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    Lüroth's theorem, and the Lüroth problem in higher dimension

    T-IV2LUROTH

    In dimension one, unirational means rational: every intermediate field $k\subsetneq L\subseteq k(t)$ is again $k(u)$.Suggest a slogan

    (Lüroth)

    Let \(k = \kbar\) and let \(L\) be a field with \(k \subsetneq L \subseteq k(t)\). Then \(L\) is purely transcendental: \(L = k(u)\) for some \(u\).

    \(L\) is a subfield of a finitely generated field of transcendence degree \(1\), so \(\trdeg_k L \leq 1\), and \(\trdeg_k L = 1\) because \(L \neq k\) and \(k\) is algebraically closed, so any element of \(L \setminus k\) is transcendental over \(k\). Such an \(L\) is itself finitely generated over \(k\), hence \(L = k(Y)\) for a unique smooth projective curve \(Y\).

    The inclusion \(L \subseteq k(t) = k(\PP^1)\) is a finite extension of function fields, so it is induced by a finite morphism \(f : \PP^1 \to Y\). A finite morphism of curves never decreases the genus, so \(g(Y) \leq g(\PP^1) = 0\) and \(g(Y) = 0\). A genus-\(0\) curve over \(k = \kbar\) is \(\PP^1\), so \(Y \cong \PP^1\) and \(L \cong k(u)\).

    (What the proof actually rests on)

    The one input is \(g(X) \geq g(Y)\) for a finite morphism \(X \to Y\) of curves, and that is Riemann–Hurwitz read backwards: for separable \(f\) of degree \(n\), \[ 2 g_X - 2 = n(2 g_Y - 2) + \deg R, \qquad \deg R \geq 0 , \] so \(g_X < g_Y\) is impossible. The inseparable case is not an exception but a separate mechanism: factor \(f\) through its purely inseparable part, which preserves the genus, and apply the above to the separable part. This is the role of the inseparable factorization in PR-IV2INSEP.

    Algebraic closedness enters twice. It makes every element of \(L \setminus k\) transcendental, and it turns \(g(Y) = 0\) into \(Y \cong \PP^1\) — over a general field a genus-\(0\) curve is a conic, which is \(\PP^1\) only when it has a rational point. The purely field-theoretic Lüroth theorem needs neither: over any field \(k\), every intermediate field \(k \subsetneq L \subseteq k(t)\) is \(k(u)\). The geometric proof above buys the statement back with the curve theory instead of with field theory.

    (The Lüroth problem)

    Restated as geometry, Lüroth says: a curve dominated by \(\PP^1\) is rational, that is, unirational implies rational in dimension \(1\). Asking the same in higher dimension is the Lüroth problem, and the answer splits by dimension and by characteristic.

    In dimension \(2\) over \(k = \kbar\) of characteristic \(0\), unirational implies rational. The proof runs through Castelnuovo’s criterion, that \(q=P_2=0\) forces rationality; the characteristic hypothesis enters in deducing these vanishings from unirationality. Castelnuovo’s criterion itself holds in every characteristic, by Zariski and Mumford. What fails in characteristic \(p\) is the other half: a unirational surface need not satisfy \(q = P_2 = 0\). Zariski’s surfaces, purely inseparable covers of \(\PP^2\), are unirational and not rational, so the dimension-\(2\) Lüroth statement is false as soon as \(p > 0\) even though the criterion it invokes is not.

    In dimension \(3\) it is false even over \(\CC\). The three standard counterexamples are the smooth cubic threefold, non-rational because its intermediate Jacobian is not a Jacobian of a curve; the smooth quartic threefold, non-rational because its group of birational self-maps is too small; and the Artin–Mumford double solid, non-rational because it has torsion in \(H^3\), which is a birational invariant. All three are unirational. Each counterexample is distinguished by a specific birational obstruction: the intermediate Jacobian, birational automorphism group, or torsion in \(H^3\).

    State Lüroth's theorem.
    Prove it using the genus.
    Where does the proof use that $k$ is algebraically closed?
    Is the analogue true for subfields of $k(t_1, t_2)$? Of $k(t_1,t_2,t_3)$?

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