Math Genius: Is the sheaf of non-vanishing holomorphic functions soft?

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It is obvious that the sheaf of holomorphic functions is not soft, not every holomorphic function on a closed set (appropriately defined) can be extended to be an entire function. What about the sheaf of non-vanishing holomorphic functions $mathcal{O}_X^{ast}$? I assume that it is not soft either, but I’d like an example.

The case of non-vanishing holomorphic functions is really only easier. For instance, consider the function $f(z)=z$, defined on a non-discrete closed set that does not contain $0$. Since a holomorphic function on a connected set is determined by its values on any non-discrete set, the only possible extension to all of $mathbb{C}$ is $f(z)=z$, but this is not a non-vanishing entire function.

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Math Genius: Clarification of Proof of Nagell-Lutz in Silverman-Tate

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I’m working through the proof of the Nagell-Lutz Theorem, specifically Section 2.4 of Silverman-Tate, wherein they prove that Points of Finite Order have Integer Coordinates. I follow most of the proof just fine, but I’m getting a little stuck on the part where they try to show that $C(p^nu)$ is a group. Here $C(p^nu)$ is the set of rational points $(x,y)$ on some elliptic curve with $ord_p(x) le -2nu$ and $ord_p(y) le -3nu$.

We convert to a new coordinate system via $t = frac{x}{y}$ and $s = frac{1}{y}$. They say then that $(t,s)in C(p^nu)$ if and only if $t in p^nu R$ and $s in p^{3nu}R$ (where $R$ is the ring of rationals with non-negative $p$-valuation). I agree so far.

However, in the next paragraph (p. 51), they claim that to show $(t,s)in C(p^nu)$, we need only show that $t in p^nu R$, and seem to ignore $s$ altogether. This is used multiple times later on in the proof, but I don’t see how this condition is sufficient.

I’ve tried showing that $t in p^nu R$ implies $s in p^{3nu}R$ by factoring the new equation for the elliptic curve as follows (here $a,b,cinmathbb{Z}$).
$$ s=t^3 + at^2s+bts^2 + cs^3 $$
$$ t^3 = s(1 – at^2 – bts – cs^2) $$
Since $ord_p(t^3) ge 3nu$, if I can show that $ord_p(1 – at^2 – bts – cs^2) = 0$, I would get my result. But I can’t see how to do that.

Any suggestions?

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Math Genius: Show that $phi^{-1}((y_1-a_1,ldots,y_n-a_n))=(x_1-b_1,ldots,x_m-b_m)$ for points in affine varieties $psi((a_1,ldots,a_n))=(b_1,ldots,b_m)$

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Let $A=k[x_1,ldots,x_m]$ and $B=k[y_1,ldots,y_n]$ over some algebraically closed field $k$. Let $psi : mathbb{A}^nrightarrow mathbb{A}^m$ be the morphism corresponding to a $k$-algebra homomorphism $phi: Arightarrow B$. Let $a=(a_1,ldots,a_n)in mathbb{A}^n$ and $psi(a)=(b_1,ldots,b_m)inmathbb{A}^m$. I want to show that $phi^{-1}((y_1-a_1,ldots,y_n-a_n))=(x_1-b_1,ldots,x_m-b_m)$.

I know that since $phi$ is a $k$-algebra homomorphism and $A$ and $B$ are finitely generated $k$-algebra, then the preimage of a maximal ideal is again a maximal ideal. I also know that by Hilberts Nullstellensatz, then a point $(a_1,ldots,a_n)$ in a affine variety corresponds to the maximal ideal $(a_1-p_1,ldots,x_n-a_n)$. And by how $psi$ is defined on the point $a$, then this seems to be natural that $phi^{-1}((y_1-a_1,ldots,y_n-a_n))=(x_1-b_1,ldots,x_m-b_m)$. However I was wondering if there were a more rigorously way of doing it or seeing it. My argument seems a little loose.

Your argument is the right idea. Let me try to give you some help.

A $k$-point in $Bbb A^n$ is the same as a map from $Bbb A^0to Bbb A^n$ picking out the point we care about. On the coordinate algebra side, this is represented by the map $k[y_1,cdots,y_n]to k$ given by $y_imapsto a_i$, and the preimage of the maximal ideal $(0)subset k$ is exactly the maximal ideal $(y_1-a_1,cdots,y_n-a_n)$.

Composing this with our map $Bbb A^nto Bbb A^m$, we get a map $Bbb A^0to Bbb A^m$ which corresponds to $k[x_1,cdots,x_m]to k$. The map $Bbb A^0to Bbb A^m$ picks out $(b_1,cdots,b_m)$, so we have that our map on coordinate rings $k[x_1,cdots,x_m]to k$ is given by $x_jmapsto b_j$, and the preimage of the maximal ideal $(0)subset k$ is exactly the maximal $(x_1-b_1,cdots,x_m-b_m)$.

Now look at the sequence of maps $k[x_1,cdots,x_m]to k[y_1,cdots,y_n]to k$. The preimage of $(0)$ along the first map is $(y_1-a_1,cdots,y_n-a_n)$, and the preimage of $(0)$ along the composite is $(x_1-b_1,cdots,x_m-b_m)$. But this means that the preimage of $(y_1-a_1,cdots,y_n-a_n)$ is $(x_1-b_1,cdots,x_m-b_m)$ by, for instance, the second-to-last line of this answer.

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Math Genius: Ideal of the twisted cubic

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The twisted cubic is the image of the morphism $phi : mathbb{P}^1 to mathbb{P}^3 , (x:y) mapsto (x^3:x^2 y:x y^2:y^3)$, it is given by $X = V(ad-bc,b^2-ac,c^2-bd)$. Now I would like to compute $I(X)$, which equals by the Nullstellensatz the radical of the ideal $I := (ad-bc,b^2-ac,c^2-bd) subseteq k[a,b,c,d]$. I think that that $I$ is already a radical ideal, even a prime ideal. Namely, I suspect that
$$phi^* : Q:=k[a,b,c,d]/I to k[s,t] , a mapsto s^3, b mapsto s^2 t , c mapsto s t^2 , d mapsto t^3$$
is an injection. If this is true: How can we prove that? I’ve already tried to find a $k$-basis of the quotient, but this turned out to be a big mess. Even the representation of the quotient as a monoid algebra doesn’t seem to help. Another idea is the following: A formal manipulation of generators and relations implies $Q_a cong k[a,b]_a$. Thus it suffices to prove that $Q to Q_a$ is injective, i.e. that $a$ is not a zero divisor.

Here is a purely algebraic proof that $I(X)=I$.
It is of course sufficient to prove that $I(X) subset I$ and for that it suffices to prove that every homogeneous polynomial $P(a,b,c,d)$
which vanishes on $X$ is in $I$.

Any homogeneous polynomial $P(a,b,c,d)in k[a,b,c,d]$ can be written $$P(a,b,c,d)=R(a,d) +S(a,d)b+T(a,d)c+i(a,b,c,d) $$ for some polynomials $R,S,Tin k[a,d]$
and a polynomial $iin I$
The easy proof is by induction on the degree of $P$ and I’ll leave it to you.

Now back to our problem. If now that homogeneous $P$ is in $ I(X)$ , we write it as in the lemma and get by using that $P$ vanishes on $X$ that for all $(x:y)in mathbb P^1_k$ $$0=P(x^3,x^2y,xy^2,y^3)=R(x^3,y^3) +S(x^3,y^3)x^2y+T(x^3,y^3)xy^2+0 $$
By considering exponents modulo $3$ for $x$ and $y$, we see that no cancellation occurs, hence that $R=S=T=0$ and thus $P=iin I$ as required.

The ideal $I$ is prime if and only if its associated projective scheme $V_p(I)subset mathbb P^3_k$ is integral.
This in turn can be checked on the four standard affine open subschemes covering $mathbb P^3_k$.
For example in the affine open subscheme (isomorphic to $mathbb A^3_k$) $U_dsubset mathbb P^3_k$ corresponding to $d=1$, the scheme $V_p(I)cap U_d$ is defined by the ideal $(a-bc,b^2-ac,c^2-b)$ which is trivially prime in $k[a,b,c]$.
Three similar calculations will imply that indeed the original ideal $I$ is prime.

There are several algorithms that can compute whether your ideal $I$ is prime or not, most of the algorithms use Groebner basis. I gave your ideal to a maple package, namely PrimDecomp, obtainable from and it says that your ideal is prime.

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Math Genius: Ramifications indices invariant under algebraic closure + question about ramification and covering

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We are considering functions fields of transcendance degree equal to $1$ (or smooth curves, it’s the same), over a perfect field if necessary, or even a finite fields, let’s say $K(X)$ and $K(Y)$ (corresponding to some curves $X$ and $Y$), with
: $K(Y) hookrightarrow K(X)$ (corresponding to a morphism $X rightarrow Y$). Let $P$ be a place of $K(Y)$, and $P’$ a place of $K(X)$ above $P$. We may assume that the curves are geometrically irreducible, as we can consider $overline{K}(X)$ and $overline{K}(Y)$, and : $overline{K}(Y) hookrightarrow overline{K}(X)$ (corresponding to $X_overline{K}$ and $Y_overline{K}$).
My question is then : is it true that the ramification indices of $P’/P$ is the same over $K$ and $overline{K}$ ? I mean, the place $P$ and $P’$ decomposed into multiples places when we switch to $overline{K}$, and those corresponding to $P’$ are dividing those corresponding to $P$ (stop me if it’s wrong ?), and then, are the ramifications indices the same than $e(P’/P)$ ?

My second question, related to the first one, is : given an hyperelliptic curve $X$ over $K$, and then :
$ X rightarrow mathbb{P}^1$ the covering of degree $2$ corresponding to the hyperelliptic curve, I read somewhere that the ramifications points are the locus of $mathbb{P}^1$ where there is only $1$ antecedent. I think this follows from the formula : $sum_{P’/P} e(P’/P) = 2$, but this is valid only if $K$ is algebraically closed, right ? Otherwise, we could have, for example, $e(P’/P)=1$ and $f(P’/P)=2$, and then no ramification but a residue fields of degree $2$. we should switch $K$ to $overline{K}$ (if the first assertion of my post is valid), right ? (and then, assume the curves are geometrically irreducible ?)

Thank you for the answers !

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Math Genius: Field of rational functions at a non singular point

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Let $mathbb{k}$ be an algebraically closed field, and let $psubset mathbb{k}[x_1,…,x_n]$ be a prime ideal contained in the ideal $(x_1,…,x_n)$. Suppose $0$ is a nonsingular point of the variety definted by $p$. Let $A=frac{mathbb{k[x_1,…,x_n]}}{p}$ and let $mathbb{K}$ denote the field of fractions of $A$. Suppose $r$ is the transcendence degree of $mathbb{K}/mathbb{k}$. Then, can we find a set $S={i_1,…,i_r}$ consisting of $r$ different elements such that ${x_{i_1},…,x_{i_r}}$ is a transcedence basis of $mathbb{K}$ over $mathbb{k}$, and for each $1leq ileq n$ such that $inotin S$, there exists a polynomials $phi_iinmathbb{k}[x_{i_1},…,x_{i_r}, y]$ such that $phi_i(x_{i_1},…,x_{i_r}, x_i)=0$ in $mathbb{K}$, but $frac{partial phi_i}{partial y}(0)ne0$.

What I am trying to do is to find something of a “basis” of $mathbb{K}/mathbb{k}$, where $x_{i_1},…,x_{i_r},phi_{i_{r+1}},…,phi_{i_n}$ defines $mathbb{K}$ and the transcendence basis, $x_1,…,x_{i_r}$ helps give the tangent space of the variety at the point and the polynomials $phi_{i_{r+1}},…,phi_{i_n}$ helps give a Jacobian matrix that has a nonsingular, diagonal submatrix.

For example: let $p=(x_1+x_2+x_3^4)subset mathbb{k}[x_1,x_2,x_3]$. Now, if I take $S={1,2}$, then we wont be able to find find a $phi_3$ which satisfies out desired conditions. This is because $frac{partial f}{partial x_3}(0)=0$ (here $f=x_1+x_2+x_3^4$). But, if I take $S={1,3}$ and $phi_2=x_1+x_2+x_3^4$, then it will satisfy our requirements.

In the characteristic $0$ situation, I strongly believe it should work. Although I feel it should work in the prime characteristic case, I don’t feel as strongly.

Could someone help me prove or disprove it!!

Thanks in advance.

Let $f_1,cdots,f_s$ be a set of generators of $P$. Let $J:=begin{pmatrix} frac{partial f_i}{partial x_j}(0)end{pmatrix}_{i,j}$ be the Jacobian matrix at the origin. The assumption that $V(P)$ is smooth at the origin means that this matrix is of rank $n-r$, and the tangent space to $V(P)$ at the origin is exactly the kernel of this matrix. Pick a $(n-r)times (n-r)$ minor of this matrix with nonzero determinant, and let $S$ (defined in your question) be the set of indices of the columns not in this minor.

Now I claim that we can find a bijection between rows $i$ in this minor and columns $j$ in this minor so that $frac{partial f_i}{partial x_j}(0)neq 0$: if not, then by the definition of the determinant, we have that the determinant of this minor vanishes, contradiction. By an application of a linear transformation, we may assume that the linear terms of these $f_i$ are exactly $x_i$ and nothing else. This is almost the statement you want – the $f_i$ form the required diagonal minor inside the Jacobian, and I would argue that if this is really your goal, then requiring the $phi_i$ be polynomials in exactly one more indeterminant is unnecessary and in fact not always possible (see the end of the post for a counterexample).

This also shows that the $x_s$ for $sin S$ form a transcendence basis. $K$ is generated as a field over $k$ by the images of all $x_i$, and it is of transcendence degree $r$. The $x_t$ for $tnotin S$ are all algebraic over $k(x_s)$: we can use elimination theory on the $f_i$ to construct the relevant polynomials in $k(x_s)[y]$ which vanish on $x_i$. So our extension $ksubset K$ can be written $ksubset k(x_s)subset K$, and as transcendence degree adds over extension, we see $ksubset k(x_s)$ must have transcendence degree $r$, which means that the $x_s$ form a transcendence basis.

Here’s why you might not be able to get $phi_i$ the way you want. Such a $phi_i$ would be a polynomial vanishing on the projection of $V(P)$ to the hyperplane spanned by $x_{i_1},cdots,x_{i_r}$ and $x_i$. It may happen that $V(P)$ is smooth at the origin, but any such projection is singular at the origin, which could imply that there’s no possible choice of $phi_i$ meeting the condition on the first derivative with respect to $y$.

Consider the curve $X$ in $Bbb A^3$ given parametrically by $$tmapsto(frac12t(t-2)(t-3),frac{-1}{2}t(t-1)(t-3),frac16t(t-1)(t-2)).$$ This is an irreducible closed curve which is smooth at the origin and passes through $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. It’s three projections to the planes $x=0$, $y=0$, and $z=0$ are the curves cut out by the irreducible polynomials $$2y^3+18y^2z-2y^2+54yz^2-21yz=54z^2-54z^3,$$ $$x^3-9x^2z-x^2+27xz^2+6xz=27z^3-27z^2,$$ and $$2x^3+6x^2y-2x^2+6xy^2-5xy=2y^2-2y^3$$ respectively, all of which are singular at the origin. So in this case there are no choices of $phi_i$ which meet both your criteria.

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Math Genius: Is the Scheme-theoretic sum of two Cartier divisors on X proper over S again proper?

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Let $X$ be a scheme over $S$. Suppose $D$ and $D^prime$ are two closed subschemes of $X$ which are both proper over $S$. So is the scheme-theoretic sum of $D$ and $D^prime$ again proper over $S$.

If the above does not hold generally, let us suppose moreover that $D$ and $D^prime$ are both effective Cartier Divisors, i.e., both are locally principal closed subschemes. In this case is the sum of the two divisors proper over $S$?

Well, recently I am reading Katz and Mazur’s book $Arithmetic space Moduli space of space Elliptic space Curves$ and the question arises in the study of effective Cartier divisors of smooth curves over S. If you have any other advice in studying this topic, please point out directly. I will be grateful!


This is just an expansion of my comment above. All intersections and unions are scheme-theoretic.

One has $f:Y=Dcup D’to S$, where $D,D’$ are closed subschemes of $Y$. One wants to show $f$ is proper if $f_{|D}, f_{|D’}$ are proper. Proper just means universally closed. So, take a morphism $Tto S$ and pull back to get $g:Ytimes_S T=Dtimes_S Tcup D’times_S Tto T$. If $Zsubset Ytimes_S T$ is a closed subset, then $Z=(Zcap Dtimes_S T)cup (Zcap D’times_S T)$ and $g(Z)=g_{|Dtimes_S t}(Zcap Dtimes_S T)cup g_{|D’times_S T}(Zcap D’times_S T)$. Since both the right hand terms are closed in $T$, since by assumption, you are done.

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Math Genius: Adjunction formula projective hypersurface interpretation

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I was reading a proof that all degree 4 hypersurfaces $X$ in $P^3(mathbb{C})$ are K3 surfaces. When they were showing that the canonical bundle $omega_X$ of $X$ is trivial, they suddenly used the fact that $omega_X cong mathcal{O}_X(-n-1+d)$ and after some Googling, if found that this was an application of the adjunction formula. However, I’m having trouble understanding what actually the object $mathcal{O}_X(-n-1+d)$ is well defined. Because this is the sheaf of holomorphic functions on $X$ but how can you define that on a natural number?

The notation $mathcal{O}_X(n)$ (for any $nin mathbb{Z}$) does not mean the evaluation of $mathcal{O}_X$ at $n$ (which does not make sense), but the $n$th Serre twist of $mathcal{O}_X$.

Precisely, a projective variety has a “twisting sheaf” $mathcal{O}_X(1)$, and for any sheaf $mathcal{F}$ of $mathcal{O}_X$-modules, its $n$th twist (for $n$ nonnegative) is
$$mathcal{F}(n) = mathcal{F} otimes mathcal{O}_X(1)otimes dots otimes mathcal{O}_X(1)$$
When $n$ is negative you have to use $mathcal{O}_X(-1)$, which is dual to $mathcal{O}_X(1)$.

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Math Genius: inflection points of a projective conic

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Suppose $C={x=(x_{0},x_{1},x_{2}) in mathbb{P}^{2}| f(x)=x_{1}^{2}-x_{0}x_{2}=0}$. I want to find inflection points of $C$. So, I need to find $P in C subset mathbb{P}^2: H(f)(P)=0$, where $H_{ij}(f)=(frac{partial^{2} f}{partial x_{i} partial x_{j}})$ and $H(f)=det H_{ij}(f)$.

The Hessian matrix:
$$ H_{ij} = begin{pmatrix} 0 & 0 & -1 \ 0 & 2 & 0 \ -1 & 0 & 0 end{pmatrix}$$
Then $det H_{ij}=-2$. What is wrong?

Thank you in advance!

Nothing is wrong here. An inflection point $p$ on a curve $C$ is a nonsingular point where the tangent line to $p$ meets $C$ with multiplicity at least three. By Bezout, the sum of the intersection multiplicities of a conic and a line in $Bbb P^2$ is two, so no conic in $Bbb P^2$ can have an inflection point.

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Math Genius: Relation between local ring of regular variety and local ring of regular subvariety

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Let $X$ be a regular affine variety and $Y subseteq X$ be a regular subvariety such that $dim(X)-dim(Y)=1$. If $y in Y,$ can we say $frac{mathcal{O}_{X,y}}{<c>} cong mathcal{O}_{Y,y} $ for some $c in (mathfrak{m}_{X,y}setminus mathfrak{m}^2_{X,y})$, where $mathfrak{m}_{X,y}$ is the unique maximal ideal of $mathcal{O}_{X,y}$?

I proved that $frac{mathcal{O}_{X,y}}{<c>}$ is regular ring and it’s dimension is $dim(X)−1$ when $c$ is in $mathfrak{m}_{X,y}setminus mathfrak{m}^2_{X,y}$. I don’t know how to proceed from this part.
I appreciate any help.

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