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Can you recover the characteristic of a perfect field from the category of smooth projective geometrically connected varieties over it?

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    $\begingroup$ There is a final object, and the covariant Yoneda functor of that object recovers the set of closed points. Now we can formulate homogeneous objects. There is a unique homogeneous variety where Aut is triply transitive. Consider the p-torsion in the stabilizer of an ordered pair of distinct points. $\endgroup$ Jul 5, 2020 at 15:36
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    $\begingroup$ Jason, does this work for fields that are not algebraically closed? $\endgroup$
    – Angelo
    Jul 5, 2020 at 17:38
  • $\begingroup$ I was assuming that the field is algebraically closed . . . $\endgroup$ Jul 5, 2020 at 19:02
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    $\begingroup$ Anyway, it's a really nice argument. $\endgroup$
    – Angelo
    Jul 5, 2020 at 20:32
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    $\begingroup$ In the algebraically closed case, here is a variant bypassing the use of $\mathbb{P}^1$: $k$ has characteristic $p>0$ iff there is a group object $E$ such that the kernel of multiplication by $l$ on $E(k)$ has order $l^2$ for each prime $l\neq p$, and smaller order if $l=p$. $\endgroup$ Jul 6, 2020 at 7:03

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Correction. As correctly noted by Remy van Dobben de Bruyn, there is a mistake in Lemma 4. What follows is a corrected argument, with the original (mistaken) post appended below the corrected argument.

Let $k$ be a perfect field. Denote by $\mathbf{V}$ the category of $k$-schemes that are smooth, projective and geometrically connected.

Proposition. A non-final object $Y$ of $\mathbf{V}$ is a curve if and only if the natural action of the symmetric group $\mathfrak{S}_2$ on $Y\times_{\text{Spec}\ k} Y$ admits a categorical quotient in $\mathbf{V}$.

Proof. For a curve, the geometric quotient is smooth, hence it is an object of $\mathbf{V}$.

Let $Y$ be a non-final object that is not a curve. Denote the dimension by $n\geq 2$. In the category of $k$-schemes, there is a categorical quotient $Y_2$ (even a geometric quotient) of the action of $\mathfrak{S}_2$ on $Y\times_{\text{Spec}\ k}Y$, and it is singular along the image of the diagonal, i.e., the embedding dimension is strictly larger than $2n$.

There is a closed immersion of $Y_2$ into a projective space, and this is an object of $\mathbf{V}$. Thus, if there is a categorical quotient $Z$, it factors this closed immersion. In particular, the morphism from $Y_2$ to $Z$ is also a closed immersion of $k$-schemes.

Since the embedding dimension of $Y$ is strictly larger than $2n$, also $Z$ has dimension strictly larger than $2n$. Thus, the closed immersion is not surjective on closed points. For any closed point of $Z$ that is not in the image of $Y_2$, consider the blowing up $\widetilde{Z}$ of the categorical quotient at that closed point. Since the field is perfect, $\widetilde{Z}$ is still a smooth $k$-scheme (this can fail for the blowing up at a closed point with inseparable residue field).

The closed immersion from $Y_2$ to $Z$ factors through the morphism $\widetilde{Z}\to Z$. Of course the identity map on $Z$ does not factor through $\widetilde{Z}$. This contradicts that $Z$ is a categorical quotient. QED

Now we repeat the last part of the argument from the original post.

Lemma. A $k$-curve $Y$ in $\mathbf{V}$ is isomorphic to $\mathbb{P}^1_k$ if and only if every $k$-curve in $\mathbf{V}$ admits a nonconstant morphism to $Y$.

Proof. This follows from the fact that $\mathbb{P}^1_k$ admits no nonconstant morphism to a curve of positive genus, and every $k$-curve of genus $0$ with a $k$-point is isomorphic to $\mathbb{P}^1_k$. QED

Corollary. The field $k$ is uniquely determined by the category $\mathbf{V}$.

Proof. The proposition and the lemma together establish that there is a characterization of the object $\mathbb{P}^1_k$ in $\mathbf{V}$ in purely categorical terms. The automorphism group of $(\mathbb{P}^1_k,0,\infty)$ is $k^\times$. The automorphism group of $(\mathbb{P}^1,\infty)$ is a semidirect product of $k^\times$ and the additive group $k$. With the structure of both the multiplicative group of $k$ and the additive group of $k$, we can recover the field $k$. QED

Original post. There is a mistake in Lemma 4, as pointed out by Remy van Dobben de Bruyn.

I am writing an answer to address the case of a field $k$ that is not necessarily algebraically closed. Denote by $\mathbf{V}$ the category of $k$-schemes that are smooth, projective and geometrically connected.

Lemma 1. The $k$-scheme $\text{Spec}\ k$ is a final object in $\mathbf{V}$.

Proof. In fact this is a final object in the larger category of all $k$-schemes. QED

Definition 2. A morphism in $\mathbf{V}$ is constant if it factors through a morphism from the domain to a final object.

Definition 3. For an object $Z$ of $\mathbf{V}$, an ordered pair $(f,g)$ of nonconstant morphisms, $f:X\to Z$ and $g:Y\to Z$, is cofinite if every ordered pair $(u,v)$ of morphisms $u:W\to X$, $v:W\to Y$ with $f\circ u$ equal to $g\circ v$ is a pair of constant morphisms.

Lemma 4. A non-final object of $\mathbf{V}$ is a $k$-curve if and only if there exists no cofinite pair of nonconstant morphisms to the object.

Proof. This is a straightforward application of Bertini-type theorems. QED

Lemma 5. A $k$-curve $Y$ in $\mathbf{V}$ is isomorphic to $\mathbb{P}^1_k$ if and only if every $k$-curve in $\mathbf{V}$ admits a nonconstant morphism to $Y$.

Proof. This follows from the fact that $\mathbb{P}^1_k$ admits no nonconstant morphism to a curve of positive genus. QED

Proposition 6. The field $k$ is uniquely determined by the category $\mathbf{V}$.

Proof. The lemmas establish that there is a characterization of the object $\mathbb{P}^1_k$ in $\mathbf{V}$ in purely categorical terms. The automorphism group of $(\mathbb{P}^1_k,0,\infty)$ is $k^\times$. The automorphism group of $(\mathbb{P}^1,\infty)$ is a semidirect product of $k^\times$ and the additive group $k$. With the structure of both the multiplicative group of $k$ and the additive group of $k$, we can recover the field $k$. QED

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    $\begingroup$ In Lemma 5, two words should be said about genus $0$ curves without rational points. (Do rational points play a role in Def 3/Lemma 4 as well?) $\endgroup$ Jul 5, 2020 at 23:54
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    $\begingroup$ @R.vanDobbendeBruyn You are correct! That lemma was wrong :( I am revising my answer now . . . $\endgroup$ Jul 6, 2020 at 13:37

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