Local Fields, Hensel's First Lemma

Hensel's First Lemma

Hensel (biography) developed three lemmas that build upon each other, just as Sylow is known for his three theorems. Of course the sylow theorems are theorems, not lemmas. The distinction is somewhat subjective. The sylow theorems can be used directly, to prove things about finite groups, while hensel's lemmas are used to prove other theorems, which are then used to prove results in algebraic number theory. Apparently Hensel's results are one step back, so they are called lemmas. No matter - a rose by any other name would smell as sweet. And these lemmas are indeed lovely.

Let R be a commutative ring, with an ideal H, such that H² = 0. The idempotents in R correspond 1-1 with the idempotents in R/H.

Clearly idempotents in R are idempotent in R/H. Suppose e and f are idempotents in R that map to the same idempotent in R/H. In other words, f = e+b for some b in H. Write (e+b)² = e+b. Since b² = 0, we have b*(1-2e) = e²-e. The right side is 0, so if 2e-1 is a unit, then b is also 0. This forces e = f, and the map from idempotents of R into the idempotents of R/H is injective. We only need show 2e-1 is a unit.

Let e ∈ R be idempotent in R/H. Thus e²-e = v, for v in H. Consider the element 2e-1, and square it to get 4e²-4e+1, or 4v+1. Since v is in H, v² = 0, and (4v+1)*(4v-1) = -1. Thus 2e-1 is a unit in R. Idempotents upstairs map uniquely into idempotents downstairs.

To show the map is surjective, let e be idempotent in R/H, so that e²-e = v, for some v in H. Set b = -v/(2e-1).

(e+b)² =
e² - 2ev/(2e-1) + v²/(2e-1)² =
e² - 2ev/(2e-1) { v² is 0 }
e+v - 2ev/(2e-1) =
e - v/(2e-1) =
e + b

That completes the correspondence.

Filtered Ring

Let the powers of an ideal H descend within a ring R. The ring R/Hⁿ is a quotient ring of R/Hⁿ⁺¹, via reduction mod Hⁿ. Note that Hⁿ squared becomes 0 in the quotient ring. Thus the idempotents of R/Hⁿ correspond to the idempotents of R/Hⁿ⁺¹.

By induction, the idempotents of R/Hⁿ, for each n, correspond to the idempotents of R/H.

Let S be the completion of R. If R is a dvr, R and S are both integral domains, with no idempotents, and there isn't much to talk about. But R could be any ring, whence its completion is the inverse limit, as described in the previous section.

Let e be an idempotent of S, which maps to a consistent set of idempotents in the quotient rings R/Hⁿ, terminating in an idempotent e′ in R/H. Let f be another idempotent, different from e. Thus f differs in the n^th entry, for some n. This becomes a different idempotent in R/Hⁿ, which leads to a different idempotent f′ in R/H. The idempotents of S map 1-1 into the idempotents of R/H.

To complete the correspondence, start with an idempotent e₁ in R/H. Pull this back to an idempotent e₂ in R/H². By induction, find an idempotent e_n in R/Hⁿ. This defines an entry e in S. Multiply e*e in S, and you multiply e_n*e_n in each ring R/Hⁿ. The result is always e_n, hence e*e = e, and e is idempotent. The correspondence is complete.

If R/H has no idempotents, e.g. when H is prime, then S has no idempotents either.