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 $p \in \langle f_1, h \rangle$.