- Let X be an algebraic variety over a field K. Find all finite field extensions L of K in which X has a rational point.
- Let X be an algebraic variety over a field K. Suppose L is an extension field of K such that the set of L-rational points of X, denoted X(L), is nonempty. Assume furthermore that L is minimal with respect to this property. Thus if F (not equal to L) is an intermediate field contained between K and L then X(F) is empty. Then L is called a minimal splitting field for X. Find all minimal splitting fields for a given variety X/K.
- Fix a number field K and a positive integer m. Does
there exist a genus one curve over K whose minimal splitting
field is of degree m over K? This is question was partially
resolved in [Lang-Tate, 1958].
- Does period=index for H^1 of an abelian variety over a number
field? ANSWER: NO. Cassels found a counterexample in the 60's.
- Let E be a genus one curve over a field K. Let L_1 and L_2 be extensions of K in which E has a rational point. Using the Riemann-Roch theorem one sees that E has a point in an extension of degree d=gcd([L_1:K],[L_2:K]). Now suppose L is a Galois extension of K which contains both L_1 and L_2. Does there exist a subfield of L of degree d over which E has a rational point? [I doubt it but I haven't quite put together a counterexample yet. It could be true though.]

If X is the curve x^2+y^2=-1 and K=Q the field of rational numbers then the problem is to characterize all number fields L in which x^2+y^2=-1 has a solution.

The problem can be rephrased in terms of finding so-called minimal splitting fields.

- (Communicated by H.W. Lenstra, proposed by June) Let R be a finite ring (i.e. R has finitely many elements) and fix finite R-modules M and N. Let M^n denote the direct sum of M with itself n times. Suppose that there is some positive integer n along with an R-module injection M^n ---> N^n. Question: Does it follow that there exists an R-module injection from M into N? The answer is well known in the special case when M^n is isomorphic to N^n. In this case the Krull-Schmidt theorem implies that the indecomposable factors of M and N are the same. Since R is finite this is enough to show that M is isomorphic to N. See Lang's Algebra.