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[parent] Kummer's lemma (Theorem)

The following result is a key ingredient in the proof of Fermat's last theorem for regular primes. More concretely, the lemma is needed to show the so-called second case of Fermat, i.e. $x^p+y^p=z^p$ does not have any non-trivial solutions in $\Ints$ with $p>2$ a regular prime and $p|xyz$ . It is due to Ernst Kummer, thus the name.

Theorem 1 (Kummer's Lemma)   Let $p>2$ be a prime, let $\zeta_p$ be a primitive $p$ th root of unity and let $K=\Rats(\zeta_p)$ be the corresponding cyclotomic field. Let $E$ be the group of algebraic units of the ring of integers $\mathcal{O}_K$ . Suppose that $p$ is a regular prime. If a unit $\epsilon\in E$ is congruent modulo $p$ to a rational integer, then $\epsilon$ is the $p$ th power of another unit also $E$ .

For a proof, see [Washington], Theorem 5.36. The reader may also be interested in generalizations due to [Washington 1992] and [Ozaki 1997].

Bibliography

Ozaki 1997
Ozaki, M., Kummer's lemma for $\Ints_p$ -extensions over totally real number fields, Acta Arith. 81 (1997), no. 1, 37-44.
Washington
Washington L. C., Introduction to Cyclotomic Fields, Second Edition, Springer-Verlag, New York.
Washington 1992
Washington, L. C., Kummer's lemma for prime power cyclotomic fields, J. Number Theory 40 (1992), no. 2, 165-173.




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Cross-references: theorem, rational integer, congruent, unit, ring of integers, algebraic units, group, cyclotomic field, root of unity, primitive, prime, solutions, regular primes, Fermat's last theorem, proof, key

This is version 3 of Kummer's lemma, born on 2006-05-26, modified 2006-09-26.
Object id is 7929, canonical name is KummersLemma.
Accessed 1464 times total.

Classification:
AMS MSC11D41 (Number theory :: Diophantine equations :: Higher degree equations; Fermat's equation)
 14H52 (Algebraic geometry :: Curves :: Elliptic curves)
 11F80 (Number theory :: Discontinuous groups and automorphic forms :: Galois representations)

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algebraic unit by pahio on 2006-05-27 04:20:09
Hi, what means the term "algebraic unit"? I think it has not been defined in PM.

Jussi
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