Generators of a certain ideal - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T15:06:34Z http://mathoverflow.net/feeds/question/87429 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/87429/generators-of-a-certain-ideal Generators of a certain ideal Pierre-Yves Gaillard 2012-02-03T13:06:21Z 2012-02-04T10:02:18Z <p>In view of Mariano Suárez-Alvarez's answer I see how badly phrased my question was, and decided to rewrite it. The drawback is that some comments of Martin Brandenburg are now incomprehensible, but I thought it would suffice to say here that Martin made some legitimate constructive criticisms to the original wording of the question. By the way, thank you also to Vladimir Dotsenko for his comments.</p> <p>Let $K$ be a commutative ring, and let $X_1,\dots,X_n$ be indeterminates. Here $n$ is an integer $\ge3$. For $1\le i &lt; j\le n$ put $$x_{ij}:=\frac{1}{X_i-X_j}$$ and let $Y_{ij}$ be an indeterminate. Let $I$ be the kernel of the $K$-algebra morphism $$\varepsilon:K[(Y_{ij})]\to K[(x_{ij})],\quad Y_{ij}\mapsto x_{ij}.$$</p> <blockquote> <p>Is $I$ finitely generated? If it is, can one give an explicit finite set of generators?</p> </blockquote> <p>Note that the identity $$\frac{1}{a-b}\ \frac{1}{a-c}+\frac{1}{b-a}\ \frac{1}{b-c}+\frac{1}{c-a}\ \frac{1}{c-b}=0.$$ shows that $I$ is nonzero.</p> <p>(I put the homological algebra tag because the ultimate goal is to know whether there is a functorial free resolution of $K[(x_{ij})]$, viewed as a $K[(Y_{ij})]$-module, and, if it exists, what can be said about it.)</p> <p>The question had been posted before on Mathematics Stack Exchange (<a href="http://math.stackexchange.com/questions/104548/generators-of-a-certain-ideal" rel="nofollow">link</a>).</p> http://mathoverflow.net/questions/87429/generators-of-a-certain-ideal/87464#87464 Answer by Mariano Suárez-Alvarez for Generators of a certain ideal Mariano Suárez-Alvarez 2012-02-03T18:20:55Z 2012-02-03T19:11:28Z <p>A polynomial $f\in K[\underline Y]$ is in the kernel of your map iff $f$ is zero in the quotient <code>$$\frac{k[X,Y]}{\bigl((X_i-X_j)Y_{i,j}-1:1\leq i&lt;j\leq n\bigr)}.$$</code>In other words, your kernel is the intersection of the ideal in the denominator with the ring $k[Y]$, <code>$$\ker\varepsilon=k[Y]\cap\bigl((X_i-X_j)Y_{i,j}-1:1\leq i&lt;j\leq n\bigr).$$</code>This intersection is generated by the elements of a Groebner basis which only contain $Y$s, assuming you are using a monomial order which <em>eliminates</em> the $X$s; this is explained in the book by Cox, Little and O'Shea, for example. </p> <p>Doing small examples shows that </p> <blockquote> <p>$(\star)$ the intersection is generated by all polynomials of the form <code>$$Y_{i,j} Y_{i,k}+ Y_{j,k}Y_{j,i}+Y_{k,i}Y_{k,j}$$</code> with $i$, $j$ and $k$ distinct. (I am identifying $Y_{i,j}$ with $-Y_{j,i}$ here when $i\neq j$)</p> </blockquote> <p>Ordering the variables as in $$X_1,X_2,X_3,X_4,Y_{1,2},Y_{1,3},Y_{1,4},Y_{2,3},Y_{2,4},Y_{3,4}$$ for $n=4$ we find the Groebner basis $$\begin{array}{l} Y_{2,3} Y_{2,4}+Y_{3,4} Y_{2,4}-Y_{2,3} Y_{3,4} \\ Y_{1,3} Y_{1,4}+Y_{3,4} Y_{1,4}-Y_{1,3} Y_{3,4} \\ Y_{1,2} Y_{1,4}+Y_{2,4} Y_{1,4}-Y_{1,2} Y_{2,4} \\ Y_{1,2} Y_{1,3}+Y_{2,3} Y_{1,3}-Y_{1,2} Y_{2,3} \\ X_3 Y_{3,4}-X_4 Y_{3,4}-1 \\ X_2 Y_{2,4}-X_4 Y_{2,4}-1 \\ X_2 Y_{2,3}-X_3 Y_{2,3}-1 \\ X_1 Y_{1,4}-X_4 Y_{1,4}-1 \\ X_1 Y_{1,3}-X_3 Y_{1,3}-1 \\ X_1 Y_{1,2}-X_2 Y_{1,2}-1 \end{array}$$ The same pattern is seen for all $n$. It is very easy to see that all these polynomials are in $((X_i-X_j)Y_{i,j}-1:1\leq i&lt;j\leq n)$, and it should not be difficult to show that they are a Groebner basis in general. I expect checking that the above claim $(\star)$ can actually be proved without much pain.</p>