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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.

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  • $\begingroup$ This is the Schanuel conjecture $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Nov 23, 2016 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
    Commented Nov 23, 2016 at 6:31
  • $\begingroup$ @Amdeberhan Yes, you are right. $\endgroup$
    – joaopa
    Commented Nov 23, 2016 at 6:36
  • $\begingroup$ Oh I see now. Well, then... $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Nov 23, 2016 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
    Commented Nov 24, 2016 at 14:05

3 Answers 3

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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
    Commented Nov 23, 2016 at 6:46
  • $\begingroup$ Take $e^z$ and $e^{2z}$. Then, are these algebraically dependent or independent as "functions"? $\endgroup$
    – T. Amdeberhan
    Commented Nov 23, 2016 at 6:49
  • $\begingroup$ dependent, since on $\mathbb C(z)$ they are roots of the polynomial $P(X^2,Y)$ $\endgroup$
    – joaopa
    Commented Nov 23, 2016 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
    Commented Nov 23, 2016 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
    Commented Nov 23, 2016 at 7:16
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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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a)

Consider the values of your functions $e^{a_1z},...,e^{a_nz}$ at $z=1$.

Lindemann-Weierstrass theorem:
"If $a_1,...,a_n$ are algebraic numbers that are $\mathbb{Q}$-linearly independent, then $e^{a_1},...,e^{a_n}$ are $\mathbb{Q}$-algebraically independent."

With help of Lindemann-Weierstrass theorem, we find that the function values of your functions are $\mathbb{Q}$-algebraically independent at $z=1$. That there exists at least one place where your function values are $\mathbb{Q}$-algebraically independent means that your functions are $\mathbb{Q}$-algebraically independent.

b)

Lang, S.: Introduction to transcendental numbers. Addison-Wesley, 1966, p. 8:

"Let $\beta_1,...,\beta_m$ complex numbers, linearly independent over the rationals. Then the functions $$e^{\beta_1t},...,e^{\beta_mt}$$ are algebraically independent over the complex numbers."

For $\beta_1,...,\beta_m$ $\mathbb{Q}$-linearly independent algebraic numbers, the prerequisites are met, and your functions are $\mathbb{C}$-algebraically independent.

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