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6h
answered Examples of fields with characteristic $2$.
6h
comment Jacobian of a bijective mapping?
I don't understand the question. What are $X$ and $Y$? Are they vector spaces? If so, what does "smooth" mean? If not, what does "linear" mean?
6h
answered Correspondence between prime and maximal ideals
6h
revised Example of a Tensor Product of Modules with Non-Decomposable Elements
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6h
revised Example of a Tensor Product of Modules with Non-Decomposable Elements
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6h
revised Example of a Tensor Product of Modules with Non-Decomposable Elements
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6h
answered Example of a Tensor Product of Modules with Non-Decomposable Elements
17h
comment An algebra exercise
Do you know how to classify permutations up to conjugacy in terms of their cycle types?
17h
comment Tensor product of two simple modules
The answer is no. I give a counterexample for $k$ not algebraically closed here: math.stackexchange.com/questions/1615711/…
20h
awarded  Enlightened
23h
awarded  Nice Answer
1d
awarded  Nice Answer
1d
comment universal coefficient theorem for mod p cohomology
Just apply the universal coefficient theorem for cohomology and see what you get.
1d
comment Correct definition of model category
Isn't there also some condition about having finite / all limits / colimits?
1d
comment What properties of busy beaver numbers are computable?
Again, you're not using the right quantifiers. Since $BB(n)$ grows arbitrarily large, for fixed $m$ there will always be some $n_0$ such that $BB(n) > m$ for $n > n_0$. So for any fixed $m$, reducing $\bmod m$ necessarily causes you to lose information about the busy beaver numbers in such a way that you can't conclude that the sequence $BB(n) \bmod m$ is uncomputable from this argument. (Again, independently, your argument also can't work because it's possible to choose a dumb encoding relative to which $BB(n) \equiv 0 \bmod m$, in which case the sequence is trivially computable.)
1d
comment On the meaning of the word “generic” in Lie Algebra (or otherwise)
@Hamed: any finite-dimensional vector space $V$ can be regarded as an affine space in the sense of algebraic geometry. The closed subsets in the topology are zero sets of polynomials, meaning elements of the symmetric algebra $S(V^{\ast})$. The condition is just that some such set exists; it doesn't specify what it is (but if you work through an example you should be able to write down what the set ends up being explicitly; probably it is the vanishing locus of a determinant or a set of minors or something like that).
1d
answered Is every monomorphism an injection?
1d
comment $\mathbb{R}^3$ not diffeomorphic to $\mathbb{R}^3\setminus \{0\}$
@monoid: I think it can be interpreted as describing the electric field of a point charge at the origin. I don't know if that's the sort of thing you have in mind.
1d
comment What properties of busy beaver numbers are computable?
Your argument doesn't have the right quantifiers in it. For fixed $m$, knowing $BB(n) \bmod m$ for all $n$ does not tell you $BB(n)$ for all $n$, so you can't use the fact that $BB(n)$ is uncomputable to conclude that $BB(n) \bmod m$ is uncomputable. In fact, 1) any finite sequence of $BB(n)$s is computable in a tautological sense, and 2) as pointed out in the comments, $BB(n) \bmod m$ is highly sensitive to details of encodings: you might choose dumb encodings with the property that it's always $0$, for example, and so trivially computable.
1d
comment Action via automorphism
The collection of automorphisms of $N$ forms a group $\text{Aut}(N)$. A group action of $A$ on $N$ is a group homomorphism $A \to \text{Aut}(N)$.