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[parent] continuation of exponent (Definition)

Theorem. Let $ K/k$ be a finite field extension and $ \nu$ an exponent valuation of the extension field $ K$. Then there exists one and only one positive integer $ e$ such that the function

\begin{displaymath} (1) \qquad\qquad\qquad \nu_0(x)\, := \, \begin{cases} & \inf... ... & \frac{\nu(x)}{e} \;\; \mbox{when }\; x \neq 0, \end{cases}\end{displaymath}
defined in the base field $ k$, is an exponent of $ k$.

Proof. The exponent $ \nu$ of $ K$ attains in the set $ k\!\smallsetminus\!\{0\}$ also non-zero values; otherwise $ k$ would be included in $ \mathcal{O}_\nu$, the ring of the exponent $ \nu$. Since any element $ \xi$ of $ K$ are integral over $ k$, it would then be also integral over $ \mathcal{O}_\nu$, which is integrally closed in its quotient field $ K$ (see theorem 1 in ring of exponent); the situation would mean that $ \xi \in \mathcal{O}_\nu$ and thus the whole $ K$ would be contained in $ \mathcal{O}_\nu$. This is impossible, because an exponent of $ K$ attains also negative values. So we infer that $ \nu$ does not vanish in the whole $ k\!\smallsetminus\!\{0\}$. Furthermore, $ \nu$ attains in $ k\!\smallsetminus\!\{0\}$ both negative and positive values, since $ \nu(a)\!+\!\nu(a^{-1}) = \nu(aa^{-1}) = \nu(1) = 0$.

Let $ p$ be such an element of $ k$ on which $ \nu$ attains as its value the least possible positive integer $ e$ in the field $ k$ and let $ a$ be an arbitrary non-zero element of $ k$. If

$\displaystyle \nu(a) = m = qe+r \quad (q,\,r \in \mathbb{Z},\;\; 0 \leqq r < e),$
then $ \nu(ap^{-q}) = m-qe = r$, and thus $ r = 0$ on grounds of the choice of $ p$. This means that $ \nu(a)$ is always divisible by $ e$, i.e. that the values of the function $ \nu_0$ in $ k\!\smallsetminus\!\{0\}$ are integers. Because $ \nu_0(p) = 1$ and $ \nu_0(p^l) = l$, the function attains in $ k$ every integer value. Also the conditions
$\displaystyle \nu_0(ab) = \nu_0(a)+\nu_0(b), \quad \nu_0(a+b) \geqq \min\{\nu_0(a),\,\nu_0(b)\}$
are in force, whence $ \nu_0$ is an exponent of the field $ k$.

Definition. Let $ K/k$ be a finite field extension. If the exponent $ \nu_0$ of $ k$ is tied with the exponent $ \nu$ of $ K$ via the condition (1), one says that $ \nu$ induces $ \nu_0$ to $ k$ and that $ \nu$ is the continuation of $ \nu_0$ to $ K$. The positive integer $ e$, uniquely determined by (1), is the ramification index of $ \nu$ with respect to $ \nu_0$ (or with respect to the subfield $ k$).

Bibliography

1
S. BOREWICZ & I. SAFAREVIC: Zahlentheorie. Birkhäuser Verlag. Basel und Stuttgart (1966).



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Other names:  prolongation of exponent
Also defines:  induce, continuation, continuation of the exponent, ramification index of the exponent

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theorems on continuation (Theorem) by pahio
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Cross-references: subfield, ramification index, exponent of the field, divisible, field, vanish, negative, contained, mean, ring of exponent, quotient field, integrally closed, integral, ring of the exponent, base field, function, integer, positive, extension field, exponent valuation, finite field extension
There are 111 references to this entry.

This is version 3 of continuation of exponent, born on 2008-04-16, modified 2008-05-22.
Object id is 10509, canonical name is ContinuationOfExponent.
Accessed 568 times total.

Classification:
AMS MSC: 11R99 (Number theory :: Algebraic number theory: global fields :: Miscellaneous)
 12J20 (Field theory and polynomials :: Topological fields :: General valuation theory)
 13A18 (Commutative rings and algebras :: General commutative ring theory :: Valuations and their generalizations)
 13F30 (Commutative rings and algebras :: Arithmetic rings and other special rings :: Valuation rings)
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