Schanuel's Conjecture

Schanuel's Conjecture

Definitions and mathematical statement

Numbers $z_1,\ldots,z_n$ are ${\mathbb{Q}}$-linearly independent when $\sum_jq_jz_j=0$ with $q_j\in{\mathbb{Q}}$ forces all $q_j=0$. The transcendence degree of a field extension is the maximum size of an algebraically independent subset. Schanuel's assertion is \[ \dim_{{\mathbb{Q}}}\operatorname{span}_{{\mathbb{Q}}}\{z_1,\ldots,z_n\}=n \quad\Longrightarrow\quad \operatorname{trdeg}_{{\mathbb{Q}}}{\mathbb{Q}}(z_1,\ldots,z_n,e^{z_1},\ldots,e^{z_n})\ge n. \] It is quantified over every $n\ge1$ and every such complex tuple; the $z_j$ themselves need not be algebraic.

Formulation and concept references: [S1].

Short English statement

Do linearly independent complex numbers, together with their exponentials, always contain at least as much algebraic independence as Schanuel predicts?

Research attempt

Status and scope.

This is an unresolved attempt, dated 27 September 2026. The aim is to identify exactly what a reduction to independent inputs accomplishes, prove the algebraic-input case using a named classical theorem, and test a proposed induction on exponential towers. The elementary reductions below are not claimed to be new. None gives the missing transcendence estimate for arbitrary complex inputs.

The external theorem used is Lindemann--Weierstrass: exponentials of distinct algebraic numbers are linearly independent over the algebraic numbers. A primary proof is [E1]. Ax's theorem [E2] concerns exponentiation in differential fields. Its functional scope is different from the numerical assertion here; no specialization principle is assumed. The power-tower reference [S1] is retained as context, not used to turn a conditional conclusion into a theorem.

1. Removing dependent coordinates without losing exponentials.

For a finite tuple $z=(z_1,\ldots,z_m)$, put \[ K(z)=\mathbb Q(z_1,\ldots,z_m,e^{z_1},\ldots,e^{z_m}),\qquad \delta(z)=\operatorname{trdeg}_{\mathbb Q}K(z) -\dim_{\mathbb Q}\langle z_1,\ldots,z_m\rangle. \] Choose a basis $b_1,\ldots,b_r$ of this rational span from among the $z_i$. For a common positive denominator $D$ there are integers $a_{ij}$ with $Dz_i=\sum_j a_{ij}b_j$. Hence \[ (e^{z_i})^D=\prod_{j=1}^r(e^{b_j})^{a_{ij}}. \tag{28.1} \] The right side belongs to $K(b)$, including when some exponents are negative, because exponentials never vanish. Each $z_i$ already belongs to $K(b)$ and each $e^{z_i}$ is algebraic over it. Conversely $K(b)\subseteq K(z)$ because the basis was selected from the tuple. Thus $K(z)/K(b)$ is algebraic and \[ \operatorname{trdeg}_{\mathbb Q}K(z) =\operatorname{trdeg}_{\mathbb Q}K(b),\qquad \delta(z)=\delta(b). \tag{28.2} \] There is no logarithm branch choice in this proof: $28.1$ follows from the single-valued exponential addition law. The assertion for independent tuples is therefore equivalent to $\delta(z)\geq0$ for every finite tuple, including dependent ones. Rank zero consists of zero inputs and gives $K(z)=\mathbb Q$.

This reformulation is useful bookkeeping, but it supplies no lower bound on the transcendence degree. In particular, selecting a rational basis does not make its entries algebraic. Any argument applying Lindemann--Weierstrass immediately after this selection would be using an unproved hypothesis.

2. Two unconditional subcases.

For one nonzero complex number $z$, the conjecture follows at once. If $z$ is transcendental, the field already contains one transcendental element. If $z$ is algebraic, Lindemann--Weierstrass applied to $0,z$ makes $e^z$ transcendental. This establishes the entire rank-one case, not just real or algebraic rank-one inputs.

Now suppose $b_1,\ldots,b_r$ are algebraic and rationally independent. Let $P\in\overline{\mathbb Q}[X_1,\ldots,X_r]$ be a nonzero polynomial, written with its distinct monomials as \[ P(X)=\sum_{\nu\in A}c_\nu X_1^{\nu_1}\cdots X_r^{\nu_r}, \qquad c_\nu\ne0. \] At the exponentials its value is \[ P(e^{b_1},\ldots,e^{b_r}) =\sum_{\nu\in A}c_\nu\exp(\nu_1b_1+\cdots+\nu_rb_r). \tag{28.3} \] The exponents are algebraic and distinct: equality for two indices would give the nontrivial rational relation $(\nu-\nu')\cdot b=0$. Lindemann--Weierstrass therefore makes $28.3$ nonzero. Consequently $e^{b_1},\ldots,e^{b_r}$ are algebraically independent, and $\operatorname{trdeg}_{\mathbb Q}K(b)=r$ since the inputs are algebraic. Together with $28.2$, this proves the rank formulation for every tuple of algebraic numbers. This familiar corollary is a genuine subcase, with its dependence on the classical theorem explicit.

The exact point of failure for transcendental inputs is visible in $28.3$: its exponents need no longer be algebraic. Their distinctness survives, but distinct complex exponents alone give no nonvanishing theorem for their exponential values. For example $0$ and $2\pi i$ are distinct and both exponentials equal $1$. That example is a stress test of the proposed use of the theorem, not a counterexample to Schanuel: the input $2\pi i$ itself contributes transcendence.

3. What a smallest counterexample would look like.

Suppose, conditionally, that the conjecture fails, and let $r$ be its smallest failing rank. The rank-one argument gives $r\geq2$. Choose an independent counterexample $b_1,\ldots,b_r$. Each $(r-1)$-coordinate subtuple satisfies the conjecture by minimality, so its field has transcendence degree at least $r-1$. Its field is a subfield of $K(b)$, whereas failure says the latter degree is less than $r$. Therefore necessarily \[ \operatorname{trdeg}_{\mathbb Q}K(b)=r-1,\qquad \delta(b)=-1. \tag{28.4} \] More generally every rational subspace of the span of dimension less than $r$ has nonnegative predimension, by choosing a basis and using minimality. Thus a least-rank obstruction would be a deficit of exactly one, invisible on all lower-dimensional rational subspaces. This narrows a hypothetical obstruction but does not rule it out. In particular, transcendence degree need not increase when a new pair $(b_r,e^{b_r})$ is adjoined; proving that it must increase is precisely where an induction would require new input.

One can also see why merely counting $2r$ generators is ineffective. A field generated by many named numbers can have much smaller transcendence degree because of polynomial relations. Rational independence excludes linear relations between the inputs, not arbitrary relations mixing inputs and outputs. Equation $28.4$ does not supply an algorithm for recognizing that mixing from decimal approximations.

4. A tower induction, with its hypothesis exposed.

Define real numbers $a_0=1$ and $a_{j+1}=e^{a_j}$. Assume Schanuel's conjecture. Then $a_1,\ldots,a_m$ are algebraically independent for every $m\geq1$. Here is the full induction. For $m=1$, $a_1=e$ is transcendental by the preceding theorem. Suppose $a_1,\ldots,a_{m-1}$ are algebraically independent. Any rational relation among $1,a_1,\ldots,a_{m-1}$ is a polynomial relation in the latter $m-1$ numbers, so all its coefficients vanish. Apply Schanuel to that $m$-tuple. The resulting field is exactly \[ \mathbb Q(1,a_1,\ldots,a_{m-1}, e,e^{a_1},\ldots,e^{a_{m-1}}) =\mathbb Q(a_1,\ldots,a_m). \] It has transcendence degree at least $m$, and it is generated by $m$ numbers, so those generators are algebraically independent.

This is a conditional consequence, not an induction proving Schanuel. Already the step from $e$ to $(e,e^e)$ invokes the conjecture on $(1,e)$, whose second input is transcendental. Positivity and rapid growth of the tower do not justify that invocation. The same distinction appears in the tuple $(1,i\pi)$: it is rationally independent by real and imaginary parts, and Schanuel would give transcendence degree at least two for $\mathbb Q(i\pi,e,-1)$. After adjoining the algebraic number $i$, this is equivalent to algebraic independence of $e$ and $\pi$. Their separate transcendence does not imply that conclusion.

5. A functional argument cannot be specialized for free.

It is possible to prove directly that the functions $t,e^t$ are algebraically independent over $\mathbb C$. Suppose $\sum_{j=0}^M P_j(t)e^{jt}=0$ with polynomial coefficients and $P_M\ne0$. Divide by $e^{Mt}$ and take real $t\to+\infty$. Every term with $j<M$ tends to zero, forcing the nonzero polynomial $P_M(t)$ to tend to zero, which is impossible. Thus the meromorphic function field $\mathbb C(t,e^t)$ has transcendence degree two over $\mathbb C$. In particular, $\mathbb Q(t,e^t)$ has transcendence degree two over $\mathbb Q$ as well, so the next comparison can keep the base field fixed.

At $t=1$, however, the corresponding numerical field $\mathbb Q(1,e)$ has transcendence degree only one over $\mathbb Q$. At $t=0$ both values are algebraic. These observations do not contradict the numerical conjecture; they show that independence of functions is not a license to retain the same degree at a prescribed point. Ax's differential theorem [E2] should therefore not be inserted at the missing numerical induction step. A valid specialization theorem would need additional arithmetic hypotheses and a proof that the chosen point satisfies them.

Verification and remaining gap.

The companion exact-arithmetic checks use the coefficient vectors of $\sqrt2,\sqrt3,(\sqrt2+\sqrt3)/2$. They confirm rational rank two and the cleared relation $2z_3-z_1-z_2=0$, which gives $(e^{z_3})^2=e^{z_1}e^{z_2}$. The independence of $\sqrt2,\sqrt3$ used to interpret these vectors is elementary: a nontrivial rational dependence would make $\sqrt{3/2}$ rational, contradicting the parity of prime valuations in a rational square. The script also checks distinct exponent vectors for all pairs of monomial indices between zero and eight in each coordinate. These finite checks test denominator handling and monomial bookkeeping only; the general injectivity proof is the argument in $28.3$. No finite search or numerical precision is reported as a transcendence test.

The outcome is an exact rank reduction, the full rank-one and algebraic-input subcases, a description of any least-rank failure, and explicit conditional consequences. The first unresolved step is a lower bound for $\operatorname{trdeg}\mathbb Q(b,e^b)$ when the independent inputs include transcendental numbers and their known transcendence does not already reach their rational rank. Neither the basis reduction, the tower construction, nor differential algebra fills that step. Schanuel's conjecture remains UNRESOLVED.

Sources

• S1 Schanuel's Conjecture and the Transcendence of Power Towers. https://www.mdpi.com/2227-7390/9/7/717. Formulation source.
• S2 Combining the conjectures of Schanuel and Zilber-Pink. https://content.ems.press/assets/public/full-texts/serials/rlm/36/3/14299588/online/10.4171-rlm-1086.pdf. Further reference; not a proof endorsement.
• S3 Schanuel Property for Elliptic and Quasi-Elliptic Functions. https://webusers.imj-prg.fr/~michel.waldschmidt/articles/pdf/SchanuelPropertyAlmostAll.pdf. Further reference; not a proof endorsement.

• E1 F. Beukers, J. P. Bézivin, and P. Robba, An Alternative Proof of the Lindemann--Weierstrass Theorem, American Mathematical Monthly 97 $1990$, no. 3, 193--197. https://doi.org/10.2307/2324683. Primary proof of the classical nonvanishing theorem used in the algebraic-input case. Full-text scan inspected at https://datuan5pdes.wordpress.com/wp-content/uploads/2016/12/10-23072324683.pdf. Consulted 27 September 2026.
• E2 James Ax, On Schanuel's conjectures, Annals of Mathematics 93 $1971$, no. 2, 252--268. https://annals.math.princeton.edu/1971/93-2/p03. Primary research reference for the differential-algebraic setting; the publisher's abstract was inspected to check scope. No numerical specialization theorem is inferred. Consulted 27 September 2026.

Schanuel's Conjecture

Definitions and mathematical statement

Numbers $z_1,\ldots,z_n$ are ${\mathbb{Q}}$-linearly independent when $\sum_jq_jz_j=0$ with $q_j\in{\mathbb{Q}}$ forces all $q_j=0$. The transcendence degree of a field extension is the maximum size of an algebraically independent subset. Schanuel's assertion is \[ \dim_{{\mathbb{Q}}}\operatorname{span}_{{\mathbb{Q}}}\{z_1,\ldots,z_n\}=n \quad\Longrightarrow\quad \operatorname{trdeg}_{{\mathbb{Q}}}{\mathbb{Q}}(z_1,\ldots,z_n,e^{z_1},\ldots,e^{z_n})\ge n. \] It is quantified over every $n\ge1$ and every such complex tuple; the $z_j$ themselves need not be algebraic.

Formulation and concept references: [S1].

Short English statement

Do linearly independent complex numbers, together with their exponentials, always contain at least as much algebraic independence as Schanuel predicts?

Research attempt

Current frontier.

The algebraic-input case is the Lindemann--Weierstrass theorem, but the entry quantifies arbitrary complex inputs. Waldschmidt's December 2025 manuscript proves the stronger independence of all $2n$ numbers for almost every complex tuple, and explicitly distinguishes this from the universal conjecture. [S3], Sections 1--2. Pila's recent paper combines Schanuel and multiplicative Zilber--Pink into an equivalent joint conjecture; it is not an unconditional proof of either. [S2], [E1]. Bertolin's June 2026 revision proposes further semi-elliptic analogues and proves special motivic cases; those hypotheses cannot be substituted for arbitrary complex exponential tuples. [E2]. The catalogue's MDPI page could not be retrieved in this run; the explicit statement supplied in the attachment and its matching formulation in Waldschmidt were used, not an invented replacement.

Attack route.

The first route is a minimal counterexample and rational change of basis: perhaps all proper rational subspaces would force incompatible algebraic constraints. The second is functional transcendence followed by specialization, suggested by the abundance of generic tuples. Its missing step is preventing transcendence degree from dropping at a specified arithmetic point. I prove the exact minimal-counterexample constraints, then test whether generic analytic independence actually provides that prevention. It does not. The deductions are recorded without claiming historical novelty for elementary predimension or specialization arguments.

Attempt.

For a finite-dimensional ${\mathbb{Q}}$-subspace $W\subset{\mathbb{C}}$ with basis $z=(z_1,\ldots,z_n)$, put \[ F_z={\mathbb{Q}}(z_1,\ldots,z_n,e^{z_1},\ldots,e^{z_n}),\qquad d(W)=\operatorname{trdeg}_{{\mathbb{Q}}}F_z-\dim_{{\mathbb{Q}}}W. \] The definition of $d(W)$ must first be shown independent of its basis.

Lemma 28.1 (basis invariance up to algebraic extension). If $w=Az$ for $A\in\operatorname{GL}_n({\mathbb{Q}})$, the compositum $F_zF_w$ is algebraic over each of $F_z$ and $F_w$. Consequently their transcendence degrees agree.

Proof. Choose a positive integer $D$ clearing the denominators of $A$. Each $w_i$ lies in $F_z$, and \[ (e^{w_i})^D=\prod_j(e^{z_j})^{DA_{ij}}\in F_z. \] Negative exponents are allowed because exponentials are nonzero. Thus $e^{w_i}$ is algebraic over $F_z$, as an actual complex root of the displayed equation. No arbitrary branch of logarithm is chosen. Applying the same argument to $A^{-1}$ proves the reverse algebraicity.$\square$

Lemma 28.2 (the form of a smallest counterexample). If the conjecture fails, choose a counterexample with $n$ minimal. Then $n\ge2$, \begin{equation}\label{wsref-0-72-32-38-3a-6d-69-6e-69-6d-61-6c} \operatorname{trdeg}_{{\mathbb{Q}}}F_z=n-1, \end{equation} and for every rational hyperplane $U\subset W$, the field generated by any basis of $U$ and its exponentials has transcendence degree $n-1$. In a basis extending that basis of $U$, the full field is algebraic over the hyperplane field.

Proof. For $n=1$, a transcendental $z\ne0$ already supplies transcendence degree at least one; for algebraic nonzero $z$, the Lindemann theorem supplies transcendental $e^z$. This is the one-dimensional case of the classical result discussed in [S3], Section 2. Hence $n\ge2$. Extend a basis of $U$ to a basis of $W$. By minimality, its first $n-1$ entries with their exponentials generate a field of transcendence degree at least $n-1$. The full field has degree at most $n-1$ by failure of the target and Lemma 28.1. Both degrees are therefore exactly $n-1$. Since the fields are finitely generated, equality of transcendence degrees makes the full extension algebraic. This holds for every $U$, not only the coordinate hyperplanes. $\square$

For $n=2$, this is quite restrictive: each nonzero rational direction $az_1+bz_2$ must individually yield a field of transcendence degree one, and the entire two-direction field still has degree one. But it is not a contradiction. Many distinct generators can live in the same transcendence-degree-one field: for instance ${\mathbb{Q}}(t)={\mathbb{Q}}(t+a)$ for every rational $a$ and transcendental $t$. That example does not implement the required exponential identities; it shows why field-degree counting alone cannot replace them. The missing step is a rigidity theorem for the actual simultaneous additive and exponential relations in (r28:minimal), not a further application of the tower formula.

Attempt to use generic analytic independence. Here is a self-contained version of the generic argument, consistent with the stronger almost-everywhere result of [S3].

Lemma 28.3 (no polynomial identity of the full exponential graph). For every nonzero $P\in{\mathbb{Q}}[X_1,\ldots,X_n,Y_1,\ldots,Y_n]$, the entire function \[ z\longmapsto P(z,e^{z_1},\ldots,e^{z_n}) \] is not identically zero.

Proof. Collect monomials in $Y$ to obtain $\sum_{a\in A}P_a(z)e^{a\cdot z}$, where $A\subset{\mathbb{Z}}_{\ge0}^n$ is finite and each $P_a$ is nonzero. Choose a real vector $v$ so that all real numbers $a\cdot v$ are distinct and none of the polynomials $P_a(tv)$ is identically zero. Such vectors exist: the first failures form finitely many hyperplanes, and for each nonzero $P_a$ one nonzero homogeneous part gives a proper algebraic exceptional set. A finite union of these proper real algebraic sets cannot fill ${\mathbb{R}}^n$. Restrict to $z=tv$. If the function vanished identically, divide by the exponential with largest exponent $a\cdot v$ and let real $t\to+\infty$. The coefficient polynomial of that term would tend to zero, since all other polynomial terms are exponentially damped. A nonzero polynomial cannot do so. This proves the assertion.$\square$

A nonzero holomorphic function on ${\mathbb{C}}^n$ has a Lebesgue-null zero set. For clarity, this follows inductively from the one-variable isolated-zero property and Fubini: expand in the last variable, choose a nonzero holomorphic coefficient in the other variables, exclude its null zero set, and then each remaining fiber has only a discrete zero set. Using countably many bounded regions handles unbounded space. There are countably many rational polynomials $P$. Thus outside a null set none of the polynomial relations in Lemma 28.3 holds, and \begin{equation}\label{wsref-0-72-32-38-3a-67-65-6e-65-72-69-63} \operatorname{trdeg}_{{\mathbb{Q}}}{\mathbb{Q}}(z_1,\ldots,z_n,e^{z_1},\ldots,e^{z_n})=2n \quad\hbox{for almost every }z\in{\mathbb{C}}^n. \end{equation} Such tuples are automatically ${\mathbb{Q}}$-linearly independent. This generic conclusion is substantially stronger numerically than the target's bound $n$, but weaker in its quantifier: it excludes an unspecified null set, which can contain all the arithmetic points of interest.

Why specialization does not finish the proof. The same exponential-polynomial argument shows that $t$ and $e^t$ are algebraically independent as functions over ${\mathbb{Q}}$. But specializing to $t=1$ yields the field ${\mathbb{Q}}(1,e)$, of transcendence degree exactly one, not two. Thus even a perfectly valid functional independence theorem can lose transcendence degree at a particular algebraic specialization. This example is compatible with Schanuel for $n=1$; it is a counterexample to the proposed no-drop inference, not to Schanuel.

A more concrete dependency warning comes from $(z_1,z_2)=(1,i\pi)$. These are ${\mathbb{Q}}$-linearly independent, because one is real and the other nonzero purely imaginary. Their exponentials are $e$ and $-1$, and \[ \operatorname{trdeg}_{{\mathbb{Q}}}{\mathbb{Q}}(1,i\pi,e,-1) =\operatorname{trdeg}_{{\mathbb{Q}}}{\mathbb{Q}}(\pi,e). \] Adjoining $i$ is algebraic, which justifies the equality. The target therefore implies algebraic independence of $e$ and $\pi$, not merely their individual transcendence. A proposed elementary specialization lemma that settles every tuple would have to establish this consequence too. Numerical recognition of a few alleged polynomial relations cannot supply a proof of either independence or its negation.

Stress test.

Lemma 28.1 uses powers of actual exponentials and is insensitive to logarithm branches; replacing it by arbitrary chosen logarithms would require an additional $2\pi i$ term. Lemma 28.2 invokes minimality only for strictly smaller linearly independent tuples, not for an unproved same-dimensional case. It proves algebraicity of the missing direction over each hyperplane field, not linearity or equality of those fields. Neither conclusion gives a differential equation: the numbers in the target are fixed constants, and differentiating them as if they were a variable analytic family is illegitimate.

The generic proof explicitly permits exceptional zeros of each entire function. Analytic continuation says a function vanishing on an open set vanishes identically; it does not say it cannot vanish at the particular point supplied by a hypothetical counterexample. Pila's joint-conjecture formulation and Bertolin's motivic program articulate additional deep structure rather than remove this obstacle by specialization. [E1], [E2]. No numerical experiment is offered as a transcendence certificate in this entry.

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Outcome.

Outcome: Exploratory approach with unresolved gap.

The attempt isolates a minimal counterexample's exact transcendence degree and its algebraic dependence over every rational hyperplane, and proves why generic functional independence cannot rule it out. These are useful constraints and explicit failure tests, not a new transcendence theorem for a previously unresolved arithmetic tuple. The unresolved task is a rigidity or specialization argument using the exponential structure that excludes all such minimal configurations. No complete proof or counterexample candidate for the universal conjecture has been obtained.

Sources

• S1 Schanuel's Conjecture and the Transcendence of Power Towers. https://www.mdpi.com/2227-7390/9/7/717. Catalogue formulation source.
• S2 Combining the conjectures of Schanuel and Zilber-Pink. https://content.ems.press/assets/public/full-texts/serials/rlm/36/3/14299588/online/10.4171-rlm-1086.pdf. Catalogue research/status reference; not a proof endorsement.
• S3 Schanuel Property for Elliptic and Quasi-Elliptic Functions. https://webusers.imj-prg.fr/~michel.waldschmidt/articles/pdf/SchanuelPropertyAlmostAll.pdf. Catalogue research/status reference; not a proof endorsement.
• E1 Jonathan Pila, Combining the conjectures of Schanuel and Zilber--Pink, Rendiconti Lincei--Matematica e Applicazioni 36 $2025$, 653--670, DOI 10.4171/RLM/1086, abstract and Sections 1--3. https://doi.org/10.4171/RLM/1086. Bibliographic completion of [S2]; PDF bears 2026 copyright. Equivalence of joint conjectures, not their proof. Consulted 27 September 2026.
• E2 Cristiana Bertolin, A conjecture in Schanuel style for 1-motives, arXiv:2509.08700v3, 29 June 2026. https://arxiv.org/abs/2509.08700v3. Abstract and stated special-case scope consulted 27 September 2026; not used as an unconditional premise of a notebook lemma.