Chapter 4 Transcendence Results

Chapter 4

Transcendence results

We recall some basic deﬁnitions.

We call α ∈ C transcendental if it is not algebraic, i.e., if it is not a zero of a non-zero polynomial from Q[X].

We call numbers α1, . . . , αn ∈ C algebraically independent if they are algebraically independent over Q, i.e., if there is no non-zero polynomial P ∈ Q[X1,...,Xn] such that P (α1, . . . , αn) = 0. According to Proposition 3.8, this is equivalent to α1, . . . , αn being algebraically independent over Q. Given a subset S of C, we deﬁne the transcendence degree of S, notation trdeg S, to be the maximal number t such that S contains t algebraically independent elements.

4.1 The transcendence of e.

z P∞ n We deﬁne as usual e = n=0 z /n! for z ∈ C. Further, Q = {α ∈ C : α algebraic over Q}. Theorem 4.1 (Hermite, 1873). e is transcendental.

We assume that e is algebraic. This means that there are q0, q1, . . . , qn ∈ Z with

n (4.1) q0 + q1e + ··· + qne = 0, q0 6= 0.

Under this hypothesis, we construct M ∈ Z with M 6= 0 and |M| < 1 and obtain

45 a contradiction. We need some auxiliary results. Of course we have to use certain properties of e. We use that (ez)0 = ez.

Let f ∈ C[X] be a polynomial. For z ∈ C we deﬁne Z z (4.2) F (z) := ez−uf(u)du. 0 Here the integration is over the line segment from 0 to z. We may parametrize this line segment by u = tz, 0 6 t 6 1. Thus, Z 1 F (z) = ez(1−t)f(zt)zdt. 0 Lemma 4.2. Suppose f has degree m. Then

m ! m X X F (z) = ez f (j)(0) − f (j)(z). j=0 j=0

Proof. Repeated integration by parts.

Corollary 4.3. Let f be as in Lemma 4.2. Then

n m X X (j) q0F (0) + ··· + qnF (n) = − qaf (a). a=0 j=0

Proof. Clear.

Our aim is to show that for a suitable choice of f, the quantity M := q0F (0) + ··· + qnF (n) is a non-zero integer with |M| < 1. Note that Corollary 4.3 gives an identity with an analytic expression on the left-hand side, and an algebraic expression on the right-hand side. We prove that M is a non-zero integer by analyzing the right-hand side, and |M| < 1 by analyzing the left-hand side. For the latter, we need the following simple estimate.

Lemma 4.4. Let f ∈ C[X] be any polynomial and let F be given by (4.2). Then For z ∈ C we have |z| |F (z)| 6 |z| · e · sup |f(u)|. u∈C, |u|6|z|

46 Proof. We have Z 1 Z 1 z(1−t) |z| |F (z)| 6 |e f(zt)z|dt 6 e |z| · |f(zt)|dt 0 0 |z| 6 |z| · e · sup |f(u)|. u∈C, |u|6|z|

Let p be a prime number, which is chosen later to be suﬃciently large to make all estimates work. We take

1 p−1 p (4.3) f(X) := (p−1)! · X {(X − 1)(X − 2) ··· (X − n)} . In this case,

n n np+p−1 X X X (j) (4.4) M = qaF (a) = − qaf (a). a=0 a=0 j=0 Lemma 4.5. We have

f (p−1)(0) = {(−1)nn!}p ;(4.5) f (j)(a) = 0 for a = 0, . . . , n, j = 0, . . . , p − 1, (a, j) 6= (0, p − 1);(4.6) (j) (4.7) f (a) ≡ 0 (mod p) for a = 0, . . . , n, j > p.

Proof. In general, if g is a polynomial of the shape (X − a)rh with a ∈ C, h ∈ C[X], then g(m)(a) = 0 for m = 0, . . . , r − 1 and g(r)(a) = r!h(a). This implies (4.5), (4.6). r To prove (4.7), observe that for any g = crX + ··· + c0 ∈ C[X] and all j > 0 we have

1 (j) r r−j r−1 r−j−1 (4.8) j!g = cr j X + cr−1 j X + ··· + cj. In particular, since (p − 1)!f ∈ Z[X] and the binomial coeﬃcients are integers, we (j) (j) have for j > p, a = 0, . . . , n that (p − 1)!f /j! ∈ Z[X], and so f (a)/p ∈ Z. This implies at once (4.7).

Lemma 4.6. Assume that p > |q0n|. Then M is a non-zero integer.

(p−1) Proof. From (4.5) it follows that the term q0f (0) is an integer not divisible by (j) p, while all other terms qaf (a) in the right-hand side of (4.4) are integers that are either 0 or divisible by p. Hence M is an integer not divisible by p.

47 Lemma 4.7. For p suﬃciently large, we have |M| < 1.

Proof. By Lemma 4.4, we have for a = 0, . . . , n,

|a| |F (a)| 6 a · e · sup |f(u)|. |u|6a

For a, b = 0, . . . , n, and u ∈ C with |u| 6 a we have |u − b| 6 |u| + |b| 6 2n. Hence (2n)np+p−1 cp sup |f(u)| 6 6 , |u|6a (p − 1)! (p − 1)! say, where c is a constant independent of p, a, b. This implies

n n ! X X cp |M| |q F (a)| q · a · ea . 6 a 6 a (p − 1)! a=0 a=0 cp For p suﬃciently large this is < 1, since for any c > 1, (p−1)! → 0 as p → ∞.

Summarizing, our assumption that e is algebraic implies that there is a quantity M, which is by Lemma 4.6 an non-zero integer, and by Lemma 4.7, of absolute value < 1. Since this is absurd, e must be transcendental.

4.2 The Lindemann-Weierstrass theorem

Lindemann proved in 1882 that eα is transcendental for algebraic α, and Weierstrass proved in 1885 that if α1, . . . , αn are algebraic numbers that are linearly independent over Q, then eα1 , . . . , eαn are algebraically independent over Q. The following result, due to A. Baker, is in fact equivalent to the Lindemann-Weierstrass Theorem.

Theorem 4.8. Let α1, . . . , αn, β1, . . . , βn ∈ Q. Suppose that α1, . . . , αn are pairwise distinct, and that β1, . . . , βn 6= 0. Then

α1 αn β1e + ··· + βne 6= 0.

Before proving this theorem, we state some corollaries.

Corollary 4.9. (i) Let α ∈ Q be non-zero. Then eα is transcendental. (ii) π is transcendental.

48 Proof. (i) Suppose that eα =: β is algebraic. Then it follows that 1 · eα − β · e0 = 0, contradicting Theorem 4.8. (ii) Suppose that π is algebraic. Then πi is algebraic. But eπi = −1 is not transcendental, contradicting (i).

Corollary 4.10. (i) Let α ∈ Q and α 6= 0. Then sin α, cos α, and tan α are transcendental. (ii) Let α ∈ Q and α 6= 0, 1. Then log α is transcendental (for any choice of log α).

Exercise 4.1. Prove Corollary 4.10.

Corollary 4.11. Let α1, . . . , αn ∈ Q. Then the following two assertions are equivalent: (i) α1, . . . , αn are linearly independent over Q; (ii) eα1 , . . . , eαn are algebraically independent.

Exercise 4.2. Prove Corollary 4.11.

We start with some preliminary comments on the proof of Theorem 4.8.

Our proof of the transcendence of e was by contradiction: we assumed that q0 + n q1e+···+qne = 0 for certain rational integers q0, . . . , qn, and constructed from this a non-zero integer M with |M| < 1. To prove the Lindemann-Weierstrass Theorem, α0 αn we may again proceed by contradiction and assume that β0e +···+βne = 0. By following the transcendence proof of e, but replacing 0, 1, , . . . , n by α1, . . . , αn and q , . . . , q by β , , . . . , β , we obtain a non-zero algebraic integer M, not necessarily in 0 n 1 n √ 1 Q, such that |M| < 1. This is, however, not a contradiction. For instance, 2 (1− 5) is an algebraic integer of absolute value < 1. Instead, we use the following observation. Let α be in the ring of integers of an algebraic number ﬁeld K. Recall that the characteristic polynomial of α is Q χα,K (X) = σ(X − σ(α)) where the product is over all embeddings σ : K,→ C of

K. This polynomial has its coeﬃcients in Z. So in particular the norm NK/Q(α) = Q σ σ(α) is in Z. If α 6= 0, we have NK/Q(α) 6= 0, hence |NK/Q(α)| > 1. So from the assumption that Theorem 4.8 is false we have to construct somehow a non-zero algebraic integer whose norm has absolute value < 1. If we just follow the transcendence proof of e without any modiﬁcations, we only get a non-zero algebraic integer M with |M| < 1, but we can not show that its norm has absolute value < 1.

49 A priori, for some embedding σ, the quantity |σ(M)| may be very large, and then

|NK/Q(M)| may be > 1.

Pn αi To circumvent this, we deduce from the expression i=1 βie a new expression Pt γi i=1 δie , where the γi, δi satisfy certain symmetry conditions. These symmetry Pt γi conditions allow to construct, under the hypothesis i=1 δie = 0, a non-zero algebraic integer having a norm with absolute value < 1. Thus, we obtain a weaker version of the Lindemann-Weierstrass Theorem, which asserts that under the said Pt γi symmetry conditions, i=1 δie 6= 0. But as will be seen, this weaker version implies the general Lindemann-Weierstrass Theorem. Theorem 4.12 (“Weak Lindemann-Weierstrass Theorem”). Let L ⊂ C be a normal algebraic number ﬁeld. Let γ1, . . . , γt, δ1, . . . , δt ∈ L such that

γ1, . . . , γt are distinct, δ1 ··· δt 6= 0, and suppose moreover, that each τ ∈ Gal(L/Q) permutes the pairs (γ1, δ1),..., (γt, δt). Then γ1 γt δ1e + ··· + δte 6= 0.

We say that τ permutes the pairs (γ1, δ1),..., (γt, δt) if (τ(γ1), τ(δ1)),..., (τ(γt), τ(δt)) is a permutation of (γ1, δ1),..., (γt, δt). We ﬁrst prove the implication Theorem 4.12=⇒Theorem 4.8. After that, we prove Theorem 4.12.

Theorem 4.12 =⇒ Theorem 4.8. Assume that Theorem 4.8 is false. This means that there are α1, . . . , αn, β1, . . . , βn ∈ Q such that, α1, . . . , αn are distinct, β1, . . . , βn are non-zero, and α1 αn β1e + ··· + βne = 0. We derive from this a contradiction to Theorem 4.12.

Let L be the number ﬁeld generated by α1, . . . , αn, β1, . . . , βn and their conjugates. Then L is a normal number ﬁeld. Let

Gal(L/Q) = {τ1, . . . , τd}.

Recall that if γ ∈ L, then the set {τ1(γ), . . . , τd(γ)} contains all conjugates of γ. Clearly, d Y τi(α1) τi(αn) τi(β1)e + ··· + τi(βn)e = 0. i=1

50 By expanding the product, we get n n X X τ1(αi1 )+···+τd(αid ) (4.9) ··· τ1(βi1 ) ··· τd(βid ) · e = 0.

i1=0 id=1 Each τ ∈ Gal(L/Q) permutes the pairs τ1(αi1 ) + ··· + τd(αid ), τ1(βi1 ) ··· τd(βid ) , since ττ1, . . . , ττd is a permutation of τ1, . . . , τd.

The exponents τ1(αi1 )+···+τd(αid ) need not be distinct. We group together the terms with equal exponents. Let γ1, . . . , γs be the distinct values among τ1(αi1 ) +

··· + τd(αid ) (0 6 i1, . . . , in 6 d), and for k = 1, . . . , s, denote by Jk the set of tuples (i1, . . . , id) such that

τ1(αi1 ) + ··· + τd(αid ) = γk. Then (4.9) becomes

s X γk X (4.10) δke = 0, where δk = τ1(βi1 ) ··· τd(βid ).

k=1 (i1,...,ik)∈Jk

Notice that each τ ∈ Gal(L/Q) permutes the pairs (γ1, δ1),..., (γs, δs). A priori, all coefficients δk might be 0. However, we show that there is a tuple (i1, . . . , ik) such that τ1(αi1 ) + ··· + τd(αid ) is different from all the other exponents. Thus, for some k, the set Jk has cardinality 1, and δk 6= 0. Define a total ordering on C by setting θ < ζ if Re θ < Re ζ or if Re θ = Re ζ and Im θ < Im ζ. This ordering has the property that if θi, ζi are complex numbers Pr Pr with θi < ζi for i = 1, . . . , r, then j=1 θj < j=1 ζj.

Since α1, . . . , αd were assumed to be distinct, for each τ ∈ Gal(L/Q), the numbers τ(α1), . . . , τ(αd) are distinct. Hence for each k ∈ {1, . . . , d}, there is an index ik such that τk(αik ) > τk(αj) for j 6= ik. This implies

τ1(αi1 ) + ··· + τd(αid ) > τ1(αj1 ) + ··· + τd(αjd )

for all tuples (j1, . . . , jd) 6= (i1, . . . , id), and so τ1(αi1 ) + ··· + τd(αid ) is distinct from the other exponents.

Assume without loss of generality that δ1, . . . , δt are the non-zero numbers among δ1, . . . , δs. Then (4.10) becomes

t X γk (4.11) δke = 0. k=1

51 By construction, the numbers γ1, . . . , γt are distinct algebraic numbers. Further, δ1, . . . , δt are non-zero. As observed before, each τ ∈ Gal(L/Q) permutes the pairs (γ1, δ1),..., (γs, δs) from (4.10). But then τ permutes also the pairs with δk 6= 0, i.e., τ permutes (γ1, δ1),..., (γt, δt). Now Theorem 4.12 implies that (4.11) is false. Thus, our assumption that Theorem 4.8 is false leads to a contradiction.

Proof of Theorem 4.12. We follow the transcendence proof of e, with the necessary modiﬁcations. Before proceeding, we observe that there is no loss of generality to assume that δ1, . . . , δt are algebraic integers. Indeed, there is a positive m ∈ Z such that mδ1, . . . , mδt are algebraic integers (e.g, we may take for m the product of the denominators of β1, . . . , βn), and clearly, the conditions and conclusion of Theorem 4.12 are unaﬀected if we replace δi by mδi for i = 1, . . . , t.

Let γ1, . . . , γt be distinct algebraic numbers and δ1, . . . , δt non-zero algebraic integers from the normal number ﬁeld L, such that each τ ∈ Gal(L/Q) permutes the pairs (γ1, δ1),..., (γt, δt). Assume that

γ1 γt (4.12) δ1e + ··· + δte = 0.

Let again p be a prime number. Further, let l be a positive rational integer such that lγ1, . . . , lγt are all algebraic integers (e.g., the product of the denominators of γ1, . . . , γt). For k = 1, . . . , t, deﬁne

t 1 tp p−1 Y p fk(X) := (p−1)! · l (X − γk) (X − γj) , j=1

Z z z−u Fk(z) := e fk(u)du, 0

Mk := δ1Fk(γ1) + ··· + δtFk(γt). We prove the following:

1) For each τ ∈ Gal(L/Q) we have τ(M1) ∈ {M1,...,Mt};

2) for suﬃciently large p, M1 is a non-zero algebraic integer;

3) |Mk| < 1 for k = 1, . . . , t and suﬃciently large p.

52 From 1) and 3) it follows that |NL/Q(M1)| < 1. But this contradicts 2). Lemma 4.13. (i) We have

t tp−1 X X (m) Mk = − δjfk (γj) for k = 1, . . . , t. j=1 m=0

(ii) For each τ ∈ Gal(L/Q) we have τ(M1) ∈ {M1,...,Mt}.

Pt γj Proof. (i) This follows at once from Lemma 4.2 and our assumption j=1 δje = 0. (ii) Let τ ∈ Gal(L/Q). Then there is a permutation τ ∗ of 1, . . . , t such that

(τ(γk), τ(δk)) = (γτ ∗(k), δτ ∗(k)) for k = 1, . . . , t.

By applying τ to the coeﬃcients of f1, we obtain

t t tp p−1 Y p tp Y p l (X − τ(γ1)) (X − τ(γj)) = l (X − γτ ∗(1)) (X − γτ ∗(j)) = fτ ∗(1). j=1 j=1

Hence

t tp−1 1 X X (m) τ(M ) = − τ(δ )f (τ(γ )) 1 (p − 1)! j τ ∗(1) j j=1 m=0 t tp−1 1 X X (m) = − δ ∗ f (γ ∗ ) = M ∗ . (p − 1)! τ (j) τ ∗(1) τ (j) τ (1) j=1 m=0

Given two algebraic numbers α, β and m ∈ Z>0, we write α ≡ β (mod m) if (α − β)/m is an algebraic integer. Lemma 4.14. let k ∈ {1, . . . , t}. Then

( t )p (p−1) tp Y (4.13) f1 (γ1) = l (γ1 − γk) , k=2 (j) (4.14) f1 (γm) = 0 for m = 1, . . . , t, j = 0, . . . , p − 1, (m, j) 6= (1, p − 1), (j) (4.15) f1 (γm) ≡ 0(mod p) for m = 1, . . . , t, j > p.

53 Proof. The proofs of (4.13) and (4.14) are completely analogous to those of (4.5) and (4.6) in Lemma 4.5. We prove only (4.15). Let m ∈ {1, . . . , t} and j > p. Deﬁne

t 1 p−1 Y p g(X) := f1(X/l) = (p−1)! · l(X − lγ1) (X − lγk) . k=2

Then (p − 1)!g has algebraically integral coeﬃcients. Using (4.8), one easily shows (j) that the coeﬃcients of (p − 1)!g1 /j! are algebraic integers. Hence for j > p, (j) g (lγj)/p is an algebraic integer, and therefore,

f (j)(γ ) ljg(j)(lγ ) 1 m = m p p is an algebraic integer. This implies at once (4.15).

Lemma 4.15. For p suﬃciently large, M1 is a non-zero algebraic integer.

Proof. An application of Lemma 4.14 gives

t p t Y M1 ≡ −δ1A (mod p) with A := l (γ1 − γk). k=2

Both δ1,A are algebraic integers, hence M1 is an algebraic integer. We prove that p for suﬃciently large p, δ1A /p is not an algebraic integer. Then necessarily, M1 6= 0. p Assume that δ1A /p is an algebraic integer. Let b = NL/Q(δ1), B = NL/Q(A). p p d Then b, B ∈ Z, and the norm NL/Q(δ1A /p) = bB /p is in Z, where d = [L : Q]. But this is impossible if p > |bB|.

Lemma 4.16. For p suﬃciently large we have |Mk| < 1 for k = 1, . . . , t.

Exercise 4.3. Prove this lemma.

Thus our assumption that Theorem 4.12 is false implies the Lemmas 4.13, 4.15, 4.16, and these together give a contradiction.

54 4.3 Other transcendence results

We give an overview of some other transcendence results, without proof. Recall that the logarithm on C is a multi-valued function: the logarithm of a non-zero complex number z is defined by log z := log |z| + i arg z, where log |z| is the usual real logarithm of the positive number |z| and arg z is any choice for the argument of z. In the theorems below, we fix a real ϕ0, and take arg z ∈ [ϕ0, ϕ0 + 2π). Having b b log a fixed ϕ0, we have fixed the argument, the logarithm, and also a := e for a, b ∈ C z P∞ n with a 6= 0, where e := n=0 z /n!. The theorems hold for any choice of ϕ0. Theorem 4.17 (Gel’fond, Schneider, 1934). Let α, β ∈ Q with α 6= 0, 1, β 6∈ Q. Then αβ is transcendental.

Corollary 4.18. Let α ∈ Q with α 6∈ Qi. Then eπα is transcendental.

Proof. Choose log(−1) = πi. Then eπα = e−iα log(−1) = (−1)−iα.

Corollary 4.19. Let α1, α2, β1, β2 ∈ Q be non-zero. Assume that log α1, log α2 are linearly independent over Q. Then

β1 log α1 + β2 log α2 6= 0.

Proof. Suppose β1 log α1 + β2 log α2 = 0. Put γ := −β2/β1. Then by assumption, γ 6∈ Q, and log α2 γ log α1 γ α2 = e = e = α1 , contradicting Theorem 4.17.

In 1966, A. Baker proved the following far-reaching generalization.

Theorem 4.20 (A. Baker, 1966). Let α1, . . . , αn, β1, . . . , βn ∈ Q be non-zero. As- sume that log α1,..., log αn are linearly independent over Q.

Then β1 log α1 + ··· + βn log αn is transcendental.

Deﬁnition. We say that non-zero complex numbers α1, . . . , αn are multiplicatively dependent if there are x1, . . . , xn ∈ Z, not all 0, such that

x1 xn α1 ··· αn = 1.

Otherwise, α1, . . . , αn are called multiplicatively independent.

55 Corollary 4.21. Let n > 1. Let α1, . . . , αn, β1, . . . , βn, γ ∈ Q be such that

α1, . . . , αn 6= 0, α1, . . . , αn are multiplicatively independent, n (β1, . . . , βn) 6∈ Q .

γ β1 βn Then e α1 ··· αn is transcendental.

γ β1 βn Proof. Suppose that δ := e α1 ··· αn is algebraic. We have

elog δ = eγ+β1 log α1+···+βn log αn and so, for some k ∈ Z,

(4.16) log δ = γ + β1 log α1 + ··· + βn log αn + 2kπi.

Note that log(−1) = (2l + 1)πi for some l ∈ Z depending on the choice of the Pn 2k log. Then log δ − i=1 βi log αi − 2l+1 log(−1) = γ ∈ Q. Hence by Theorem 4.20, log δ, log α1,..., log αn and log(−1) are linearly dependent over Q, that is, there are x1, . . . , xn, xn+1, xn+2 ∈ Z, not all 0, such that

(4.17) x1 log α1 + ··· + xn log αn + xn+1 log(−1) + xn+2 log δ = 0.

Eliminating log δ from (4.16) and (4.17), we get

2k xn+2γ+(xn+2β1+x1) log α1+···+(xn+2βn+xn) log αn+(xn+2· 2l+1 +xn+1) log(−1) = 0.

In (4.17), at least one of x1, . . . , xn, xn+2 is 6= 0, since log(−1) 6= 0. Since (β1, . . . , βn) 6∈ n Q we have xn+2βi + xi 6= 0 for at least one i ∈ {1, . . . , n}. Applying again Theorem 4.20, we infer that there are y1, . . . , yn+1 ∈ Z, not all 0, such that

y1 log α1 + ··· + yn log αn + yn+1 log(−1) = 0.

In fact, at least one of y1, . . . , yn must be 6= 0. Now we get

2y1 2yn 2y1 log α1+···+2yn log αn −2yn+1 log(−1) −2yn+1(2l+1)πi α1 ··· αn = e = e = e = 1, contrary to our assumption.

In fact, the conditions that the αi be multiplicatively independent, and the βi not all in Q are needed only if γ = 0. These conditions can be dropped if we assume that γ 6= 0.

56 Exercise 4.4. Let α1, . . . , αn, β1, . . . , βn, γ ∈ Q,and suppose that γ, α1, . . . , αn 6= 0. γ β1 βn Using Corollary 4.21, prove that e α1 ··· αn is transcendental.

There is a far reaching conjecture, due to Schanuel, which implies all results mentioned before and much more.

Schanuel’s Conjecture. (1960’s) Let x1, . . . , xn be any (not necessarily algebraic) complex numbers which are linearly independent over Q. Then

x1 xn trdeg(x1, . . . , xn, e , . . . , e ) > n.

We give some examples of known cases. Examples. 1. Let x ∈ C∗. Then either x is transcendental, or x is algebraic x x and then by Lindemann’s Theorem, e is transcendental. Hence trdeg(x, e ) > 1. Schanuel’s Conjecture is still open for n > 2.

2. Let α1, . . . , αn ∈ Q and suppose they are linearly independent over Q. By Corollary 4.11 (a consequence of the Lindemann-Weierstrass Theorem), eα1 , . . . , eαn α1 αn are algebraically independent. Hence trdeg(α1, . . . , αn, e , . . . , e ) = n. We deduce some consequences of Schanuel’s Conjecture which are still open. Conjecture. e and π are algebraically independent.

Proof under assumption of Schanuel’s Conjecture. The transcendence degree of a set of numbers does not change if some algebraic numbers are added to or removed from it. Moreover, the transcendence degree of this set does not change if we multiply its elements with non-zero algebraic numbers. So by Schanuel’s conjecture, trdeg(e, π) = trdeg(e, πi) = trdeg(1, πi, e, eπi) = 2.

Conjecture. Let α1, . . . , αn ∈ Q such that α1, . . . , αn 6= 0 and log α1,..., log αn are linearly independent over Q. Then log α1,..., log αn are algebraically independent.

Proof under assumption of Schanuel’s Conjecture. We have

trdeg(log α1,..., log αn) = trdeg(log α1,..., log αn, α1, . . . , αn) = n.

57 The above conjecture implies that for every non-zero polynomial P ∈ Q[X1,...,Xn] we have P (log α1,..., log αn) 6= 0. Baker’s Theorem 4.20 implies that this holds for linear polynomials P ∈ Q[X1,...,Xn], but even for quadratic polynomials P this is still open. For instance, the above conjecture implies that log 2 · log 3 is transcendental, but as yet not even this very special case could be proved. We mention some other consequences of Schanuel’s conjecture, the deduction of which is left as an exercise.

Exercise 4.5. Deduce the following from Schanuel’s conjecture: (i) Let α ∈ Q, α 6∈ iQ. Then π and eπα are algebraically independent. β β2 βd−1 (ii) Let α, β ∈ Q with α 6∈ {0, 1} and β of degree d > 2. Then α , α , . . . , α are algebraically independent. ∞ xn−1 e (iii) Deﬁne the sequence {xn}n=1 by x1 = e and xn = e for n > 2, i.e., x2 = e , ee x3 = e , etc. Then x1, . . . , xN are algebraically independent for every N > 1. (iv) Let α ∈ Q \{0, 1}. Then log α, log log α are algebraically independent. (v) Let p, q be two distinct prime numbers. Then for every irrational x ∈ R, at least one of the numbers px, qx is transcendental.

Remark. The following has been proved.

π 1 In 1996, Nesterenko proved (among other things), that π, e and Γ( 4 ) are al- R ∞ x−1 −t gebraically independent. Recall that Γ(x) = 0 t e dt for x > 0, that Γ(n) = 1 √ (n − 1)! for every positive integer n, and that Γ( 2 ) = π. For α, β as in (ii), Diaz proved in 1989 that

β β2 βd−1 trdeg(α , α , . . . , α ) > [(d + 1)/2] where [x] is the largest integer 6 x. This settles (ii) for d = 3. In the 1960’s, Lang and Ramachandra independently proved (among other things) that if p1, p2, p3 are three distinct primes and x an irrational real, then at least one x x x of the numbers p1, p2, p3 is transcendental.

58 4.4 Siegel’s Lemma

In the next section we will prove a special case of the Gel’fond-Schneider Theorem, that is that if α, β are real algebraic numbers with α 6= 0, 1, β 6∈ Q, then αβ is transcendental. In the present section we develop a tool which is very important in Diophantine approximation, that is the so-called Siegel’s Lemma, which was formally stated for the ﬁrst time by Siegel in 1929, but known before. Essentially, it states that under certain hypotheses, a system of M homogeneous linear equations in N unknowns

a11x1 + ··· + a1N xN = 0 . . (4.18) . .

aM1x1 + ··· + aMN xN = 0 has a non-trivial solution in integer coordinates, the absolute values of which are not too large.

Theorem 4.22 (Siegel’s Lemma). Assume that N > M > 0, A > 1, and

aij ∈ Z, |aij| 6 A for i = 1,...,M, j = 1,...,N.

N Then (4.18) has a solution x = (x1, . . . , xN ) ∈ Z \{0} with

M/(N−M) max |xi| 6 (NA) . 16i6N

N PN Proof. For i = 1,...,M, x ∈ Z , put li(x) := j=1 aijxj and let

N N X X −Ci := min(aij, 0),Di := max(aij, 0). j=1 j=1

Notice that Ci + Di 6 NA. Let B be a positive integer, and let SB be the set of N y = (y1, . . . , yN ) ∈ Z with 0 6 yj 6 B for j = 1,...,N. For each y ∈ SB we have

−CiB 6 li(y) 6 DiB for i = 1,...,M.

N Notice that SB has cardinality (B + 1) . Further, if y runs through SB, then (l1(y), . . . , lM (y)) runs through a set of cardinality at most

M Y M (CiB + DiB + 1) 6 (NAB + 1) . i=1

59 We choose B such that (B + 1)N > (NAB + 1)M . Then by the box principle, there are distinct y1, y2 ∈ SB with li(y1) = li(y2) for i = 1,...,M. Take x = y1 − y2. Then x satisﬁes (4.18) and |xi| 6 B for i = 1,...,N. We ﬁnish our proof by showing that the choice B = [(NA)M/(N−M)] is valid. Indeed, with this choice of B we have (B + 1)N−M > (NA)M , hence

(B + 1)N > (NA(B + 1))M > (NAB + 1)M .

We need a generalization where the coeﬃcients aij are algebraic integers instead of just rational integers. To deduce this, we need some facts from algebraic number theory, which we recall without proof.

Let K be an algebraic number ﬁeld of degree d. Then its ring of integers OK is a free Z-module of rank d, that is, there are ω1, . . . , ωd ∈ OK such that every element β of OK can be expressed uniquely as

(4.19) β = x1ω1 + ··· + xdωd with x1, . . . , xd ∈ Z.

We call {ω1, . . . , ωd} an integral basis of K.

Let σ1, . . . , σd be the embeddings of K in C. Then

σi(β) = x1σi(ω1) + ··· + xdσi(ωd) for i = 1, . . . , d, or b = Ωx

T T where b := (σ1(β), . . . , σd(β)) , x = (x1, . . . , xd) , and   σ1(ω1) ··· σ1(ωd)  . .  Ω :=  . .  . σd(ω1) ··· σd(ωd)

2 The matrix Ω is invertible; in fact, in fact it can be shown that DK := (det Ω) is a non-zero rational integer, the discriminant of K. We deﬁne the house of an algebraic integer α by

α := max(|α(1)|,..., |α(m)|),

60 where α(1), . . . , α(m) are the conjugates of α. If α belongs to an algebraic number ﬁeld K of degree d and σ1, . . . , σd are the embeddings of K in C, then each of the conjugates of α occurs d/m times among σ1(α), . . . , σd(α). Hence we have also α = maxi |σi(α)|. Lemma 4.23. There is an eﬀectively computable number C(K) depending on K and ω1, . . . , ωd, such that if β ∈ OK then for the rational integers x1, . . . , xd given by (4.19) we have max |xi| 6 C(K) · β . 16i6d

−1 Pd Proof. By expanding x = Ω b we get xi = j=1 θijσj(β) for i = 1, . . . , d, where −1 Ω = (θij). Apply the triangle inequality.

We consider again systems (4.18), but now the coeﬃcients aij are from OK .

Corollary 4.24. Let [K : Q] = d, let M,N be integers with N > dM > 0, let A be a real > 1, and suppose that

aij ∈ OK , aij 6 A for i = 1, . . . , M, j = 1,...,N.

N Then (4.18) has a solution x = (x1, . . . , xN ) ∈ Z \{0} such that

dM/(N−dM) (4.20) max |xi| 6 (C(K)NA) , 16i6N where C(K) is the constant from Lemma 4.23.

Pd Proof. In general, if we write β ∈ OK as k=1 ykωk with yk ∈ Z, then β = 0 if and only if yk = 0 for k = 1, . . . , d. By Lemma 4.23 we have

d X aij = ωkbikj with bikj ∈ Z, |bikj| 6 C(K) · A. k=1

PN Pd PN N Then j=1 aijxj = k=1 ωk j=1 bikjxj for x ∈ Z , hence (4.18) holds for some x ∈ ZN if and only if

bik1x1 + ··· + bikN xN = 0 for i = 1, . . . , M, k = 1, . . . , d.

Notice that this is a system of dM equations in N unknowns. By Theorem 4.22 this system has a non-trivial solution x ∈ ZN with (4.20).

61 4.5 The Gel’fond-Schneider Theorem

We prove the following theorem.

Theorem 4.25. Let α, β be real algebraic numbers such that α > 0, α 6= 1 and β 6∈ Q. Then αβ is transcendental.

Here αβ = eβ log α with the usual natural logarithm for positive real numbers. The proof in the case that α, β are not both real or α < 0 goes along the same lines, but with additional complications. Gel’fond and Schneider independently proved the above theorem, in the general case where α, β may be complex, with somewhat diﬀerent proofs. We follow Schneider’s proof.

We assume that γ := αβ is algebraic. Let K = Q(α, β, γ), d = [K : Q]. Recall that there are positive integers m1, m2, m3 such that m1α, m2β, m3γ are algebraic integers. Then with m = m1m2m3, mα, mβ, mγ are algebraic integers.

Let D1,D2,L be parameters, which will be chosen optimally later. In what follows, c1, c2,... will be constants depending only on α, β, γ and L, and independent of D1,D2,L.

2 Lemma 4.26. Assume that D1D2 > 2dL , L > 5. Then there are integers aij (i = 0,...,D1 − 1, j = 0,...,D2 − 1), not all zero, such that the function

D1−1 D2−1 X X i jz (4.21) F (z) = aijz α i=0 j=0 has zeros a + bβ with a, b = 1,...,L, and such that

(4.22) |aij| 6 exp c1(D1 log L + D2L) (i = 0,...,D1 − 1, j = 0,...,D2 − 1).

Proof. We have to ﬁnd aij ∈ Z, not all zero, such that F (a + bβ) = 0 for a, b = 1,...,L. Using αa+bβ = αaγb, this translates into a system of L2 linear equations in

D1D2 unknowns,

D1−1 D2−1 X X i aj bj aij(a + bβ) α γ = 0 (a, b = 1,...,L). i=0 j=0

62 To apply Siegel’s Lemma we want all coeﬃcients of this system of equations to be algebraic integers. To this end, we multiply the equations with mD1+2LD2 and obtain

D1−1 D2−1 X X D1+2LD2 i aj bj (4.23) aijm (a + bβ) α γ = 0 (a, b = 1,...,L). i=0 j=0

We estimate the houses of the coeﬃcients mD1+2LD2 (a + bβ)iαajγbj of this system. Let σ : K,→ C be an embedding of K. Then for the image under σ of a typical coeﬃcient we have

|mD1+2LD2 (a + bσ(β))iσ(α)ajσ(γ)bj|

D1+2LD2 D1 LD2 LD2 6 m L(1 + |σ(β)|) (1 + |σ(α)|) (1 + |σ(γ)|) 6 exp c2(D1 log L + D2L) where the constant c2 has been chosen large enough in terms of m, σ(α), σ(β), σ(γ). By making c2 even larger, we can make it independent of σ, so that the houses of the coeﬃcients of system (4.23) are all bounded above by exp c2(D1 log L + D2L) . Now Corollary 4.24 implies that system (4.23) has a solution in integers aij, not all zero, such that

2 2 c2(D1 log L+D2L)dL /(D1D2−dL ) |aij| 6 C(K)D1D2e 6 exp c1(D1 log L + D2L) ,

2 choosing c1 suﬃciently large. Here we have used our assumption D1D2 > 2dL .

2 We now choose the parameters D1,D2,L such that D1D2 = 2dL and D1 = D2L (to make D1 log L and D2L about equal), i.e. √ √ 3/2 1/2 (4.24) D1 = 2d · L ,D2 = 2dL . Then the estimate in Lemma 4.26 becomes

3/2 (4.25) |aij| 6 exp c3L log L .

We note that F (z) is a so-called exponential polynomial, i.e., a function of the shape r X γkz E(z) = pk(z)e , k=1 where the pk(z) are non-zero polynomials, and the γk distinct numbers. We need a simple result on the number of zeros of such a function.

63 Lemma 4.27. Assume that the γk and the coeﬃcients of the pk are all reals. Put Pr M := k=1(1 + deg pk). Then E(z) has at most M zeros in R.

Exercise 4.6. Prove this lemma. Hint. Proceed by induction on M. Apply Rolle’s Theorem, which asserts that if G is a diﬀerentiable real function and a, b are reals with a < b and G(a) = G(b) = 0, then there is c with a < b < c and G0(c) = 0.

Notice that we can apply this lemma to our above function F (z), thanks to our assumption that α, β are real. Thus, this lemma implies that F (z) has at 2 2 most D1D2 = 2dL zeros. We know already that F (z) has the L zeros a + bβ (1 6 a, b 6 L). These zeros are all different, since β 6∈ Q. We briefly sketch the idea how to derive a contradiction from this. Details are provided later. Here it is important that we have some freedom to choose the parameters D1,D2,L introduced above. Thus, we can choose L sufficiently large to make all estimates work. We show that for all sufficiently small and all sufficiently large L, we have 1+ 2(1+) F (a + bβ) = 0 for all integers a, b with 1 6 a, b 6 L . Thus, F has at least L zeros. But for L sufficiently large this is larger than dL2, and so this contradicts Lemma 4.27.

1+ To prove that F (a+bβ) = 0 for all integers a, b with 1 6 a, b 6 L , we proceed 1+ as follows. The number A := mD1+2D2[L ]F (a + bβ) is an algebraic integer. Using an analytic argument, we show that |A| is very small. Further, by a trivial estimate we show that |σ(A)| is not too large for the embeddings σ of K in C not equal to Q the identity. It will follow that |NK/Q(A)| = | σ σ(A)| < 1, where the product is over all embeddings, the identity included. But then, F (a + bβ) = A = 0, since the norm of a non-zero algebraic integer is a non-zero element of Z. We now work out the details. We need a few facts from complex analysis. Recall that an entire function is a function f : C → C that is everywhere analytic, i.e., 0 f(w)−f(z) f (z) = limw→z w−z exists for every z ∈ C. The following two lemmas are standard, and their proofs can be found in any textbook on complex analysis.

Lemma 4.28. Let f be an entire function and a ∈ C a zero of f. Then the function f(z) 0 g deﬁned by g(z) = z−a if z 6= a and g(a) = f (a) is also entire.

64 Lemma 4.29 (Maximum Modulus Principle). Let f be an entire function. For R > 0, deﬁne

|f|R := sup |f(z)|. |z|=R

Then for every z ∈ C with |z| 6 R we have |f(z)| 6 |f|R, i.e., |f(z)| attains its maximum on the disk |z| 6 R on the boundary of that disk.

As a consequence of these two lemmas we obtain the following:

Lemma 4.30. Let f be an entire function and a1, . . . , ar distinct zeros of f. Let R,T be reals such that |ai| 6 R for i = 1, . . . , r and T > 2R. Then r |f(z)| 6 |f|T 4R/T for all z ∈ C with |z| 6 R.

Proof. Let g(z) = f(z)/(z − a1) ··· (z − ar). By Lemma 4.28, g(z) is entire. Let z ∈ C with |z| 6 R. On the one hand, by Lemma 4.29, r Y r r |f(z)| 6 |g(z)| (|z| + |ai|) 6 |g(z)|(2R) 6 |g|T (2R) , i=1 on the other hand, we have for w ∈ C with |w| = T ,

|f(w)| r |g(w)| = 6 |f(w)| · (2/T ) , |w − a1| · · · |w − ar| 1 r since |w − ai| > |w| − |ai| > T − R > 2 T . Hence |g|T 6 |f|T (2/T ) . Our lemma follows.

Proof of Theorem 4.25. Let 0 < < 1/8. Put R := (1 + |β|)L1+,T := (1 + |β|)L1+2. Our intention is to show that if L is sufficiently large, then F (a + bβ) = 0 for all 1+ 2(1+) integers a, b with 1 6 a, b 6 L . This yields L distinct zeros of F , since β 6∈ Q. Assuming L is sufficiently large, this contradicts Lemma 4.27, since by that 2 lemma, F cannot have more than D1D2 = 2dL zeros. 1+ Choose integers a, b with 1 6 a, b 6 L . We first estimate |F (a + bβ)| by applying Lemma 4.30. A simple application of the triangle inequality gives

D1−1 D2−1 3/2 X X i T j c3L log L D1 TD2 |F |T 6 |aij|T |(1 + |α|) 6 D1D2e T (1 + |α|) . i=0 j=0

65 √ √ 3/2 1/2 1+2 Using our choices D1 = 2dL , D2 =√ 2dL , T = (1 + |β|)T , we see that (3/2)+2 the term with exponent TD2 = (1 + |β|) 2dL dominates. We thus obtain

(3/2)+2 |F |T 6 exp c4L . As a consequence,

(3/2)+2 − L2 |F (a + bβ)| 6 exp c4L · (4L ) .

1+ We multiply F (a + bβ) with mD1+2D2[L ] to obtain an algebraic integer, i.e.,

D1−1 D2−1 1+ D1+2D2[L ] X X i a b j A := m aij(a + bβ) (α γ ) . i=0 j=0 Now clearly, (3/2)+2 2 |A| 6 exp c5L − L log L . We estimate the absolute values of the other conjugates of A. Let σ be an embedding of K = Q(α, β, γ) in C. Then by the triangle inequality,

D1−1 D2−1 1+ D1+2D2[L ] X X i a b j |σ(A)| 6 m |aij|(a + b|σ(β)|) (|σ(α)| |σ(γ)| ) i=0 j=0 1+ 1+ 3/2 1+ D2L D1+2D2[L ] c3L log L L 6 D1D2m e (1 + |σ(β)|) (1 + |σ(α)|)(1 + |σ(γ)|)

(3/2)+ 6 exp c6L . So altogether we have an upper bound for |A| and upper bounds for |σ(A)| for the other d − 1 embeddings σ of K not equal to the identity. By taking their product we obtain

(3/2)+2 (3/2)+ 2 |NK/Q(A)| 6 exp c5L + (d − 1)c6L − L log L . Now we choose L large enough so that the exponent becomes negative; this is possible since the term L2 log L grows more fastly than the other terms in the exponent as 2(1+) 2 L → ∞. Further, we assume L > 2dL = D1D2. With such a choice of L, we get |NK/Q(A)| < 1. Since the norm of a non-zero algebraic integer is a non-zero integer, this must imply A = 0, or equivalently, F (a + bβ) = 0. This holds for all 1+ integers a, b with 1 6 a, b 6 L . As explained above, this contradicts Lemma 4.27. Our proof of Theorem 4.25 is complete.