Mixing
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Is it true that the generators of a commutative ring is finite?
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So I am thinking about the structure of a commutative ring $R$. Give a family of principal ideals $A={(x_{i})}$, and if $R=bigcup_{i}(x_{i})$, is it true that there are finite number of principal ideals in $A$ such that $(1)=(x_{1},x_{2},...,x_{n})$? Actually, this comes from the statement that "Affine schemes are quasi-compact". Can anyone help? Thanks!
abstract-algebra ideals
edited Jan 25 at 10:26
user26857
39.4k124183
asked Jan 25 at 9:09
Yuyi ZhangYuyi Zhang
15117
$endgroup$
add a comment |
$begingroup$
So I am thinking about the structure of a commutative ring $R$. Give a family of principal ideals $A={(x_{i})}$, and if $R=bigcup_{i}(x_{i})$, is it true that there are finite number of principal ideals in $A$ such that $(1)=(x_{1},x_{2},...,x_{n})$? Actually, this comes from the statement that "Affine schemes are quasi-compact". Can anyone help? Thanks!
abstract-algebra ideals
edited Jan 25 at 10:26
user26857
39.4k124183
asked Jan 25 at 9:09
Yuyi ZhangYuyi Zhang
15117
$endgroup$
$begingroup$
Are you sure you mean $cup_i(x_i)$, and not the ideal generated by the union (equivalently, by elements $x_i$? In both cases the statement is true, but from the given context I presume you mean the latter.
$endgroup$
– Wojowu
Jan 25 at 9:34
add a comment |
$begingroup$
So I am thinking about the structure of a commutative ring $R$. Give a family of principal ideals $A={(x_{i})}$, and if $R=bigcup_{i}(x_{i})$, is it true that there are finite number of principal ideals in $A$ such that $(1)=(x_{1},x_{2},...,x_{n})$? Actually, this comes from the statement that "Affine schemes are quasi-compact". Can anyone help? Thanks!
abstract-algebra ideals
edited Jan 25 at 10:26
user26857
39.4k124183
asked Jan 25 at 9:09
Yuyi ZhangYuyi Zhang
15117
$endgroup$
So I am thinking about the structure of a commutative ring $R$. Give a family of principal ideals $A={(x_{i})}$, and if $R=bigcup_{i}(x_{i})$, is it true that there are finite number of principal ideals in $A$ such that $(1)=(x_{1},x_{2},...,x_{n})$? Actually, this comes from the statement that "Affine schemes are quasi-compact". Can anyone help? Thanks!
abstract-algebra ideals
abstract-algebra ideals
edited Jan 25 at 10:26
user26857
39.4k124183
asked Jan 25 at 9:09
Yuyi ZhangYuyi Zhang
15117
edited Jan 25 at 10:26
user26857
39.4k124183
asked Jan 25 at 9:09
Yuyi ZhangYuyi Zhang
15117
edited Jan 25 at 10:26
user26857
39.4k124183
edited Jan 25 at 10:26
user26857
39.4k124183
edited Jan 25 at 10:26
user26857
39.4k124183
39.4k124183
asked Jan 25 at 9:09
Yuyi ZhangYuyi Zhang
15117
asked Jan 25 at 9:09
Yuyi ZhangYuyi Zhang
15117
asked Jan 25 at 9:09
Yuyi ZhangYuyi Zhang
15117
15117
$begingroup$
Are you sure you mean $cup_i(x_i)$, and not the ideal generated by the union (equivalently, by elements $x_i$? In both cases the statement is true, but from the given context I presume you mean the latter.
$endgroup$
– Wojowu
Jan 25 at 9:34
add a comment |
$begingroup$
Are you sure you mean $cup_i(x_i)$, and not the ideal generated by the union (equivalently, by elements $x_i$? In both cases the statement is true, but from the given context I presume you mean the latter.
$endgroup$
– Wojowu
Jan 25 at 9:34
$begingroup$
Are you sure you mean $cup_i(x_i)$, and not the ideal generated by the union (equivalently, by elements $x_i$? In both cases the statement is true, but from the given context I presume you mean the latter.
$endgroup$
– Wojowu
Jan 25 at 9:34
$begingroup$
Are you sure you mean $cup_i(x_i)$, and not the ideal generated by the union (equivalently, by elements $x_i$? In both cases the statement is true, but from the given context I presume you mean the latter.
$endgroup$
– Wojowu
Jan 25 at 9:34
add a comment |
1 Answer
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If $R = bigcup_i (x_i)$, then $1in (x_i)$ for some $i$, so one of the principal ideals must be $(1)=R$.
answered Jan 25 at 9:20
SladeSlade
25.3k12665
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add a comment |
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$begingroup$
If $R = bigcup_i (x_i)$, then $1in (x_i)$ for some $i$, so one of the principal ideals must be $(1)=R$.
answered Jan 25 at 9:20
SladeSlade
25.3k12665
$endgroup$
add a comment |
$begingroup$
If $R = bigcup_i (x_i)$, then $1in (x_i)$ for some $i$, so one of the principal ideals must be $(1)=R$.
answered Jan 25 at 9:20
SladeSlade
25.3k12665
$endgroup$
add a comment |
$begingroup$
If $R = bigcup_i (x_i)$, then $1in (x_i)$ for some $i$, so one of the principal ideals must be $(1)=R$.
answered Jan 25 at 9:20
SladeSlade
25.3k12665
$endgroup$
If $R = bigcup_i (x_i)$, then $1in (x_i)$ for some $i$, so one of the principal ideals must be $(1)=R$.
answered Jan 25 at 9:20
SladeSlade
25.3k12665
answered Jan 25 at 9:20
SladeSlade
25.3k12665
answered Jan 25 at 9:20
SladeSlade
25.3k12665
answered Jan 25 at 9:20
SladeSlade
25.3k12665
25.3k12665
add a comment |
add a comment |
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$begingroup$
Are you sure you mean $cup_i(x_i)$, and not the ideal generated by the union (equivalently, by elements $x_i$? In both cases the statement is true, but from the given context I presume you mean the latter.
$endgroup$
– Wojowu
Jan 25 at 9:34