# Jack Huizenga

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 Name Jack Huizenga Member for 2 years Seen 9 mins ago Website Location Chicago Age 28
I'm a postdoc at the University of Illinois at Chicago working on algebraic geometry.
 1d comment Examples of intersections of two hypersurfaces with high-dimensional singular locusThe intersection of two smooth hypersurfaces $X,Y$ is singular at a point $p$ if and only if $X$ and $Y$ fail to be transverse at $p$. It should be easy to show that the singular locus of the intersection can have any codimension you want. As you should interpret the intersection scheme-theoretically, it is even possibly for the intersection to be everywhere non-reduced, in which case you should view it as being singular everywhere. (For example, this is the case for the two hypersurfaces $z=0$ and $z=x^2$ in 3-space, whose intersection should be viewed as a double line in the plane $z=0$). May14 comment An example of a tensor product consisting of only simple tensors?@David: there won't be any interesting surjective maps from a field to another ring. I think your argument is fine Chris, even if this question would be better suited to math.stackexchange.com. May13 comment Is a complete intersection satisfying Jacobian matrix smooth criterion a smooth variety?This is really differential geometry, not algebraic geometry (although it is true in the algebraic setting as well, when the proof is written appropriately). Your $r$ functions give a map $f:(\mathbb{C}^*)^n \to \mathbb{C}^r$. The variety $X$ is $f^{-1}(0)$, and the condition on the matrix says that $0$ is a regular value of $f$. So $X$ is a smooth manifold. Near any point $p\in X$, $X$ is the transverse intersection of the $r$ hypersurfaces $f_1=f_2=\cdots f_r = 0$, and these hypersurfaces are smooth at $p$. May10 revised Surfaces in $\mathbb{P}^3$ with isolated singularitiesdeleted 2 characters in body May2 revised Is it true that $p^2+1$ is square free if $p>7$ is a Mersenne primeadded 8 characters in body Apr28 comment Hilbert function of (weighted) projectivized tangent conesThe tangent cone is a local intrinsic construction of an abstract variety, and doesn't "see" any embedding into another space. Subvarieties of weighted projective spaces still have projectivized tangent cones, which are subschemes of the projectivized Zariski tangent space (which is an ordinary projective space). Apr28 comment Hilbert function of (weighted) projectivized tangent conesThe answer to your first question is yes; this is just saying that the degree and dimension of the tangent cone is $\mu$ and $r-1$, respectively, which can be found in many books, e.g. Harris' "First Course" book. For your second question, I don't know what you would mean by a "weighted projective tangent cone." The ring $\oplus m^i/m^{i+1}$ is canonically $\mathbb{Z}$-graded. Apr25 comment Punctual Hilbert SchemesCan you give an example of a special ring for which this fails? (Even in the dimension 2 case might be useful to see what goes wrong) Apr25 comment An expression with an alternating trilinear form, written in terms of the determinant and a symmetric bilinear formThere would have to be some relation between $f$ and $\sigma$. As it stands, if $f$ is nonzero and $\sigma$ is zero there is no hope. Apr16 awarded ● Nice Answer Apr16 comment Algebraic machinery for algebraic geometryI of course agree with your opinion here Dmitry. Once you've been working in the subject for a while you start to see why some of the nitpicky details are important, but it's certainly possible to go a long way without, say, ever just studying Hartshorne for 2 years. And at least for certain people, it's a lot more interesting to start doing geometry than to bash your head against concepts like flatness (which I'm not discrediting--it's extremely important, but at least in nice cases more intuitive than Hartshorne might have you believe) before having any idea what it might be good for. Apr16 answered Does there exist an infinite non-commutative division ring with finite center? Apr15 revised Are these two definitions of $\mathcal{O}(1)$ over a ruled surface closely related?added 16 characters in body Apr13 comment Gluing free modules to get a finitely generated free moduleAre you basically just asking for generalizations of the Quillen-Suslin theorem to other rings? Apr4 revised Hilbert scheme of points on a surface as moduli space of semistable sheavesadded 9 characters in body Apr4 accepted Hilbert scheme of points on a surface as moduli space of semistable sheaves Apr4 revised Hilbert scheme of points on a surface as moduli space of semistable sheavesadded 15 characters in body Apr4 answered Hilbert scheme of points on a surface as moduli space of semistable sheaves Mar8 comment euler class of the normal bundle and self intersection number@Allen: Fair enough. I suppose I saw the tag algebraic-geometry and responded hastily. Mar8 comment euler class of the normal bundle and self intersection numberYour statement that $T_X|_S = T_S \oplus N_{S/X}$ is not true. There is an exact sequence relating these things, but it generally does not split. Mar7 comment Is the universal closed subscheme reduced?$Z$ is certainly reduced. In fact, it is smooth at any point lying over a point in the Hilbert scheme corresponding to a collection of distinct points, so it is generically smooth (and generically reduced). But it is also flat over the Hilbert scheme, so there are no embedded components, and it is reduced. (In fact, I believe it is actually smooth, but this is slightly more subtle.) I don't see, however, why knowing $Z$ is reduced gives what you want; reduced schemes have lots of nonreduced subschemes. Mar3 comment Higher dimensional Bezout via Hilbert polynomials: a reference.It's easy to prove this if you know that every component of a hyperplane section of an irreducible variety has codimension at most 1. Obviously this is easier to show from some foundations than others, but it's probably a more important fact than Bezout. Feb16 comment $G=\langle a\rangle H$ for subgroup $H$More generally, this is obviously false for any non-cyclic group with a unique maximal proper subgroup. Feb12 accepted On the construction of the varieties parametrizing special linear series on a curve Feb12 comment On the construction of the varieties parametrizing special linear series on a curveIn this context "Canonical blowup" is just a name. It seems best to work directly with the definition of $\tilde X_k(\phi)$ in section 2.4, as a degeneracy locus of bundle maps. At least in nice cases it is also possible to construct $\tilde X_k(\phi)$ as some iterated blowup of subvarieties (I don't think it's good enough to just blow up $X_{k-1}$--you then also have to blow up the proper transform of $X_{k-2}$, etc.), but if $\phi$ isn't very nice (for instance if $X_k(\phi)$ has the wrong dimension) this might not work. Feb12 answered On the construction of the varieties parametrizing special linear series on a curve Feb2 comment Irreducibility of fibers vs. irreducibility of fibered productPardon my ignorance, but how do the notions of "fiber over the generic point" and "general fiber" relate to one another? For example, if we look at a 2:1 branched cover $\mathbb{P}^1\to \mathbb{P}^1$, then is it not the case that the general fiber is reducible (generally consisting of 2 points) but the fiber over the generic point (which is the general point of $\mathbb{P}^1$) is irreducible? (Even if this is the case, its possible Alexander cares about the fiber over the generic point instead of the fiber over a general point. Feb1 revised Irreducibility of fibers vs. irreducibility of fibered productadded 28 characters in body Feb1 answered Irreducibility of fibers vs. irreducibility of fibered product Jan30 comment Class of integrable 0/1-functions “with no null sets.”On the Wikipedia page for the Lebesgue density theorem (which is very closely related to what you ask about) it is noted that if $E$ and its complement both have positive measure then there are always points where the "approximate density" is strictly between $0$ and $1$. It is then always possible change $E$ by either including or removing such points. Dec11 comment A Question About Free ResolutionsNo. The relations among the generators will typically satisfy further relations, which necessitates taking longer resolutions. Dec10 accepted Does a weaker condition than vanishing derivative imply a function being constant? Dec10 revised Does a weaker condition than vanishing derivative imply a function being constant? deleted 1105 characters in body Dec10 revised Does a weaker condition than vanishing derivative imply a function being constant? added 1105 characters in body Dec10 answered Does a weaker condition than vanishing derivative imply a function being constant? Dec3 revised Necessary and sufficient conditions for a sum of idempotents to be idempotentadded 140 characters in body Dec3 revised Necessary and sufficient conditions for a sum of idempotents to be idempotentadded 4 characters in body Dec3 revised Necessary and sufficient conditions for a sum of idempotents to be idempotentadded 113 characters in body; added 38 characters in body; added 6 characters in body Dec3 answered Necessary and sufficient conditions for a sum of idempotents to be idempotent Dec2 accepted is there some condition I can impose on families of curves on a surface such that the second Ext between the ideal sheaf and the structure sheaf is zero? Dec1 comment is there some condition I can impose on families of curves on a surface such that the second Ext between the ideal sheaf and the structure sheaf is zero?In general yes, see the edit. Dec1 revised is there some condition I can impose on families of curves on a surface such that the second Ext between the ideal sheaf and the structure sheaf is zero?added 425 characters in body Dec1 answered is there some condition I can impose on families of curves on a surface such that the second Ext between the ideal sheaf and the structure sheaf is zero? Nov26 comment Does Physics need non-analytic smooth functions?Keep in mind that you can use Taylor's theorem instead of a full Taylor series to work with smooth functions which are not necessarily analytic. So long as you only care about finitely many terms of the series, this usually does the trick. Nov26 comment Genus of non-complete intersectionsNo. Think of it this way: being a complete intersection actually specifies a resolution of the ideal of the curve (via the Koszul complex). From this information the Euler characteristic of the structure sheaf can be determined, which gives the genus. More generally, if only generators of the ideal are known, the full structure of the resolution can be mysterious. If you know the full resolution you can determine the arithmetic genus (and hence the geometric genus since the curve is smooth), but the genus depends intimately on the resolution, i.e. on the "relations" between the $F_i$.