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Let $S=k[x_1,\dots,x_n]$ be a polynomial ring, and $A:=k[x^{u^{(1)}}, \dots x^{u^{(l)}}]$ a monomial subalgebra, generated by monomials $x^{u^{(i)}} = \prod_{j=1}^n x_j^{u^{(i)}_{j}}$ with $u^{(i)} \in \mathbf{N}^n$. What are sufficient criteria for the inclusion $A \to S$ to split? What is the splitting? Since S is normal, and direct summands of normal rings are normal, normality of A is necessary.

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You want the inclusion to split as a map of rings? (the zero you put there is off, then :) ) –  Mariano Suárez-Alvarez Oct 29 '11 at 9:53
    
In characteristic zero, normality is also enough. The norm map on quotient fields restricts to a splitting of rings. Of course this has nothing to do with the special form of $A$. –  Graham Leuschke Oct 29 '11 at 10:50
    
I guess my previous comment only applies if $S$ is a finitely generated $A$-module, that is, some power of each $x_j$ is in $A$. –  Graham Leuschke Oct 29 '11 at 13:40
    
@Mariano: fixed, thanks. –  Thomas Oct 29 '11 at 13:52
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1 Answer

up vote 1 down vote accepted

This part answers when $A$ (the monomial algebra) is a direct summand of the polynomial ring as an $A$-module.

Normality is also sufficient, also this doesn't depend on the field $k$. This is essentially exercise 6.1.10 in "Cohen-Macaulay rings" by Bruns and Herzog.

Let $S\subset T$ be affine semigroups. Call $S$ a full subsemigroup if $S= T\cap \mathbb ZS$. It is not hard to show that if $S$ is a positive affine semigroup then it is normal if and only if it is isomorphic to a full subsemigroup of $\mathbb N^n$, for some $n\geq 0$.

The claim in question follows from the general fact that if $S$ is full in $T$ then $k[S]$ is a direct $k[S]$-summand of $k[T]$. The proof is straightforward, denote by $W$ the $k$-algebra spanned by elements in $T-S$, then clearly $k[T]=k[S]\oplus W$ as vector spaces. To show that $W$ is in fact a $k[S]$ module, it is enough to check that if $\alpha\in S$ and $\beta\in T-S$ then $\alpha+\beta\in T-S$. If we assume for the sake of contradiction that $\alpha'=\alpha+\beta\in S$, then $\beta=\alpha'-\alpha\in T$ which implies $\beta\in S$ since $S$ is full.


This part answers when the inclusion $A\to k[x_1,\dots,x_n]$ splits as a map of rings.

I hope I haven't made any silly mistakes, but the final answer seems to be a very restricted class of monomial algebras. Let $I=\lbrace i_1,\dots i_m\rbrace \subset \lbrace 1,2,\dots,m\rbrace$ and let $J$ be its complement. I claim that all $A$ in our problem are generated by monomials of the form $x_{i_r}f_r$ where $f_r$ are monomials in variables indexed by $J$ (for arbitrary choices of $I$ and $f_r$'s).

First, to see that the inclusion splits for such monomial algebras notice that we have the map $k[x_1,\dots,x_n]\to A$ defined by $x_{i_r}\to x_{i_r}f_r$ for all $1\le r\le m$ and $x_i\to 1$ for all $i\in J$.

Now suppose we have a monomial algebra $A=k[x^{\alpha_1},\dots,x^{\alpha_m}]$ and the inclusion $i:A\to k[x_1,\dots,x_n]$. In the conditions of our problem we have a map $q:k[x_1,\dots,x_n]\to A$ so that $qi=\text{id}_A$. (Here I am denoting $x^{\alpha_i}=x_1^{\alpha_{i1}}\cdots x_n^{\alpha_{in}}$.)

Our conclusion follows because we have $x^{\alpha_r}=\prod q(x_i)^{\alpha_{ri}}$, so that for every $r$ we have for some $i$, $\alpha_{ri}=1$ and $q(x_i)=\beta_r x^{\alpha_r}$ for some $\beta_r\in k$ (because $x^{\alpha_r}$ is a generator so it cannot have a non-trivial factorization). The rest of the variables with $\alpha_{ri}\geq 1$ are sent to scalars. therefore each $x^{\alpha_r}$ factors as $x_{i_r}f_r$ where all $i_r$'s are distinct and all $f_r$'s are monomials in the rest of the variables.

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Thanks! In most cases being a $k[S]$-module direct summand is what one is interested in. I originally though of a direct summand of rings, is that a meaningful question at all? –  Thomas Oct 30 '11 at 18:07
    
I added what I think answers the question for splitting as a map of rings. –  Gjergji Zaimi Nov 8 '11 at 0:22
    
Looks alright. Thanks. –  Thomas Nov 23 '11 at 16:47
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