Mixing
Suppose $A,~B$ are not zero. Prove or disprove $Aoplus B$ is not isomorphic to $A$
Clash Royale CLAN TAG#URR8PPP
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It seems like a very stupid question, but I just cannot solve it...
Suppose $A,~B$ are not zero. Prove or disprove $Aoplus B$ is not isomorphic to $A$.
Furthermore, we can ask the similar question:
Suppose $A,~I$ are not zero. Prove or disprove $Aoplus I$ is not isomorphic to $A$.
Here $A$ is a ring and $I$ is an ideal of $A$. $Aoplus I$ is the direct sum of $A$-modules.
I encounter this problem when I was trying to solve the ex 2.28 on Atiyah-Macdonald:
If a local ring is absolutely flat, then it is a field.
Take $0neq x$ and we have a maximal ideal $(x)subsetmathfrakm$. According to some results from Atiyah-Macdonald, we can find an idempotent element $ein mathfrakm$ such that $(e)=(x)$ and $(e)oplus (1-e)=A$. Also, the Jacobson radical of $A$ is $mathfrakm$ so $1-e$ is unit. Then we have
$$(e)oplus A=A$$
I want to conclude $e=0$ so that $x=0$. Then $mathfrakm$ has to be $0$. Hence $A$ is a field.
But I don't know how to deduce $e=0$ from $(e)oplus A=A$, which is intuitively true to me...
abstract-algebra commutative-algebra
asked Jul 18 at 10:19
Aolong Li
848415
add a comment |Â
up vote
4
down vote
favorite
It seems like a very stupid question, but I just cannot solve it...
Suppose $A,~B$ are not zero. Prove or disprove $Aoplus B$ is not isomorphic to $A$.
Furthermore, we can ask the similar question:
Suppose $A,~I$ are not zero. Prove or disprove $Aoplus I$ is not isomorphic to $A$.
Here $A$ is a ring and $I$ is an ideal of $A$. $Aoplus I$ is the direct sum of $A$-modules.
I encounter this problem when I was trying to solve the ex 2.28 on Atiyah-Macdonald:
If a local ring is absolutely flat, then it is a field.
Take $0neq x$ and we have a maximal ideal $(x)subsetmathfrakm$. According to some results from Atiyah-Macdonald, we can find an idempotent element $ein mathfrakm$ such that $(e)=(x)$ and $(e)oplus (1-e)=A$. Also, the Jacobson radical of $A$ is $mathfrakm$ so $1-e$ is unit. Then we have
$$(e)oplus A=A$$
I want to conclude $e=0$ so that $x=0$. Then $mathfrakm$ has to be $0$. Hence $A$ is a field.
But I don't know how to deduce $e=0$ from $(e)oplus A=A$, which is intuitively true to me...
abstract-algebra commutative-algebra
asked Jul 18 at 10:19
Aolong Li
848415
1
Perhaps "Suppose $A$ and $B$ are not zero (modules). $Aoplus B$ is not isomorphic to $A$" would be clearer. @Arthur
– Michael Burr
Jul 18 at 10:31
@Arthur Thanks for pointing out the potential confusion. Fixed!
– Aolong Li
Jul 18 at 10:36
@MichaelBurr Thanks for your suggestion! Fixed!
– Aolong Li
Jul 18 at 10:36
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
It seems like a very stupid question, but I just cannot solve it...
Suppose $A,~B$ are not zero. Prove or disprove $Aoplus B$ is not isomorphic to $A$.
Furthermore, we can ask the similar question:
Suppose $A,~I$ are not zero. Prove or disprove $Aoplus I$ is not isomorphic to $A$.
Here $A$ is a ring and $I$ is an ideal of $A$. $Aoplus I$ is the direct sum of $A$-modules.
I encounter this problem when I was trying to solve the ex 2.28 on Atiyah-Macdonald:
If a local ring is absolutely flat, then it is a field.
Take $0neq x$ and we have a maximal ideal $(x)subsetmathfrakm$. According to some results from Atiyah-Macdonald, we can find an idempotent element $ein mathfrakm$ such that $(e)=(x)$ and $(e)oplus (1-e)=A$. Also, the Jacobson radical of $A$ is $mathfrakm$ so $1-e$ is unit. Then we have
$$(e)oplus A=A$$
I want to conclude $e=0$ so that $x=0$. Then $mathfrakm$ has to be $0$. Hence $A$ is a field.
But I don't know how to deduce $e=0$ from $(e)oplus A=A$, which is intuitively true to me...
abstract-algebra commutative-algebra
asked Jul 18 at 10:19
Aolong Li
848415
It seems like a very stupid question, but I just cannot solve it...
Suppose $A,~B$ are not zero. Prove or disprove $Aoplus B$ is not isomorphic to $A$.
Furthermore, we can ask the similar question:
Suppose $A,~I$ are not zero. Prove or disprove $Aoplus I$ is not isomorphic to $A$.
Here $A$ is a ring and $I$ is an ideal of $A$. $Aoplus I$ is the direct sum of $A$-modules.
I encounter this problem when I was trying to solve the ex 2.28 on Atiyah-Macdonald:
If a local ring is absolutely flat, then it is a field.
Take $0neq x$ and we have a maximal ideal $(x)subsetmathfrakm$. According to some results from Atiyah-Macdonald, we can find an idempotent element $ein mathfrakm$ such that $(e)=(x)$ and $(e)oplus (1-e)=A$. Also, the Jacobson radical of $A$ is $mathfrakm$ so $1-e$ is unit. Then we have
$$(e)oplus A=A$$
I want to conclude $e=0$ so that $x=0$. Then $mathfrakm$ has to be $0$. Hence $A$ is a field.
But I don't know how to deduce $e=0$ from $(e)oplus A=A$, which is intuitively true to me...
abstract-algebra commutative-algebra
asked Jul 18 at 10:19
Aolong Li
848415
edited Jul 18 at 10:35
asked Jul 18 at 10:19
Aolong Li
848415
asked Jul 18 at 10:19
Aolong Li
848415
asked Jul 18 at 10:19
Aolong Li
848415
848415
1
Perhaps "Suppose $A$ and $B$ are not zero (modules). $Aoplus B$ is not isomorphic to $A$" would be clearer. @Arthur
– Michael Burr
Jul 18 at 10:31
@Arthur Thanks for pointing out the potential confusion. Fixed!
– Aolong Li
Jul 18 at 10:36
@MichaelBurr Thanks for your suggestion! Fixed!
– Aolong Li
Jul 18 at 10:36
add a comment |Â
1
Perhaps "Suppose $A$ and $B$ are not zero (modules). $Aoplus B$ is not isomorphic to $A$" would be clearer. @Arthur
– Michael Burr
Jul 18 at 10:31
@Arthur Thanks for pointing out the potential confusion. Fixed!
– Aolong Li
Jul 18 at 10:36
@MichaelBurr Thanks for your suggestion! Fixed!
– Aolong Li
Jul 18 at 10:36
1
1
Perhaps "Suppose $A$ and $B$ are not zero (modules). $Aoplus B$ is not isomorphic to $A$" would be clearer. @Arthur
– Michael Burr
Jul 18 at 10:31
Perhaps "Suppose $A$ and $B$ are not zero (modules). $Aoplus B$ is not isomorphic to $A$" would be clearer. @Arthur
– Michael Burr
Jul 18 at 10:31
@Arthur Thanks for pointing out the potential confusion. Fixed!
– Aolong Li
Jul 18 at 10:36
@Arthur Thanks for pointing out the potential confusion. Fixed!
– Aolong Li
Jul 18 at 10:36
@MichaelBurr Thanks for your suggestion! Fixed!
– Aolong Li
Jul 18 at 10:36
@MichaelBurr Thanks for your suggestion! Fixed!
– Aolong Li
Jul 18 at 10:36
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
The statement doesn't hold in general because it can happen like this: Let $A=prod_i=1^infty F$ and consider the ideal $I$ which is the set of elements that are zero outside of the first coordinate. Then $Acong Aoplus I$.
There are much easier ways to prove an absolutely flat local ring is a field. For one thing, a local ring only has trivial idempotents, so if your ideal $I=(e)$, then $I=0$ or $I=A$. The ring only has trivial ideals then, and hence it is a field.
answered Jul 18 at 10:36
rschwieb
100k1193227
Thank you so much!
– Aolong Li
Jul 18 at 10:40
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
The statement doesn't hold in general because it can happen like this: Let $A=prod_i=1^infty F$ and consider the ideal $I$ which is the set of elements that are zero outside of the first coordinate. Then $Acong Aoplus I$.
There are much easier ways to prove an absolutely flat local ring is a field. For one thing, a local ring only has trivial idempotents, so if your ideal $I=(e)$, then $I=0$ or $I=A$. The ring only has trivial ideals then, and hence it is a field.
answered Jul 18 at 10:36
rschwieb
100k1193227
Thank you so much!
– Aolong Li
Jul 18 at 10:40
add a comment |Â
up vote
4
down vote
accepted
The statement doesn't hold in general because it can happen like this: Let $A=prod_i=1^infty F$ and consider the ideal $I$ which is the set of elements that are zero outside of the first coordinate. Then $Acong Aoplus I$.
There are much easier ways to prove an absolutely flat local ring is a field. For one thing, a local ring only has trivial idempotents, so if your ideal $I=(e)$, then $I=0$ or $I=A$. The ring only has trivial ideals then, and hence it is a field.
answered Jul 18 at 10:36
rschwieb
100k1193227
Thank you so much!
– Aolong Li
Jul 18 at 10:40
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
The statement doesn't hold in general because it can happen like this: Let $A=prod_i=1^infty F$ and consider the ideal $I$ which is the set of elements that are zero outside of the first coordinate. Then $Acong Aoplus I$.
There are much easier ways to prove an absolutely flat local ring is a field. For one thing, a local ring only has trivial idempotents, so if your ideal $I=(e)$, then $I=0$ or $I=A$. The ring only has trivial ideals then, and hence it is a field.
answered Jul 18 at 10:36
rschwieb
100k1193227
The statement doesn't hold in general because it can happen like this: Let $A=prod_i=1^infty F$ and consider the ideal $I$ which is the set of elements that are zero outside of the first coordinate. Then $Acong Aoplus I$.
There are much easier ways to prove an absolutely flat local ring is a field. For one thing, a local ring only has trivial idempotents, so if your ideal $I=(e)$, then $I=0$ or $I=A$. The ring only has trivial ideals then, and hence it is a field.
answered Jul 18 at 10:36
rschwieb
100k1193227
answered Jul 18 at 10:36
rschwieb
100k1193227
answered Jul 18 at 10:36
rschwieb
100k1193227
answered Jul 18 at 10:36
rschwieb
100k1193227
100k1193227
Thank you so much!
– Aolong Li
Jul 18 at 10:40
add a comment |Â
Thank you so much!
– Aolong Li
Jul 18 at 10:40
Thank you so much!
– Aolong Li
Jul 18 at 10:40
Thank you so much!
– Aolong Li
Jul 18 at 10:40
add a comment |Â
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1
Perhaps "Suppose $A$ and $B$ are not zero (modules). $Aoplus B$ is not isomorphic to $A$" would be clearer. @Arthur
– Michael Burr
Jul 18 at 10:31
@Arthur Thanks for pointing out the potential confusion. Fixed!
– Aolong Li
Jul 18 at 10:36
@MichaelBurr Thanks for your suggestion! Fixed!
– Aolong Li
Jul 18 at 10:36