Coleman norm contraction 2026-10-14
For odd , the Coleman norm operator satisfies the displayed inclusion for every . Write for multiplication by on the finite free module over . Its trace is , where is the Coleman trace operator. In the linear term is therefore divisible by , and all higher terms are divisible by . Congruence reflection for Frobenius substitution gives the assertion. A norm-fixed element of lies in for every and hence equals .
Past exam of the mathematics course of the University of Cambridge 2005 iii Paper 30 2 Solution Created 2026-10-03 Updated 2026-10-14
Let be the matrix of multiplication by in the basis from Question 1. Define the Coleman norm operator by the norm for a finite free ring extension:The entries lie in , and is an isomorphism, so this determines uniquely. If is a unit, is invertible and its determinant is a unit. Also , giving . Splitting the conjugates, or diagonalizing multiplication after adjoining the roots of unity and inverting , gives the equivalent formulaIn particular, this definition has the required target , rather than merely a ring with extra roots of unity in its coefficients.
For the requested ring congruence, observe that the reduction of is the injective map . If , its reduction first gives . Then ; repeat to obtain . Applying this to provesThis congruence reflection for Frobenius substitution is valid for all , without assuming is a unit.
A further property needed for the fixed points is Coleman norm contraction. If with , expand the determinant:Here is the coefficient of degree in . The trace formula from Question 1 makes the linear term divisible by ; every later term is divisible by and . Reflecting the resulting ring congruence through givesIf and , iteration gives for every . Since , .
To identify all fixed points, we also need for every unit . In characteristic , the extension has degree and its norm for a finite free ring extension is . Indeed, after passing to fraction fields it is a purely inseparable field extension of degree , so multiplication has only the eigenvalue , with multiplicity , after splitting. Its determinant is . Thereforeand congruence reflection for Frobenius substitution gives .
Now choose any lift of a given and put . Multiplicativity givesThus is a Cauchy sequence for the finer of the topologies on integral formal power series and converges to a unit reducing to . The Coleman norm operator is continuous: multiplication matrices and their determinants are continuous, and the finite-free coordinate decomposition and its inverse preserve the -adic filtration defining the finer of the topologies on integral formal power series. Hence . If two fixed units reduce to , their quotient is fixed and belongs to , so it is by the preceding argument. This proves uniqueness, independence of the initial lift, and multiplicativity of the Coleman norm-fixed lift. In particular,Every unit consequently has a unique factorization with , giving the canonical Coleman decomposition of power-series units:The final assertion as printed is false. The original PDF identifies a ring with a product of unit groups. With multiplication, has a zero and is not a group. With addition, it has exponent , whereas has a nonidentity element of order : because is odd and . Thus neither interpretation makes the printed assertion true. The two boxed isomorphisms above are the valid natural conclusions; they do not require guessing which symbols the author intended to replace.