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I have a silly question. I am clearly missing a subtlety with the use of Hensel's Lemma.

Suppose I have a variety $S/\mathbb{Q}$. I wish to determine all of its $\mathbb{Q}_2$ points. I search $\mod 4$, and find exactly one point, let's call it $P$.

I apply Hensel's Lemma to lift $P$ to a unique $P'$ in $\mathbb{Q}_2$. Now I know $S(\mathbb{Q}_2)=\{P'\}$.

The thing about $P$ is that it was also a $\mathbb{Z}$ point of $S$, so Hensel lifting was easy. It looks something like $P'=(1+O(2^{\infty}):0+O(2^{\infty}):1+O(2^\infty))$.

Now here is my question. Suppose I want to determine $S(\mathbb{Q})$. If $Q\in S(\mathbb{Q})$ then in particular $\tilde{Q}\in S(\mathbb{Q}_2)$ where $\tilde{Q}$ is the image of $Q$ in $\mathbb{Q}_2$. There is only one point in $S(\mathbb{Q}_2)$ so I know it must equal $P'=(1+O(2^{\infty}):0+O(2^{\infty}):1+O(2^\infty))$. That must mean that the only rational solution is $(1:0:1)$.

What's wrong here? Surely this can't be right.

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First of all, to make sense of "reduction mod $4$" you need to assume that $S$ is defined over $\Bbb Z$. Formally, the points of $S$ modulo $4$ are none other than $S(\Bbb Z / 4 \Bbb Z) = \mathrm{Hom}_{\mathrm{Spec} \, \Bbb Z}(\mathrm{Spec} \, \Bbb Z / 4 \Bbb Z, S)$, which would not make sense if $S$ was only defined over $\Bbb Q$. It does not make sense to talk about the integer points of a variety over $\Bbb Q$ a priori: the equations $x-3=0$ and $2x-3=0$ define isomorphic (zero-dimensional) varieties over $\Bbb Q$, but not over $\Bbb Z$ (the first one has a $\Bbb Z$-point, whereas the second one does not).

Assume $S$ is defined over $\Bbb Z$, then. Without further assumptions, it may very well happen that Hensel's lemma does not apply. Consider for instance the (zero-dimensional) variety defined over $\Bbb Z$ by $x^2-8$. There is exactly one solution modulo $4$, but there is none over $\Bbb Q_2$ (such an $x$ would need to have $2$-adic valuation $3/2$). What is missing is a smoothness assumption. You might have seen that for a single variable polynomial $f$, a solution $\alpha$ modulo $p$ is required to be a single root in order to lift to $\Bbb Z_p$; requiering $S$ to be smooth is the natural generalization of this condition to higher-dimensional varieties (technically we only need to assume that $S$ is smooth over $\Bbb Z[1/2]$ but I will ignore that).

Assume $S/\Bbb Z$ is smooth. Then indeed, any $\Bbb Z / 4 \Bbb Z$-point lifts to a $\Bbb Z_2$-point. But uniqueness only holds in the zero-dimensional case, and fails in general. For instance, take $S = \Bbb A^1$ the affine line over $\Bbb Z$, which is smooth. Then $S(\Bbb Z / 4 \Bbb Z) = \Bbb Z / 4 \Bbb Z$ has four points, each of which lifts to infinitely many points in $S(\Bbb Z_2) = \Bbb Z_2$. Hence as soon as you are working with positive-dimensional varieties, you may have a single point modulo $4$, and yet lots of them over $\Bbb Z_2$.

If you are working with a smooth, zero-dimensional scheme $S / \Bbb Z$ defined by monic polynomials, then having a single solution modulo $4$ does mean having a single solution over $\Bbb Z_2$, and these actually constitute all solutions over $\Bbb Q_2$! As $\Bbb Q$-points inject inside $\Bbb Q_2$-points, you have found that your variety is in fact made of a single point over $\Bbb Q$!

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    $\begingroup$ You are absolutely right, I mistakenly thought Hensel lifting a point on a variety would give a unique point. Of course there are many possible ways of performing the Hensel lift (my variety is positive dimensional). Thank you! $\endgroup$ Commented 14 hours ago

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