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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- …
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 …
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 …
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 …
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 …
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 $A$, …
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 …
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 …
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 …
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 …
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 …