Proposition If \( I + J = \langle 1 \rangle \), then \( I \cap J = I J \)
Proof.
- \( \forall t \in I \cap J , t \in I \land t \in J \). \( I + J = \langle 1 \rangle \implies \exists i_1 , j_1 \in I, J : i_1+j_1=1 \). Note, \( t=i_1t+j_1t \in IJ \). So, \( I \cap J \subseteq I J \)
- \( \forall t \in I J, \exists i_k , j_k \in I, J : t= \sum {i_kj_k}
\). Note, for all \( k
\), \( i_k , j_k \in I, R \implies i_kj_k \in I \). Since \( I \) is closed under
addition, \( t= \sum {i_kj_k} \in I \).
Similarly, \( t \in J \). Thus, \( t \in I \cap J \) and \( I \cap J \supseteq I J \)
Overall, \( I \cap J = I J \)
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