the monomial orders can be extended? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T15:15:26Z http://mathoverflow.net/feeds/question/87994 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/87994/the-monomial-orders-can-be-extended the monomial orders can be extended? willy 2012-02-09T15:00:02Z 2012-02-09T21:56:03Z <p>Let $k[x_0,\cdots,x_N]$ be the polynomial ring. Suppose that we have the monomial orders on $k[x_0,...,x_m]$ and $k[x_m,...,x_N]$ then is there a monomial order on $k[x_0,\cdots,x_N]$ such that this order induces given monomial orders on $k[x_0,\cdots,x_m]$ and $k[x_m,\cdots,x_N]$? I think it may be possible, but i dont know how to construct the monomial order formally.</p> http://mathoverflow.net/questions/87994/the-monomial-orders-can-be-extended/88008#88008 Answer by Alberto García-Raboso for the monomial orders can be extended? Alberto García-Raboso 2012-02-09T16:51:01Z 2012-02-09T21:56:03Z <p>It is a result of Lorenzo Robbiano's (<a href="http://www.ams.org/mathscinet-getitem?mr=826583" rel="nofollow">Term orderings on the polynomial ring</a>; see also section 1.2 of Greuel and Pfister's <a href="http://books.google.com/books?id=7PMdgAPjscsC&amp;lpg=PR3&amp;dq=A%2520singular%2520introduction%2520to%2520commutative%2520algebra&amp;pg=PR3#v=onepage&amp;q&amp;f=false" rel="nofollow">A Singular Introduction to Commutative Algebra</a>) that every monomial ordering on a polynomial ring in $n$ variables can be obtained from a matrix $A \in \mathrm{GL}(n, \mathbb{R})$ in the following way: <code>$$x^\alpha &gt;_A x^\beta \qquad\Leftrightarrow\qquad A\alpha &gt; A\beta$$</code> where you arrange the exponents as a column vector and the ordering on the right-hand side is the lexicographical ordering on $\mathbb{R}^n$.</p> <p>Notice that different matrices can give rise to the same monomial ordering. E.g., there are only two different monomial orderings on $k[x]$: the lexicographical ($x > 1$) and the negative lexicographical ($1 &lt; x$). Any 1x1 matrix with positive entry will give the first one; the second one can be obtained from any 1x1 matrix with negative entry.</p> <p>Choose matrices <code>$$A = (a_{ij})_{i,j = 0, \ldots, m} \in \mathrm{GL}(m+1, \mathbb{R})$$</code> and <code>$$B = (b_{ij})_{i,j = m, \ldots, N} \in \mathrm{GL}(N-m+1, \mathbb{R})$$</code> that give rise to your original monomial orderings $>_A$ on $k[x_0, \ldots, x_m]$, and $>_B$ on $k[x_m, \ldots, x_N]$, respectively. The signs of $a_{mm}$ and $b_{mm}$ determine the restrictions of $>_A$ and $>_B$ to $k[x_m]$. If you want to "glue" this orderings then it is necessary that these coincide, i.e., that the signs of $a_{mm}$ and $b_{mm}$ be the same. I claim that this condition is also sufficient.</p> <p>We thus want to construct a matrix <code>$$C = (c_{ij})_{i,j = 0, \ldots, N} \in \mathrm{GL}(N+1, \mathbb{R})$$</code> Start by scaling $B$ in such a way that $a_{mm} = b_{mm}$ (you can always multiply a matrix by a positive number without changing the associated ordering). Define then</p> <ul> <li>$c_{ij} = a_{ij}$ for $0 \leq i, j \leq m$,</li> <li>$c_{ij} = b_{ij}$ for $m \leq i, j \leq N$,</li> <li>$c_{ij} = 0$ otherwise.</li> </ul> <p>If this matrix is invertible, then it defines an ordering $>_C$ on $k[x_0, \ldots, x_N]$ satisfying your requirements.</p> <p>Let me denote by $A_i$ the columns of $A$. Since $A$ is invertible, these are $m+1$ linearly independent vectors in $\mathbb{R}^{m+1}$. The same can be said about the columns $B_i$ of $B$ as vectors in $\mathbb{R}^{N-m+1}$. Let now $C_i$ be the columns of $C$, and suppose they satisfy a linear relation $$\lambda_0 C_0 + \ldots + \lambda_N C_N = 0$$ in $\mathbb{R}^{N+1}$. Projecting onto the first $m+1$ coordinates yields $$\lambda_0 A_0 + \ldots + \lambda_m A_m = 0$$ which implies $\lambda_0 = \cdots = \lambda_m = 0$. Similarly, projection onto the last $N-m+1$ coordinates results in the equation $$\lambda_m B_m + \ldots + \lambda_N B_N = 0$$ forcing the vanishing of the remaining $\lambda$'s and proving that the columns of $C$ are linearly independent. </p>