Why are solutions $x=(x_1,\ldots,x_n)$ of a $p$-adic variety with one of the $x_i$ invertible called primitive solutions?

Question

In Serre's arithmetic he calls a solution $(x_1,\ldots,x_n)$ of a $p$-adic variety $$ X = \mathbb{V}(f^{(1)},\ldots,f^{(k)}) \subset (\mathbb{Z}_p)^n $$ primitive if one of the $x_i$ is invertible. Where does this terminology come from and why should I care about the special case of primitive zeros?

Answer 1

Let $K$ be a field. Then "the $K$-valued points of projective $n$-space", denoted $\mathbb{P}^n(K)$, is equal to the set of $(n+1)$-tuples $(x_0, x_1, \ldots,x_n)$, with at least one coordinate nonzero, modulo the equivalence relation of scaling: for nonzero $t$ in $K$, we identify $(x_0, x_1, \ldots, x_n)$ with$$t(x_0, x_1, \ldots, x_n) = (tx_0, tx_1, \ldots, tx_n).$$We write $[x_0, x_1, \ldots, x_n]$ for the equivalence class of $(x_0, \ldots, x_n)$ and refer to $(x_0, x_1, \ldots, x_n)$ as "homogeneous coordinates" for the projective point $[x_0, x_1, \ldots, x_n]$. One particular thing to note is that the set $K^n$ embeds naturally in $\mathbb{P}^n(K)$ via $(x_1, \ldots, x_n)$ goes to $[1, x_1, \ldots, x_n]$, and in a sense $\mathbb{P}^n(K)$ is a kind of completion of $K^n$.

Just as one considers "closed algebraic" subsets of $K^n$ determined by roots of polynomials $f(x_1, \ldots, x_n)$, one considers "closed algebraic" subsets of $\mathbb{P}^n(K)$ determined by roots of homogeneous polynomials $F(x_0, x_1, \ldots, x_n)$. The word "homogeneous" means that each summand of $F$ has the same total degree. If that degree is $d$, then$$F(tx_0, tx_1, \ldots, tx_n) = t^d F(x_0, x_1, ..., x_n),$$and therefore if$$[x_0, x_1, \ldots, x_n] = [x'_0, x'_1, \ldots, x'_n]$$in $\mathbb{P}^n(K)$ then$$F(x_0, x_1, \ldots, x_n) = 0$$if and only if$$F(x_0', x_1',\ldots, x_n') = 0.$$So the question "is $F = 0$?" is well-defined on projective points. Moreover, given $f(x_1, \ldots, x_n)$, one can "complete" it to a homogeneous polynomial $F(x_0, x_1, \ldots, x_n)$ by adding extra powers of $x_0$ as necessary (typical example: $1 + x_1 + x_1x_2$ becomes $x_0^2 + x_0 x_1 + x_1 x_2$). And under the embedding of $K^n$ into $\mathbb{P}^n(K)$ that I described above, the algebraic set $f = 0$ in $K^n$ embeds into its closure $F = 0$ in $\mathbb{P}^n(K)$.

Algebraic geometry starts off with "affine" algebraic varieties, built out of equations $f$ in a fixed $K^n$, but eventually it generalizes to "global" algebraic varieties that are patched together out of such things, and the very first examples are projective varieties. One can patch $\mathbb{P}^n(K)$ out of $n+1$ copies of $K^n$, and the same goes for algebraic subsets.

Anyway, that is the context. Let me return to your question. Suppose $R$ is a domain and $K$ is its fraction field. Given a projective point $[x_0, x_1, \ldots, x_n]$, we can write$$x_i = {{y_i}\over{z_i}}$$ with $y_i$, $z_i$ in $R$ (and $z_i$ nonzero), and then multiplying the whole tuple by the scalar $z_0z_1\ldots z_n$ we can arrange that all the $x_i$ lie in $R$. But can we do better? When $R$ is a UFD, then there is a well-defined greatest common divisor $d$ of the $x_i$, and we can multiply the whole tuple by the scalar $1/d$ to arrange that the $x_i$ have no common factor. This gives a "primitive" representative of the equivalence class. When furthermore $R$ is a PID (such as $\mathbb{Z}$ or a DVR like $\mathbb{Z}_p$), for primitive $[x_0, \ldots, x_n]$ one will even have$$Rx_0 + Rx_1 + \ldots + Rx_n = R$$as ideals.