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bio website math.sunysb.edu/~jstarr/…
location Stony Brook University, Stony Brook, NY USA
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1d
comment cycle class as Chern class
I am afraid that I do not see how $[Z]$, defined with those funny factors of $\sqrt{-1}$ and $\pi$, can possibly be an element of $H^{2p}(X,\mathbb{Q})$, i.e., with $\mathbb{Q}$-coefficients.
1d
answered On transforming pair of bivariate polynomials to pair of univariate polynomials by applying polynomial map
1d
awarded  Nice Answer
1d
comment sufficient condition of complete intersection
Like Daniel, I have trouble making sense of this. Perhaps by "complete intersection", you mean the local commutative algebra definition, i.e., for every closed point $x\in X$, the local ring $\mathcal{O}_{X,x}$ is a local complete intersection ring. But then, as Daniel says, it follows directly from your hypothesis.
Dec
15
awarded  Enlightened
Dec
15
awarded  Nice Answer
Dec
15
comment The cohomology of an $S_{3}$ cover of an elliptic curve ramified in one point
What precisely do you mean "ramified in one point"? Do you mean that there is a unique point in $E$ over which the cover is branched? By my computation, lying over this unique point in $E$ there are either $2$ or $3$ ramification points in $C$.
Dec
15
comment A perfect domain that is not integrally closed?
It appears to me that your ring is seminormal. Is every perfect integral domain a seminormal ring?
Dec
15
revised Fano variety of lines on the Segre and the Grassmannian
added 1224 characters in body
Dec
15
comment Fano variety of lines on the Segre and the Grassmannian
You are correct. I made a mistake. I will fix it now.
Dec
15
answered Fano variety of lines on the Segre and the Grassmannian
Dec
15
comment Fano variety of lines on the Segre and the Grassmannian
@DanielLitt: "... are you sure these 'virtual' counts imply anything ...?" Yes, the virtual count equals the actual number if the set is "transversal". Since the empty set is transversal, a nonzero virtual count implies that there exists at least one line. Since that is all that Landsberg asked, dhy has answered Landsberg's question.
Dec
14
comment Codimension in zero and positive characteristic
@gio: If the polynomials $F_i$ are homogeneous, then the semicontinuity theorem holds. This was also Allen Knutson's point. I suggest you look up in a textbook "semicontinuity" (for dimension of fibers) and "properness". Since semicontinuity is a result on the domain of the morphism, and you want a result on the target, you need to use properness.
Dec
13
revised Push-forward of a quasi-coherent graded algebra under a proper map
added 175 characters in body
Dec
13
answered Push-forward of a quasi-coherent graded algebra under a proper map
Dec
13
comment personal relationships
This does not sound appropriate to me.
Dec
13
comment Codimension in zero and positive characteristic
@AllenKnutson: "Is this a properness over $\text{Spec} \mathbb{Z}$ issue?" Yes, precisely. The example above only works because the induced morphism from the zero scheme of $F_0$, $F_1$, $F_2$ to $\text{Spec} \mathbb{Z}$ is not proper.
Dec
13
comment Torsors and twists of algebraic groups
@Mostafa, Regarding your new question, "Is the automorphism group ... always affine?", the answer is "no". For $G=\mathbb{G}_m\times \mathbb{G}_m$, the automorphism group has countably many connected components via the action of $\mathbf{SL}_{2}(\mathbb{Z})$ on $G$. So it makes sense to restrict to the connected component of the identity.
Dec
13
revised Codimension in zero and positive characteristic
added 238 characters in body
Dec
13
answered Codimension in zero and positive characteristic