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An undergraduate at UCLA.


Aug
10
answered Is the sphere $S^n$ always arcwise connected?
Jul
5
comment Axioms of associative algebra?
Oh, also, lay off lang, get Artin, it's far better.
Jul
5
comment Axioms of associative algebra?
No, and in general it is not. It is best to think of such a structure as a vecotorspace (and if your more inclined, a module) that just happens to have a notion of multiplication. Then the first axiom just expresses that the actions respect each other. An example you could look at are of-course function algebras, fields over their characteristic, or one of the classics, the quaternions (which are not commutative!). Algebras are actually very nice, for example, Commutative Rings are just $Z$-algebras (you can define algebras independent of rings, so this is actually meaning-ful). Hope this help
Jul
2
awarded  Curious
Jul
1
comment Holomorphically simply connected implies simply connected
The idea would be that if the set is not simply connected, then there is a loop that cannot be contracted. Not being contractable means there is a hole in the inside of the loop, so that we can use the standard $1/z$ trick to get a holomorphic (on our set) function that has nonzero integral, and thus isn't simply connected. Notably, this is a proof by contradiction. I don't know of a constructive proof or method, and I would be surprised if one existed.
Jul
1
comment How to solve this simple linear equation
Felt like this was too short for an answer. We have that $2x+(5+3)=2x+8=46+3x$ so that subtracting $2x$, $2x+8-2x=46+3x-2x$ which is the same as $2x-2x+8=(3-2)x+46$ or $8=x+46$ or thus subtracting $46$, $x=-38$.
Jul
1
comment Is set of even integers an integral domain?
I would say no, but then again, that is just a matter of taste and what you tend to use rings for.
Jul
1
comment For vector spaces $V,W$, probe that $\Lambda (V)\otimes\Lambda (W) \cong \Lambda (V\oplus W)$
This can be shown degreewise by showing that $\Lambda^i(V\oplus W)=\bigoplus \Lambda^j(V)\otimes \Lambda^{i-j}(W)$ and this is so by writing a basis. If you need universal properties, you can get linear maps $\Lambda(V)\to \Lambda(V\oplus W)$ (by the universal property of $\Lambda(V)$) and likewise for $W$ and then by linearlity (since tensorproduct is the coproduct) to a map from $\Lambda(V)\otimes \Lambda(W)\to \Lambda(V\oplus W)$. Now use the universal property to get a map backwards, and use uniqueness to get that they are inverses. Drawing all the diagrams is a good exercise.
Jun
25
accepted What are necessary and sufficient conditions for the product of spheres to be paralellizable?
May
13
answered Homomorphism from $\mathbb{Z}\oplus \mathbb{Z}_2$ to $\mathbb{Z}$.
Apr
20
awarded  Notable Question
Apr
17
accepted Conditions on a $1$-form in $\mathbb{R}^3$ for there to exist a function such that the form is closed.
Mar
31
awarded  Nice Question
Mar
31
comment If the tensor product of two modules is free of finite rank, then the modules are finitely generated and projective
This is a truly beautiful way to prove this. Thanks!
Mar
31
accepted If the tensor product of two modules is free of finite rank, then the modules are finitely generated and projective
Mar
31
revised If the tensor product of two modules is free of finite rank, then the modules are finitely generated and projective
added 47 characters in body
Mar
31
asked If the tensor product of two modules is free of finite rank, then the modules are finitely generated and projective
Mar
13
revised Conditions on a $1$-form in $\mathbb{R}^3$ for there to exist a function such that the form is closed.
edited tags
Mar
13
comment Conditions on a $1$-form in $\mathbb{R}^3$ for there to exist a function such that the form is closed.
@SanathDevalapurkar A smooth function, sorry
Mar
13
revised Conditions on a $1$-form in $\mathbb{R}^3$ for there to exist a function such that the form is closed.
edited tags