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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. –  martini May 12 '12 at 11:32
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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. –  David Wheeler May 12 '12 at 11:33
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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? –  molan May 12 '12 at 11:33
    
@David Wheeler: Arh, of course it's supposed to be the intersection of the ideals. –  Fritz Langahn May 12 '12 at 12:14
    
Thanks, David, now that I think about it, the interpretation of $(a)$ as the smallest ideal really helps me out. –  Fritz Langahn May 12 '12 at 12:24

1 Answer 1

Clearly, every element in $$ \{ra + as + z\cdot a + \sum_{i=1}^m r_ias_i : r,a,r_i,s_i \in R, z\in \mathbb Z \}$$ must lie in every ideal that contains $a$, since each of $ra$, $as$, $z\cdot a$, and each $r_ias_i$ will like in an ideal that contains $a$. Thus, this set is contained in the intersection.

In order to show that this set equals the intersection, we need to show that this set is in fact an ideal that contains $a$, because this will establish that it is one of the sets being intersected.

That the set contains $a$ follows by setting $r=s=m=0$ and $z=1$.

Now let $ra+as+z\cdot a + \sum_{i=1}^m r_ias_i$ and $ta+au+w\cdot a + \sum_{j=1}^n t_jau_j$ be two elements of the set. Then their difference is $$(r-t)a + a(s-w) + (z-w)\cdot a + \sum x_kay_k$$ where $k$ ranges from $1$ to $n+m$, $x_k=r_k$ and $y_k=s_k$ for $k=1,\ldots,m$, and $x_k=t_{k-m}$, $y_k = u_{k-m}$ for $k=m+1,\ldots,n$. Thus, the set is a subgroup.

Multiplying $ra + as + z\cdot a + \sum_{i=1}^m r_ias_i$ by $x$ on the left gives $$(xr+z\cdot x)a + a0 + 0\cdot a + \sum_{i=0}^m t_iau_i$$ where $t_0=x$, $u_0 = s$, $t_i=xr_i$ for $i=1,\ldots,m$, and $u_i=s_i$ for $i=1,\ldots,m$. This is an element of the set.

Similarly, if we multiply by $x$ on the right we get $$0a + a(sx+z\cdot x) + \sum_{i=0}^m v_iaw_i,$$ where $v_0 = r$, $w_0=x$, $v_i=r_i$ for $i=1,\ldots,m$, and $w_i = s_ix$ for $i=1,\ldots,m$. Hence, it also lies in the set.

Thus, the set in question is an ideal that contains $a$, hence it contains $(a)$. Since it is contained in the intersection, we are done.

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