Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This question comes from the explicit construction of a smooth projective model of a hyperelliptic curve. Nevertheless it is fully elementary and, to me, more interesting than hyperelliptic curves.

Notations. In the following, $k$ will always denote a commutative ring with $1$. "Graded $k$-algebra" will always mean a $k$-algebra graded by $\mathbb N$. Whenever $S$ is a graded $k$-algebra and $d$ is a positive integer, we denote by $S^{(d)}$ the graded $k$-algebra whose $i$-th graded component is $S^{di}$ (this means the $di$-th graded component of $S$) for each $i\in\mathbb N$. The algebra structure on $S^{(d)}$ is inherited from $S$ (since $S^{(d)}$ is a subset of $S$ and easily seen to be a subalgebra). Whenever we speak of "standard polynomial algebras", we mean polynomial $k$-algebras with standard grading (i. e., any indeterminate has degree $1$).

First, here is the fact that helps us construct that projective model:

Theorem 1. Let $e$ be a positive integer. Let $k\left[X,Y\right]$ and $k\left[X_0,X_1,...,X_e\right]$ be standard polynomial algebras. Then, the $k$-algebra homomorphism

$k\left[X_0,X_1,...,X_e\right]\to k\left[X,Y\right]^{(e)},$

$X_i\mapsto X^iY^{e-i}$

is graded and surjective. Its kernel is the ideal generated by terms of the form $X_iX_j-X_{i+1}Y_{j-1}$ for $i$ and $j$ satisfying $0\leq i < j-1 < j \leq n$.

This is easily proven combinatorially, by constructing a basis of $k\left[X,Y\right]^{(e)}$ of monomials and lifting it to a basis of $k\left[X_0,X_1,...,X_e\right]$ of monomials.

The natural questions are now:

Question 2. Theorem 1 gives $k\left[X,Y\right]^{(e)}$ as a graded quotient $k$-algebra of $k\left[X_0,X_1,...,X_e\right]$. Can we similarly represent $k\left[X,Y,Z\right]^{(e)}$ or $k\left[Y_1,Y_2,...,Y_n\right]^{(e)}$ ?

Question 3. Now assume that we grade a polynomial algebra $k\left[T_1,T_2,...,T_e\right]$ in a nonstandard way, i. e., we have $\deg T_i=\alpha_i$ for some positive integers $\alpha_i$. (The ground ring $k$ is still in the $0$-th component.) What is a necessary and sufficient condition on $d$ for the $k$-algebra $k\left[T_1,T_2,...,T_e\right]^{(d)}$ to be generated by its degree-$1$ component (as an algebra over $k$)? Clearly, a necessary condition is for $d$ to be divisible by all $\alpha_i$, but I can't see whether it is sufficient.

Elementary reformulation: If $\alpha_1$, $\alpha_2$, ..., $\alpha_e$ are positive integers, then what conditions do we have to impose on a positive integer $d$ in order for the following to hold: Whenever $\beta_1$, $\beta_2$, ..., $\beta_e$ are nonnegative integers satisfying $d\mid\alpha_1\beta_1+\alpha_2\beta_2+...+\alpha_e\beta_e$, there exist nonnegative integers $\gamma_1\leq \beta_1$, $\gamma_2\leq \beta_2$, ..., $\gamma_e\leq \beta_e$ such that $\alpha_1\gamma_1+\alpha_2\gamma_2+...+\alpha_e\gamma_e = d$.

On the one hand, this looks like elementary number theory; on the other it reminds me of combinatorial facts like the one claiming that a regular bipartite graph can be factored into perfect matchings. None of these helps me proving or disproving the natural conjecture (that the condition is that $d$ is divisible by all $\alpha_i$), though...

share|improve this question
    
Question 2: On the geometric side, you are asking about the $e$th Veronese embedding of $n-1$ dimensional projective space. I am happy to elaborate in person. –  Steven Sam Nov 16 '11 at 2:06
    
Thanks, Steven. We'll see whether I can hijack tomorrow's combinatorics preseminar with this question. –  darij grinberg Nov 16 '11 at 6:49
    
Hyperelliptic curves are actually pretty interesting. –  JSE Nov 17 '11 at 2:37

1 Answer 1

Question 2: The following map defines a surjective $k$-algebra homomorphism: $$\varphi: k[X_{i_1,...,i_n} \mid i_1 + ... + i_n = e] \to k[Y_1,...Y_n],\quad X_{i_1,...,i_n} \mapsto Y_1^{i_1} \cdots Y_n^{i_n}.$$ For, let non-negative rational integers $j_1,...,j_n$ be given, those sum is $de$ and let $I_p =(i_{p1},...,i_{pn})$ be non-negative rational integers such that $i_{p1} + ... + i_{pn} = e$. Because of $$Y_1^{j_1} \cdots Y_n^{j_n}\overset{!}{=}\varphi(\prod_{p=1}^d X_{I_p}) =\prod_{p=1}^d (Y_1^{i_{p1}} \cdots Y_n^{i_{pn}}) = (Y_1^{\sum_{p=1}^d i_{p1}}) \cdots (Y_n^{\sum_{p=1}^d i_{pn}})$$ we want to solve $$\sum_{p=1}^d i_{p,q} = j_q,\quad (q=1,...,n).$$ In case $d=1$ choose $i_{1q} = j_q$. Assume the equation is solvable for $d-1$. Choose $0 \le i_{d,q} \le j_q$ such that $i_{d1} + ... + i_{dn} = e$ (possible since $j_1 +...+j_n = de \ge e$). Then the linear system above is equivalent to $$\sum_{p=1}^{d-1} i_{p,q} = j_q - i_{d,q},\quad (q=1,...,n)$$ which is solvable by induction hypothesis.

share|improve this answer
    
Thanks, but what interests me is the kernel of your $\varphi$. –  darij grinberg Nov 16 '11 at 6:48

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.