Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 24 / 4

Put $f=[k_L:k]$. For an unramified extension the ramification index is one and $[L:K]=f$. The residue $\bar{\alpha }$ has degree $f$ over $k$. Thus $1,\alpha ,\dots ,{\alpha }^{f-1}$ are linearly independent over $K$: a nonzero linear dependence can be divided by a coefficient of least valuation and reduced to a nonzero relation among the reduced powers. Hence $[K(\alpha ):K]\ge f=[L:K]$, and $\alpha$ generates $L$.

The same argument gives the integral statement, which is stronger than merely generating the field. WriteIf $m={\min }_j{v}_K(c_j)$, divide by ${\pi }_K^m$. Since ${\pi }_K$ remains a uniformizer in the unramified extension and the reduced powers form a basis of $k_L/k$, the resulting sum has nonzero residue. Therefore $v_L(x)=m$. In particular $x$ is integral if and only if every coefficient is integral. This proves unramified residue generator integral basis:

For the finite separable field extension define the inverse different, or codifferent, byIt is a fractional ideal of ${\mathcal{O}}_L$. The different ideal is its inverse fractional ideal.

Suppose the integer ring is generated by $\alpha$, with monic minimal polynomial of an algebraic element $g$ of degree $d$. We prove the derivative formula using the coefficient pairing for a monogenic algebra. If ${\alpha }_1,\dots ,{\alpha }_d$ are its distinct conjugate polynomial roots, Lagrange interpolation for a polynomial $h$ of degree less than $d$ givesComparing coefficients of $T^{d-1}$ yieldsFor any integral polynomial, first reduce modulo $g$; the coefficients stay integral because $g$ is monic over ${\mathcal{O}}_K$.

It follows that $g^{\prime }(\alpha {)}^{-1}{\mathcal{O}}_L$ lies in the codifferent. To prove equality, consider the matrixAll entries are integral. Entries with $i+j<d-1$ are zero, and those with $i+j=d-1$ are one. Reversing the column order makes the matrix triangular with diagonal ones; thus its determinant is a unit of ${\mathcal{O}}_K$.

Write an arbitrary $x\in L$ as $x=g^{\prime }(\alpha {)}^{-1}{\sum }_j{b}_j{\alpha }^j$ with $b_j\in K$. Membership in the codifferent says $A(b_j{)}_j\in {\mathcal{O}}_K^d$. Since $A$ is invertible over ${\mathcal{O}}_K$, this forces every $b_j$ integral. ThereforeSeparability ensures that $g^{\prime }(\alpha )\ne 0$; the trace pairing calculation also proves that no extra factor has been overlooked.

For $L={\mathbb{Q}}_p({\zeta }_p)$, put $\pi ={\zeta }_p-1$. The polynomialis Eisenstein at $p$. It gives $[L:{\mathbb{Q}}_p]=e=p-1$, total ramification, and $v_L(\pi )=1$. Also ${\mathcal{O}}_L={\mathbb{Z}}_p[\pi ]$: in the power basis, terms $c_i{\pi }^i$ have valuations $(p-1)v_p(c_i)+i$ in different residue classes modulo $p-1$, so cannot cancel at the minimum. Integrality forces all $c_i$ integral.

Use part (b) with ${\Phi }_p(T)$ and differentiate the quotient at ${\zeta }_p$:The root of unity is a unit, while $v_L(p)=p-1$ and $v_L(\pi )=1$. Hence the cyclotomic local different exponent isThis includes $p=2$: that extension is trivial and the exponent is zero.

For $m=p^2-1$, construct an unramified quadratic extension $E/{\mathbb{Q}}_p$ by lifting an irreducible quadratic polynomial over ${\mathbb{F}}_p$. Its residue field is ${\mathbb{F}}_{p^2}$. Each of its nonzero residue elements is a simple polynomial root of $T^m-1$, so Hensel's lemma lifts all $m$ polynomial roots into $E$. A primitive residue element lifts to a primitive mth polynomial root: its reduction already has order $m$, and its mth power is one.

Such a primitive polynomial root cannot generate a proper subfield of $E$, since its residue has order $p^2-1>p-1$ and therefore generates the quadratic residue extension. Thus ${\mathbb{Q}}_p({\zeta }_m)=E$ is unramified of degree two. Part (a) gives ${\mathcal{O}}_E={\mathbb{Z}}_p[{\zeta }_m]$. The reduced minimal polynomial of an algebraic element has degree two and is separable, so its derivative at ${\zeta }_m$ is a unit. Part (b) givesBoth answers use normalized discrete valuations on the top field.