Reconstruction from Milnor K-theory modulo the characteristic

Proof. Choose a transcendence basis t_1,\ldots,t_d over \kappa, and write F_0=\kappa(t_1,\ldots,t_d). Perfectness of \kappa gives F_0^p=\kappa(t_1^p,\ldots,t_d^p). The monomials t_1^{i_1}\cdots t_d^{i_d} with 0\leq i_j<p form a basis of F_0 over F_0^p: group polynomial exponents modulo p, and for rational functions use a/b=ab^{p-1}/b^p. Linear independence follows by clearing denominators and separating the exponent classes. Thus [F_0:F_0^p]=p^d.

The extension F/F_0 is finite, and Frobenius induces an isomorphism of extensions F/F_0\simeq F^p/F_0^p. Computing [F:F_0^p] in the two towers through F_0 and F^p gives

Cancellation proves the assertion without a separability assumption.□

Proof. Since s^p\in C, the minimal polynomial of s over M divides T^p-s^p=(T-s)^p. If its degree were 1\leq e<p, it would be (T-s)^e, and its coefficient -es of T^{e-1} would force s\in M. Thus every strict adjunction has degree p. When [F:C] is finite, successive adjunctions must terminate, so they extend any p-independent list to a generating list. If its length is e, the tower formula gives [F:C]=p^e, and the usual basis of each simple extension gives the asserted monomial basis.

Define the D_j on these monomials and extend C-linearly. When two monomials are multiplied, reducing an exponent by p only removes a factor in C and leaves its value in \Fp unchanged. This proves the product rule. The operators commute because they are diagonal on the same basis. On C(s_j) their eigenvector assertions follow from the distinct eigenvalues 0,1,\ldots,p-1 on 1,s_j,\ldots,s_j^{p-1}. Finally, the expansions of an element in both C(s_j) and C(s_k) can have only their constant term in common.□

Proof. Suppose instead that s\notin C and t\notin C(s). By Lemma 2.2, the pair s,t is p-independent. Consider

Each logarithmic derivative is additive on products and vanishes on C^\times. Hence \beta is a well-defined \Fp-bilinear map V_F\times V_F\to F. For f\ne0,1, the identities D_s(1-f)=-D_s f and D_t(1-f)=-D_t f give \beta([f],[1-f])=0. Thus \beta descends through the Steinberg relations to an \Fp-linear map W_F\to F. But D_s s=s, D_t t=t, and D_s t=D_t s=0, so this map sends \{s,t\}_F to 1. The symbol cannot vanish.□

Proof. Put q=-dt/c. The hypotheses give q\ne0,1, while [q]=[t] and [1-q]=[c+dt], since the omitted factors lie in C^\times. Therefore the identity is the Steinberg relation \{q,1-q\}_F=0.□

Proof. Choose Z in the intersection (3.1). Extend X,Y to a p-basis and let D_X,D_Y be the commuting Euler derivations of Lemma 2.2. We first translate the two descriptions of Z into differential equations. Since U and V are not scalar multiples of monomials in X and Y, respectively, the eigenvector description in that lemma gives

Define

Here A_X\in C(X)\setminus C and A_Y\in C(Y)\setminus C: if, for example, B_X\in C, then D_XU=B_X^{-1}U, making U an eigenvector and hence a scalar multiple of a monomial in X. In particular, A_X,A_Y,B_X,B_Y are all nonzero.

The derivations

satisfy \delta X=X, \delta Y=Y, \delta'U=U, and \delta'V=V. Thus \delta kills Y/X, whereas \delta' kills V/U; both kill C. The two expressions for Z therefore give

Set

Dividing the preceding equations by Z and subtracting yields

Both q_X and q_Y are nonzero: if one vanished, these two equations would force the other A to vanish.

Since \delta commutes with D_X and D_Y, the equation \delta Z=Z implies that q_X,q_Y are killed by \delta:

By (3.3), A_X/A_Y=-q_Y/q_X, so \delta(A_X/A_Y)=0. Since D_YA_X=D_XA_Y=0, we obtain

The left side lies in C(X) and the right side in C(Y). Their intersection is C, by the p-monomial basis. Consequently A_X and A_Y are eigenvectors of their respective Euler derivations with the same eigenvalue. The distinct eigenvalues on 1,X,\ldots,X^{p-1}, and on 1,Y,\ldots,Y^{p-1}, now give

for c_X,c_Y\in C^\times and one integer m, 1\le m<p. The eigenvalue cannot be zero because A_X,A_Y\notin C.

It remains to integrate (3.5) inside the one-variable subfields. Put n=p-m. The identity 1+c_XX^m=B_X\ne0 ensures that all terms in

are defined. Thus D_X kills U^n/(X^n+c_XX^p)\in C(X)^\times. Its kernel on C(X) is C, so this quotient belongs to C^\times. Since X^p\in C^\times, we have U^n\in C+CX^n with both coefficients nonzero. Replacing X,U,c_X,D_X by Y,V,c_Y,D_Y proves the second inclusion.

Finally, suppose U^r=h+jX^r with h,j\in C and 1\le r<p. Both coefficients must be nonzero: otherwise r[U]\in\Fp[X], contrary to [U]\notin\Fp[X]. Taking logarithmic derivatives gives

The last expression is a nonzero scalar multiple of X^{p-r}, since X^p\in C^\times. Comparison with (3.5) and linear independence of the p monomials forces p-r=m, hence r=n. The argument for the inclusion involving V is identical.□

Proof. Represent r by an integer 1\le r<p and take x\in F\setminus F^p. The points [1],[x],[x+1] lie on one line. Their images under multiplication by r are [1],[x^r],[(x+1)^r]. The first two are distinct, so the asserted line containment implies

If 1<r<p, the binomial expansion has nonzero coefficient r at x^{r-1}. This contradicts the linear independence of 1,x,\ldots,x^{p-1} over F^p from Lemma 2.2.□

Proof. Put H=K^p and C=L^p. We first find one scalar for which lines map into lines. Only after this construction will we apply it to the inverse.

A connected graph of independent image pairs. On the nonzero elements of V_K, join [x] and [y] by an edge when

This condition is independent of the chosen representatives. Since X,Y\notin C, the fields C(X) and C(Y) have degree p over C. An edge therefore means that X,Y are p-independent. The graph is complete between distinct parts of the partition by the fields C(X). It has at least two parts: by Lemma 2.1, [L:C]\ge p^2, so for any X\notin C there is an element of L\setminus C(X), and \Theta is surjective. Thus every vertex has a neighbor, and any two vertices are joined by a path of length at most two.

An affine configuration along an edge. Fix an edge [x],[y]. The elements 1,x,y are linearly independent over H. Indeed, a dependence would give y=c+dx with c,d\in H and d\ne0, since x,y\notin H. If c=0, the two vertices would coincide. Otherwise Lemma 2.4 gives \{x,y\}_K=0, and compatibility followed by Lemma 2.3 gives Y\in C(X), again contradicting the edge condition.

For arbitrary a,b\in H^\times, set

These are nonzero: u,v are nonzero because x,y\notin H, and z\ne0 because x,y are H-linearly independent. Moreover,

For example, [u]=i[x] with 0\le i<p would give x+a=h x^i for some h\in H^\times, contrary to the linear independence of 1,x,\ldots,x^{p-1} and the condition a\ne0. Figure 1 shows the incidence that explains the two expressions for z.

Lemma 2.4, applied also to z/x=b-a(y/x) and z/u=b-a(v/u), gives

Write [U]=\Theta([u]), [V]=\Theta([v]), and [Z]=\Theta([z]). Compatibility and Lemma 2.3 applied to the first two symbols show that

Linearity and injectivity transport (4.2) to [U]\notin\Fp[X] and [V]\notin\Fp[Y]. In particular U,V\notin C. Also Y/X\notin C, since X,Y are p-independent, and V/U\notin C: otherwise U would belong to C(X)\cap C(Y)=C. We may therefore apply Lemma 2.3 to the last two symbols in (4.3). The result is

All hypotheses of Proposition 3.1 now hold. It gives a common integer 1\le n<p such that

and either inclusion uniquely determines n.

Propagation of the scalar. Fix representatives for all vertices and their images. For a vertex [x], fix one neighbor [y] and one b\in H^\times. As a\in H^\times varies, Proposition 3.1 applies to (4.1). The second inclusion in (4.4), whose pair Y,V stays fixed, forces the same n for every a. To compare choices of neighbor and b, specialize to a=1, keeping X and a representative U of \Theta([x+1]) fixed. Uniqueness for the first inclusion shows that these choices cannot change n. Thus each vertex has a well-defined scalar. Adjacent vertices have the same scalar by (4.4). Connectedness consequently supplies a single n for all vertices and all a\in H^\times.

Set \Psi=n\Theta, with n viewed in \Fp^\times. The line joining [1] and [x] consists of those two points and [x+a] for a\in H^\times. The first inclusion in (4.4) says exactly that its image is contained in the line joining [1] and [X^n]. Every projective line is a multiplicative translate of one through [1]: multiply the line joining [s],[t] by s^{-1}. Multiplication by a nonzero field element is linear over the field of p-th powers, and \Psi respects the multiplicative group law. Thus \Psi maps every line into a line.

Equality of lines and uniqueness. Scalar multiplication preserves compatibility, since it scales the tensor square by the nonzero scalar n^2. Hence \Psi^{-1} is compatible. Apply the construction just completed, with the fields reversed, to obtain r\in\Fp^\times for which r\Psi^{-1} maps lines into lines. Its composition with \Psi is multiplication by r on V_K and has the same line-containment property. Lemma 4.1 forces r=1. Thus both \Psi and its inverse map lines into lines. If \Psi(\ell)\subseteq\ell', choose a line \ell'' containing \Psi^{-1}(\ell'). Then

Two projective lines cannot properly contain one another, so \ell=\ell'' and both inclusions are equalities. Applying \Psi gives \Psi(\ell)=\ell'.

Finally, if m\Theta also sends lines onto lines, then (m\Theta)\circ\Psi^{-1} is multiplication by mn^{-1} on V_L and preserves lines. Another application of Lemma 4.1 gives m=n.□

Proof. The inverse also maps lines onto lines: a target line is the image of the line joining the preimages of any two distinct points on it. Projective spans are obtained by repeatedly joining pairs of points by lines, so \psi and its inverse preserve spans. Minimal spanning sets therefore correspond, and \dim_H E=\dim_C E'=N+1 for some N\ge2. Choose a basis e_0,\ldots,e_N of E and representatives e'_0,\ldots,e'_N of their images. They form a basis of E'. Keeping e'_0 fixed, scale each e'_i, i\ge1, so that

This is possible because [e_0+e_i] lies on the line joining [e_0] to [e_i] and differs from both endpoints.

The hyperplanes spanned by e_1,\ldots,e_N and by e'_1,\ldots,e'_N correspond. On their affine complements use the coordinates

The induced affine bijection preserves parallelism: parallel lines have the same point on the corresponding hyperplane at infinity. On the ith coordinate axis it is a bijection \sigma_i:H\to C fixing 0 and 1. Varying only the jth input coordinate moves along a line with point at infinity [e_j], whose image has point at infinity [e'_j]. Hence all other output coordinates are unchanged. It follows that the affine map is

In each plane of two coordinate axes, the diagonal through (0,0) and (1,1) maps to the corresponding diagonal. Thus all \sigma_i equal one bijection \sigma:H\to C. In the first two coordinates, the line (t,t+a) is parallel to this diagonal and passes through (0,a). Its image therefore gives

The line (t,at) passes through (0,0) and (1,a), so its image gives

Since \sigma(1)=1, it is a field isomorphism.

Define S(\sum_{i=0}^N t_i e_i)=\sum_{i=0}^N\sigma(t_i)e'_i. This is an additive, \sigma-semilinear bijection inducing \psi on the affine chart. Every point at infinity is the direction of an affine line through the origin, so it induces \psi there as well.

For uniqueness, suppose that T is another semilinear lift, possibly with a different field isomorphism. For every nonzero v\in E write T(v)=c_vS(v) with c_v\in C^\times. If v,w are independent, additivity applied to v+w and independence of S(v),S(w) give c_v=c_w=c_{v+w}. Two dependent nonzero vectors can each be compared to a vector outside their common line. Thus c_v is a single scalar c, and T=cS. Comparing the images of av then shows that the two field isomorphisms also agree.□

Proof. Perfectness gives the inclusion from right to left. If f\in F\setminus\kappa, algebraic closedness makes f transcendental over \kappa. Complete f to a transcendence basis f,t_2,\ldots,t_d, and put B=\kappa(t_2,\ldots,t_d) and E=B(f). The extension F/E is finite; write D=[F:E]. For every m\ge1, the polynomial T^{p^m}-f is Eisenstein at the prime f of B[f], and is therefore irreducible over E. If f=g^{p^m} with g\in F, it follows that

This fails for all sufficiently large m, so f does not belong to the intersection.□

Proof of Theorem 1.1. A field isomorphism \alpha:K\to L sends K^p to L^p and Steinberg pairs to Steinberg pairs. It therefore induces an element of \Isom_{\mathrm M}(V_K,V_L), and the canonical map in the theorem is well defined.

For surjectivity, take \Theta\in\Isom_{\mathrm M}(V_K,V_L). Proposition 4.2 gives n\in\Fp^\times such that \Psi=n\Theta maps lines onto lines. Put H=K^p and C=L^p. By Lemma 2.1, \dim_H K and \dim_C L are at least p^2\ge4, so Proposition 5.1 gives a semilinear lift S:K\to L. Since \Psi([1])=[1], scale S to arrange S(1)=1. For each f\in K^\times, the two semilinear bijections

induce the same projective map: the equality is precisely \Psi([fw])=\Psi([f])+\Psi([w]) in the additive notation for V_L. Their values at 1 agree, so the uniqueness assertion in Proposition 5.1 makes them equal. Thus S is multiplicative, and hence a field isomorphism. It maps every K^{p^m} onto L^{p^m}, so Lemma 5.2 gives S(k)=l. Its induced map is n\Theta, proving surjectivity onto the scalar quotient.

For injectivity, suppose that field isomorphisms \alpha,\beta:K\to L induce maps with \beta_*=r\alpha_* for r\in\Fp^\times. Both preserve projective lines, so multiplication by r on V_L also preserves lines. Lemma 4.1 gives r=1. The maps \alpha and \beta are therefore semilinear lifts of the same projective bijection. Proposition 5.1 makes them scalar multiples, and \alpha(1)=\beta(1)=1 makes them equal.□