Problem:
Solution:
(i) $ \implies $ (ii)
If $ f_i \in I $, $ \forall i $, then $ \sum\limits_{i = 1}^{s}h_i f_i \in I $ because $ I $ is an ideal, therefore $ \langle f_1, \cdots f_s \rangle \subset I $.
(ii) $ \implies $ (i)
If $ \langle f_1, \cdots f_s \rangle \subset I $, then $ \sum\limits_{i = 1}^{s}h_i f_i \in \langle f_1, \cdots f_s \rangle \subset I $ because $ \langle f_1, \cdots f_s \rangle $ is an ideal. For $ k \in [1, s] $, set $ h_k = 1 $ and $ h_i = 0 $ for all $ i \ne k $, we get (i).
Solution:
(i) $ \implies $ (ii)
If $ f_i \in I $, $ \forall i $, then $ \sum\limits_{i = 1}^{s}h_i f_i \in I $ because $ I $ is an ideal, therefore $ \langle f_1, \cdots f_s \rangle \subset I $.
(ii) $ \implies $ (i)
If $ \langle f_1, \cdots f_s \rangle \subset I $, then $ \sum\limits_{i = 1}^{s}h_i f_i \in \langle f_1, \cdots f_s \rangle \subset I $ because $ \langle f_1, \cdots f_s \rangle $ is an ideal. For $ k \in [1, s] $, set $ h_k = 1 $ and $ h_i = 0 $ for all $ i \ne k $, we get (i).
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