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First of all, a happy new year. Be it better than 2015, healthy, wealthy, fruitful and cross-fertilizing for you, familly and friends.

In order to cope with families of solutions of evolution equations, I had to prove the following lemma

Lemma: Let $Z=\{z_n\}_{n\in \mathbb{N}}$ be a set of indeterminates, then $[e^{z_0},e^{z_1}]$ is algebraically independent on $\mathbb{C}[Z]$ within $\mathbb{C}[[Z]]$. In other words, if a finitely supported sum $$ \sum_{n,m}P_{n,m}[(z_i)_{i\geq 0}]e^{nz_0}e^{mz_1} $$
is zero, then every polynomial $P_{n,m}\in \mathbb{C}[Z]$ is zero.

I cannot imagine it is unknown among specialists. My question is :

Does someone have a reference for this property ?

Thanks in advance.

On request, a proof below (I have withdrawn the - too long - previous one, using "orders of infinity".)

All relies on the following proposition which is characteristic free.

Proposition Let $(\mathcal{A},d)$ be a commutative differential ring without zero divisor, and $R=ker(d)$ be its subring of constants. Let $z\in \mathcal{A}$ such that $d(z)=1$ and $S=\{e_\alpha\}_{\alpha\in I}$ be a set of eigenfunctions of $d$ all different ($I\subset R$) i.e.

  1. $e_\alpha\not=0$
  2. $d(e_\alpha)=\alpha e_\alpha\ ;\ \alpha\in I$

Then the family $(e_\alpha)_{\alpha\in I}$ is linearly free over $R[z]$ (the subring generated by $R\cup \{z\}$, see remark).

From this, one can show the

Corollary Let $\mathcal{A}$ be a $\mathbb{Q}$-algebra (associative, commutative and unital) and $z$ an indeterminate, then $\{z,e^z\}\subset \mathcal{A}[[z]]$ are algebraically independent over $\mathcal{A}$.

Remark If $\mathcal{A}$ is a $\mathbb{Q}$-algebra or only of characteristic zero, i.e. $$ n1_\mathcal{A}=0\Rightarrow n=0 $$ then $d(z)=1$ implies that $z$ is transcendent over $R$. This is never the case in characteristic $p$ where $z^p$ is a constant. End of remark

One finishes the job proving, by successive extensions, that the sets
$$ \{e^{z_0},e^{z_1},\cdots e^{z_n},z_0,z_1,\cdots ,z_n\} $$
are algebraically independent over $\mathcal{A}$. Which is stronger than the desired result.

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  • $\begingroup$ I believe you can prove it by looking at the Wronskian and seeing that it doesn't vanish, at least this how it was shown to me in classes in ODE where the two exponent are solutions to the second order ode $a\ddot{x}+b\dot{x}+cx=0$ under certain condtions on the equation. $\endgroup$ – Alan Jan 4 '16 at 10:17
  • $\begingroup$ @Alan Does the Wronskian prove "algebraic" or linear independence only ? moreover I am chasing (I have one proof) for algebraic independence over $\mathbb{C}[Z]$ and not $\mathbb{C}$ (rather references) $\endgroup$ – Duchamp Gérard H. E. Jan 4 '16 at 10:57
  • $\begingroup$ It proves that they aren't linear dependent which I am not sure, but I believe you can use it also for algebraic dependence. I mean algebraic dependence means there aren't coefficients $a_0,a_1 \in \mathbb{C}$ s.t $a_0e^{z_0}+a_1 e^{z_1} = 0$. I see my mistake you look at polynomials in $Z$. I still believe you can construct an ODE for it and use the wronskian, but I am not sure how exactly. $\endgroup$ – Alan Jan 4 '16 at 11:14
  • $\begingroup$ @Alan [I mean algebraic dependence means there aren't coefficients]--->no, it is much stronger than that, I will make this precise in my question. Thank you for interaction. $\endgroup$ – Duchamp Gérard H. E. Jan 4 '16 at 12:21
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    $\begingroup$ Take z_i = R^{N^i} with N a sufficiently large positive integer so that for each m,n all the monomials in P_{m,n} become of distinct degrees in R, which is a real number that we're going to take to infinity. Then you get a sum of terms of the form R^c e{bR + aR^2} with a,b,c\geq 0. Taking the lexicographically maximal (a,b,c) you get that (const) R^c e^{bR + aR^2} = O(R^{c-1} e^{bR + aR^2}) if c > 0, so c = 0 if const\neq 0. Similarly b = 0 and a = 0 and you conclude. Hope I haven't made a mistake! $\endgroup$ – alpoge Jan 4 '16 at 20:08
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First of all, I am not technically answering your question because you ask specifically for a reference. However, here is a nice algebraic proof.

The result should follow immediately from the following lemma, assumed known for $k = \mathbb C$.

Lemma 1. Let $k$ be a field of characteristic $0$. Then the elements $z, e^z \in k[[z]]$ are algebraically independent over $k$.

Proof. If they are algebraically dependent, there exists a finitely generated (over $\mathbb Q$) subfield $k'\subseteq k$ over which they are algebraically dependent. But any finitely generated field of characteristic $0$ embeds into $\mathbb C$, contradicting the fact that the lemma is true over $\mathbb C$. $\square$

(This type of argument is known in algebraic geometry as the Lefschetz principle. I believe the first place where it appeared was Lefschetz's Algebraic Geometry textbook from the fifties, but I have not been able to verify that it didn't appear somewhere before. See also this answer on MO for a survey.)

Recall the following lemmata from transcendence theory.

Lemma 2. Let $\Omega$ be a big overfield, and let $k \subseteq \Omega$ be some small base field. Let $\alpha_1, \ldots, \alpha_n \in \Omega$. Then $\alpha_1, \ldots, \alpha_n$ are algebraically independent over $k$ if and only if $\operatorname{tr.deg} (k(\alpha_1,\ldots,\alpha_n)/k) = n$.

(Reference: this follows immediately from Theorem VIII.1.1 of Lang's Algebra.)

Lemma 3. Let $k \subseteq l \subseteq m$ be a tower. Then $\operatorname{tr.deg} (m/k) = \operatorname{tr.deg} (m/l) + \operatorname{tr.deg} (l/k)$.

(Reference: exercise in [loc. cit.].)

Proposition. The elements $z_0, \ldots, z_n, e^{z_0}, e^{z_1}$ are algebraically independent.

Proof. By Lemma 2, we have to prove that $\operatorname{tr.deg} (\mathbb C(z_0, \ldots, z_n, e^{z_0}, e^{z_1})/\mathbb C) = n+3$. We clearly have $$\operatorname{tr.deg}(\mathbb C(z_2,\ldots,z_n)/\mathbb C) = n-1,$$ since the $z_i$ are assumed to be algebraically independent. Letting $k = \mathbb C(z_2,\ldots,z_n)$, Lemma 1 tells us that $$\operatorname{tr.deg}(k(z_1, e^{z_1})/k) = 2.$$ Hence, by Lemma 3, we get $$\operatorname{tr.deg}(\mathbb C(z_1,\ldots,z_n,e^{z_1})/\mathbb C) = n+1.$$ Another application of Lemma 1 and Lemma 3 gives the result. $\square$

The result now follows, as you noted, because an algebraic dependence is defined over a finitely generated subextension.

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    $\begingroup$ I will accept your contribution as a fine answer after inspection (I have Lang's textbook at home). Thank you for having understood - and mentioned - that it was overall a reference request (+1 as a reaction for now). $\endgroup$ – Duchamp Gérard H. E. Jan 5 '16 at 5:26
  • $\begingroup$ [But any finitely generated field of characteristic 0 embeds into $\mathbb{C}$] I cannot believe this, dear correspondent. For example, I doubt $\mathbb{C}(X)$ can be embedded in $\mathbb{C}$. Can we fix ? $\endgroup$ – Duchamp Gérard H. E. Jan 5 '16 at 8:20
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    $\begingroup$ @DuchampGérardH.E. The field $\mathbb{C}(X)$ is not finitely generated over the prime field $\mathbb{Q}$. I advise you to search for "Lefschetz principle". $\endgroup$ – Jason Starr Jan 5 '16 at 11:53
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    $\begingroup$ But yes, $\mathbb C(X)$ does embed in $\mathbb C$. Every field of characteristic $0$ and cardinality at most $2^\omega$ does. More generally, if $K$ is an uncountable algebraically closed field, and $k$ a field of the same characteristic such that $|k|\le|K|$, then $k$ embeds in $K$. This all follows from the fact that two algebraically closed fields of the same characteristic and transcendence degree are isomorphic. $\endgroup$ – Emil Jeřábek Jan 5 '16 at 12:58
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    $\begingroup$ Right; often when something is said to be finitely generated without mention of over what it is finitely generated, this means it's finitely generated over the minimal thing possible. Likewise, a finitely generated ring (commutative with unit) would be a ring that is finitely generated as $\mathbb Z$-algebra, etc. Sometimes an author might write absolutely finitely generated to avoid any confusion. I have clarified this in my answer. $\endgroup$ – R. van Dobben de Bruyn Jan 5 '16 at 17:06
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Here is an analytic argument. Suppose that

$$ \sum_{j,k=0}^N P_{j,k}(z_0,z_1,z_2, \dotsc,z_n)e^{jz_0}e^{k z_1}=0. $$

Fix $z_2,\dotsc, z_n$ and $\newcommand{\ii}{\boldsymbol{i}}$ $\newcommand{\bR}{\mathbb{R}}$ let $z_0=\ii t_0$, $z_1=\ii t_1$, $\ii=\sqrt{-1}$, $t_0,t_1\in\bR$. We obtain an equality of the form $(\vec{z}=(z_2,\dotsc, z_n)$)

$$\sum_{j,k}P_{j,k}(t_0,t_1,\vec{z}) e^{\ii (jt_0+kt_1)}=0, $$

for all $t_0,t_1\in\bR$. Fix $t_1$. The functions $t^me^{\ii jt}$, $m,n\in\mathbb{Z}_{\geq 0}$ are linearly independent.(You can use Wronskians to prove this.) We deduce that

$$\sum_k P_{j,k}(t_0,t_1,\vec{z})e^{\ii k t_1}=0, $$

for all $j$, $t_0,t_1\in\bR$, $\vec{z}\in\mathbb{C}^{n-2}$. The same argument implies

$$ P_{j,k}(t_0,t_1,\vec{z})=0, $$

for all $j,k$, $t_0,t_1\in\bR$, $\vec{z}\in\mathbb{C}^{n-2}$.

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    $\begingroup$ Thank you but I wanted is overall a reference request (and - very secondarily - an elegant proof), I'l have a look, +1 for the contribution however. $\endgroup$ – Duchamp Gérard H. E. Jan 5 '16 at 6:34

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