アルティン・ハッセの指数関数
The coefficients of "E"("x") are rational. We can use either formula for "E"("x") to prove that, unlike "e", all of its coefficients are "p"-integral; in other words, the denominators of the coefficients of "E"("x") are not divisible by "p". A first proof uses the definition of "E"("x") and Dwork's lemma, which says that a power series "f"("x") = 1 + ... with rational coefficients has "p"-integral coefficients if and only if "f"("x")/"f"("x") ≡ 1 mod "pZ<nowiki></nowiki>"x"<nowiki></nowiki>. When "f"("x") = "E"("x"), we have "f"("x")/"f"("x") = "e", whose constant term is 1 and all higher coefficients are in "pZ. A second proof comes from the infinite product for "E"("x"): each exponent -μ("n")/"n" for "n" not divisible by "p" is a "p"-integral, and when a rational number "a" is "p"-integral all coefficients in the binomial expansion of (1 - "x") are "p"-integral by "p"-adic continuity of the binomial coefficient polynomials "t"("t"-1)...("t"-"k"+1)/"k"! in "t" together with their obvious integrality when "t" is a nonnegative integer ("a" is a "p"-adic limit of nonnegative integers) . Thus each factor in the product of "E"("x") has "p"-integral coefficients, so "E"("x") itself has "p"-integral coefficients.