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Non-commutative rings and algebras, non-associative algebras, universal algebra and lattice theory, linear algebra, semigroups. For questions specific to commutative algebra (that is, rings that are assumed both associative and commutative), rather use the tag ac.commutative-algebra.

1 vote

GCD computation for multiple polynomials and degree of Bezout coefficients

You can just recursively compute the gcd of $t\ge 2$ polynomials by $$ \gcd(P_1, P_2, \dots , P_t) = \gcd( P_1, \gcd(P_2, \dots , P_t)), $$ starting with $\gcd(P_{t-1},P_t)$, $\gcd(P_{t-2}, \gcd(P_{t- …
Dietrich Burde's user avatar
2 votes

Matrix Inverse with Same Principal Minors

Let $A=(a_{ij})$. Then we obtain a system of polynomial equations in the variables $a_{ij}$, given by determinant conditions $\det A(\alpha)=\det A^{-1}(\alpha)$. If one determines a Groebner basis f …
Dietrich Burde's user avatar
20 votes

Are there only finitely many associative algebras of fixed dimension?

Even for commutative associative algebras it is not true. The article of Björn Poonen "Isomorphism types of commutative algebras of finite rank over an algebraically closed field" gives a classificati …
Dietrich Burde's user avatar
1 vote

Efficient Algorithm for Matrix Version of Waring's Problem

The result mentioned above is Waring's problem for matrices (respectively for algebraic number fields). The result is: Theorem (Katre, Kuhle 1990): Let $R$ be an order in an algebraic number field $K …
Dietrich Burde's user avatar
3 votes

degree of polynomial in Gröbner basis

In general, there is the following result on upper bounds for the degree of elements in the (reduced) Groebner basis: Let $G$ be a reduced Groebner basis of an ideal $I=\langle f_1,\ldots, f_r\rangle …
Dietrich Burde's user avatar
7 votes
1 answer
646 views

Existence of a generalized matrix inverse over an arbitrary field?

Let $A\in M_n(K)$ be a square matrix over a field $K$. The notion of inverse matrix was generalized by Moore and Penrose for real and complex matrices (also called pseudo-inverse $A^{\dagger}$ of $A$, …
Dietrich Burde's user avatar
14 votes
1 answer
1k views

Examples of polynomial rings $A[x]$ with relatively large Krull dimension

If $A$ is a commutative ring we have the estimate $$ \dim (A)+1 \le \dim (A[x])\le 2\dim (A)+1 $$ for the Krull dimension, with $\dim (A)+1 = \dim (A[x])$ for Noetherian rings. I am looking for nice …
Dietrich Burde's user avatar
23 votes
3 answers
2k views

Why is $(\mathbb{Z}/3\mathbb{Z})^3$ not a class group of an imaginary quadratic number field ?

In general, it seems not known which finite abelian groups are class groups of quadratic number fields. For imaginary quadratic number fileds, I read that $(\mathbb{Z}/3\mathbb{Z})^3$ is the smallest …
Dietrich Burde's user avatar
4 votes
Accepted

"as close to being semisimple as it can possibly be."

In the semisimple case it is really easy to calculate $ext_A^i(M,N)$, $i\ge 1$, with the above assumptions. It is zero. For Koszul rings this is almost true, i.e., $ext^i(M,N)$ is concentrated in degr …
Dietrich Burde's user avatar
2 votes

Why/when classification of simple objects is "simple" ? E.g. (unknown) classification of sim...

In general, the classification of finite-dimensional simple non-associative algebras is difficult, i.e., not known, even for algebraically closed fields of characteristic zero. I have tried to start …
Dietrich Burde's user avatar
6 votes

Poincaré duality for (co)homology of Lie algebras?

As far as I know, a generalisation of Poincare duality for Lie algebra cohomology over rings is given in M. Hazewinkel, "A duality theorem for the cohomology of Lie algebras" Math. USSR-Sb. , 12 (19 …
Dietrich Burde's user avatar