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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.

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You need at least a compatibility condition, namely that the orders induced on $k[x_m]$ coincide. –  Alberto García-Raboso Feb 9 '12 at 15:40
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It is a result of Lorenzo Robbiano's (Term orderings on the polynomial ring; see also section 1.2 of Greuel and Pfister's A Singular Introduction to Commutative Algebra) 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: $$ x^\alpha >_A x^\beta \qquad\Leftrightarrow\qquad A\alpha > A\beta $$ where you arrange the exponents as a column vector and the ordering on the right-hand side is the lexicographical ordering on $\mathbb{R}^n$.

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 ($1 < x$). Any 1x1 matrix with positive entry will give the first one; the second one can be obtained from any 1x1 matrix with negative entry.

Choose matrices $$ A = (a_{ij})_{i,j = 0, \ldots, m} \in \mathrm{GL}(m+1, \mathbb{R}) $$ and $$ B = (b_{ij})_{i,j = m, \ldots, N} \in \mathrm{GL}(N-m+1, \mathbb{R}) $$ 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.

We thus want to construct a matrix $$ C = (c_{ij})_{i,j = 0, \ldots, N} \in \mathrm{GL}(N+1, \mathbb{R}) $$ 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

  • $c_{ij} = a_{ij}$ for $0 \leq i, j \leq m$,
  • $c_{ij} = b_{ij}$ for $m \leq i, j \leq N$,
  • $c_{ij} = 0$ otherwise.

If this matrix is invertible, then it defines an ordering $>_C$ on $k[x_0, \ldots, x_N]$ satisfying your requirements.

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.

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@willy: if my post answers your question, you should accept it. –  Alberto García-Raboso Feb 9 '12 at 21:56
    
The matrix C which you make is not invertible in general. and moreover the C does not induce the order on $k[x_m,\cdots,x_N]$ what i given. finally, $a_{mm}$ and $b_{mm}$ may be zeros –  willy Feb 20 '12 at 10:00
    
@willy: $a_{mm}$ cannot be zero, for then $>_A$ would induce the zero order on $k[x_m]$, which is not an order. Equivalently, $a_{mm} = 0$ would say that be saying the vector $x_m$ is an eigenvalue of $A$ with eigenvalue zero, contradicting the invertibility of $A$. –  Alberto García-Raboso Feb 20 '12 at 13:24
    
Consider the permutation matrix $A$, this matrix can have the $a_mm=0$, but this matrix defines the order on $k[x_0,\cdots,x_m]$, moreover $a_mm=0$ does not imply the matrix is not invertiable, for example, the matrix (0,1:1,0) is invertible and this matrix defines the lexicographic order on $k[x_0,x_1]$ such that $x_1>x_0$. –  willy Feb 21 '12 at 17:50
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