Hilbert's Tenth Problem over C(t) with named t

Hilbert's Tenth Problem over C(t) with named t

Definitions and mathematical statement

Let ${\mathbb{C}}(t)$ be the field of rational functions in an indeterminate $t$ with complex coefficients. The input is a finite polynomial \[ P(t,X_1,\ldots,X_n)\in{\mathbb{Z}}[t,X_1,\ldots,X_n], \] encoded by its integer coefficients and exponents. Decide whether there exist $x_1(t),\ldots,x_n(t)\in{\mathbb{C}}(t)$ with $P(t,x_1(t),\ldots,x_n(t))=0$ as a rational-function identity. The named element $t$ is part of the language, so this is not simply the algebraic-closedness problem for ${\mathbb{C}}$.

The target asks for one algorithm halting correctly for every such input, or an undecidability proof. The possible solutions may use arbitrary complex constants; those constants are not supplied as arbitrary uncomputable input data. The catalogue links this to the corresponding existential theory, with the language conventions of the cited source retained.

Formulation and concept references: [S1].

Short English statement

Can an algorithm decide whether a polynomial equation with integer coefficients and a named variable t has solutions in the complex rational-function field?

Research attempt

GPT-6 Astra Ultra; 27 September 2026. Status: UNRESOLVED.

The input has integer coefficients and a named transcendental element $t$, whereas a solution may use arbitrary complex constants. The latter permission does not prevent exact searches of bounded degree. I make that observation effective, prove the one-unknown case decidable, and identify the precise missing degree bound for the general problem. Nothing below gives either a decision algorithm in all numbers of unknowns or an undecidability reduction.

An exact test for solutions of bounded degree.

Fix an input $P(t,X_1,\ldots,X_n)$ and a nonnegative integer $D$. Represent a proposed solution by \[ X_i=A_i(t)/B_i(t),\qquad \deg A_i,\deg B_i\le D, \quad B_i\ne0. \] For each $i$, choose a possible denominator degree $d_i\in\{0,\ldots,D\}$ and normalize its leading coefficient to one. Write \[ B_i=t^{d_i}+\sum_{j<d_i}b_{ij}t^j,\qquad A_i=\sum_{j=0}^D a_{ij}t^j. \] The coefficients $a_{ij},b_{ij}$ are unknown complex numbers. There are finitely many choices of the tuple $(d_i)$. Let $e_i=\deg_{X_i}P$ and multiply the substituted expression by $\prod_i B_i^{e_i}$. The result is a polynomial in $t$ whose coefficients are polynomials over $\mathbb Z$ in the new unknown constants. Its being identically zero is a finite system of polynomial equations.

Groebner-basis ideal membership decides whether that system has a complex solution: by the weak Nullstellensatz it has one exactly when its ideal is proper. The normalization has already excluded zero denominators, so no extra interpretation of undefined rational expressions is needed. A numerator and denominator need not be coprime; allowing redundancy does not introduce false solutions. Conversely every bounded-degree rational solution has one of these normalized forms. Thus this is a terminating, exact bounded-degree test, not a numerical search for approximately vanishing coefficients.

There is a further consequence about constants. A proper ideal over $\mathbb Q$ remains proper over $\overline{\mathbb Q}$ and has a point there; equivalently one can apply the Nullstellensatz after extending to the algebraic closure. Properness over $\mathbb C$ and over $\overline{\mathbb Q}$ agrees by faithful scalar extension. Hence, for these inputs, \[ P=0\text{ has a solution in }\mathbb C(t) \quad\Longleftrightarrow\quad P=0\text{ has one in }\overline{\mathbb Q}(t). \tag{1} \] This does not restrict all solutions to algebraic constants. It asserts that one such solution exists whenever any complex-constant solution exists, because a finite coefficient variety is enough at its degree.

What the search proves, and what it lacks.

Running the bounded test for $D=0,1,2,\ldots$ semidecides positive instances. Every rational function has finite numerator and denominator degrees, so a genuine solution is eventually detected. A negative input need not produce any stopping signal. In particular, "enumerate all degrees and decide each one" is not a decision procedure.

The missing ingredient can be expressed as an exact equivalence. There is a decision algorithm for the stated problem if and only if there is a computable function $B(P)$ such that every soluble input has a solution with all numerator and denominator degrees at most $B(P)$. One direction follows by the bounded test. For the other, assuming a decision algorithm, first decide solubility. On a negative input return zero; on a positive input run the degree search until its first success and return that degree. This constructs a total computable bound. The equivalence does not provide the function. A bound on one chosen geometric presentation which is not effectively obtainable from $P$ would not settle the algorithmic question.

A complete subcase: one unknown.

Write \[ P(t,X)=a_m(t)X^m+\cdots+a_1(t)X+a_0(t), \qquad a_m\ne0. \] The zero polynomial is soluble, and a nonzero polynomial independent of $X$ is not. If $a_0=0$, then $X=0$ is a solution. Otherwise a rational root $A/B$, in lowest terms in the unique factorization domain $\mathbb C[t]$, satisfies \[ a_m A^m+a_{m-1}A^{m-1}B+\cdots+a_0B^m=0. \] Reduction modulo $A$ gives $A\mid a_0B^m$, whence $A\mid a_0$. Reduction modulo $B$ gives $B\mid a_m A^m$, whence $B\mid a_m$. Therefore \[ \deg A\le\deg a_0,\qquad \deg B\le\deg a_m. \tag{2} \] Taking $D=\max(\deg a_0,\deg a_m)$ in the bounded test decides this entire one-unknown subcase, in arbitrary degree in $X$. Unknown complex constant factors cause no problem because the coefficient test already handles them. This proof uses unique factorization only in the one-variable coefficient ring; it does not supply analogous bounds for coupled rational functions in several unknowns.

Stress tests on tempting field arguments.

The equation $X^2=t$ has no solution in $\mathbb C(t)$. The valuation at $t=0$ of a square is even, whereas that of $t$ is one. Algebraic closure of the constant field therefore does not make its rational function field algebraically closed. On the other hand, \[ X=(t+1)/2,\qquad Y=(t-1)/(2i) \] satisfy $X^2+Y^2=t$. The same obstruction does not survive an arbitrary increase in the number of unknowns. The displayed identity was checked by exact expansion, not by evaluating at finitely many values of $t$.

Another trap is to transfer the usual integer sum-of-squares encoding unchanged: over $\mathbb C$, $U^2+V^2=0$ does not imply $U=V=0$. For example $(1,i)$ is a counterexample. The finite coefficient systems used above are internal to a valid algebraically closed field algorithm; I have not claimed they can be replaced by a sum of squares in a reduction to the original single-equation format.

Source scope and outcome.

[S1] discusses Hilbert's tenth problem across different fields. The named parameter and computable coefficient ring are retained throughout this attempt; an algorithm allowed arbitrary unencoded complex coefficients as input would be a different formulation. Likewise a negative result for a different constant field or for functions of several independent parameters would require a valid reduction before it answered this case. The proved results are $1$, bounded-degree decidability, positive semidecidability, the computable-bound equivalence, and the one-unknown algorithm. The first unresolved step is an effective degree bound in arbitrary numbers of unknowns, or an obstruction proving that none exists. Neither is established here, so the requested full decision problem remains UNRESOLVED.

Sources

• S1 B. Poonen, Hilbert's tenth problem over rings of number-theoretic interest, Arizona Winter School notes, 2003. https://math.mit.edu/~poonen/papers/aws2003.pdf. Formulation source.
• E1 Bjorn Poonen, Hilbert's tenth problem over rings of number theoretic interest, Arizona Winter School notes $2003$. https://math.mit.edu/~poonen/papers/aws2003.pdf. Primary author notes corresponding to [S1], consulted for coefficient-field and function-field distinctions on 27 September 2026. The degree-bound reduction in this attempt is proved explicitly.