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Results tagged with ra.rings-and-algebras
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user 4790
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
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14
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What is the minimal number of symmetric generators of the full matrix algebra?
It is a well-known result of Burnside that a set of (say, real) $n \times n$ matrices generate the full matrix algebra if and only if they do not have a common non-trivial invariant subspace of $\math …
- 27k
7
votes
Accepted
Order of ring automorphisms of localizations of polynomial rings over finite fields
Every such automorphism is contained in the automorphism group of the field or rational functions $F(t)$ over $F$, which equals $\mathrm{PGL}_2(F)$, and so is a finite group.
[Edit:] upon further ref …
- 27k
14
votes
Accepted
non-isomorphic stably isomorphic fields
I don't think that there are any really easy examples. In the famous paper of Beauville, Colliot-Thélène, Sansuc and Swinnerton-Dyer "Variétés stablement rationnelles non rationnelles" they construct …
- 27k
9
votes
Accepted
Does $S$ being a free rank-$n$ $R$-algebra imply that $S/R$ is free rank $n-1$?
This is not true in general. For example, assume that $P$ is a projective module on $R$ that is not free, but such that $P \oplus R$ is free (there are many such examples). Set $S= R \oplus P$, and gi …
- 27k
3
votes
An algebra constructed from symmetric differences
This is the group algebra of the additive group $(\mathbb Z/2\mathbb Z)^S$, hence it is the product of $2^{|S|}$ copies of $\mathbb C$.
- 27k
13
votes
Consequences of not requiring ring homomorphisms to be unital?
Of course, if we assume that the ring have a unit, then there is absolutely no reason not to assume that homomorphisms preserve it (or are there books that do that?)
My impression is that there has b …
- 27k
11
votes
Accepted
$K_{0}(R) =\mathbb{Z}$ but some f.g. projective not stably free?
If $R$ is a commutative ring with $K_{0}(R)=\mathbb{Z}$, then $\mathop{\rm Spec} R$ is connected, because otherwise $R$ would split as a product, and $K_{0}(R)$ would contain a copy of $\mathbb{Z} \op …
- 27k