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For some ring $R$ (no $1$ or commutativity necessary) and $a \in R$, we defined the principal ideal $(a)$ by

$$ (a) := \bigcap \{ I : a \in I \subseteq R, I \text{ is an ideal}\}.$$

Now as a homework question, we shall show that $(a)$ always is of the form $$ (a) = \{ra + as + z\cdot a + \sum_{i=1}^m r_ias_i : r,a,r_i,s_i \in R, z\in \mathbb Z \},$$

where $z\cdot a$ is defined as repeated addition of $a$ with itself. Of course, any element of this form is contained in $(a)$, but I'm clueless as how to show the opposite inclusion. Could you give any hints on how to arrive at that any $x\in (a)$ can be written in this special sum form?

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    Hint: Show that the right hand side (the set of sums) is an ideal containing $a$, then by definition $(a) \subseteq $ the right hand side.2012-05-12
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    Are you sure you mean union? Because R is always an ideal of R, and a is certainly in R. I believe what you are after is the SMALLEST ideal containing a. So it suffices to show that each term of your sum in the second definition must lie in (a). That is, suppose J is an ideal containing a, and prove (a) is contained in J.2012-05-12
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    In fact, let $S$ be a subset of a ring $R$. Then the idea generated by $S$ is exactly $\sum\limits_{s\in S}Rs+\sum\limits_{s\in S}sR+\sum\limits_{s\in S}RsR$ where the sum is finite. If $R$ is commutative, then it will be $\sum\limits_{s\in S}sR$ where the sum is finite. Now can you check this?2012-05-12
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    @David Wheeler: Arh, of course it's supposed to be the **intersection** of the ideals.2012-05-12
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    Thanks, David, now that I think about it, the interpretation of $(a)$ as *the smallest ideal* really helps me out.2012-05-12

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