why is ideal $I_1, I_2, \dots, I_n$ contained in $I_1, I_2, \dots, I_{n-1}$

Question

Let $A$ be the ring, and $I_1, I_2, \dots, I_n$ are the ideals. I need to show that $I_1 I_2 \dots I_n \subset I_1 I_2\dots I_{n-1}$. I know that $I_1 I_2 \dots I_n \subset I_1 \cap I_2 \cap \dots \cap I_n$, and also $I_1 I_2 \dots I_{n-1} \subset I_1 \cap I_2 \cap \dots \cap I_{n-1}$. Clearly, $I_1 \cap I_2 \cap \dots \cap I_{n} \subset I_1 \cap I_2 \cap \dots \cap I_{n-1}$, but how can I conclude that $I_1 I_2 \dots I_n \subset I_1 I_2\dots I_{n-1}$?

Answer 1

You can prove this easily by showing set inclusion. Take an element in $I_1 I_2 \dots I_n$. This set is the ideal generated by the elements $\prod_{k=1}^n i_k$ where $i_k \in I_k$. Hence any element here will look like the linear combinations of such elements, say $\sum_{m=1}^{m_0} \alpha_m(\prod_{k=1}^n .i_{k_m})$. This element can be written as $\sum_{m=1}^{m_0} (\alpha_m.i_{n_m})(\prod_{k=1}^{n-1}i_{k_m})$ which belongs in $I_1 I_2\dots I_{n-1}$. Hence $I_1 I_2 \dots I_n \subset I_1 I_2\dots I_{n-1}$.

Answer 2

Since $I_1I_2 \ldots I_{n - 1}$ is an ideal, we have

$I_1I_2 \ldots I_{n - 1}A \subset I_1I_2 \ldots I_{n - 1}; \tag 1$

this actually follows simply from the fact that

$I_{n - 1}A \subset I_{n - 1}; \tag 2$

now since

$I_n \subset A, \tag 3$

we further have

$I_1I_2 \ldots I_{n - 1} I_n \subset I_1I_2 \ldots I_{n - 1}A \subset I_1I_2 \ldots I_{n - 1}. \tag 4$