or as Weil would say "$k(x)$ is regular over $k$") then $k(x) \otimes\_k k(y)$ is a domain and hence "independence over $k$" in such cases is precisely the condition that $k(x) \otimes\_k k(y) \rightarrow \mathbf{K}$ is injective. For example: taking $\mathbf{C}$ to be the universal domain, the generic points $\pi$ and $e$ of the affine line over $\mathbf{Q}$ are independent over $\mathbf{Q}$ if and only if $\pi$ and $e$ are algebraically independent over $\mathbf{Q}$ (the latter property being an unsolved problem), whereas the pair of generic points $\pi^5 - 2$ and $\pi^3 + 7 \pi - 4$ are certainly not independent over $\mathbf{Q}$. The notion of "independence" (over $k$) for a finite set of points is defined similarly using the tensor product over $k$ of several subfields of $\mathbf{K}$.


So in modern terms, the lemma of Nagata is saying that if $C$ is a degree-$d$ curve inside the projective plane over $\mathbf{K}$ (I really mean $\mathbf{K}$ and not the prime field) and if $x\_1, \dots, x\_{16}$ are $\mathbf{K}$-points of $C$ that lie in the usual affine plane with coordinates that are collectively algebraically independent over the prime field then the multiplicities of $C$ at the points $x\_i$ have sum at most $4d$. Of course, one has to look at Nagata's stuff to decide if his curves are meant to be geometrically reduced or geometrically irreducible, etc. 


