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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.
12
votes
Accepted
Urge/reason for inventing interior product ( Grassmann algebra )
Here is an elementary motivation for interior products .
Suppose $V$ is an $n$-dimensional vector space .
To every non-zero vector $ v\in V$ you can associate the complex $$ 0\to V\to...\to \La …
3
votes
Isomorphism problem for finite dimensional central division algebras over a function field i...
Dear Albert, the key theoretical tool in your problem is the theorem (due independently to Auslander-Brumer and Fadeev) relating the Brauer group of a field $k$ and that of its rational function field …
7
votes
Superfluous definitions
A Lie subgroup of a Lie group is usually defined as a subgroup which is also a submanifold.
But actually any closed subgroup of a Lie group is automatically a manifold, hence a Lie subgroup.
Similarly …
26
votes
What makes a theorem *a* "nullstellensatz."
What I find intriguing is that the Nullstellensatz is underappreciated in the sense that many people appeal to a variation of it without saying (or realizing) they do.
For example, Hadamard's lemma …
7
votes
Alternating forms as skew-symmetric tensors: some inconsistency?
Dear Paul, first of all let me congratulate you for the extremely clear formulation of your interesting question (which is not silly at all, contrary to what you say): +1.
The source of your trouble …
5
votes
reduced ⊗ reduced = reduced; what about connected?
Ad (1): It is perfectly true that the tensor product of two reduced algebras over a perfect field is reduced. You can find a proof in Bourbaki's Algebra (Chapter V; §15; 4,5), but of course this vener …
12
votes
Are quotients of polynomial rings almost UFDs?
No, quotients of polynomial rings are definitely not "almost UFDs".
Any finitely generated ring over $K$ is such a quotient and this means a lot of non UFDs.
Said differently, any algebraic variety in …