Primitive idempotents that cannot be homogenized in Koszul algebras
Henry Zweiman
September 9, 2026
Abstract
We construct primitive idempotents that no algebra automorphism can carry to a homogeneous idempotent in a locally finite standard graded algebra with semisimple degree zero. The algebra is a matrix algebra over a two-dimensional domain defined by two quadrics, and is Koszul. This gives a negative answer to Dramburg’s Question 6.2, including over algebraically closed fields of characteristic zero. The same fixed algebra contains such an idempotent congruent to a constant matrix modulo any prescribed power of the positive-degree ideal. Each is conjugate to that constant matrix after completion. The proof combines an explicit matrix with a line-module obstruction, detected by its two trivializations over a Veronese ring and a polynomial quotient. Counterexamples exist over every field, in matrix size two outside characteristic two and size three in characteristic two.
1. The question and the construction
A nonnegatively graded -algebra is standard graded if it is generated by and , locally finite if each is finite dimensional, and semiconnected if is semisimple. Dramburg asks whether every complete set of primitive orthogonal idempotents of such an algebra can be carried to its degree-zero parts by an automorphism [3, Question 6.2]. His earlier question explicitly includes the possibility of assuming that is algebraically closed of characteristic zero [4]. We give counterexamples satisfying all these conditions, with the additional Koszul property.
The relation between idempotent matrices and projective modules is classical. Petersson describes line-module obstructions to automorphisms of two-by-two matrix algebras [6, Sections 5–6]. Kanwar, Leroy and Matczuk study conjugacy in polynomial and power-series extensions, and give a quaternion-polynomial example using Ojanguren–Sridharan [5, p. 9 of the author preprint]. De Jong also describes nonhomogeneous idempotents arising from line bundles in a Laurent extension [2]. Our contribution is an explicit split construction over every ground field, with a quadratic complete-intersection center, and an arbitrarily deep finite-jet failure in one fixed Koszul algebra. The underlying projective-module method is not new.
Fix a field and put
For , define four elements of by
and set
Write for the standard matrix units. In let
The notation for the zero block is omitted when .
Theorem 1.
Let be an integer whose image in is nonzero. For the ring and matrices above, the following statements hold.
is a two-dimensional standard graded domain and a quadratic complete intersection. The algebra is locally finite, standard graded and Koszul, with .
For every , the elements form a complete set of primitive orthogonal idempotents. Their degree-zero parts are .
No -algebra automorphism of carries to any homogeneous idempotent. In particular, no automorphism carries the displayed complete system to its degree-zero parts, even allowing a permutation.
We have . Nevertheless and are conjugate by a unit congruent to modulo in the -adic completion. Their images are conjugate in every quotient , .
Taking when , and otherwise, gives a counterexample over every field. In particular, works over . The theorem concerns arbitrary algebra automorphisms, not just conjugation by units or automorphisms fixing the center pointwise.
identifies with the indicated subring of . The two defining quadrics form a Gröbner basis. The ring is Koszul.
Proof.
Use lexicographic order . The leading monomials and are relatively prime, so Buchberger’s criterion gives the asserted Gröbner basis. Its standard monomials are
Their images in have exponent pairs
The first exponent modulo determines , since the four possibilities are . The two exponents then determine . Thus the standard monomials have distinct images, proving injectivity and the domain assertion.
Formula (4) also makes a free module of rank four over , with basis . Consequently , and the defining ideal has height two. Two generators of height two in a polynomial ring form a regular sequence, so this is a complete intersection. Finally, a standard graded algebra with a quadratic Gröbner basis is Koszul; see, for example, [1, Section 1].
Work henceforth in , using the identifications of Lemma 2. Introduce
Thus is the fourth Veronese subring. Give its degree-four monomials degree one. Let be the -span in of all its monomials having -exponent at least two. This is an ideal. In the quotient only the monomials and , , remain. Hence
The symbol in denotes the residue class of .
The inclusion followed by the quotient to sends to zero and to . It therefore agrees with the composition , where kills . We will use this commutative square of ring maps. We do not need flatness of over , a computation of , or a description of the full normalization of .
Both and have unit group . For , this follows because it is a connected positively graded domain: the highest nonzero degrees of two nonzero elements add under multiplication, so a product equal to forces both elements to have degree zero. The assertion for is immediate.
3. An explicit line-module obstruction
For an indeterminate , consider the matrix
Its determinant is . Direct multiplication gives
Put . The identities
show that (8), evaluated at , is precisely and has entries in . Since is injective, is idempotent over .
Set . The determinant and trace of are and . Over every residue field its image has dimension one; thus and are finitely generated projective modules of rank one. Here a line module means a projective module of rank one.
Proposition 3.
For every and every positive integer nonzero in , the line module is not free. The line modules , , are pairwise nonisomorphic.
Proof.
After base change to , the module has the basis vector
Indeed, conjugation in (8) identifies the image with the span of the first column of an invertible matrix. After base change to , the idempotent becomes , so has basis .
These identifications are compatible over . Since in , we have
Tensor powers of these basis vectors consequently differ by .
Suppose and choose a global basis. Over its expression is for some ; over it is for some . These scalars must be units because both expressions are bases. They lie in , by the unit computations above. Compatibility over gives
The left side is . The elements and are linearly independent over , and in , a contradiction.
Likewise, an isomorphism becomes multiplication by units in the two chosen trivializations. Compatibility implies an equality
Comparing the constant terms gives , and then in , so .
All base changes used here are applied to direct summands of free modules, and to a hypothesized isomorphism. They require no flatness assumption.
In characteristic zero, each has infinite order in . We have not computed this Picard group, and need no assertion about its other elements.
4. Excluding arbitrary automorphisms
The following matrix-unit argument is a useful form of the classical projective-module obstruction. It avoids an assumption that an automorphism fixes the center.
Lemma 4.
Let be a commutative ring and let be an idempotent whose image is a line module. If a ring automorphism of carries to , then .
Proof.
Let be such an automorphism and set . These matrices satisfy the matrix-unit identities, and . They yield a direct sum
Multiplication by is an -linear isomorphism from to , with inverse multiplication by . Thus . Taking the top exterior power gives .
The matrices act -linearly regardless of the action of on scalar matrices. In particular, the proof applies to automorphisms acting nontrivially on the center.
The ring statements follow from Lemma 2. The entrywise grading on is standard and locally finite, and is semisimple. For the Koszul assertion, take a linear graded projective resolution of over and apply the exact graded Morita functor
It gives a linear resolution of the simple -module , regarded as an -module. Taking copies gives a linear resolution of as a left -module, which is the Koszul condition with semisimple degree zero.
The identities for give the stated orthogonal decomposition of . Every displayed summand has image of rank one over . For a line module one has , as can be checked after localization. Since is a domain, this endomorphism ring has no nontrivial idempotents. Hence each displayed matrix idempotent is primitive. The degree-zero assertions follow directly from (2).
The image of is isomorphic to . Proposition 3 says that is not free. Lemma 4 therefore excludes an automorphism taking to . A nonzero homogeneous idempotent in a nonnegative grading has degree zero, since its square has twice its degree. A primitive idempotent in has rank one: any larger rank splits into orthogonal constant idempotents in . Every constant rank-one idempotent is conjugate over to . Composing with that constant conjugation proves the third assertion, including the statement about permutations of the complete system.
The four elements have respective degrees for . Thus , where equality uses the standard grading. Put and
Then , by idempotence. Also . Therefore the geometric series for converges in the -adic completion, and becomes a finite inverse in every quotient . In each such ring , proving the final assertion.
Corollary 5.
For the fixed algebra in Theorem 1 and every , there is a primitive idempotent congruent to modulo whose orbit under all algebra automorphisms contains no homogeneous idempotent. The family consists of pairwise distinct inner-conjugacy classes.
Proof.
Choose with . For the last assertion, an inner conjugation between and would induce an -linear isomorphism between their images and , contradicting Proposition 3 when .
5. Scope and consequences
The construction gives a negative answer to the homogenization question even for a prime algebra finite over its center, with center a two-dimensional quadratic complete-intersection domain. The obstruction is algebraic: completion removes it, and no fixed finite order of agreement with the degree-zero part detects it. This does not assert that the lie in distinct orbits under all automorphisms; we only need to exclude the homogeneous orbit for each of them.
There is no conflict with the uniqueness theorem for standard semiconnected locally finite gradings [3, Theorem 1.2]. That theorem compares gradings already given on an algebra. The primitive systems here cannot be made homogeneous by choosing another grading of that class: a graded isomorphism from such a grading to the present one would carry to a homogeneous idempotent, contrary to Theorem 1.
The center in (1) is not seminormal: belongs to its fraction field, since , and while . This failure provides the nonconstant nilpotent unit in the comparison ring . The example therefore does not settle analogous questions restricted to normal centers or to algebras of finite global dimension. These additional restrictions require separate arguments.
References
[1] A. Conca, Koszul algebras and Gröbner bases of quadrics, arXiv:0903.2397v1 (2009), Section 1. https://arxiv.org/abs/0903.2397.
[3] D. Dramburg, Isomorphisms of graded semiconnected algebras, arXiv:2609.03288v1 (2026), especially Definition 1.1, Theorem 1.2 and Question 6.2. https://arxiv.org/abs/2609.03288.
[4] D. Dramburg, Are idempotents in a nonnegatively graded algebra conjugate to homogeneous idempotents?, MathOverflow question 479788, September 29, 2024. https://mathoverflow.net/q/479788.