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The fixed point theorem fails without completeness: the additive group acts on the affine line by translations

Statement refuted

Over an algebraically closed field $k$ , every nonempty finite-type $k$ -scheme with an action of a smooth connected solvable affine algebraic group $G$ has a $k$ -point fixed by $G (k)$ . In other words, the completeness hypothesis in the Borel fixed point theorem (Borel fixed point theorem for complete schemes) can be dropped.

The witness is the following. Let $k$ be a field and let $X = A_{k}^{1} = Spec k [x]$ (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution, Affine schemes and their coordinate rings), with the action of $G_{a}$ on $X$ given on $R$ -points by $a \cdot z = z + a$ (translation). Then $X$ is nonempty of finite type over $k$ with an action of the smooth connected unipotent group $G_{a}$ (Unipotent algebraic groups and unipotent representations, Upper unitriangular groups are unipotent, and the additive group is U_2), so $G = G_{a}$ is smooth connected solvable, but $X$ is not complete and the action has no fixed point: for every $z \in X (k)$ and every $a \neq 0$ in $k$ , $a \cdot z \neq z$ . Hence the completeness hypothesis cannot be dropped, even for the smallest positive-dimensional smooth connected solvable group.

Facts & Assumptions

Given: A field $k$ , the additive group $G_{a} = Spec k [t]$ with $Δ (t) = t \otimes 1 + 1 \otimes t$ , and $X = Spec k [x] = A_{k}^{1}$ .

[F1]

$G_{a}$ is the group scheme with $G_{a} (R) = (R, +)$ for every $k$ -algebra $R$ , and it is a smooth connected unipotent group; the map $a \mapsto (\begin{matrix} 1 & a \\ 0 & 1 \end{matrix})$ identifies it with $U_{2}$ . (The upper unitriangular group scheme U_n and its coordinate ring, Upper unitriangular groups are unipotent, and the additive group is U_2, Unipotent algebraic groups and unipotent representations)

[F2]

An action of a group scheme $G$ on a scheme $X$ is a morphism $α : G \times_{k} X \to X$ satisfying the usual identities; on $R$ -points it gives an action of the abstract group $G (R)$ on $X (R)$ . (Algebraic group actions, orbit maps, orbit subschemes and scheme-theoretic stabilizers)

[F3]

$X = A_{k}^{1}$ is not complete. After base change to $A_{t}^{1}$ , its projection has the closed subset $Z = V (t x - 1) \subseteq Spec k [t, x]$ . Its image is exactly $D (t) \subseteq Spec k [t]$ : the quotient ring is $k [t, t^{- 1}]$ , and a prime lifts precisely when it does not contain $t$ . This image contains the generic prime $(0)$ and excludes the closed prime $(t)$ , so is not closed. The structure morphism is therefore not universally closed and hence not proper or complete. (Proper morphisms, Complete varieties)

[F4]

The fixed point theorem is cited only as the contrast with the present computation; the verification below is a direct computation over $k$ and uses no choice principle, so no assumption of the Axiom of Choice is made in this counterexample. (Borel fixed point theorem for complete schemes)

Proof

Given: A field $k$ , $G = G_{a}$ , and $X = Spec k [x]$ .

1.1F1F2

The morphism $α : G_{a} \times_{k} X \to X = Spec k [x]$ , dual to $k [x] \to k [t] \otimes_{k} k [x] = k [t, x]$ , $x \mapsto x + t$ , defines an action: on $R$ -points it is $(a, z) \mapsto z + a$ , and the identities $0 \cdot z = z$ and $a \cdot (b \cdot z) = z + b + a = (a + b) \cdot z$ hold in every $k$ -algebra $R$ . Hence $G_{a}$ acts on $X$ by translation, algebraically.

1.2F1F3

The group $G_{a}$ is smooth connected unipotent by [F1], and is commutative since addition commutes on every algebra-valued point; its commutator is the identity, so its derived series terminates after one step and it is solvable. The scheme $X$ is nonempty of finite type over $k$ , and it is not complete by [F3].

2.1step 1.1

The action has no fixed point: for $z \in X (k) = k$ and $a \in k$ , the equation $a \cdot z = z$ reads $z + a = z$ , i.e. $a = 0$ . Hence for every $z \in X (k)$ and every $a \neq 0$ the translate differs from $z$ , and $X (k)$ contains no point fixed by all of $G_{a} (k)$ .

3.1F4step 1.2step 2.1∎

Therefore the statement refuted is false: the action of the smooth connected solvable group $G_{a}$ on the nonempty finite-type scheme $A_{k}^{1}$ has no fixed point, so the completeness hypothesis in the Borel fixed point theorem is indispensable even in this minimal example, in contrast with [F4].

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Sources

J. S. Milne, Algebraic Groups (corrected 2022 printing, Cambridge University Press) (standard reference, not scraped)