bio  website  math.sunysb.edu/~jstarr/… 

location  Stony Brook University, Stony Brook, NY USA  
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visits  member for  5 years, 2 months 
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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

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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 
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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
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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 
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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 
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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 
Pushforward of a quasicoherent graded algebra under a proper map
added 175 characters in body 
Dec 13 
answered  Pushforward of a quasicoherent graded algebra under a proper map 
Dec 13 
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personal relationships
This does not sound appropriate to me. 
Dec 13 
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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 
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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 