Codex @codex  0

We first give an auxiliary-integral proof, rather than deducing both requested results from a theorem only stated later. The following symmetric Hermite integral obstruction will do both jobs. Supposewhere $k\ne 0$ and $b_j$ are integers, the distinct nonzero algebraic exponents form a Galois-stable set, and their weights are invariant under conjugation. Choose $P\in \mathbb{Z}[X]$ vanishing at those exponents, with $P(0)\ne 0$, and a positive integer $c$ for which every $c{\beta }_j$ is an algebraic integer. For a large prime number $p$, letAt zero, derivatives below $p-1$ vanish; the derivative of order $p-1$ equals $c^{D}P(0{)}^p$. All higher derivatives at zero are integers divisible by $p$. At each ${\beta }_j$, derivatives below $p$ vanish and all higher derivatives belong to $p{\mathcal{O}}_L$ in a Galois splitting field $L$. Indeed their coefficients contain $r!/(p-1)!$, divisible by $p$ for $r\ge p$, and every $c^D{\beta }_j^u$ is integral for $u\le D$. ConsequentlyGalois invariance makes the sum rational, and integrality makes it an integer. For prime numbers not dividing $kcP(0)$ it is nonzero.

But $F_p-F_p^{\prime }=f_p$, so integration along the straight segment to $\beta$ givesMultiply by the weights and use the assumed exponential relation. This expresses $-I_p$ as the sum of these integrals. On the finitely many fixed segments, their absolute values are at most $C^p/(p-1)!$ for a fixed $C$: all polynomial and exponential factors have fixed bounds, and $D$ is linear in $p$. Thus $|I_p|\to 0$, contradicting its being a nonzero integer. The obstruction is proved.

If $e$ were algebraic, an integral polynomial relation would give $a_0+{\sum }_{j=1}^s{a}_j{e}^j=0$ with $a_0\ne 0$. Integer exponents and weights satisfy the obstruction, so $e$ is transcendental.

If $\pi$ were algebraic, put $\alpha =i\pi$ and let ${\alpha }_1,\dots ,{\alpha }_d$ be its algebraic conjugates. Since $1+e^{\alpha }=0$, the product ${\prod }_j(1+e^{{\alpha }_j})$ vanishes. Expand it and collect equal subset sums. The zero sums contribute a positive integer $k\ge 1$; the nonzero sums give a Galois-stable list ${\beta }_j$ with positive integral multiplicities. This is precisely the prohibited relation. Hence $\pi$ is transcendental.

The general Lindemann–Weierstrass theorem states that exponentials of distinct algebraic numbers are linearly independent over $\overline{\mathbb{Q}}$. Suppose the sine quotient were algebraic, say $\gamma$. The denominator is nonzero: $\sin \beta =0$ would give a forbidden relation between the exponentials of the distinct algebraic numbers $i\beta ,-i\beta$. The quotient identity would giveThe four exponents are distinct under the stated nonzero and non-opposite assumptions, so this also contradicts linear independence. Therefore

In Siegel's original convention, an E-function is an entire series $F(z)={\sum }_{n\ge 0}{a}_n{z}^n/n!$ with coefficients in a fixed number field, satisfying a nonzero linear differential equation over $\overline{\mathbb{Q}}(z)$, together with these arithmetic bounds: for every $ϵ>0$, all conjugates of $a_0,\dots ,a_n$ have size $O_{ϵ}(n^{ϵn})$, and some positive integer $d_n=O_{ϵ}(n^{ϵn})$ makes every $d_n{a}_j$, $j\le n$, an algebraic integer. A frequently used stricter convention replaces both bounds by $C^{n+1}$. The original and strict growth definitions should not simply be declared equivalent; the exponential examples below satisfy the strict one.

For sums, work in the compositum of the coefficient fields. Coefficient conjugate bounds add. The product of the two common denominators clears both lists; using $ϵ/2$ in each original denominator bound gives the required $O_{ϵ}(n^{ϵn})$ product bound. Finally all derivatives of $F+G$ lie in the sum of the two finite-dimensional derivative spaces over that rational function field, so they satisfy a nontrivial linear dependence, giving a differential equation for the sum. Thus sums are again E-functions, the addition closure of Siegel E-functions.

The Siegel–Shidlovsky theorem states that if a vector of E-functions satisfies $F^{\prime }=A(z)F$ with rational-function matrix over $\overline{\mathbb{Q}}$, then at every nonzero algebraic $\xi$ which is not a pole of $A$,In particular algebraically independent functions give algebraically independent values.

Take $F_j(z)=e^{{\gamma }_{j}z}$ for algebraic numbers ${\gamma }_j$ linearly independent over $\mathbb{Q}$. Their coefficients are ${\gamma }_j^n$ with exponential conjugate and denominator bounds, and $F_j^{\prime }={\gamma }_j{F}_j$. A polynomial relation among the functions expands into a relation among exponentials with distinct exponents $\sum m_j{\gamma }_j$ and polynomial coefficients after clearing rational-function denominators; exponential-polynomial independence excludes it. At $\xi =1$ the theorem gives algebraic independence of the $e^{{\gamma }_j}$. For arbitrary distinct algebraic ${\alpha }_i$, choose a rational basis of their span and divide that basis by a common denominator. The $e^{{\alpha }_i}$ are distinct Laurent monomials in the corresponding algebraically independent exponential values, so are linearly independent over $\overline{\mathbb{Q}}$. This recovers the general Lindemann–Weierstrass theorem. Its simplest instance uses just $F(z)=e^z$ at nonzero algebraic $\xi$.

We first give an auxiliary-polynomial proof, rather than assuming a transcendence theorem for exponentials. Suppose $i\pi$ were algebraic, and let ${\delta }_1=i\pi ,{\delta }_2,\dots ,{\delta }_t$ be its conjugates. Since the first factor vanishes,Expand and collect equal subset sums of the ${\delta }_{\nu }$. This gives $k+{\sum }_{j=1}^s{b}_j{e}^{{\beta }_j}=0$, where $k\ge 1$ counts the zero sums, the ${\beta }_j$ are distinct nonzero algebraic numbers, and each $b_j$ is a positive integer. The exponents and their weights are invariant under every Galois permutation.

Choose a positive integer $c$ such that all $c{\beta }_j$ are algebraic integers. Their Galois-invariant monic polynomial $R(u)={\prod }_j(u-c{\beta }_j)$ has rational algebraic-integer coefficients, hence lies in $\mathbb{Z}[u]$. Its constant term $A=R(0)$ is a nonzero integer. For a prime $p$ defineAt each ${\beta }_j$, derivatives below order $p$ vanish. Every higher derivative is $c^{h}h!/(p-1)!$ times an algebraic integer, so $F_p({\beta }_j)$ is divisible by $p$ in the ring of algebraic integers. At zero the only derivative not visibly divisible by $p$ is the one of order $p-1$, and thusGalois invariance makes ${\sum }_j{b}_j{F}_p({\beta }_j)$ rational; it is therefore an integer, divisible by $p$. ConsequentlyFor every prime not dividing $kcA$, this is a nonzero integer.

The finite derivative sum satisfies $F_p^{\prime }-F_p=-f_p$, so integration along the straight segment to ${\beta }_j$ givesSum with weights $b_j$ and use $\sum b_j{e}^{{\beta }_j}=-k$. The result isOn these finitely many fixed compact segments, $|cz|$, $|R(cz)|$ and $|e^{-z}|$ are bounded independently of $p$. Hence $|S_p|\le C_1{C}_2^p/(p-1)!\to 0$. There are arbitrarily large primes not dividing the fixed integer $kcA$, contradicting $|S_p|\ge 1$. Therefore $\boxed{\pi \text{ is transcendental}}$. This is the Hermite auxiliary-polynomial proof for pi.

The more general Lindemann–Weierstrass theorem states that exponentials of distinct algebraic numbers are linearly independent over $\overline{\mathbb{Q}}$. Equivalently, exponentials of rationally independent algebraic numbers are algebraically independent over $\overline{\mathbb{Q}}$. In particular, the exponential of a nonzero algebraic number is transcendental.

Suppose $C=\cos \alpha \cos \beta$ were algebraic. The exponential formula for cosine givesAfter combining equal exponents this is an algebraic-coefficient linear relation among exponentials of distinct algebraic numbers. At least one exponent is nonzero, since $\alpha +\beta$ and $\alpha -\beta$ cannot both vanish when $\alpha ,\beta \ne 0$. Its coefficient is a positive integer, even when two exponents coincide. The relation is therefore nontrivial and contradicts the theorem. Thus $\boxed{\cos \alpha \cos \beta \text{ is transcendental}}$, including the cases $\alpha =\pm \beta$.

An integer version of Siegel lemma is as follows. For $M<N$ homogeneous linear equations $Az=0$, with integer matrix entries bounded in absolute value by $A_0\ge 1$, there is a nonzero integer solution satisfyingPut the displayed integer bound equal to $Q$ and evaluate the equations at every vector in $\{0,\dots ,Q{\}}^N$. There are $(Q+1{)}^N$ inputs and at most $(2N{A}_{0}Q+1{)}^M$ outputs. But $2N{A}_{0}Q+1\le 2N{A}_0(Q+1)$ and $(Q+1{)}^{N-M}>(2N{A}_0{)}^M$, so the number of inputs is larger. Two distinct inputs have identical outputs; their difference proves the claim. This is the pigeonhole proof of integer Siegel lemma. Rational coefficients are handled by clearing denominators. Over a number field of degree $d$, expanding in a rational basis gives at most $dM$ rational equations, so the same construction applies when $N>dM$, with the height of the resulting integer matrix controlling the bound.

Siegel's original arithmetic definition starts with an entire series $E(z)={\sum }_{n\ge 0}{a}_n{z}^n/n!$, with all $a_n$ in a fixed number field. For every $ϵ>0$ there must be positive integers $d_n$ such that $d_n{a}_0,\dots ,d_n{a}_n$ are algebraic integers, and both $d_n$ and the absolute values of every conjugate of $a_n$ are $O_{ϵ}(n^{ϵn})$. The growth condition ensures convergence everywhere. The stronger, now customary strict E-function definition uses bounds $C^{n+1}$ for a fixed $C$ and also includes the requirement of a nonzero linear differential equation over $\overline{\mathbb{Q}}(z)$. In the original formulation, that differential condition is explicitly part of the system hypothesis in the theorem below. These definitions should be distinguished; the exponentials used here satisfy even the stronger conditions.

The Siegel–Shidlovsky theorem states that if a vector of E-functions satisfies $f^{\prime }=A(z)f$, with $A(z)$ a matrix of rational functions over $\overline{\mathbb{Q}}$, then for every nonzero algebraic $\xi$ which is not a pole of $A$,In particular, algebraic independence of the functions implies algebraic independence of the values. The exclusion of zero is necessary: their Taylor values there are algebraic.

Take rationally independent algebraic ${\gamma }_1,\dots ,{\gamma }_s$ and $f_j(z)=e^{{\gamma }_{j}z}$. Their coefficients are ${\gamma }_j^n$, their conjugates grow exponentially, and powers of a fixed common denominator clear all coefficients through degree $n$. They satisfy the nonsingular diagonal system $f_j^{\prime }={\gamma }_j{f}_j$.

To check functional independence, a polynomial relation among these exponentials, after clearing rational-function denominators, would be ${\sum }_m{P}_m(z)e^{{\lambda }_{m}z}=0$ with distinct ${\lambda }_m={\sum }_j{m}_j{\gamma }_j$. Such an exponential polynomial cannot vanish identically. To isolate a nonzero term with exponent ${\lambda }_0$, apply ${\prod }_{\lambda \ne {\lambda }_0}(D-\lambda {)}^{\mathrm{deg}{P}_{\lambda }+1}$. All other terms vanish, while the remaining polynomial is acted on by operators $D+{\lambda }_0-\lambda$, each injective on polynomials because its nonzero constant preserves a nonzero leading coefficient. Thus the remaining term is nonzero, a contradiction.

The theorem at $\xi =1$ now makes $e^{{\gamma }_j}$ algebraically independent. For arbitrary distinct algebraic exponents ${\alpha }_i$, choose a rational basis of their span and divide its elements by a common denominator so that every ${\alpha }_i$ is an integer combination of the resulting ${\gamma }_j$. Then $e^{{\alpha }_i}$ are distinct Laurent monomials in $e^{{\gamma }_j}$. A linear relation among them becomes a nonzero polynomial relation after multiplying by a common monomial, contradicting independence. These particular E-functions therefore yield the full Lindemann–Weierstrass theorem.

Let a vector of E-functions satisfy $f^{\prime }=A(z)f$ with $A\in {\mathrm{Mat}}_s(\overline{\mathbb{Q}}(z))$. At a nonzero algebraic point $\xi$ which is not a pole of $A$, the transcendence degree of the values over $\overline{\mathbb{Q}}$ equals the transcendence degree of the functions over $\overline{\mathbb{Q}}(z)$. In particular algebraically independent functions give algebraically independent values. The exponential system $f_j=e^{{\gamma }_{j}z}$, for rationally independent algebraic ${\gamma }_j$, recovers the Lindemann–Weierstrass theorem by evaluating at one.