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842112
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location USA
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visits member for 2 years, 6 months
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PhD

Interests:

Ring and module theory

Clifford algebra/Geometric algebra

Mathematical physics

Applications of abstract algebra


2d
awarded  Necromancer
2d
revised Is there a better way to solve this problem in Linear Algebra?
added 157 characters in body
2d
comment Is there a better way to solve this problem in Linear Algebra?
Dear @JessicaB : Oh. Well, I doubt anyone else is going to understand that. I guess I'll just transfer the remark into the answer then. It is next to incomprehensible as it stands
2d
comment Why is axiom of choice needed? (Equivalent conditions for Noetherian)
@user71815 Somewhat similarly to the above, you'll need a choice function to prove that the "all submodules f.g." is equivalent to the ascending chain condition. Suppose $N$ is a submodule of $M$ which isn't finitely generated. Then $N\setminus N'$ is nonempty for any f.g. submodule $N'<N$ and you need to pick some element out of this nonempty set to make a second f.g. module $N''$ over $N'$. A choice function will enable the existence of these elements, and then you can lift a strictly ascending chain out of the poset of finitely generated submodules of $N$.
2d
comment Ellipse Problem
Just general writing advice: using three "of"'s in quick succession makes the sentence hard to understand. Sometimes it pays to write a little more to make a sentence easier to digest :) That was what I tried to do with my revision. I hope this is ok...
2d
revised Ellipse Problem
grammar, more readable statement of question
2d
comment how to determine the right side of the dice
For general information (but it is really a nitpick) the singular form of "dice" in English is "die." Not trying to be morbid, that's really what it is :)
2d
comment Is there a better way to solve this problem in Linear Algebra?
This probably needs to be improved to be more explicit. It looks like the comment you're talking about has disappeared.
2d
comment Proof vector x + ⃗y = ⃗x + ⃗z then ⃗y = ⃗z
Transitivity? Either you mean something else or the equivalence relation that you're thinking of does not jump to my mind... It looks like injectivity of a map given by adding $x$, to me.
2d
awarded  Necromancer
Oct
19
comment Is product of prime ideals prime?
It's straightforward to prove that if the product is prime, then one of the prime ideals contains the other. Thus is clearly not the case here, so the strategy won't work.
Oct
19
comment Proving commutativity
Dear @MartinBrandenburg: Given a choice between A) (maybe) preventing a single user from successfully fishing a full answer and B) keeping our body of answers as organized as possible, I feel B is the clear winner. The chances of A being successful already seem incredibly slim in a case like this. Regards.
Oct
18
comment $R/J(R)$ not semisimple Artinian
@user26857 looks good: thanks.
Oct
18
comment Proving ring $R$ with unity is commutative if $(xy)^2 = x^2y^2$
Might contain logic that you like : math.stackexchange.com/q/971443/29335
Oct
18
comment Proving ring $R$ with unity is commutative if $(xy)^2 = x^2y^2$
Since "abelian" has other meanings for rings, it's more standard to call them "commutative."
Oct
18
revised Abstract algebra book with real life applications
added 11 characters in body
Oct
18
revised $R/J(R)$ not semisimple Artinian
added 224 characters in body
Oct
18
comment $R/J(R)$ not semisimple Artinian
@karparvar You mean Jacobson semisimple, I guess. Sure, it's trivial to give it a nonzero radical. I edited some in.
Oct
17
comment prime ring and the proof of it's property
If you included your partial work, we could probably coach you along. The problem pretty much solves itself.
Oct
17
revised $R/J(R)$ not semisimple Artinian
edited tags