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...

  • $\begingroup$ 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. $\endgroup$ – Steven Sam Nov 16 '11 at 2:06
  • $\begingroup$ Thanks, Steven. We'll see whether I can hijack tomorrow's combinatorics preseminar with this question. $\endgroup$ – darij grinberg Nov 16 '11 at 6:49
  • $\begingroup$ Hyperelliptic curves are actually pretty interesting. $\endgroup$ – JSE Nov 17 '11 at 2:37

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.

  • $\begingroup$ Thanks, but what interests me is the kernel of your $\varphi$. $\endgroup$ – darij grinberg Nov 16 '11 at 6:48

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