Reputation
1,344
Next privilege 2,000 Rep.
Edit questions and answers
Badges
1 8 21
Newest
 Yearling
Impact
~28k people reached

Nov
20
comment Generators of a finitely generated free module over a commutative ring
@MakotoKato, I'm not so sure if the uniqueness of rank for commutative rings can be proved without choice. The proof I know uses it. So it may not be a drawback here.
Nov
20
comment Generators of a finitely generated free module over a commutative ring
Note that this would not be a contradiction if $A$ was not commutative.
Nov
20
comment Free Module $R^{n}$ with zero divisors
Why do you think there should be no torsion elements in a free module (over a ring with zero divisors)? The axioms for a free module are clearly satisfied by $R^n$.
Nov
20
comment How to show that a form on $\mathbb{C}$ defines a holomorphic $1$-form on $\mathbb{C}/\Gamma$?
Where are you stuck? How did you define holomorphic 1-forms on $\mathbb{C}/\Gamma$ (a Riemann surface)? This is what you have to check. Do you know what the complex charts are?
Nov
19
comment $\mathbb{Z}_{11} [x]/\langle x^2-2\rangle $ and $\mathbb{Z}_{11} [x]/\langle x^2-3\rangle$ are not isomorphic
You don't have to be sorry, but please don't use the imperative "show" either. How far do you get yourself on this problem? Where are you stuck?
Nov
19
comment Question of Hartshorne book's Proposion II.(2.6)
Could you please elaborate which part exactly you do not understand? In your quote there are a lot of different things happening. Surely you don't want us to explain what a local ring is.
Nov
19
comment If $A$ is singular, is $A^3+A^2+A$ singular?
At least if $p$ does not have a constant term.
Nov
17
comment Reading circle in mathematics?
Thank you for your ideas. It seems like this is the way to go, although it probably will never be as interactive as in other disciplines. I also like the suggestion to do basic number theory, one can never have enough experience with that.
Nov
17
answered Are all projection maps in a categorical product epic?
Nov
17
comment Should every line be infinite in both two directions?
It depends on what you want. If you consider, e.g., $[0,1]$ as a subspace in its own right, then the line segment $[0,1]$ (in $\mathbb{R}$) is really a line in the subspace $[0,1]$: It goes 'from one end of the space to the other'. So if you want to consider the situation relative to the ambient space, you could add the axiom. If you care about definitions without reference to the embedding, then you should probably not.
Nov
17
comment Binary notation in Magma
Please let me know if you try this and it's actually faster.
Nov
17
comment Should every line be infinite in both two directions?
A line segment/ray does not fit the definition of a line (an induced line) in your link. It does not consist of all points that satisfy the condition.
Nov
17
answered Binary notation in Magma
Nov
15
comment Is there no solution to the blue-eyed islander puzzle?
In a way, the speech act ensures that everyone has the same information or at least can do the same deductions.
Nov
15
comment Is there no solution to the blue-eyed islander puzzle?
The new information is burrowed in the chain of thought inferences the people do (precisely at the end - the $n=1$ case). The people on the island are quick and logical thinkers, they are not mind-readers. For sure, it does not have to be an outsider who speaks, it could as well be one of the persons on the island. That is why there is this law prohibiting discussing eye-colors.
Nov
15
comment Is there no solution to the blue-eyed islander puzzle?
Why would it not have a solution? You stated the perfectly valid proof (that the 100 blue-eyed people kill themselves after 100 days) yourself.
Nov
15
comment Help with nilradicals
The nilradical contains all the nilpotent elements of a ring $R$, that is the $x\in R$ for which some power $x^n$ will be $0$. Take for example $6\in Z/12$. Is $6^n=0$ for some $n$, modulo $12$? You can just go through the elements of $Z/12$ to get an intuition.
Nov
15
revised Vakil's Foundations of Algebraic Geometry, Exercise 5.5.E
added 3 characters in body
Nov
15
comment Vakil's Foundations of Algebraic Geometry, Exercise 5.5.E
It sure is. However, for the silly sake of tidiness, my solution follows the path the hint suggests (without shortcut). :)
Nov
15
comment Vakil's Foundations of Algebraic Geometry, Exercise 5.5.E
Yup, thank you. I have fixed it.