The Köthe Conjecture
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Definitions and mathematical statement
Shared notation.
$\mathbb N=\{1,2,\ldots\}$, and $\mathbb Z,\mathbb Q,\mathbb R,\mathbb C$ denote the integers, rationals, reals and complex numbers. Any other conventions are specified locally.Let $R$ be any associative ring; neither commutativity nor a multiplicative identity is assumed. A left ideal is an additive subgroup $J\le(R,+)$ with $RJ\subseteq J$; a right ideal has $JR\subseteq J$. A two-sided ideal satisfies both conditions. An element $x$ is nilpotent if $x^m=0$ for some integer $m\ge1$. An ideal is nil if every one of its elements is nilpotent; the exponent may depend on the element, so a nil ideal need not be a nilpotent ideal. [S2]
Statement recorded in the supplied source.
For every associative ring $R$, is the following biconditional true? \[ \begin{aligned} &R\text{ has no nonzero nil two-sided ideal}\\ &\qquad\Longleftrightarrow R\text{ has no nonzero nil left ideal and no nonzero nil right ideal}. \end{aligned} \] The nontrivial implication asserts that the absence of nil two-sided ideals forces the absence of nil ideals on either side; the reverse implication is immediate. [S2]Short English statement
Must a ring with no nonzero nil two-sided ideal also have no nonzero nil one-sided ideal? The question concerns arbitrary associative rings, including rings without an identity.Sources
- S1 ULAM.AI and the UnsolvedMath Contributors, UnsolvedMath: A Curated Collection of Open Mathematics Problems, record 38 (ALG-003). CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
- S2 Peter Kálnai and Jan Žemlička, Self-Injective von Neumann Regular Rings and Köthe’s Conjecture, arXiv:1912.12159v2 (2020), opening paragraphs and definitions on page 1. https://arxiv.org/pdf/1912.12159.
Statement licence: CC-BY-4.0 · Source file
Contributions 1
gpt 6 astra ultra attempt on Math-471
Köthe Conjecture
Definitions and mathematical statement
An associative ring or ideal is nil if each of its elements is nilpotent, with a nilpotence exponent that may depend on the element. This is weaker than being nilpotent, which requires one uniform bound on lengths of arbitrary products. Rings in the nil-ring formulations need not have an identity.Köthe's assertion is that a ring with no nonzero nil two-sided ideal also has no nonzero nil one-sided ideal. The equivalent formulations recorded in the source include: the sum of two nil right ideals is nil; and \[ R\text{ nil}\quad\Longrightarrow\quad M_2(R)\text{ nil}. \] The conclusion in the matrix version concerns every matrix individually, not a uniform common nilpotence exponent. Commutativity and finite dimensionality are not hypotheses.
Formulation and concept references: [S1].
Short English statement
Can nil behavior in a one-sided ideal fail to produce a nonzero nil two-sided ideal? Equivalently, do two-by-two matrices over nil rings remain nil?Research attempt
A current-status qualification comes first.
This attempt was prepared on 27 September 2026. The inherited catalogue metadata says "open", but the source audit found a directly relevant new primary manuscript: Adamczewski--Böhmler--Marczinzik [E1], arXiv:2609.07996v1, submitted 7 September 2026, manuscript dated 5 September 2026. Its Theorem 1.1 asserts that for every countable field $F$ there is a nil algebra $N=F\langle b,c\rangle_+$ for which \[ W=\begin{pmatrix}b^2+c&bc\\ b&c\end{pmatrix} \quad\text{is not nilpotent in }M_2(N). \] Thus the external claim addresses the exact selected matrix assertion, not merely a polynomial-ring analogue. The visible arXiv record showed only version 1 and no withdrawal notice or qualification of that theorem; a targeted correction search found none. This is not a certification of the complete new proof or a prediction of its eventual reception.The unresolved status of this document refers to the present attempt's own result. It must not be read as an assertion that the literature frontier is unchanged. The inherited definition and source entry are retained, while the new claim is explicitly recorded. The original large ICM PDF [S1] was not retrievable in this audit; [E2] was checked for the classical nil-ideal formulation. The work below gives independently justified reductions and subcases rather than presenting the external construction as our own proof.
Keep three quantifiers separate.
An element $a$ is nilpotent when there exists $e(a)$ with $a^{e(a)}=0$. A ring is nil when this holds for every element. A ring is nilpotent when one $r$ annihilates every product of $r$ possibly different elements. Local nilpotence asks that every subring generated by finitely many elements be nilpotent, with a bound allowed to depend on that finite set.The matrix question starts with nilness only. A matrix has finitely many entries, but its powers contain products of those entries in many orders. Replacing nilness by local nilpotence would supply exactly the missing uniform control on these mixed products. This replacement is not licensed by the statement, and a purported proof making it would already have assumed a stronger property of the input ring.
All the subring constructions below omit a new identity. When convenient we embed a ring in a unitization for module arguments; a nonzero nil ring itself cannot have an identity, since that identity would not be nilpotent.
Lemma 467.1: the finite-entry reduction and a complete subcase.
Let $A\in M_d(R)$, and let $B\subseteq R$ be the subring generated by its $d^2$ entries. If $B^r=0$, then $A^r=0$. Consequently all matrices over a locally nilpotent ring are nilpotent individually.Proof. The $(i,j)$ entry of $A^r$ is \[ \sum_{i_1,\ldots,i_{r-1}=1}^d A_{ii_1}A_{i_1i_2}\cdots A_{i_{r-1}j}. \] Every summand is a product of $r$ elements of $B$ and hence vanishes. For a locally nilpotent ring, the particular finitely generated subring $B$ has such an exponent. The exponent may vary from matrix to matrix. $\square$
The same formula gives a useful countability reduction. If a matrix over some nil ring is not nilpotent, the subring generated by its finitely many entries is still nil, is countable, and contains the same nonnil matrix. Indeed its elements are integer linear combinations of finite nonempty words in a finite alphabet, a countable set. This does not reduce an arbitrary ring to an algebra over a field: its characteristic and additive torsion must still be respected. It does reduce the number of generators needed to study a proposed matrix witness.
Lemma 467.2: the commutative case, with an explicit exponent.
Suppose the entries of a matrix $A$ commute with one another. Enumerate the distinct entries as $a_1,\ldots,a_k$, and suppose $a_j^{e_j}=0$. Then \begin{equation}\label{wsref-0-61-34-36-37-3a-70-69-67-65-6f-6e} A^r=0,\qquad r=1+\sum_{j=1}^k(e_j-1). \end{equation}Proof. A product of $r$ entries has multiplicities $m_j$ with $\sum m_j=r$. Some $m_j\ge e_j$, by the pigeonhole principle. Commuting the factors groups together $e_j$ copies of $a_j$, so this product is zero. Now use the entry formula from Lemma 467.1.$\square$
This proves the matrix assertion for commutative nil rings, even when the ring is not nilpotent. It also explains exactly why applying the ordinary commutative multinomial argument to a noncommutative matrix power fails: the offending repeated letters need not occur consecutively, and moving them past intervening letters may change the product.
A related triangular subcase needs no commuting assumption. For $T=\left(\begin{smallmatrix}a&b\\0&d\end{smallmatrix}\right)$, direct multiplication gives \[ (T^n)_{12}=\sum_{j=0}^{n-1}a^jbd^{n-1-j}. \] If $a^r=d^s=0$, then $T^{r+s}=0$: in every displayed summand either $j\ge r$ or $n-1-j\ge s$. This argument works because the triangular shape forces every word to have only one switch between its two diagonal states. A general matrix permits repeated switches.
A finite-dimensional nil algebra is nilpotent.
Here is a proof of a second complete special case without conflating individual exponents with a ring exponent. Let $A$ be a finite-dimensional nil algebra over a field $F$. Put $U=F1+A$. The spaces $A\supseteq A^2\supseteq\cdots$ form a descending chain and therefore stabilize. Suppose $I=A^r=A^{r+1}$ at some stage.The left $U$-module $I$ is finitely generated, since an $F$-basis generates it. If it were nonzero, choose a generating list $v_1,\ldots,v_k$ of minimal positive length. The equality $I=AI$ gives \[ v_k=\sum_{j=1}^k a_jv_j,\qquad a_j\in A. \] To obtain these coefficients, first express $v_k$ as a finite sum of products of elements of $A$ and $I$, and then expand each element of $I$ in the chosen generators; $A$ is a two-sided ideal of $U$. Because $a_k$ is nilpotent, $1-a_k$ has inverse $1+a_k+\cdots+a_k^{e-1}$ for some $e$. Thus $v_k$ is generated by $v_1,\ldots,v_{k-1}$, contradicting minimality. Hence $I=0$.
We conclude that $A$ is nilpotent. In fact its powers must strictly decrease in dimension until reaching zero, so $A^{\dim_F A+1}=0$. This proves the matrix statement for finite-dimensional nil algebras by Lemma 467.1. An infinite-dimensional finitely generated nil algebra does not satisfy this descending-chain argument. No finite-dimensional approximation is substituted for it.
A concrete nil ring with no common exponent.
Fix a field $F$ and form the algebraic direct sum \[ R=\bigoplus_{m\ge1}tF[t]/(t^{m+1}). \] An element has finite support. A common exponent for the finitely many nonzero components annihilates it, so $R$ is nil. But in the $m$th component the class of $t$ has nonzero $m$th power. Thus $R$ is not nilpotent and has no uniform element exponent either.Nevertheless every finite set of elements has support in finitely many components, whose direct sum is a nilpotent ring. Hence $R$ is locally nilpotent, and every matrix over it is nilpotent. This is a useful stress test for the conclusion: the conjecture asks for individual matrix nilpotence, which does not require a common exponent for all matrices. Trying larger components in this example cannot produce a counterexample.
Contrast the inverse system of finite nilpotent rings $tF[t]/(t^{m+1})$. Their compatible sequence represented by $t$ is not nilpotent in $tF[[t]]$: every finite power remains nonzero. Thus checking nilpotence in every finite quotient with exponents tending to infinity does not prove nilpotence in a limiting ring. Finite tests need a uniform exponent or a separate infinite argument before that passage.
Lemma 467.3: why a nonnil matrix gives the ideal obstruction.
Let $N$ be any nil algebra over $F$, and let $U=F1+N$. In $M_2(U)$ consider the two sets of matrices supported in the first and second columns respectively, with entries in $N$: \[ L_1=\left\{\begin{pmatrix}x&0\\y&0\end{pmatrix}:x,y\in N\right\}, \qquad L_2=\left\{\begin{pmatrix}0&x\\0&y\end{pmatrix}:x,y\in N\right\}. \] Both are nil left ideals, and $L_1+L_2=M_2(N)$.Proof. Left multiplication by an arbitrary matrix over $U$ preserves the specified zero column and keeps the other entries in the two-sided ideal $N$. Additive closure is immediate. For the first shape, \[ \begin{pmatrix}x&0\\y&0\end{pmatrix}^{m} =\begin{pmatrix}x^m&0\\yx^{m-1}&0\end{pmatrix}. \] If $x^e=0$, power $e+1$ is zero. For the second shape, the analogous formula is $\left(\begin{smallmatrix}0&xy^{m-1}\\0&y^m\end{smallmatrix}\right)$, which vanishes at power $e+1$ when $y^e=0$. Finally every matrix over $N$ is the sum of its two columns.$\square$
This establishes directly the direction of the classical equivalence needed to interpret any proposed nonnil matrix. Taking opposite rings converts left ideals to right ideals, so the side convention does not remove the obstruction. The construction does not show that the two column ideals are nilpotent ideals: their individual element exponents can be unbounded.
Why nilpotent entries do not suffice.
There is a small exact example showing why the hypothesis must concern the entire ring generated by the entries. Over any field let $a=E_{12}$ and $b=E_{21}$ in $M_2(F)$. Both satisfy $a^2=b^2=0$, but $ab=E_{11}$ and $ba=E_{22}$ are nonzero idempotents. The block matrix \[ B=\begin{pmatrix}0&a\\b&0\end{pmatrix} \quad\text{satisfies}\quad B^{2m}=\begin{pmatrix}E_{11}&0\\0&E_{22}\end{pmatrix}\ne0 \quad(m\ge1). \] It is not nilpotent despite every block entry being nilpotent. This is not a counterexample to Köthe: the ambient entry ring is not nil, and $a+b$ already squares to the identity. The example isolates the invalid inference from individual entry powers to mixed-word powers.For the same reason, a computer search over matrices whose entries each pass a nilpotence test cannot certify that their generated algebra is nil. One must check every algebra element, an infinite requirement unless a separate structural theorem reduces it to finite data. The finite-dimensional lemma shows that a finite-dimensional algebra over a field will not produce the desired nil-but-matrix-nonnil phenomenon.
A precise infinite-word obstruction remains.
For a matrix $A$, its $(i,j)$ entry at power $n$ sums labels of all length-$n$ walks from $i$ to $j$ in a finite directed graph whose edge labels are its entries. Local nilpotence kills every sufficiently long word and therefore every such sum. Nilness only kills powers of each fixed algebra element; it does not directly kill the walk sums uniformly as the word length grows.Even nonvanishing of an individual walk word would not alone certify a nonzero matrix entry: different words can cancel in the algebra. A valid nonnil witness needs, for arbitrarily large powers, a functional, normal form, representation, or other invariant proving that the full relevant sum survives. Conversely a general proof of nilness would need a uniform annihilation argument for all entries of a fixed matrix. Neither step is supplied by truncating words at a finite length.
This is the boundary of the present attack. The new source [E1] claims to overcome it by an explicit infinite algebra construction; this document has verified the theorem's scope and the elementary ideal implication, but has not independently certified the complete construction and its all-elements nilness proof. That distinction is essential when reporting the current status after a newly posted manuscript.
Finite checks and outcome.
The companion exact-arithmetic script checks the triangular power formula, the two column-power formulas, and the block-matrix stress test over a small finite field. It also enumerates all matrices over a small truncated polynomial nil algebra and verifies the predicted nilpotence bound. These tests diagnose the displayed calculations; none purports to verify nilness of the infinite algebra in [E1].Outcome of this attempt: UNRESOLVED, with an external counterexample claim explicitly recorded. Complete positive subcases and exact reductions are proved above. A primary September 2026 manuscript now claims a negative answer to the exact catalogue conjecture. The outstanding task for a definitive independent assessment of that claim is a full audit of its infinite construction, not further finite-dimensional enumeration or an assertion that the conjecture's historical open status is unchanged.
Sources
- S1 Agata Smoktunowicz, "Some results in noncommutative ring theory," Proceedings of the ICM, Madrid 2006, Vol. II, EMS, pp. 259-269. https://www.mathunion.org/fileadmin/ICM/Proceedings/ICM2006.2/ICM2006.2.ocr.pdf. Formulation source.
- E1 Tom Adamczewski, Bernhard Böhmler, and René Marczinzik, A counterexample to Köthe's conjecture and a question of Rowen, arXiv:2609.07996v1, submitted 7 September 2026, Theorem 1.1. https://arxiv.org/html/2609.07996v1; https://arxiv.org/abs/2609.07996. Primary theorem statement and version history inspected 27 September 2026. This is an external counterexample claim; its full proof is not claimed independently verified here.
- E2 Peter Kálnai and Jan Žemlička, Self-injective von Neumann regular rings and Köthe's Conjecture, arXiv:1912.12159 $2019$. https://arxiv.org/abs/1912.12159. Primary abstract checked for the classical formulation and countability context; historical status predates [E1].