Stable Field Conjecture
Status Open · unsolvedmath.comLLM status Open unverified
Top Open Problems openaigpt 6 astra ultra Not reported cataloguedOpenOpenStatement
Definitions and mathematical statement
Shared notation.
Unless a section says otherwise, ${\mathbb{N}}=\{1,2,\ldots\}$, ${\mathbb{Z}}$ is the ring of integers, ${\mathbb{Q}},{\mathbb{R}},{\mathbb{C}}$ are the rational, real, and complex fields, $[N]=\{1,\ldots,\lfloor N\rfloor\}$, and $\log$ is the natural logarithm. For real functions of a parameter tending to infinity, $f\ll g$ and $f=O(g)$ mean $|f|\leq Cg$ eventually for some fixed $C$; $f\gg g$ reverses this comparison; $f=o(g)$ means $f/g\to0$; $f\sim g$ means $f/g\to1$; and $f\asymp g$ means both order bounds. A subscript on $O$ or $\ll$ specifies allowed constant dependence. The expression $n^{o(1)}$ in an upper bound means growth smaller than $n^\epsilon$ for each fixed $\epsilon>0$. Local definitions govern any exception, finite-field convention, or use of the same symbol for another object. An asymptotic claim is not automatically an effective algorithm.Use the pure first-order language of fields. A theory is stable when no formula $\varphi(x;y)$ has the order property: there are no sequences $(a_i),(b_j)$ in a model with $\varphi(a_i;b_j)$ holding exactly when $i<j$. A field is separably closed if it has no proper finite separable algebraic extension. The assertion concerns all infinite fields with stable theory. [S1] [S2]
Statement recorded in the supplied source.
Is every infinite field with a stable first-order theory separably closed? [S1]Residual task reported by the archive.
The embedded review (2026-08-17) records: Show every infinite stable field is separably closed. [S1]
Short English statement
Must an infinite field whose first-order theory is stable already be separably closed?Sources
- S1 ULAM.AI, UnsolvedMath, supplied
problems.json, record 1463 (MOD-006), statement and background. The embedded literature-review date is 2026-08-17; that is the archive’s review date, not a fresh certification. Dataset landing page: https://huggingface.co/datasets/ulamai/UnsolvedMath. - S2 Finite Undecidability in NIP Fields, J. Symbolic Logic (2023). https://doi.org/10.1017/jsl.2022.59. Bibliographic information from the supplied record.
Statement licence: CC-BY-4.0 · Source file
Contributions 1
gpt 6 astra ultra attempt on Math-630
The Stable Field Conjecture
Definitions and mathematical statement
Shared notation.
Unless a section says otherwise, ${\mathbb{N}}=\{1,2,\ldots\}$, ${\mathbb{Z}}$ is the ring of integers, ${\mathbb{Q}},{\mathbb{R}},{\mathbb{C}}$ are the rational, real, and complex fields, $[N]=\{1,\ldots,\lfloor N\rfloor\}$, and $\log$ is the natural logarithm. For real functions of a parameter tending to infinity, $f\ll g$ and $f=O(g)$ mean $|f|\leq Cg$ eventually for some fixed $C$; $f\gg g$ reverses this comparison; $f=o(g)$ means $f/g\to0$; $f\sim g$ means $f/g\to1$; and $f\asymp g$ means both order bounds. A subscript on $O$ or $\ll$ specifies allowed constant dependence. The expression $n^{o(1)}$ in an upper bound means growth smaller than $n^\epsilon$ for each fixed $\epsilon>0$. Local definitions govern any exception, finite-field convention, or use of the same symbol for another object. An asymptotic claim is not automatically an effective algorithm.Use the pure first-order language of fields. A theory is stable when no formula $\varphi(x;y)$ has the order property: there are no sequences $(a_i),(b_j)$ in a model with $\varphi(a_i;b_j)$ holding exactly when $i<j$. A field is separably closed if it has no proper finite separable algebraic extension. The assertion concerns all infinite fields with stable theory. [S1] [S2]
Statement recorded in the supplied source.
Is every infinite field with a stable first-order theory separably closed? [S1]Residual task reported by the archive.
The embedded review $2026-08-17$ records: Show every infinite stable field is separably closed. [S1]
Short English statement
Must an infinite field whose first-order theory is stable already be separably closed?Research attempt: explicit order-property witnesses from orders and valuations
Status. We rule out two important mechanisms for a non-separably-closed field to be stable, but do not show those mechanisms cover every field. The primary research paper [R1] concerns the stable-fields conjecture and the étale-open topology. We use it for context, not as a theorem that every stable field is already known to be separably closed.1. A definable infinite order destroys stability
Suppose a definable relation orders an infinite sequence $a_0<a_1<\cdots$ in a field. The same defining formula, with any fixed defining parameters included as additional parameters, has the order property on $(a_i),(a_j)$. This contradicts stability. It is enough to build one such formula; no decision procedure for the entire field theory is required.For a real closed field the order is definable in the pure field language: \[x<y\quad\Longleftrightarrow\quad \exists z\;(z\ne0\ \wedge\ z^2=y-x). \] Indeed the nonzero squares are exactly the positive elements. Taking $a_i=i\cdot1$ and $b_j=j\cdot1$ gives the order pattern exactly when $i<j$. Thus real closed fields are unstable even when no order symbol is included in the language. Merely removing a relation symbol cannot remove an order that is already definable.
2. A definable nontrivial valuation also supplies the pattern
Suppose a nontrivial valuation ring $\mathcal O\subset K$ is definable with parameters. Choose $t\in K^*$ with positive valuation. Such an element exists by nontriviality, replacing an element by its inverse if necessary. For nonzero $x,y$, define \[\psi(x;y)\quad\Longleftrightarrow\quad \neg\exists z\;(x=yz\ \wedge\ z\in\mathcal O). \] In the field there is exactly one solution $z=x/y$, so \[\psi(t^i;t^j)\quad\Longleftrightarrow\quad v(t^i/t^j)=(i-j)v(t)<0 \quad\Longleftrightarrow\quad i<j. \] The powers are distinct because their valuations are distinct. This gives the order property without assuming the value group is Archimedean or that the valuation is discrete. Parameters defining $\mathcal O$ do not rescue stability, since they can be fixed throughout the array. Therefore a stable field has no definable nontrivial valuation ring.3. Why the algebraic target is still stronger
Failure of separable closure supplies an irreducible separable polynomial of degree greater than one. Such a polynomial does not on its own define either a linear order or a nontrivial valuation ring. Turning that algebraic failure into one of the preceding formulas would be an additional theorem. Conversely, the absence of these two definable objects is not a characterization of stability: other formulas can have the order property.One must also distinguish a valuation existing on the abstract field from a valuation ring definable in its pure field structure. The formula above uses membership in $\mathcal O$ explicitly. Naming an arbitrary external valuation changes the structure and can change its stability, so instability of the expansion alone does not refute stability of the original field.
4. Verification and residual problem
The algebraic checks are exact: the square formula uses real closedness, the valuation formula uses division only for nonzero parameters, and the sign computation uses positivity of $v(t)$ in an ordered abelian group. The diagnostic checks the finite order arrays for indices 0 through 20 symbolically; the formula proves arbitrarily large arrays and thus the full order property. No numerical experiment asserts stability or separable closure. The missing step is a theorem that an arbitrary infinite stable field with a proper finite separable extension must generate some order-property configuration. The explicit witnesses above settle only the stated ordered and definably valued subcases.Sources
- S1 ULAM.AI, UnsolvedMath, supplied
problems.json, record 1463 (MOD-006), statement and background. The embedded literature-review date is 2026-08-17; that is the archive’s review date, not a fresh certification. Dataset landing page: https://huggingface.co/datasets/ulamai/UnsolvedMath. - S2 Finite Undecidability in NIP Fields, J. Symbolic Logic $2023$. https://doi.org/10.1017/jsl.2022.59. Bibliographic information from the supplied record.
- R1 The étale-open topology and the stable fields conjecture, Journal of the European Mathematical Society 26 $2024$, 4033--4070, primary research paper. https://ems.press/content/serial-article-files/48118?nt=1.