Editing Lindemann–Weierstrass theorem (section)

====Equivalence of the two statements====

Baker's formulation of the theorem clearly implies the first formulation. Indeed, if <math>\alpha(1),\ldots,\alpha(n)</math> are algebraic numbers that are linearly independent over <math>\Q</math>, and

:<math>P(x_1, \ldots, x_n)= \sum b_{i_1,\ldots, i_n} x_1^{i_1}\cdots x_n^{i_n}</math>

is a polynomial with rational coefficients, then we have

:<math>P\left(e^{\alpha(1)},\dots,e^{\alpha(n)}\right)= \sum b_{i_1,\dots,i_n} e^{i_1 \alpha(1) + \cdots + i_n \alpha(n)},</math>

and since <math>\alpha(1),\ldots,\alpha(n)</math> are algebraic numbers which are linearly independent over the rationals, the numbers <math>i_1 \alpha(1) + \cdots + i_n \alpha(n)</math> are algebraic and they are distinct for distinct ''n''-tuples <math>(i_1,\dots,i_n)</math>. So from Baker's formulation of the theorem we get <math> b_{i_1,\ldots,i_n}=0</math> for all ''n''-tuples <math>(i_1,\dots,i_n)</math>.

Now assume that the first formulation of the theorem holds. For <math>n=1</math> Baker's formulation is trivial, so let us assume that <math>n>1</math>, and let <math>a(1),\ldots,a(n)</math> be non-zero algebraic numbers, and <math>\alpha(1),\ldots,\alpha(n)</math> distinct algebraic numbers such that:

:<math>a(1)e^{\alpha(1)} + \cdots + a(n)e^{\alpha(n)} = 0.</math>

As seen in the previous section, and with the same notation used there, the value of the polynomial

:<math>Q(x_{11},\ldots,x_{nd(n)},y_1,\dots,y_n)=\prod\nolimits_{\sigma\in S}\left(x_{1\sigma(1)}y_1+\dots+x_{n\sigma(n)}y_n\right),</math>

at

:<math>\left (a(1)_1,\ldots,a(n)_{d(n)},e^{\alpha(1)},\ldots,e^{\alpha(n)} \right)</math>

has an expression of the form

: <math>b(1)e^{\beta(1)}+ b(2)e^{\beta(2)}+ \cdots + b(M)e^{\beta(M)}= 0,</math>

where we have grouped the exponentials having the same exponent. Here, as proved above, <math>b(1),\ldots, b(M)</math> are rational numbers, not all equal to zero, and each exponent <math>\beta(m)</math> is a linear combination of <math>\alpha(i)</math> with integer coefficients. Then, since <math>n>1</math> and <math>\alpha(1),\ldots,\alpha(n)</math> are pairwise distinct, the <math>\Q</math>-vector subspace <math>V</math> of <math>\C</math> generated by <math>\alpha(1),\ldots,\alpha(n)</math> is not trivial and we can pick <math>\alpha(i_1),\ldots,\alpha(i_k)</math> to form a basis for <math>V.</math> For each <math>m=1,\dots,M</math>, we have

:<math>\begin{align}
\beta(m) = q_{m,1} \alpha(i_1) + \cdots + q_{m,k} \alpha(i_k), && q_{m,j} = \frac{c_{m,j}}{d_{m,j}}; \qquad c_{m,j}, d_{m,j} \in \Z.
\end{align}</math>

For each <math>j=1,\ldots, k,</math> let <math>d_j</math> be the least common multiple of all the <math>d_{m,j}</math> for <math>m=1,\ldots, M</math>, and put <math>v_j = \tfrac{1}{d_j} \alpha(i_j)</math>. Then <math>v_1,\ldots,v_k </math> are algebraic numbers, they form a basis of <math>V</math>, and each <math>\beta(m)</math> is a linear combination of the <math>v_j</math> with integer coefficients. By multiplying the relation

:<math>b(1)e^{\beta(1)}+ b(2)e^{\beta(2)}+ \cdots + b(M)e^{\beta(M)}= 0,</math>

by <math>e^{N(v_1+ \cdots + v_k)}</math>, where <math>N</math> is a large enough positive integer, we get a non-trivial algebraic relation with rational coefficients connecting <math>e^{v_1},\cdots,e^{v_k}</math>, against the first formulation of the theorem.