Mixing
Is there a ring containing ideals where the union of the ideals forms a subring but not an ideal?
$begingroup$
In general the union of two subrings is not a subring, and likewise for the union of two ideals. However, I was wondering if there exists an example of a union of ideals that is actually a subring but is not an ideal? I've tried looking around without any success so I imagine the answer is no.
abstract-algebra ring-theory ideals
asked Jan 14 at 22:53
JohnCJohnC
1168
$endgroup$
add a comment |
$begingroup$
In general the union of two subrings is not a subring, and likewise for the union of two ideals. However, I was wondering if there exists an example of a union of ideals that is actually a subring but is not an ideal? I've tried looking around without any success so I imagine the answer is no.
abstract-algebra ring-theory ideals
asked Jan 14 at 22:53
JohnCJohnC
1168
$endgroup$
add a comment |
$begingroup$
In general the union of two subrings is not a subring, and likewise for the union of two ideals. However, I was wondering if there exists an example of a union of ideals that is actually a subring but is not an ideal? I've tried looking around without any success so I imagine the answer is no.
abstract-algebra ring-theory ideals
asked Jan 14 at 22:53
JohnCJohnC
1168
$endgroup$
In general the union of two subrings is not a subring, and likewise for the union of two ideals. However, I was wondering if there exists an example of a union of ideals that is actually a subring but is not an ideal? I've tried looking around without any success so I imagine the answer is no.
abstract-algebra ring-theory ideals
abstract-algebra ring-theory ideals
asked Jan 14 at 22:53
JohnCJohnC
1168
asked Jan 14 at 22:53
JohnCJohnC
1168
asked Jan 14 at 22:53
JohnCJohnC
1168
asked Jan 14 at 22:53
JohnCJohnC
1168
asked Jan 14 at 22:53
JohnCJohnC
1168
1168
add a comment |
add a comment |
1 Answer
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$begingroup$
If ${I_j}_{jin J}$ is a non-empty family of ideals such that $I:=bigcup_{jin J}I_j$ is closed under addition (as required for a subring), then $I$ is already an ideal.
answered Jan 14 at 23:03
Hagen von EitzenHagen von Eitzen
280k23272504
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1 Answer
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1 Answer
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votes
$begingroup$
If ${I_j}_{jin J}$ is a non-empty family of ideals such that $I:=bigcup_{jin J}I_j$ is closed under addition (as required for a subring), then $I$ is already an ideal.
answered Jan 14 at 23:03
Hagen von EitzenHagen von Eitzen
280k23272504
$endgroup$
add a comment |
$begingroup$
If ${I_j}_{jin J}$ is a non-empty family of ideals such that $I:=bigcup_{jin J}I_j$ is closed under addition (as required for a subring), then $I$ is already an ideal.
answered Jan 14 at 23:03
Hagen von EitzenHagen von Eitzen
280k23272504
$endgroup$
add a comment |
$begingroup$
If ${I_j}_{jin J}$ is a non-empty family of ideals such that $I:=bigcup_{jin J}I_j$ is closed under addition (as required for a subring), then $I$ is already an ideal.
answered Jan 14 at 23:03
Hagen von EitzenHagen von Eitzen
280k23272504
$endgroup$
If ${I_j}_{jin J}$ is a non-empty family of ideals such that $I:=bigcup_{jin J}I_j$ is closed under addition (as required for a subring), then $I$ is already an ideal.
answered Jan 14 at 23:03
Hagen von EitzenHagen von Eitzen
280k23272504
answered Jan 14 at 23:03
Hagen von EitzenHagen von Eitzen
280k23272504
answered Jan 14 at 23:03
Hagen von EitzenHagen von Eitzen
280k23272504
answered Jan 14 at 23:03
Hagen von EitzenHagen von Eitzen
280k23272504
280k23272504
add a comment |
add a comment |
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