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Results tagged with ra.rings-and-algebras
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user 121
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.
3
votes
Accepted
Norm of a number in an algebraically closed field
For any norm of a quadratic extension, you can specify the Galois-fixed elements as those $x$ for which $N(x) = x^2$. If you had an invariant norm, it would imply the real-closed field is unique. H …
- 45.7k
2
votes
Explanation and Definition of Iwahori order
I can give a concrete description of David Savitt's example (following your link). Let $A = \mathbb{C}[\![ t]\!]$, the ring of formal Taylor series with complex coefficients. It has a maximal ideal …
- 45.7k
2
votes
Group laws on elliptic curves and varieties
The short answer is that the group law on the set of rational points of an elliptic curve defined by a Weierstrass equation uses all of the structure and properties of a field in an essential way.
To …
- 45.7k
5
votes
Accepted
A classification of non-associative algebras with a norm?
I don't think these objects can be classified in a manner similar to the normed unital division algebras, if you take "algebra" to mean "vector space $V$ equipped with a bilinear map $V \otimes V \to …
- 45.7k
9
votes
Accepted
A back and forth Euclidean algorithm over the integers--does it have bounded length?
We get an isomorphic problem by switching $c$ with $d$, and replacing $b$ with $-b$. Then we are considering matrices $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ in $SL_2(\mathbb{Z})$. Passage fr …
- 45.7k
2
votes
Dual of a module
Let $M = \mathbb{Z}_p$ with $T$ acting as $0$. Then (if I'm not missing anything) $X \cong \mathbb{Q}_p/\mathbb{Z}_p$, and $X/X[p^n] \cong X \neq 0$, while $M[p^n] = 0$.
- 45.7k
1
vote
What is an algebraic group over a noncommutative ring?
You seem to be asking two different questions. The first is, "how do I define the notion of algebraic group over a noncommutative ring?" The second is, "given an algebraic group (viewed as a functor …
- 45.7k
17
votes
Direct sum of Hopf algebras
The answer is "no". In the commutative case, this would be asking for the disjoint union of two group schemes to be a group scheme. Even more concretely (say, working with finite dimensional commuta …
- 45.7k
11
votes
An algebra of "integrals"
I don't see why you want $A$ to be an algebra, since the integral of 1 doesn't seem like a reasonable unit. Did you want some compatibility with higher dimensional integrals using the Fubini theorem? …
- 45.7k
6
votes
The evaluation and coevaluation maps for an object isomorphic to a dualisable object
Yes.
First, set $Y^* = X^*$.
Then, compose the evaluation and coevaluation maps of $X$ with maps like $\operatorname{id_{X^*}} \otimes \sigma$.
- 45.7k
15
votes
What is interesting/useful about big Witt Vectors?
Here is a long article by Hazewinkel and a discussion on the nLab.
The functor of taking big Witt vectors is right adjoint to the forgetful functor from lambda-rings to commutative rings. Lambda rin …
- 45.7k
3
votes
Realizing a restriction as direct/inverse image of sheaves
I'm going to make a bold guess about what you are doing. You want to view $\mathbb{C}$ as the real points of two dimensional affine space over $\mathbb{R}$, and its ring of functions can be seen as $ …
- 45.7k
3
votes
Modules over Laurent series rings
Sorry, the first version of this answer was broken in a few ways.
For your first question, it seems that there is more than one construction that specializes to what you want. For example, you can t …
- 45.7k
10
votes
Accepted
Codes, lattices, vertex operator algebras
I think the analogy you describe cannot be made precise with our current technology. For example, the word "functor" doesn't seem to have made an appearance yet in this context.
If you have a code, …
- 45.7k
6
votes
Generalization of eigenvalues/vectors to modules?
In the case of commutative rings, you can view spectra as points in the quotient of $End_R(M)$ by the conjugation action of $Aut_R(M)$. You can use the tensor product to turn this into a quotient of …
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