Reputation
4,300
Next privilege 5,000 Rep.
Approve tag wiki edits
Badges
1 5 30
Impact
~24k people reached

May
16
comment Is every fiber of a morphism between varieties of pure dimension?
In my example it is still the case: generic fiber has pure dimension 0, and the exceptional fiber has pure dimension $n-1$
May
16
answered Is every fiber of a morphism between varieties of pure dimension?
May
16
reviewed Approve Simplifying Radicals (Algebra II Basics)
Mar
30
awarded  Nice Question
Dec
20
awarded  Constituent
Dec
9
awarded  Caucus
Nov
24
comment Shortlist of problems in linear algebra
@RipanSaha Thanks for the link! I have seen that. It's nice, but it's not what I am looking for. First of all, it's pretty long. Second, it's rather computational. I am looking for a short list of problems, and not for complete beginners, but for someone who already knows some basics like row reduction, but is not familiar or is not familiar enough with more abstract "coordinate-free" approach to linear algebra.
Nov
24
comment Shortlist of problems in linear algebra
@Timbuc I agree with you and disagree at the same time. I am not looking for a problem list for complete beginners, and it doesn't mean that the problems tell you everything. It's just if you know how to solve them, it means you know enough of linear algebra.
Nov
24
asked Shortlist of problems in linear algebra
Nov
18
comment K[x]-module is k-vector space
Any matrix defines a linear transformation. And backwards, if you have a linear transformation, if you fix a basis in your vector space, it will give you a matrix, matrix of this linear transformation. So linear transformations are like matrices, but by default without given basis.
Nov
18
comment K[x]-module is k-vector space
Exactly! $x$ can go to any matrix. If you know where $x$ goes, say, to matrix $A$, can you tell where $x^2$ should then go? What about $x^3$?
Nov
18
answered K[x]-module is k-vector space
Nov
11
answered Homotopy groups of pairs and homotopy fibration of inclusions
Oct
21
comment Why exactly is this injective? Algebraic Topology.
I repeat one of my previous comments. By definition of this map it comes from the natural inclusion $ker(d^i) \to A^i$. It is by definition. Otherwise it doesn't even make sense to talk about the exact sequence.
Oct
21
comment Why exactly is this injective? Algebraic Topology.
In your third line you said "show the sequence is exact". This "the" indicates you know the terms of the sequence and the maps between them. So what are the maps? How do you get the sequence? Do you understand where the sequence came from?
Oct
21
comment Why exactly is this injective? Algebraic Topology.
Where did you get the map from the third formula in your question from? By definition of this map it comes from the natural inclusion $ker(d^i) \to A^i$.
Oct
21
comment Why exactly is this injective? Algebraic Topology.
Of course the can be a non-injective map $Ker(d^i) \to A^i$. But in your case the map $Ker(d^i) \to A^i$ you are considering is by definition just the obvious inclusion.
Oct
21
answered Why exactly is this injective? Algebraic Topology.
Oct
20
answered So why isn't $\Bbb R^n = \oplus _{n = 1}^{m}\Bbb R^n$
Oct
19
comment Prove that $p_M+p_{M^\perp}=I$, where $I$ is the identity on $H$.
It is direct sum of vector spaces. Do you know what that is?