# I don't understand part of a proof of the Lindemann-Weierstrass theorem

I'm looking at this PlanetMath article on the Lindemann-Weierstrass theorem, specifically the dicussion right after equation (4). I think this is about the same as what appears below equation (5) in Baker's text, p6. There is a statement in there that I can't quite understand.

The article defines $$\alpha_1,...,\alpha_N$$ as the roots of an irreducible polynomial with integer coefficients, and integers $$\beta_1,...,\beta_N$$. It defines $$S_N$$ as the set of permutations of $$N$$ elements. It then considers the expansion. $$\prod_{\sigma \in S_N} (\beta_1 e^{\alpha_{\sigma(1)}} + ... + \beta_N e^{\alpha_{\sigma(N)}})$$ It notes that "[t]here are $$N!$$ factors in this product, so expanding the product, it is a sum of terms of the form $$e^{h_1\alpha_1 + ... h_N \alpha_N}$$ with integral coefficients, and $$h_1+...+h_N=N!$$. The set of all such exponents forms a complete set of conjugates." I'm good so far.

Then it says that "[b]y symmetry considerations, we see that the coefficients of two conjugate terms are equal." This is the part that I don't get. Any explanation of this is welcome, but the rest of the discussion might help you to see where my confusion is coming from.

It seems to me that, for each list of nonnegative integer h-values such that $$h_1+⋯+h_N=N!$$, the set of terms whose exponent has coefficients in this list (in some order) has two properties: a) each term has the same coefficient, b) the associated set of exponents, "$$R$$," is the set of roots of some rational polynomial, because $$\prod_{r \in R} (x-r)$$ is symmetric in the $$\alpha$$ values and the $$\alpha$$ values are a complete set of conjugates.

If that's true, I think it proves the bold statement above, but I'm not sure. As far as I can tell, the same exponent might appear multiple times, along with its conjugates, associated with different partitions of $$N!$$ into $$h_1,...,h_N$$. In this case I think I would need to collect terms together in order to show the truth of the bold statement.

Maybe it is easier if we remove some of the notation. Consider
$$F(a_1, ..., a_N) = \prod_{\sigma \in S_N} (\beta_1 a_{\sigma(1)} + ... + \beta_Na_{\sigma(N)})$$ and we want to know that if $$\tau \in S_N$$ (more precisely the galois group embedded in $$S_N$$) then the terms $$a_1^{h_1}...a_N^{h_N}$$ and $$a_{\tau(1)}^{h_1}...a_{\tau(N)}^{h_N}$$ have the same coefficients.

But the above polynomial $$F$$ is symmetric, that is, $$F(a_{\tau(1)}, ..., a_{\tau(N)}) = F(a_{1}, ..., a_{N})$$. This is becuase \begin{align*} F(a_{\tau(1)}, ..., a_{\tau(N)}) &= \prod_{\sigma \in S_N} (\beta_1 a_{\sigma(\tau(1))} + ... + \beta_N a_{\sigma(\tau(N))}) \\ &= \prod_{\gamma = \sigma\tau \; : \; \sigma \in S_N} (\beta_1 a_{\gamma(1)} + ... + \beta_N a_{\gamma(N)}) \\ &= \prod_{\gamma \in S_N} (\beta_1 a_{\gamma(1)} + ... + \beta_N a_{\gamma(N)}) \\ &= F(a_1, ..., a_N) \end{align*}

Hence the monomials $$a_1^{h_1}...a_N^{h_N}$$ and $$a_{\tau(1)}^{h_1}...a_{\tau(N)}^{h_N}$$ have the same coefficients.

In our case we may take $$a_i = e^{\alpha_i}$$.

• I. Thanks, that makes sense! I would like to partition your monomials into equivalence classes under this "subscript-permuting" relation. Call those equivalence classes {C_1,...C_k}, if that makes sense. If I understand correctly, you have shown that there are coefficients d_1,...d_k such that every monomial in C_i has the coefficient d_i for all i. Is that right? – capet Jun 28 at 18:33
• II. If I am correct in the above (I), then I think the following is also true: Let {C'_1,...C'_k} be analogous equivalence classes to the {C_1,...C_k} above, with a_i substituted with e^(α_i) as you described. Then for each i, the exponents of C'_i are the roots of a polynomial with rational coefficients, since all the α_i are conjugates. Is that right? – capet Jun 28 at 18:40
• @capet For (I), yes that is correct, one might view your $C_1,..., C_k$ as $S_N$ orbits of monomials. For II and III no not quite. You do not need elements to be conjugate to be the roots of a rational polynomial - only algebraic. Furthermore the exponents in C'_i are not nessicarily conjugate (but what we do want is if two exponents are conjugate they must both lie in C'_i). The point is that if $g$ is a field automorphism, then $g(h_1\alpha_1 + ... +h_N\alpha_N) = h_1 g(\alpha_1) + ... + h_N g(\alpha_N)$ and the $\alpha_i$ are all the conjugates so $g$ just permutes subscripts – Mummy the turkey Jun 28 at 21:29
• cntd... and we are then using the other definition of conjugates, as orbits under elements of the galois group of a normal closure. – Mummy the turkey Jun 28 at 21:33
• @capet I am using the different yet equivalent definitions of conjugate, see this wiki page en.wikipedia.org/wiki/Conjugate_element_(field_theory) (P.S. to typeset maths put the statement between \\$ signs) – Mummy the turkey Jun 28 at 22:03