Normality of an affine semigroup - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T17:42:38Z http://mathoverflow.net/feeds/question/13477 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/13477/normality-of-an-affine-semigroup Normality of an affine semigroup unknown (google) 2010-01-30T16:06:21Z 2010-02-06T18:26:22Z <p>An affine monoid is a finitely generated commutative submonoid of $\mathbb Z^k$ for some positive integer k. Let S be an affine monoid and let G(S) be the group generated by S. We say the monoid S is normal if and only if for all $g \in G(S)$ and $n \in \mathbb N \setminus {0}$, $ng \in S$ implies $g \in S$. </p> <p>Let $S$ be the submonoid generated by the finite set <code>$T= \{(p_0,p_1, \cdots ,p_{n-1}) \in (\mathbb Z_{\geq 0})^n: \sum_{i=0}^{n-1}p_i=n \,\, and \,\, \sum_{i=0}^{n-1}i.p_i \cong 0 \pmod n\}$</code>. Now my question is how to prove that $S$ is normal ?</p> http://mathoverflow.net/questions/13477/normality-of-an-affine-semigroup/13879#13879 Answer by yanzhang for Normality of an affine semigroup yanzhang 2010-02-02T22:33:00Z 2010-02-02T22:54:25Z <p>I worked on this for a bit with Ricky Liu, who came up with this very quick solution:</p> <p>Take your set T. Suppose $\sum p_i = kn$, where $k \geq 2$. Create the following set T': let $i$ appear $p_i$ times. This creates a set $T'$ with at least $2n$ elements (by your first constraint and $k \geq 2$), whose sum is divisible by $n$ by your second constraint.</p> <p>However, by <a href="http://en.wikipedia.org/wiki/Erd%C5%91s-Ginzburg-Ziv%5Ftheorem" rel="nofollow">Erdos-Ginsberg-Ziv</a>, there's a subset of $n$ elements which add to $n$, which exactly corresponds to your generator, so we're done.</p> http://mathoverflow.net/questions/13477/normality-of-an-affine-semigroup/14410#14410 Answer by unknown (yahoo) for Normality of an affine semigroup unknown (yahoo) 2010-02-06T18:26:22Z 2010-02-06T18:26:22Z <p>Sorry, I did not get the above answer. Can somebody please explain me. </p> <p>Suppose, $q.(p_0,p_1, \cdots ,p_{n-1}) \in S$, where $q \in \mathbb N \setminus 0$ and $(p_0,p_1, \cdots ,p_{n-1}) \in G(S)$, why is it clear from the above answer that $(p_0,p_1, \cdots ,p_{n-1})$ is already in $S$. Some of the $p_i$'s can be negative as well. It is not at all clear to me. Thanks in advance.</p>