+ rI. Then, since d = c oF 0 we see that 9 ~ 9 - eX 9i ~ + Tl. + r also, we see r = Tl, as desired (by the assumption that the remainder is unique). • Case 3. d # 0 and d # c. Set h = 9 - dX 9i' Then the coefficient of X Ip(9i) in h is O. Since d # 0 we have 9 ~ h. Also, since d # c we have 9 - cXgi ~ h. So if h ~+ T2, Bueh that T2 is reduced, we get 9 ~ h ~ + T2 and 80 T2 = T, since the remainder. is unique. And 80 9 - cXgi ~ h --S+ T, as desired.

8. 1O. J = yx-x and 12 = y2 - x. J,h}. We use deglex with y> x. 10 that J ~+ 0 and J ~+ x 2 - x, the latter being reduced with respect to F. 7, F is not a Grübner basis. We can see this in another way. J,h) and J ~+ x 2 - x we have x 2 - xE (h, 12). J) = xy orlp(h) = y2. 1), Fis not a Grobner basis. 9. [x, y, z]. [x, y, z] with x < y < z. We will prove that G is a Grübner basis for J. Suppose to the contrary that there exists J E J such that lt(f) ~ (lt(91),lt(92)) = (z, y). Then, z does not divide lt(f), and y does not divide lt(f).

Let h = xy - x, 12 = x 2 - y E lQl[x, y] with the deglex term order with x < y. Let F = {h, h}. Then 8(h, 12) = xh - yh = y2 - x 2 ...!... y2 _ y, and h = y2 - Y is reduced with respect to F. Sa we add h to F, and , F' = {h, fz, h}· Then 8(h, 12) -----+ O. Now 8(h, h) = Yh - xh = 0, and F' F' 8(fz, h) = y 2h - x 2h = _y3 + x 2y -----+ x 2y - y2 -----+ O. Thus {h, fz, h} is a let F Grobner basis. 1. 8. Given F = {h, ... 1) will produce a Griibner basis Jor the ideal 1 = (h,··· ,J,). PROOF. We first need to show that this algorithm terminates.

### An Introduction to Grobner Bases (Graduate Studies in Mathematics, Volume 3) by Philippe Loustaunau, William W. Adams

by Mark

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