Введение
О алгебраической независимости экспонент линейно независимых алгебраических чисел над Q
В трансцендентной теории чисел теорема Линдемана — Вейерштрасса является результатом, широко используемым для установления трансцендентности чисел. Она формулируется следующим образом:
Иными словами, поле расширения имеет степень трансцендентности n над Q.
Эквивалентная формулировка выглядит так: . Эта эквивалентность преобразует линейную зависимость над алгебраическими числами в алгебраическую зависимость над Q, используя тот факт, что симметричный многочлен, аргументами которого являются все сопряженные друг другу числа, принимает рациональное значение. Теорема названа в честь Фердинанда фон Линдемана и Карла Вейерштрасса. Линдеманн доказал в 1882 году, что e^α трансцендентно для любого ненулевого алгебраического числа α, тем самым установив трансцендентность числа e (см. ниже). Все это впоследствии было обобщено в гипотезе Шануэля.
In other words, the extension field has transcendence degree n over
An equivalent formulation , is the following: This equivalence transforms a linear relation over the algebraic numbers into an algebraic relation over by using the fact that a symmetric polynomial whose arguments are all conjugates of one another gives a rational number. The theorem is named for Ferdinand von Lindemann and Karl Weierstrass. Lindemann proved in 1882 that e^(α) is transcendental for every non zero algebraic number α, thereby establishing that is transcendental (see below). and all of these would be further generalized by Schanuel's conjecture.
Конвенция о наименовании
Теорема также известна как теорема Эрмита — Линдемана и теорема Эрмита — Линдемана — Вейерштрасса. Шарль Эрми доказал более простую теорему, в которой показатели αi должны быть рациональными целыми числами, а линейная независимость гарантируется только над областью рациональных целых чисел; этот результат иногда называют теоремой Эрмита. Хотя это может показаться частным случаем вышеуказанной теоремы, общий результат можно свести к этому более простому случаю. Линдеманн первым допустил алгебраические числа в работе Эрмита в 1882 году. Вскоре после этого Вейерштрасс получил окончательный результат, а дальнейшие упрощения были сделаны несколькими математиками, наиболее заметно — Давидом Гильбертом и Полом Горданом.
Превышение и
Трансцендентность *e* (математической константы) и *i* являются прямыми следствиями этой теоремы. Предположим, что α – ненулевое алгебраическое число; тогда {α} – линейно независимое множество над рациональными числами, и поэтому, согласно первой формулировке теоремы, {e<sup>α</sup>} – алгебраически независимое множество; или другими словами, e<sup>α</sup> – трансцендентное. В частности, *e* трансцендентно. (Более элементарное доказательство того, что *e* является трансцендентным, приведено в статье о трансцендентных числах.) В качестве альтернативы, согласно второй формулировке теоремы, если α – ненулевое алгебраическое число, то {0, α} – множество различных алгебраических чисел, и поэтому множество линейно независимо над алгебраическими числами, и в частности e<sup>α</sup> не может быть алгебраическим, а значит, оно трансцендентно. Чтобы доказать, что *i* трансцендентно, мы докажем, что оно не является алгебраическим. Если бы *i* было алгебраическим, то и -*i* было бы также алгебраическим, а затем по теореме Линдемана — Вейерштрасса (см. тождество Эйлера) *e<sup>i</sup>* было бы трансцендентным, что является противоречием. Следовательно, *i* не является алгебраическим, что означает, что оно трансцендентно. Незначительное изменение того же доказательства покажет, что если α – ненулевое алгебраическое число, то sin(α), cos(α), tan(α) и их гиперболические аналоги также являются трансцендентными.
Модульная гипотеза
Аналог теоремы, связанной с модулярной функцией j-инвариантом, был предложен Дэниелом Бертраном в 1997 году и остаётся нерешённой проблемой. Обозначая квадрат номы через q и гипотезу – как, формулировка следующая.
Доказательство
Доказательство опирается на два предварительных леммы. Заметьте, что одной лишь леммы B уже достаточно для вывода исходного утверждения теоремы Линдемана — Вейерштрасса.
Последний шаг
Теперь мы докажем теорему: пусть a(1), ..., a(n) – ненулевые алгебраические числа, а α(1), ..., α(n) – различные алгебраические числа. Тогда предположим, что:
We will show that this leads to contradiction and thus prove the theorem. The proof is very similar to that of Lemma B, except that this time the choices are made over the a(i)'s:
For every i ∈ {1, , n}, a(i) is algebraic, so it is a root of an irreducible polynomial with integer coefficients of degree d(i). Let us denote the distinct roots of this polynomial a(i)1, , a(i)d(i), with a(i)1 = a(i). Let S be the functions σ which choose one element from each of the sequences (1, , d(1)), (1, , d(2)), , (1, , d(n)), so that for every 1 ≤ i ≤ n, σ(i) is an integer between 1 and d(i). We form the polynomial in the variables
Since the product is over all the possible choice functions σ, Q is symmetric in for every i. Therefore Q is a polynomial with integer coefficients in elementary symmetric polynomials of the above variables, for every i, and in the variables yi. Each of the latter symmetric polynomials is a rational number when evaluated in
The evaluated polynomial vanishes because one of the choices is just σ(i) = 1 for all i, for which the corresponding factor vanishes according to our assumption above. Thus, the evaluated polynomial is a sum of the form
where we already grouped the terms with the same exponent. So in the left hand side we have distinct values β(1), , β(N), each of which is still algebraic (being a sum of algebraic numbers) and coefficients The sum is nontrivial: if is maximal in the lexicographic order, the coefficient of is just a product of a(i)j's (with possible repetitions), which is non zero. By multiplying the equation with an appropriate integer factor, we get an identical equation except that now b(1), , b(N) are all integers. Therefore, according to Lemma B, the equality cannot hold, and we are led to a contradiction which completes the proof. ∎
Note that Lemma A is sufficient to prove that e is irrational, since otherwise we may write e = p / q, where both p and q are non zero integers, but by Lemma A we would have qe − p ≠ 0, which is a contradiction. Lemma A also suffices to prove that is irrational, since otherwise we may write = k / n, where both k and n are integers) and then ±i are the roots of n2x2 + k2 = 0; thus 2 − 1 − 1 = 2e0 + ei + e−i ≠ 0; but this is false. Similarly, Lemma B is sufficient to prove that e is transcendental, since Lemma B says that if a0, , an are integers not all of which are zero, then
Lemma B also suffices to prove that is transcendental, since otherwise we would have 1 + ei ≠ 0.
Мы покажем, что это приводит к противоречию и, таким образом, докажем теорему. Доказательство очень похоже на доказательство леммы B, за исключением того, что на этот раз выбор делается над a(i):
We will show that this leads to contradiction and thus prove the theorem. The proof is very similar to that of Lemma B, except that this time the choices are made over the a(i)'s:
For every i ∈ {1, , n}, a(i) is algebraic, so it is a root of an irreducible polynomial with integer coefficients of degree d(i). Let us denote the distinct roots of this polynomial a(i)1, , a(i)d(i), with a(i)1 = a(i). Let S be the functions σ which choose one element from each of the sequences (1, , d(1)), (1, , d(2)), , (1, , d(n)), so that for every 1 ≤ i ≤ n, σ(i) is an integer between 1 and d(i). We form the polynomial in the variables
Since the product is over all the possible choice functions σ, Q is symmetric in for every i. Therefore Q is a polynomial with integer coefficients in elementary symmetric polynomials of the above variables, for every i, and in the variables yi. Each of the latter symmetric polynomials is a rational number when evaluated in
The evaluated polynomial vanishes because one of the choices is just σ(i) = 1 for all i, for which the corresponding factor vanishes according to our assumption above. Thus, the evaluated polynomial is a sum of the form
where we already grouped the terms with the same exponent. So in the left hand side we have distinct values β(1), , β(N), each of which is still algebraic (being a sum of algebraic numbers) and coefficients The sum is nontrivial: if is maximal in the lexicographic order, the coefficient of is just a product of a(i)j's (with possible repetitions), which is non zero. By multiplying the equation with an appropriate integer factor, we get an identical equation except that now b(1), , b(N) are all integers. Therefore, according to Lemma B, the equality cannot hold, and we are led to a contradiction which completes the proof. ∎
Note that Lemma A is sufficient to prove that e is irrational, since otherwise we may write e = p / q, where both p and q are non zero integers, but by Lemma A we would have qe − p ≠ 0, which is a contradiction. Lemma A also suffices to prove that is irrational, since otherwise we may write = k / n, where both k and n are integers) and then ±i are the roots of n2x2 + k2 = 0; thus 2 − 1 − 1 = 2e0 + ei + e−i ≠ 0; but this is false. Similarly, Lemma B is sufficient to prove that e is transcendental, since Lemma B says that if a0, , an are integers not all of which are zero, then
Lemma B also suffices to prove that is transcendental, since otherwise we would have 1 + ei ≠ 0.
Для каждого i ∈ {1, ..., n}, a(i) является алгебраическим, поэтому это корень неприводимого многочлена с целыми коэффициентами степени d(i). Обозначим различные корни этого многочлена a(i)1, ..., a(i)d(i), где a(i)1 = a(i). Пусть S – множество функций σ, которые выбирают один элемент из каждой последовательности (1, ..., d(1)), (1, ..., d(2)), ..., (1, ..., d(n)), так что для каждого 1 ≤ i ≤ n, σ(i) – целое число между 1 и d(i). Сформируем многочлен в переменных.
We will show that this leads to contradiction and thus prove the theorem. The proof is very similar to that of Lemma B, except that this time the choices are made over the a(i)'s:
For every i ∈ {1, , n}, a(i) is algebraic, so it is a root of an irreducible polynomial with integer coefficients of degree d(i). Let us denote the distinct roots of this polynomial a(i)1, , a(i)d(i), with a(i)1 = a(i). Let S be the functions σ which choose one element from each of the sequences (1, , d(1)), (1, , d(2)), , (1, , d(n)), so that for every 1 ≤ i ≤ n, σ(i) is an integer between 1 and d(i). We form the polynomial in the variables
Since the product is over all the possible choice functions σ, Q is symmetric in for every i. Therefore Q is a polynomial with integer coefficients in elementary symmetric polynomials of the above variables, for every i, and in the variables yi. Each of the latter symmetric polynomials is a rational number when evaluated in
The evaluated polynomial vanishes because one of the choices is just σ(i) = 1 for all i, for which the corresponding factor vanishes according to our assumption above. Thus, the evaluated polynomial is a sum of the form
where we already grouped the terms with the same exponent. So in the left hand side we have distinct values β(1), , β(N), each of which is still algebraic (being a sum of algebraic numbers) and coefficients The sum is nontrivial: if is maximal in the lexicographic order, the coefficient of is just a product of a(i)j's (with possible repetitions), which is non zero. By multiplying the equation with an appropriate integer factor, we get an identical equation except that now b(1), , b(N) are all integers. Therefore, according to Lemma B, the equality cannot hold, and we are led to a contradiction which completes the proof. ∎
Note that Lemma A is sufficient to prove that e is irrational, since otherwise we may write e = p / q, where both p and q are non zero integers, but by Lemma A we would have qe − p ≠ 0, which is a contradiction. Lemma A also suffices to prove that is irrational, since otherwise we may write = k / n, where both k and n are integers) and then ±i are the roots of n2x2 + k2 = 0; thus 2 − 1 − 1 = 2e0 + ei + e−i ≠ 0; but this is false. Similarly, Lemma B is sufficient to prove that e is transcendental, since Lemma B says that if a0, , an are integers not all of which are zero, then
Lemma B also suffices to prove that is transcendental, since otherwise we would have 1 + ei ≠ 0.
Так как произведение берется по всем возможным функциям выбора σ, Q симметричен относительно для каждого i. Следовательно, Q – многочлен с целыми коэффициентами в элементарных симметричных многочленах вышеуказанных переменных для каждого i, и в переменных yi. Каждый из последних симметричных многочленов является рациональным числом при вычислении в.
We will show that this leads to contradiction and thus prove the theorem. The proof is very similar to that of Lemma B, except that this time the choices are made over the a(i)'s:
For every i ∈ {1, , n}, a(i) is algebraic, so it is a root of an irreducible polynomial with integer coefficients of degree d(i). Let us denote the distinct roots of this polynomial a(i)1, , a(i)d(i), with a(i)1 = a(i). Let S be the functions σ which choose one element from each of the sequences (1, , d(1)), (1, , d(2)), , (1, , d(n)), so that for every 1 ≤ i ≤ n, σ(i) is an integer between 1 and d(i). We form the polynomial in the variables
Since the product is over all the possible choice functions σ, Q is symmetric in for every i. Therefore Q is a polynomial with integer coefficients in elementary symmetric polynomials of the above variables, for every i, and in the variables yi. Each of the latter symmetric polynomials is a rational number when evaluated in
The evaluated polynomial vanishes because one of the choices is just σ(i) = 1 for all i, for which the corresponding factor vanishes according to our assumption above. Thus, the evaluated polynomial is a sum of the form
where we already grouped the terms with the same exponent. So in the left hand side we have distinct values β(1), , β(N), each of which is still algebraic (being a sum of algebraic numbers) and coefficients The sum is nontrivial: if is maximal in the lexicographic order, the coefficient of is just a product of a(i)j's (with possible repetitions), which is non zero. By multiplying the equation with an appropriate integer factor, we get an identical equation except that now b(1), , b(N) are all integers. Therefore, according to Lemma B, the equality cannot hold, and we are led to a contradiction which completes the proof. ∎
Note that Lemma A is sufficient to prove that e is irrational, since otherwise we may write e = p / q, where both p and q are non zero integers, but by Lemma A we would have qe − p ≠ 0, which is a contradiction. Lemma A also suffices to prove that is irrational, since otherwise we may write = k / n, where both k and n are integers) and then ±i are the roots of n2x2 + k2 = 0; thus 2 − 1 − 1 = 2e0 + ei + e−i ≠ 0; but this is false. Similarly, Lemma B is sufficient to prove that e is transcendental, since Lemma B says that if a0, , an are integers not all of which are zero, then
Lemma B also suffices to prove that is transcendental, since otherwise we would have 1 + ei ≠ 0.
Вычисленный многочлен обращается в нуль, потому что один из вариантов – это σ(i) = 1 для всех i, для которого соответствующий множитель обращается в нуль в соответствии с нашим предположением выше. Таким образом, вычисленный многочлен является суммой вида,
We will show that this leads to contradiction and thus prove the theorem. The proof is very similar to that of Lemma B, except that this time the choices are made over the a(i)'s:
For every i ∈ {1, , n}, a(i) is algebraic, so it is a root of an irreducible polynomial with integer coefficients of degree d(i). Let us denote the distinct roots of this polynomial a(i)1, , a(i)d(i), with a(i)1 = a(i). Let S be the functions σ which choose one element from each of the sequences (1, , d(1)), (1, , d(2)), , (1, , d(n)), so that for every 1 ≤ i ≤ n, σ(i) is an integer between 1 and d(i). We form the polynomial in the variables
Since the product is over all the possible choice functions σ, Q is symmetric in for every i. Therefore Q is a polynomial with integer coefficients in elementary symmetric polynomials of the above variables, for every i, and in the variables yi. Each of the latter symmetric polynomials is a rational number when evaluated in
The evaluated polynomial vanishes because one of the choices is just σ(i) = 1 for all i, for which the corresponding factor vanishes according to our assumption above. Thus, the evaluated polynomial is a sum of the form
where we already grouped the terms with the same exponent. So in the left hand side we have distinct values β(1), , β(N), each of which is still algebraic (being a sum of algebraic numbers) and coefficients The sum is nontrivial: if is maximal in the lexicographic order, the coefficient of is just a product of a(i)j's (with possible repetitions), which is non zero. By multiplying the equation with an appropriate integer factor, we get an identical equation except that now b(1), , b(N) are all integers. Therefore, according to Lemma B, the equality cannot hold, and we are led to a contradiction which completes the proof. ∎
Note that Lemma A is sufficient to prove that e is irrational, since otherwise we may write e = p / q, where both p and q are non zero integers, but by Lemma A we would have qe − p ≠ 0, which is a contradiction. Lemma A also suffices to prove that is irrational, since otherwise we may write = k / n, where both k and n are integers) and then ±i are the roots of n2x2 + k2 = 0; thus 2 − 1 − 1 = 2e0 + ei + e−i ≠ 0; but this is false. Similarly, Lemma B is sufficient to prove that e is transcendental, since Lemma B says that if a0, , an are integers not all of which are zero, then
Lemma B also suffices to prove that is transcendental, since otherwise we would have 1 + ei ≠ 0.
где мы уже сгруппировали члены с одинаковой степенью. Следовательно, в левой части у нас есть различные значения β(1), ..., β(N), каждое из которых все еще алгебраическое (будучи суммой алгебраических чисел) и коэффициенты. Сумма нетривиальна: если максимальна в лексикографическом порядке, то коэффициент при – это произведение a(i)j (с возможными повторениями), которое не равно нулю. Умножив уравнение на соответствующий целочисленный множитель, мы получим тождественное уравнение, за исключением того, что теперь b(1), ..., b(N) – все целые числа. Следовательно, согласно лемме B, равенство не может выполняться, и мы приходим к противоречию, которое завершает доказательство. ∎
We will show that this leads to contradiction and thus prove the theorem. The proof is very similar to that of Lemma B, except that this time the choices are made over the a(i)'s:
For every i ∈ {1, , n}, a(i) is algebraic, so it is a root of an irreducible polynomial with integer coefficients of degree d(i). Let us denote the distinct roots of this polynomial a(i)1, , a(i)d(i), with a(i)1 = a(i). Let S be the functions σ which choose one element from each of the sequences (1, , d(1)), (1, , d(2)), , (1, , d(n)), so that for every 1 ≤ i ≤ n, σ(i) is an integer between 1 and d(i). We form the polynomial in the variables
Since the product is over all the possible choice functions σ, Q is symmetric in for every i. Therefore Q is a polynomial with integer coefficients in elementary symmetric polynomials of the above variables, for every i, and in the variables yi. Each of the latter symmetric polynomials is a rational number when evaluated in
The evaluated polynomial vanishes because one of the choices is just σ(i) = 1 for all i, for which the corresponding factor vanishes according to our assumption above. Thus, the evaluated polynomial is a sum of the form
where we already grouped the terms with the same exponent. So in the left hand side we have distinct values β(1), , β(N), each of which is still algebraic (being a sum of algebraic numbers) and coefficients The sum is nontrivial: if is maximal in the lexicographic order, the coefficient of is just a product of a(i)j's (with possible repetitions), which is non zero. By multiplying the equation with an appropriate integer factor, we get an identical equation except that now b(1), , b(N) are all integers. Therefore, according to Lemma B, the equality cannot hold, and we are led to a contradiction which completes the proof. ∎
Note that Lemma A is sufficient to prove that e is irrational, since otherwise we may write e = p / q, where both p and q are non zero integers, but by Lemma A we would have qe − p ≠ 0, which is a contradiction. Lemma A also suffices to prove that is irrational, since otherwise we may write = k / n, where both k and n are integers) and then ±i are the roots of n2x2 + k2 = 0; thus 2 − 1 − 1 = 2e0 + ei + e−i ≠ 0; but this is false. Similarly, Lemma B is sufficient to prove that e is transcendental, since Lemma B says that if a0, , an are integers not all of which are zero, then
Lemma B also suffices to prove that is transcendental, since otherwise we would have 1 + ei ≠ 0.
Обратите внимание, что леммы A достаточно, чтобы доказать, что e иррационально, поскольку в противном случае мы можем записать e = p / q, где p и q – ненулевые целые числа, но по лемме A у нас будет qe – p ≠ 0, что является противоречием. Леммы A также достаточно, чтобы доказать, что иррационально, поскольку в противном случае мы можем записать = k / n, где k и n – целые числа), а затем ±i – корни уравнения n²x² + k² = 0; таким образом, 2 − 1 − 1 = 2e⁰ + ei + e⁻i ≠ 0; но это неверно. Аналогично, леммы B достаточно, чтобы доказать, что e трансцендентно, поскольку лемма B утверждает, что если a₀, ..., aₙ – целые числа, не все из которых равны нулю, то
We will show that this leads to contradiction and thus prove the theorem. The proof is very similar to that of Lemma B, except that this time the choices are made over the a(i)'s:
For every i ∈ {1, , n}, a(i) is algebraic, so it is a root of an irreducible polynomial with integer coefficients of degree d(i). Let us denote the distinct roots of this polynomial a(i)1, , a(i)d(i), with a(i)1 = a(i). Let S be the functions σ which choose one element from each of the sequences (1, , d(1)), (1, , d(2)), , (1, , d(n)), so that for every 1 ≤ i ≤ n, σ(i) is an integer between 1 and d(i). We form the polynomial in the variables
Since the product is over all the possible choice functions σ, Q is symmetric in for every i. Therefore Q is a polynomial with integer coefficients in elementary symmetric polynomials of the above variables, for every i, and in the variables yi. Each of the latter symmetric polynomials is a rational number when evaluated in
The evaluated polynomial vanishes because one of the choices is just σ(i) = 1 for all i, for which the corresponding factor vanishes according to our assumption above. Thus, the evaluated polynomial is a sum of the form
where we already grouped the terms with the same exponent. So in the left hand side we have distinct values β(1), , β(N), each of which is still algebraic (being a sum of algebraic numbers) and coefficients The sum is nontrivial: if is maximal in the lexicographic order, the coefficient of is just a product of a(i)j's (with possible repetitions), which is non zero. By multiplying the equation with an appropriate integer factor, we get an identical equation except that now b(1), , b(N) are all integers. Therefore, according to Lemma B, the equality cannot hold, and we are led to a contradiction which completes the proof. ∎
Note that Lemma A is sufficient to prove that e is irrational, since otherwise we may write e = p / q, where both p and q are non zero integers, but by Lemma A we would have qe − p ≠ 0, which is a contradiction. Lemma A also suffices to prove that is irrational, since otherwise we may write = k / n, where both k and n are integers) and then ±i are the roots of n2x2 + k2 = 0; thus 2 − 1 − 1 = 2e0 + ei + e−i ≠ 0; but this is false. Similarly, Lemma B is sufficient to prove that e is transcendental, since Lemma B says that if a0, , an are integers not all of which are zero, then
Lemma B also suffices to prove that is transcendental, since otherwise we would have 1 + ei ≠ 0.
Леммы B также достаточно, чтобы доказать, что трансцендентно, поскольку в противном случае у нас будет 1 + ei ≠ 0.
We will show that this leads to contradiction and thus prove the theorem. The proof is very similar to that of Lemma B, except that this time the choices are made over the a(i)'s:
For every i ∈ {1, , n}, a(i) is algebraic, so it is a root of an irreducible polynomial with integer coefficients of degree d(i). Let us denote the distinct roots of this polynomial a(i)1, , a(i)d(i), with a(i)1 = a(i). Let S be the functions σ which choose one element from each of the sequences (1, , d(1)), (1, , d(2)), , (1, , d(n)), so that for every 1 ≤ i ≤ n, σ(i) is an integer between 1 and d(i). We form the polynomial in the variables
Since the product is over all the possible choice functions σ, Q is symmetric in for every i. Therefore Q is a polynomial with integer coefficients in elementary symmetric polynomials of the above variables, for every i, and in the variables yi. Each of the latter symmetric polynomials is a rational number when evaluated in
The evaluated polynomial vanishes because one of the choices is just σ(i) = 1 for all i, for which the corresponding factor vanishes according to our assumption above. Thus, the evaluated polynomial is a sum of the form
where we already grouped the terms with the same exponent. So in the left hand side we have distinct values β(1), , β(N), each of which is still algebraic (being a sum of algebraic numbers) and coefficients The sum is nontrivial: if is maximal in the lexicographic order, the coefficient of is just a product of a(i)j's (with possible repetitions), which is non zero. By multiplying the equation with an appropriate integer factor, we get an identical equation except that now b(1), , b(N) are all integers. Therefore, according to Lemma B, the equality cannot hold, and we are led to a contradiction which completes the proof. ∎
Note that Lemma A is sufficient to prove that e is irrational, since otherwise we may write e = p / q, where both p and q are non zero integers, but by Lemma A we would have qe − p ≠ 0, which is a contradiction. Lemma A also suffices to prove that is irrational, since otherwise we may write = k / n, where both k and n are integers) and then ±i are the roots of n2x2 + k2 = 0; thus 2 − 1 − 1 = 2e0 + ei + e−i ≠ 0; but this is false. Similarly, Lemma B is sufficient to prove that e is transcendental, since Lemma B says that if a0, , an are integers not all of which are zero, then
Lemma B also suffices to prove that is transcendental, since otherwise we would have 1 + ei ≠ 0.
Соответствующий результат
Также известен вариант теоремы Линдемана — Вейерштрасса, в котором алгебраические числа заменяются трансцендентными числами Лиувиля (или, в общем случае, U-числами).