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

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.

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

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 $$a,b,cinmathbb{Z}$$).
$$s=t^3 + at^2s+bts^2 + cs^3$$
gives:
$$t^3 = s(1 – at^2 – bts – cs^2)$$
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)\$

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.

## Math Genius: Ideal of the twisted cubic

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\$.

Lemma
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 http://wwwb.math.rwth-aachen.de/~markus/ and it says that your ideal is prime.

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

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 !

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

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!!

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.

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

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!

Sincerely.

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.

## Math Genius: Adjunction formula projective hypersurface interpretation

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 $$mathcal{O}_X(1)$$.

## Math Genius: inflection points of a projective conic

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?

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

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 $$frac{mathcal{O}_{X,y}}{} cong mathcal{O}_{Y,y}$$ 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 $$frac{mathcal{O}_{X,y}}{}$$ 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.