Mixing
Two sided ideals are maximal right ideal iff they are maximal left ideal.
$begingroup$
Let $R$ be a ring with unity and $I$ be a two-sided ideal in $R$. Then $I$ is a maximal right ideal if and only if it is a maximal left ideal.
Would anyone give me an idea to prove the statement? Thanks.
abstract-algebra ring-theory maximal-and-prime-ideals
asked Jan 22 at 8:03
bozcanbozcan
26518
$endgroup$
add a comment |
$begingroup$
Let $R$ be a ring with unity and $I$ be a two-sided ideal in $R$. Then $I$ is a maximal right ideal if and only if it is a maximal left ideal.
Would anyone give me an idea to prove the statement? Thanks.
abstract-algebra ring-theory maximal-and-prime-ideals
asked Jan 22 at 8:03
bozcanbozcan
26518
$endgroup$
add a comment |
$begingroup$
Let $R$ be a ring with unity and $I$ be a two-sided ideal in $R$. Then $I$ is a maximal right ideal if and only if it is a maximal left ideal.
Would anyone give me an idea to prove the statement? Thanks.
abstract-algebra ring-theory maximal-and-prime-ideals
asked Jan 22 at 8:03
bozcanbozcan
26518
$endgroup$
Let $R$ be a ring with unity and $I$ be a two-sided ideal in $R$. Then $I$ is a maximal right ideal if and only if it is a maximal left ideal.
Would anyone give me an idea to prove the statement? Thanks.
abstract-algebra ring-theory maximal-and-prime-ideals
abstract-algebra ring-theory maximal-and-prime-ideals
asked Jan 22 at 8:03
bozcanbozcan
26518
asked Jan 22 at 8:03
bozcanbozcan
26518
asked Jan 22 at 8:03
bozcanbozcan
26518
asked Jan 22 at 8:03
bozcanbozcan
26518
asked Jan 22 at 8:03
bozcanbozcan
26518
26518
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Suppose $R/I$ only has trivial right ideals. Then it has only trivial left ideals. For if $x$ is a nonzero member of $R/I$, $x(R/I)=R/I$, and $x$ is right invertible, say by element $y$ similarly $y$ is right invertible, say by element $z$, but it is any easy exercise to prove $x=z$, so $x$ is a unit and $R/I$ is a division ring, and therefore only has trivial left and right ideals.
By a symmetric argument, the words left and right can be interchanged.
answered Jan 22 at 10:06
rschwiebrschwieb
107k12102251
$endgroup$
$begingroup$
Why is "$R/I$ only has trivial right ideals implies that it has the only trivial left ideals" true?
$endgroup$
– bozcan
Jan 22 at 13:22
$begingroup$
@bozcan Literally everything I wrote is dedicated to proving that.
$endgroup$
– rschwieb
Jan 22 at 13:57
$begingroup$
If there is any further question about that, there are numerous posts to read about it, such as math.stackexchange.com/questions/1151319/…
$endgroup$
– rschwieb
Jan 22 at 14:23
add a comment |
$begingroup$
Hint: Never mind. As rschweib points out in the comments below, this doesn't appear to lead anywhere. The only somewhat dubious advantage to be gained from trying it would appear to be in showing you that you're better off trying something else. Apologies to anyone whom I sent off on a wild goose chase.
If $ J $ is a one-sided ideal in $ R $ containing $ I $, let $ j $ be an arbirary member of $ J $, and consider the set $ K=left{,x in J mid,j,x,j in J,right} $.
- Is $ j in K $?
- Can you determine whether $ K $ is any sort of ideal ?
- Can you determine what the intersection $ Kcap I $ is ?
answered Jan 22 at 9:12
lonza leggieralonza leggiera
97728
$endgroup$
$begingroup$
Is there a typo? Because as written, $J=K$ and the hint doesn’t seem to go anywhere.
$endgroup$
– rschwieb
Jan 22 at 10:20
$begingroup$
Not a typo. A blunder, rather, in my supposed proof that $ K $ had to be two-sided. The idea was to find a description of $ J $ which it was not too difficult to show to be two-sided. If such a thing is possible, it appears to be much more difficult than I had initially hoped, and $ K $ certainly doesn't qualify.
$endgroup$
– lonza leggiera
Jan 22 at 21:33
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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$begingroup$
Suppose $R/I$ only has trivial right ideals. Then it has only trivial left ideals. For if $x$ is a nonzero member of $R/I$, $x(R/I)=R/I$, and $x$ is right invertible, say by element $y$ similarly $y$ is right invertible, say by element $z$, but it is any easy exercise to prove $x=z$, so $x$ is a unit and $R/I$ is a division ring, and therefore only has trivial left and right ideals.
By a symmetric argument, the words left and right can be interchanged.
answered Jan 22 at 10:06
rschwiebrschwieb
107k12102251
$endgroup$
$begingroup$
Why is "$R/I$ only has trivial right ideals implies that it has the only trivial left ideals" true?
$endgroup$
– bozcan
Jan 22 at 13:22
$begingroup$
@bozcan Literally everything I wrote is dedicated to proving that.
$endgroup$
– rschwieb
Jan 22 at 13:57
$begingroup$
If there is any further question about that, there are numerous posts to read about it, such as math.stackexchange.com/questions/1151319/…
$endgroup$
– rschwieb
Jan 22 at 14:23
add a comment |
$begingroup$
Suppose $R/I$ only has trivial right ideals. Then it has only trivial left ideals. For if $x$ is a nonzero member of $R/I$, $x(R/I)=R/I$, and $x$ is right invertible, say by element $y$ similarly $y$ is right invertible, say by element $z$, but it is any easy exercise to prove $x=z$, so $x$ is a unit and $R/I$ is a division ring, and therefore only has trivial left and right ideals.
By a symmetric argument, the words left and right can be interchanged.
answered Jan 22 at 10:06
rschwiebrschwieb
107k12102251
$endgroup$
$begingroup$
Why is "$R/I$ only has trivial right ideals implies that it has the only trivial left ideals" true?
$endgroup$
– bozcan
Jan 22 at 13:22
$begingroup$
@bozcan Literally everything I wrote is dedicated to proving that.
$endgroup$
– rschwieb
Jan 22 at 13:57
$begingroup$
If there is any further question about that, there are numerous posts to read about it, such as math.stackexchange.com/questions/1151319/…
$endgroup$
– rschwieb
Jan 22 at 14:23
add a comment |
$begingroup$
Suppose $R/I$ only has trivial right ideals. Then it has only trivial left ideals. For if $x$ is a nonzero member of $R/I$, $x(R/I)=R/I$, and $x$ is right invertible, say by element $y$ similarly $y$ is right invertible, say by element $z$, but it is any easy exercise to prove $x=z$, so $x$ is a unit and $R/I$ is a division ring, and therefore only has trivial left and right ideals.
By a symmetric argument, the words left and right can be interchanged.
answered Jan 22 at 10:06
rschwiebrschwieb
107k12102251
$endgroup$
Suppose $R/I$ only has trivial right ideals. Then it has only trivial left ideals. For if $x$ is a nonzero member of $R/I$, $x(R/I)=R/I$, and $x$ is right invertible, say by element $y$ similarly $y$ is right invertible, say by element $z$, but it is any easy exercise to prove $x=z$, so $x$ is a unit and $R/I$ is a division ring, and therefore only has trivial left and right ideals.
By a symmetric argument, the words left and right can be interchanged.
answered Jan 22 at 10:06
rschwiebrschwieb
107k12102251
edited Jan 22 at 13:59
answered Jan 22 at 10:06
rschwiebrschwieb
107k12102251
answered Jan 22 at 10:06
rschwiebrschwieb
107k12102251
answered Jan 22 at 10:06
rschwiebrschwieb
107k12102251
107k12102251
$begingroup$
Why is "$R/I$ only has trivial right ideals implies that it has the only trivial left ideals" true?
$endgroup$
– bozcan
Jan 22 at 13:22
$begingroup$
@bozcan Literally everything I wrote is dedicated to proving that.
$endgroup$
– rschwieb
Jan 22 at 13:57
$begingroup$
If there is any further question about that, there are numerous posts to read about it, such as math.stackexchange.com/questions/1151319/…
$endgroup$
– rschwieb
Jan 22 at 14:23
add a comment |
$begingroup$
Why is "$R/I$ only has trivial right ideals implies that it has the only trivial left ideals" true?
$endgroup$
– bozcan
Jan 22 at 13:22
$begingroup$
@bozcan Literally everything I wrote is dedicated to proving that.
$endgroup$
– rschwieb
Jan 22 at 13:57
$begingroup$
If there is any further question about that, there are numerous posts to read about it, such as math.stackexchange.com/questions/1151319/…
$endgroup$
– rschwieb
Jan 22 at 14:23
$begingroup$
Why is "$R/I$ only has trivial right ideals implies that it has the only trivial left ideals" true?
$endgroup$
– bozcan
Jan 22 at 13:22
$begingroup$
Why is "$R/I$ only has trivial right ideals implies that it has the only trivial left ideals" true?
$endgroup$
– bozcan
Jan 22 at 13:22
$begingroup$
@bozcan Literally everything I wrote is dedicated to proving that.
$endgroup$
– rschwieb
Jan 22 at 13:57
$begingroup$
@bozcan Literally everything I wrote is dedicated to proving that.
$endgroup$
– rschwieb
Jan 22 at 13:57
$begingroup$
If there is any further question about that, there are numerous posts to read about it, such as math.stackexchange.com/questions/1151319/…
$endgroup$
– rschwieb
Jan 22 at 14:23
$begingroup$
If there is any further question about that, there are numerous posts to read about it, such as math.stackexchange.com/questions/1151319/…
$endgroup$
– rschwieb
Jan 22 at 14:23
add a comment |
$begingroup$
Hint: Never mind. As rschweib points out in the comments below, this doesn't appear to lead anywhere. The only somewhat dubious advantage to be gained from trying it would appear to be in showing you that you're better off trying something else. Apologies to anyone whom I sent off on a wild goose chase.
If $ J $ is a one-sided ideal in $ R $ containing $ I $, let $ j $ be an arbirary member of $ J $, and consider the set $ K=left{,x in J mid,j,x,j in J,right} $.
- Is $ j in K $?
- Can you determine whether $ K $ is any sort of ideal ?
- Can you determine what the intersection $ Kcap I $ is ?
answered Jan 22 at 9:12
lonza leggieralonza leggiera
97728
$endgroup$
$begingroup$
Is there a typo? Because as written, $J=K$ and the hint doesn’t seem to go anywhere.
$endgroup$
– rschwieb
Jan 22 at 10:20
$begingroup$
Not a typo. A blunder, rather, in my supposed proof that $ K $ had to be two-sided. The idea was to find a description of $ J $ which it was not too difficult to show to be two-sided. If such a thing is possible, it appears to be much more difficult than I had initially hoped, and $ K $ certainly doesn't qualify.
$endgroup$
– lonza leggiera
Jan 22 at 21:33
add a comment |
$begingroup$
Hint: Never mind. As rschweib points out in the comments below, this doesn't appear to lead anywhere. The only somewhat dubious advantage to be gained from trying it would appear to be in showing you that you're better off trying something else. Apologies to anyone whom I sent off on a wild goose chase.
If $ J $ is a one-sided ideal in $ R $ containing $ I $, let $ j $ be an arbirary member of $ J $, and consider the set $ K=left{,x in J mid,j,x,j in J,right} $.
- Is $ j in K $?
- Can you determine whether $ K $ is any sort of ideal ?
- Can you determine what the intersection $ Kcap I $ is ?
answered Jan 22 at 9:12
lonza leggieralonza leggiera
97728
$endgroup$
$begingroup$
Is there a typo? Because as written, $J=K$ and the hint doesn’t seem to go anywhere.
$endgroup$
– rschwieb
Jan 22 at 10:20
$begingroup$
Not a typo. A blunder, rather, in my supposed proof that $ K $ had to be two-sided. The idea was to find a description of $ J $ which it was not too difficult to show to be two-sided. If such a thing is possible, it appears to be much more difficult than I had initially hoped, and $ K $ certainly doesn't qualify.
$endgroup$
– lonza leggiera
Jan 22 at 21:33
add a comment |
$begingroup$
Hint: Never mind. As rschweib points out in the comments below, this doesn't appear to lead anywhere. The only somewhat dubious advantage to be gained from trying it would appear to be in showing you that you're better off trying something else. Apologies to anyone whom I sent off on a wild goose chase.
If $ J $ is a one-sided ideal in $ R $ containing $ I $, let $ j $ be an arbirary member of $ J $, and consider the set $ K=left{,x in J mid,j,x,j in J,right} $.
- Is $ j in K $?
- Can you determine whether $ K $ is any sort of ideal ?
- Can you determine what the intersection $ Kcap I $ is ?
answered Jan 22 at 9:12
lonza leggieralonza leggiera
97728
$endgroup$
Hint: Never mind. As rschweib points out in the comments below, this doesn't appear to lead anywhere. The only somewhat dubious advantage to be gained from trying it would appear to be in showing you that you're better off trying something else. Apologies to anyone whom I sent off on a wild goose chase.
If $ J $ is a one-sided ideal in $ R $ containing $ I $, let $ j $ be an arbirary member of $ J $, and consider the set $ K=left{,x in J mid,j,x,j in J,right} $.
- Is $ j in K $?
- Can you determine whether $ K $ is any sort of ideal ?
- Can you determine what the intersection $ Kcap I $ is ?
answered Jan 22 at 9:12
lonza leggieralonza leggiera
97728
edited Jan 23 at 20:47
answered Jan 22 at 9:12
lonza leggieralonza leggiera
97728
answered Jan 22 at 9:12
lonza leggieralonza leggiera
97728
answered Jan 22 at 9:12
lonza leggieralonza leggiera
97728
97728
$begingroup$
Is there a typo? Because as written, $J=K$ and the hint doesn’t seem to go anywhere.
$endgroup$
– rschwieb
Jan 22 at 10:20
$begingroup$
Not a typo. A blunder, rather, in my supposed proof that $ K $ had to be two-sided. The idea was to find a description of $ J $ which it was not too difficult to show to be two-sided. If such a thing is possible, it appears to be much more difficult than I had initially hoped, and $ K $ certainly doesn't qualify.
$endgroup$
– lonza leggiera
Jan 22 at 21:33
add a comment |
$begingroup$
Is there a typo? Because as written, $J=K$ and the hint doesn’t seem to go anywhere.
$endgroup$
– rschwieb
Jan 22 at 10:20
$begingroup$
Not a typo. A blunder, rather, in my supposed proof that $ K $ had to be two-sided. The idea was to find a description of $ J $ which it was not too difficult to show to be two-sided. If such a thing is possible, it appears to be much more difficult than I had initially hoped, and $ K $ certainly doesn't qualify.
$endgroup$
– lonza leggiera
Jan 22 at 21:33
$begingroup$
Is there a typo? Because as written, $J=K$ and the hint doesn’t seem to go anywhere.
$endgroup$
– rschwieb
Jan 22 at 10:20
$begingroup$
Is there a typo? Because as written, $J=K$ and the hint doesn’t seem to go anywhere.
$endgroup$
– rschwieb
Jan 22 at 10:20
$begingroup$
Not a typo. A blunder, rather, in my supposed proof that $ K $ had to be two-sided. The idea was to find a description of $ J $ which it was not too difficult to show to be two-sided. If such a thing is possible, it appears to be much more difficult than I had initially hoped, and $ K $ certainly doesn't qualify.
$endgroup$
– lonza leggiera
Jan 22 at 21:33
$begingroup$
Not a typo. A blunder, rather, in my supposed proof that $ K $ had to be two-sided. The idea was to find a description of $ J $ which it was not too difficult to show to be two-sided. If such a thing is possible, it appears to be much more difficult than I had initially hoped, and $ K $ certainly doesn't qualify.
$endgroup$
– lonza leggiera
Jan 22 at 21:33
add a comment |
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