UTM Ideals Varieties and Algorithm - Chapter 1 Section 5 Exercise 6

Problem:

Solution:

Since $ h = GCD(f_2, \cdots, f_s) $, therefore $ f_i = hg_i $ for all $ i \in [2, s] $.

If $ p \in \langle f_2, \cdots, f_s \rangle $, then $ p = \sum\limits_{i = 2}^{s}{p_if_i} = \sum\limits_{i = 2}^{s}{p_ihg_i} = h\sum\limits_{i = 2}^{n}{p_ig_i} $, therefore $ p \in \langle h \rangle $

Now assume without proof (yet, we will do that soon) that there exists polynomial $ q_i $ such that $ h = \sum\limits_{i = 2}^{s}{q_if_i} $, then it is easy to show if $ p \in \langle h \rangle $, then $ p = rh = r\sum\limits_{i = 2}^{s}{q_if_i} = \sum\limits_{i = 2}^{n}{rq_if_i} $, and therefore $ p \in \langle f_2, \cdots, f_s \rangle $

So now we establish $ \langle h \rangle = \langle f_2, \cdots, f_s \rangle $

If $ p \in \langle f_1, h \rangle $, then $ p = af_1 + bh = af_1 + b\sum\limits_{i = 2}^{s}{q_if_i} $, therefore $ p \in \langle f_1, f_2, \cdots, f_s \rangle $.

If $ p \in \langle f_1, f_2, \cdots, f_2 \rangle $, then $ p = af_1 + c $ where $ c \in \langle f_2, \cdots, f_s \rangle = \langle h \rangle $, so we can write $ p = af_1 + rh $, so $ p \in \langle f_1, h \rangle $.