Let $f(x, y) = \sum_{m=0}^{M-1}\sum_{n=0}^{N-1} a_{m,n} x^m y^n$ and $g(x, y) = \sum_{m=0}^{M-1}\sum_{n=0}^{N-1} b_{m,n} x^m y^n$. Computing the Gröbner basis, we get an univariate polynomial $h_1(x)$ and a bivariate polynomial $h_2(x, y)$. What is the degree of $h_1(x)$? Thanks a lot.
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$\begingroup$ That depends on the chosen monomial order, and on $f$ and $g$. A lower bound is the degree of the zero polynomial. $\endgroup$– Dietrich BurdeMay 27, 2013 at 20:55
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$\begingroup$ I guess the monomial order is y > x (since (s)he eliminates y), so h1 divides the resultant of f,g w.r.t. y. $\endgroup$– pinakiMay 27, 2013 at 23:45
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In general, there is the following result on upper bounds for the degree of elements in the (reduced) Groebner basis: Let $G$ be a reduced Groebner basis of an ideal $I=\langle f_1,\ldots, f_r\rangle $. Let $d$ be the maximal degree (with respect to some monomial order) of $deg(f_1),\ldots ,deg(f_r)$. Let $n$ be the dimension of the ring (the number of variables). Then we have $$ max \lbrace deg(g) \mid g\in G \rbrace \le 2\left(\frac{d^2}{2}+d\right)^{2^{n-1}}. $$ The result is due to Thomas W. Dubé (1990).
Of course, for the equations $f(x,y)=g(x,y)=0$ we may just take the resultant $h_1(x)=res(f(y)(x),g(y)(x))$ with respect to $x$. No Groebner basis is needed then, and the degree can be estimated easily.
answered May 27, 2013 at 21:04