Math Amateur
Gold Member
MHB
- 3,920
- 48
In R. Y. Sharp "Steps in Commutative Algebra", Section 2.23 on sums of ideals reads as follows:
------------------------------------------------------------------------------
2.23 SUMS OF IDEALS. Let $$ ( {I_{\lambda})}_{\lambda \in \Lambda} $$ be a family of ideals of the commutative ring $$ R $$. We define the sum $$ {\sum}_{\lambda \in \Lambda} I_{\lambda} $$ of this family to be the ideal generated by $$ {\cup}_{\lambda \in \Lambda}I_{\lambda} $$ :
Thus $$ {\sum}_{\lambda \in \Lambda} I_{\lambda} = ( {\cup}_{\lambda \in \Lambda}I_{\lambda} ) $$
In particular if $$ \Lambda = \emptyset $$ then $$ {\sum}_{\lambda \in \Lambda} I_{\lambda} = 0 $$.
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 $$ \Lambda \ne 0 $$, an arbitrary element can be expressed in the form $$ {\sum}_{i=1}^{n} {c_{\lambda}}_i $$, where $$ n \in \mathbb{N} , {\lambda}_1, {\lambda}_2, ... \ ... , {\lambda}_n \in \Lambda $$ and $$ {c_{\lambda}}_i \in {I_{\lambda}}_i $$ for each $$ i = 1,2, ... \ ... , n $$
------------------------------------------------------------------------
Now 2.18 states the following:
Let $$ \emptyset \ne H \subseteq R $$. 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 ...
$$ (H) = \{ {\sum}_{i=1}^{n} r_ih_i \ | \ n \in \mathbb{N}, r_1, r_2, ... \ ... r_n \in R, h_1, h_2, ... \ ... h_n \in H \} $$
My problem is as follows:
Given the above, and in particular:
" ... ... an arbitrary element can be expressed in the form $$ {\sum}_{i=1}^{n} {c_{\lambda}}_i $$, where $$ n \in \mathbb{N} , {\lambda}_1, {\lambda}_2, ... \ ... , {\lambda}_1 \in \Lambda $$ and $$ {c_{\lambda}}_i \in {I_{\lambda}}_i $$ for each $$ i = 1,2, ... \ ... , n $$ ... ... "
why does each $$ {c_{\lambda}}_i \in {I_{\lambda}}_i $$ ... why for example, cannot each $$ {c_{\lambda}}_i $$ come from, say, $$ {I_{\lambda}}_1 $$To emphasize the point consider $$ I_1 \cup I_2 $$. Following the form of the expression for (H) given above we would have
$$ I_1 \cup I_2 = \{ {\sum}_{i=1}^{n} s_it_i \ | \ n \in \mathbb{N}, s_1, s_2, ... \ ... s_n \in R , \ t_1, t_2, ... \ ... t_n \in I_1 \cup I_2 \} $$
Now in this expression all of the $$ t_i \in I_1 \cup I_2 $$ could conceivably come from $$ I_1 $$ which again seems at odds with Sharp's claim above that
" ... ... an arbitrary element can be expressed in the form $$ {\sum}_{i=1}^{n} {c_{\lambda}}_i $$, where $$ n \in \mathbb{N} , {\lambda}_1, {\lambda}_2, ... \ ... , {\lambda}_1 \in \Lambda $$ and $$ {c_{\lambda}}_i \in {I_{\lambda}}_i $$ for each $$ i = 1,2, ... \ ... , n $$ ... ... "
Can someone please clarify this issue?
Peter
------------------------------------------------------------------------------
2.23 SUMS OF IDEALS. Let $$ ( {I_{\lambda})}_{\lambda \in \Lambda} $$ be a family of ideals of the commutative ring $$ R $$. We define the sum $$ {\sum}_{\lambda \in \Lambda} I_{\lambda} $$ of this family to be the ideal generated by $$ {\cup}_{\lambda \in \Lambda}I_{\lambda} $$ :
Thus $$ {\sum}_{\lambda \in \Lambda} I_{\lambda} = ( {\cup}_{\lambda \in \Lambda}I_{\lambda} ) $$
In particular if $$ \Lambda = \emptyset $$ then $$ {\sum}_{\lambda \in \Lambda} I_{\lambda} = 0 $$.
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 $$ \Lambda \ne 0 $$, an arbitrary element can be expressed in the form $$ {\sum}_{i=1}^{n} {c_{\lambda}}_i $$, where $$ n \in \mathbb{N} , {\lambda}_1, {\lambda}_2, ... \ ... , {\lambda}_n \in \Lambda $$ and $$ {c_{\lambda}}_i \in {I_{\lambda}}_i $$ for each $$ i = 1,2, ... \ ... , n $$
------------------------------------------------------------------------
Now 2.18 states the following:
Let $$ \emptyset \ne H \subseteq R $$. 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 ...
$$ (H) = \{ {\sum}_{i=1}^{n} r_ih_i \ | \ n \in \mathbb{N}, r_1, r_2, ... \ ... r_n \in R, h_1, h_2, ... \ ... h_n \in H \} $$
My problem is as follows:
Given the above, and in particular:
" ... ... an arbitrary element can be expressed in the form $$ {\sum}_{i=1}^{n} {c_{\lambda}}_i $$, where $$ n \in \mathbb{N} , {\lambda}_1, {\lambda}_2, ... \ ... , {\lambda}_1 \in \Lambda $$ and $$ {c_{\lambda}}_i \in {I_{\lambda}}_i $$ for each $$ i = 1,2, ... \ ... , n $$ ... ... "
why does each $$ {c_{\lambda}}_i \in {I_{\lambda}}_i $$ ... why for example, cannot each $$ {c_{\lambda}}_i $$ come from, say, $$ {I_{\lambda}}_1 $$To emphasize the point consider $$ I_1 \cup I_2 $$. Following the form of the expression for (H) given above we would have
$$ I_1 \cup I_2 = \{ {\sum}_{i=1}^{n} s_it_i \ | \ n \in \mathbb{N}, s_1, s_2, ... \ ... s_n \in R , \ t_1, t_2, ... \ ... t_n \in I_1 \cup I_2 \} $$
Now in this expression all of the $$ t_i \in I_1 \cup I_2 $$ could conceivably come from $$ I_1 $$ which again seems at odds with Sharp's claim above that
" ... ... an arbitrary element can be expressed in the form $$ {\sum}_{i=1}^{n} {c_{\lambda}}_i $$, where $$ n \in \mathbb{N} , {\lambda}_1, {\lambda}_2, ... \ ... , {\lambda}_1 \in \Lambda $$ and $$ {c_{\lambda}}_i \in {I_{\lambda}}_i $$ for each $$ i = 1,2, ... \ ... , n $$ ... ... "
Can someone please clarify this issue?
Peter
Last edited: