Rational solutions to $P(x,y)=0$ for $P$ reducible over ${\mathbb C}$

Question

There are facts in Mathematics that are so "obvious" and "well-known" that no-one includes a proper proof. An example is:

Theorem: If polynomial $P(x,y)$ with rational coefficients is irreducible over ${\mathbb Q}$ but not absolutely irreducible, then the equation $P(x,y)=0$ has at most finitely many rational solutions, and there is an algorithm for listing them all.

This theorem is very important and used (explicitly or implicitly) in hundreds of papers and books devoted to study rational points on curves $P(x,y)=0$, because it allows to assume without loss of generality that $P(x,y)$ is absolutely irreducible, define genus, and proceed in standard way. However, I am not able to find the full proof with all details. At best, there are short sentences like

"In this case all rational points are singular and the statement follows from Bézout's Theorem"

or

"If it not absolutely irreducible, then the absolute factors are conjugate, so any rational solution to one of them satisfies them all, so one can solve the system to find the finitely many rational points and check whether any of them are integer points."

or

"if $P(x,y)$ is irreducible over ${\mathbb Q}$ but not absolutely irreducible then the action of Galois acts transitively on the Q-irreducible components, but rational points are fixed by Galois, so X(Q) is contained in the intersection of the Q-irreducible components; thus in this case we reduce to the 0-dimensional problem."

However, I cannot find anywhere a proper proof with all details and with definitions of all concepts involved. Why exactly all rational points are singular? Bézout's Theorem is the statement about finiteness but not about algorithm for computing all points. What exactly is meant by "factors are conjugate" or "action of Galois acts transitively", and why this is true?

So, the question is to present a more detailed proof of this theorem or point me to a reference with a proper detailed proof.

Answer 1

The main idea of the proof already appears in what you've written, but here are some more details.

Factor $P(x,y) = Q_1(x,y) \cdots Q_n(x,y)$ into irreducibles, where the factorization takes place over a number field $K \supsetneq \mathbb Q$. Let $(a,b)$ be a rational solution to $P(x,y) = 0$. Then $Q_i(a,b) = 0$ for some $i$. Since $Q_i$ is not defined over $\mathbb Q$, then there is some element $\sigma \in {\rm Gal}(K/\mathbb Q)$ such that $Q_i^\sigma \ne Q_i$. (Note that $\sigma$ is just acting on the coefficients of $Q_i$.) But $P^\sigma = P$ since $P$ is defined over $\mathbb Q$, and $Q_i^\sigma$ is still an irreducible factor of $P = P^\sigma$, so $Q_i^\sigma = Q_j$ for some $j \ne i$.

Using the fact that $(a,b)$ is rational, we have

$$ Q_j(a,b) = Q_i^\sigma(a,b) = \left(Q_i(a,b)\right)^\sigma = 0^\sigma = 0, $$

so $(a,b)$ lies on the intersection $\{Q_i = 0\} \cap \{Q_j = 0\}$. The fact that $(a,b)$ is a singular point on the curve now essentially follows from the product rule from calculus.

Answer 2

This also follows from Prop. 2.3.26(i) in Bjorn Poonen's Rational Points on Varieties, where it is stated that if for a finite type $k$-scheme $X$ the set of rational points $X(k)$ is Zariski dense, then irreducibility implies geometrical irreducibility for $X$. (The set of $k$-points on a curve over $k$ being Zariski dense if and only if it's infinite.)

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Reading your question more carefully, I see that this reference does not answer all your questions. In particular it doesn't answer your question as to why the points are singular. I'd say this is because if some irreducible components $C_1,\ldots,C_i$ of a curve $C$ intersect in a point $P$, then $P$ must be a singular point, because the local ring $\mathscr{O}_{C,P}$ will not be an integral domain, much less a discrete valuation ring. [For simplicity's sake assume $C$ to be planar, and assume each $C_j$ is given by an affine equation $f_j=0$. Then $f_1 \ldots f_i = 0$ in $\mathscr{O}_{C,P}$, whereas none of the $f_j$ vanishes individually.]