Let $a_1,\ldots,a_n$ be $\mathbb Q$-linearly independant algebraic numbers. Are the functions $e^{az},\ldots,e^{a_nz}$ algebraically independent functions (over $\mathbb C(z)$ or $\mathbb Q(z)$)?

I ask this because I wonder whether the Lindemann-Weierstrass theorem is a consequence of the Siegel-Shidlovskii theorem.

Thanks in advance for any hints.

  • $\begingroup$ This is the Schanuel conjecture $\endgroup$ – მამუკა ჯიბლაძე Nov 23 '16 at 6:01
  • $\begingroup$ I suppose the OP's question is much "weaker" because he/she is asking about functions not numbers. Am I right? $\endgroup$ – T. Amdeberhan Nov 23 '16 at 6:31
  • $\begingroup$ @Amdeberhan Yes, you are right. $\endgroup$ – joaopa Nov 23 '16 at 6:36
  • $\begingroup$ Oh I see now. Well, then... $\endgroup$ – მამუკა ჯიბლაძე Nov 23 '16 at 11:00
  • $\begingroup$ A concise exposition of this can be found in Alan Baker's Transcendental number theory, chapter 11 — The Siegel-Shidlovsky theorem. In particular, I quote the last sentence of §1: “Plainly also Theorem 11.1 includes Lindemann's theorem.” $\endgroup$ – ACL Nov 24 '16 at 14:05

This can be checked by computing the Wronskian which results in the determinantal valuation $$e^{(a_1+\cdots+a_n)z}V(a_1,\dots,a_n)$$ where $V:=V(a_1,\dots,a_n)$ is the determinant of the Vandermonde matrix $$V=\prod_{1\leq i<j\leq n}(a_j-a_i).$$ So, it can only vanish when there are duplicates $a_i=a_j$ for some $i,j$. There is no need to have algebraic independence condition on the numbers. If the $a_i$'s are distinct then the Wronskian does not vanish identically, hence the functions are linearly independent.

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    $\begingroup$ For me, the Wronskian leads only to linear independances, not algebraic independance. Am I wrong? $\endgroup$ – joaopa Nov 23 '16 at 6:46
  • $\begingroup$ Take $e^z$ and $e^{2z}$. Then, are these algebraically dependent or independent as "functions"? $\endgroup$ – T. Amdeberhan Nov 23 '16 at 6:49
  • $\begingroup$ dependent, since on $\mathbb C(z)$ they are roots of the polynomial $P(X^2,Y)$ $\endgroup$ – joaopa Nov 23 '16 at 6:52
  • $\begingroup$ Then, given $e^{az}$ and $e^{bz}$, we write $e^{az}=e^{(a-b)z}e^{bz}$. Dependent? $\endgroup$ – T. Amdeberhan Nov 23 '16 at 6:54
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    $\begingroup$ The OP asked for algebraic independence. To achieve that, you need to consider an arbitrary polynomial relation $P(e^{a_1z},\dots,e^{a_n z})=0$ between these functions, and prove that $P=0$. However, expanding $P$ and using the functional equation $e^{(a+b)z}=e^{az}e^{bz}$ reduces the relation $P(e^{a_1z},\dots,e^{a_n z})=0$ to a linear relation between functions of the form $e^{bz}$, where $b$ is of the form $m_1 a_1+\dots+m_n a_n$. If $a_1,\dots,a_n$ are $\mathbf Q$-linearly independent, there is no possible cancellation, hence $P=0$ by the linear independence criterion. $\endgroup$ – ACL Nov 23 '16 at 7:16

The numbers $a_j$ do not have to be algebraic.

Theorem. If $a_j,\; 1\leq j\leq n$ are $Q$-linearly independent then $e^{a_jz}$ are algebraically independent over $C(z)$

Proof. Let $$F(x_1,\ldots,x_n)=\sum_j c_j(z)x_1^{m_{j,1}}\ldots x_n^{m_{j,n}},$$ where $c_j\in C(z)$, and $(m_{j,1}\ldots,m_{j,n})\neq(m_{k,1}\ldots,m_{k,n})$ for every pair $j\neq k$. Suppose that $$F(e^{a_1z},\ldots,e^{a_nz})=\sum_jc_j(z)e^{z\sum_km_{j,k}a_k}\equiv 0.$$ All exponentials here are distinct, because $a_j$ are $Q$-linearly independent. Then we have a contradiction from the asymptotics in the complex plane: if there is only one exponent of the largest modulus, its growth dominates the rest in certain directions. If there are several, their arguments are different and each dominates in certain direction.

Remark. Just noticed that this essentially coincides with ACL's comment:-)

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