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Apr
22
comment Is the preimage of the non-normal locus a divisor?
@RghtHndSd: This is certainly true, but "being pure" means that all components of $\nu^{-1}(Z)$ have the same dimension as $Z$. Your statement only implies that the maximal dimension among all components of $\nu^{-1}(Z)$ is equal to that of $Z$.
Apr
22
comment Every variety contains open affine normal subvariety
@DanielMcLaury: Good point, fixed.
Apr
21
comment Is the preimage of the non-normal locus a divisor?
@AsalBeagDubh: A very good nitpick, I fixed that.
Apr
20
comment Multiplicity of point as a zero of polynomial.
Also, how do you define $I_p(C,L)$?
Mar
19
comment Are unipotent algebraic groups connected?
this answer on mathoverflow strongly suggests that your argument is solid.
Mar
16
comment An isomorphism from $GL_{2}(\mathbb{F_{2}})$ to $S_{3}$
These two groups are not isomorphic. $\operatorname{GL}_3(\mathbb F_2)$ has at least $7$ elements, because there are $7$ nonzero vectors in $\mathbb F_2^3$. On the other hand, $S_3$ only contains $6$ elements.
Mar
11
comment Resolution of singularities of the determinant hypersurface
This is indeed a very nice resolution. I would have preferred an embedded one, but this is quite a nice start, so +1 and accept. Thanks!
Mar
11
comment Can a birational morphism surject from an affine to a projective variety?
+1 and accept, also thank you, this is a nice proof.
Mar
7
comment True or False: $f$ is injective if and only if $f^*$ is surjective where $f^*$ is corresponding to the pullback to $f$.
Of course, you can easily turn this into a direct proof, which is then more elegant. But it is late, and I will leave you with this.
Mar
6
comment Can a birational morphism surject from an affine to a projective variety?
@AlexYoucis: What is the argument for the fact that it has an inverse on some $U\subseteq Y$ whose complement is codim. 2? The approach sounds interesting!
Mar
4
comment Can a birational morphism surject from an affine to a projective variety?
@karl_christ: No, I mean surjective. Every fiber is nonempty.
Mar
4
comment Can a birational morphism surject from an affine to a projective variety?
@karl_christ: Such a map would only give me an isomorphism of a dense open subset $U\subseteq X$ with another dense open subset $V\subseteq Y$ of $Y$. This proves nothing.
Mar
4
comment Can a birational morphism surject from an affine to a projective variety?
@KReiser: Yea, I edited my question, this only makes sense for irreducible varieties.
Mar
3
comment Can a birational morphism surject from an affine to a projective variety?
@Ben: I am not so much interested in pathologic cases, you may safely assume that all the objects are actually interesting.
Feb
27
comment Is this graph theory problem NP-Complete?
What I ment to say is: I expect the answer to be a resounding "we still don't know", all we know that it is in P.
Feb
27
comment Is this graph theory problem NP-Complete?
Technical nag: Your problem is not a decision problem, so it would at best be NP-hard. However, you can do DFS from each vertex to find all cycles in which that vertex is contained. This sounds like polynomial time to me, and if you can prove that the problem is not NP-hard, then you would have shown P$\ne$NP.
Feb
27
comment Qing Liu's definition of an algebraic variety, a non-separated line
This is a big problem with the notion of "variety" that has bugged me for quite some time as well: There is no really good consensus about what the word precisely means.
Feb
18
comment Covers with fixed ramification
From your notation I take it that $D$ is an effective (Weyl) divisor? In other words, a codimension one subvariety? Because the multiplicities of that divisor are quite meaningless for the ramification question at hand. Also. Is $D$ base point free by any chance?
Feb
9
comment How many non-increasing sequences are there over the natural numbers?
@dtldarek: Very true - the mental image was, of course, that this is so-to-speak the "limit point" of the sequence.
Feb
9
comment Describe invariants using coaction.
@LJR - yes, indeed, I forgot to elaborate on that. Except maybe you should write $c_i$ instead of $c$, because you could have $\alpha_i(1)\ne\alpha_j(1)$. Doesn't affect the proof though.