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- Jun 22, 2012

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In R. Y. Sharp "Steps in Commutative Algebra", Section 2.23 on sums of ideals reads as follows:

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2.23 SUMS OF IDEALS. Let \(\displaystyle ( {I_{\lambda})}_{\lambda \in \Lambda} \) be a family of ideals of the commutative ring \(\displaystyle R \). We define the sum \(\displaystyle {\sum}_{\lambda \in \Lambda} I_{\lambda} \) of this family to be the ideal generated by \(\displaystyle {\cup}_{\lambda \in \Lambda}I_{\lambda} \) :

Thus \(\displaystyle {\sum}_{\lambda \in \Lambda} I_{\lambda} = ( {\cup}_{\lambda \in \Lambda}I_{\lambda} ) \)

In particular if \(\displaystyle \Lambda = \emptyset \) then \(\displaystyle {\sum}_{\lambda \in \Lambda} I_{\lambda} = 0 \).

Since an arbitrary ideal of R is closed under addition and under scalar multiplication by arbitrary elements of R, it follows from 2.18 that, in the case where \(\displaystyle \Lambda \ne 0 \), an arbitrary element can be expressed in the form \(\displaystyle {\sum}_{i=1}^{n} {c_{\lambda}}_i \), where \(\displaystyle n \in \mathbb{N} , {\lambda}_1, {\lambda}_2, ... \ ... , {\lambda}_n \in \Lambda \) and \(\displaystyle {c_{\lambda}}_i \in {I_{\lambda}}_i \) for each \(\displaystyle i = 1,2, ... \ ... , n \)

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Now 2.18 states the following:

Let \(\displaystyle \emptyset \ne H \subseteq R \). We define the ideal of R generated by H, denoted by (H) or RH or HR, to be the intersection of the family of all ideals of R which contain H.

Then Sharp shows that ...

\(\displaystyle (H) = \{ {\sum}_{i=1}^{n} r_ih_i \ | \ n \in \mathbb{N}, r_1, r_2, ... \ ... r_n \in R, h_1, h_2, ... \ ... h_n \in H \} \)

My problem is as follows:

Given the above, and in particular:

" ... ... an arbitrary element can be expressed in the form \(\displaystyle {\sum}_{i=1}^{n} {c_{\lambda}}_i \), where \(\displaystyle n \in \mathbb{N} , {\lambda}_1, {\lambda}_2, ... \ ... , {\lambda}_1 \in \Lambda \) and \(\displaystyle {c_{\lambda}}_i \in {I_{\lambda}}_i \) for each \(\displaystyle i = 1,2, ... \ ... , n \) ... ... "

why does each \(\displaystyle {c_{\lambda}}_i \in {I_{\lambda}}_i \) ... why for example, cannot each \(\displaystyle {c_{\lambda}}_i \) come from, say, \(\displaystyle {I_{\lambda}}_1 \)

To emphasize the point consider \(\displaystyle I_1 \cup I_2 \). Following the form of the expression for (H) given above we would have

\(\displaystyle I_1 \cup I_2 = \{ {\sum}_{i=1}^{n} s_it_i \ | \ n \in \mathbb{N}, s_1, s_2, ... \ ... s_n \in R , \ t_1, t_2, ... \ ... t_n \in I_1 \cup I_2 \} \)

Now in this expression all of the \(\displaystyle t_i \in I_1 \cup I_2 \) could conceivably come from \(\displaystyle I_1 \) which again seems at odds with Sharp's claim above that

" ... ... an arbitrary element can be expressed in the form \(\displaystyle {\sum}_{i=1}^{n} {c_{\lambda}}_i \), where \(\displaystyle n \in \mathbb{N} , {\lambda}_1, {\lambda}_2, ... \ ... , {\lambda}_1 \in \Lambda \) and \(\displaystyle {c_{\lambda}}_i \in {I_{\lambda}}_i \) for each \(\displaystyle i = 1,2, ... \ ... , n \) ... ... "

Can someone please clarify this issue?

Peter

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2.23 SUMS OF IDEALS. Let \(\displaystyle ( {I_{\lambda})}_{\lambda \in \Lambda} \) be a family of ideals of the commutative ring \(\displaystyle R \). We define the sum \(\displaystyle {\sum}_{\lambda \in \Lambda} I_{\lambda} \) of this family to be the ideal generated by \(\displaystyle {\cup}_{\lambda \in \Lambda}I_{\lambda} \) :

Thus \(\displaystyle {\sum}_{\lambda \in \Lambda} I_{\lambda} = ( {\cup}_{\lambda \in \Lambda}I_{\lambda} ) \)

In particular if \(\displaystyle \Lambda = \emptyset \) then \(\displaystyle {\sum}_{\lambda \in \Lambda} I_{\lambda} = 0 \).

Since an arbitrary ideal of R is closed under addition and under scalar multiplication by arbitrary elements of R, it follows from 2.18 that, in the case where \(\displaystyle \Lambda \ne 0 \), an arbitrary element can be expressed in the form \(\displaystyle {\sum}_{i=1}^{n} {c_{\lambda}}_i \), where \(\displaystyle n \in \mathbb{N} , {\lambda}_1, {\lambda}_2, ... \ ... , {\lambda}_n \in \Lambda \) and \(\displaystyle {c_{\lambda}}_i \in {I_{\lambda}}_i \) for each \(\displaystyle i = 1,2, ... \ ... , n \)

------------------------------------------------------------------------

Now 2.18 states the following:

Let \(\displaystyle \emptyset \ne H \subseteq R \). We define the ideal of R generated by H, denoted by (H) or RH or HR, to be the intersection of the family of all ideals of R which contain H.

Then Sharp shows that ...

\(\displaystyle (H) = \{ {\sum}_{i=1}^{n} r_ih_i \ | \ n \in \mathbb{N}, r_1, r_2, ... \ ... r_n \in R, h_1, h_2, ... \ ... h_n \in H \} \)

My problem is as follows:

Given the above, and in particular:

" ... ... an arbitrary element can be expressed in the form \(\displaystyle {\sum}_{i=1}^{n} {c_{\lambda}}_i \), where \(\displaystyle n \in \mathbb{N} , {\lambda}_1, {\lambda}_2, ... \ ... , {\lambda}_1 \in \Lambda \) and \(\displaystyle {c_{\lambda}}_i \in {I_{\lambda}}_i \) for each \(\displaystyle i = 1,2, ... \ ... , n \) ... ... "

why does each \(\displaystyle {c_{\lambda}}_i \in {I_{\lambda}}_i \) ... why for example, cannot each \(\displaystyle {c_{\lambda}}_i \) come from, say, \(\displaystyle {I_{\lambda}}_1 \)

To emphasize the point consider \(\displaystyle I_1 \cup I_2 \). Following the form of the expression for (H) given above we would have

\(\displaystyle I_1 \cup I_2 = \{ {\sum}_{i=1}^{n} s_it_i \ | \ n \in \mathbb{N}, s_1, s_2, ... \ ... s_n \in R , \ t_1, t_2, ... \ ... t_n \in I_1 \cup I_2 \} \)

Now in this expression all of the \(\displaystyle t_i \in I_1 \cup I_2 \) could conceivably come from \(\displaystyle I_1 \) which again seems at odds with Sharp's claim above that

" ... ... an arbitrary element can be expressed in the form \(\displaystyle {\sum}_{i=1}^{n} {c_{\lambda}}_i \), where \(\displaystyle n \in \mathbb{N} , {\lambda}_1, {\lambda}_2, ... \ ... , {\lambda}_1 \in \Lambda \) and \(\displaystyle {c_{\lambda}}_i \in {I_{\lambda}}_i \) for each \(\displaystyle i = 1,2, ... \ ... , n \) ... ... "

Can someone please clarify this issue?

Peter

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