Commutative ring and maximal ideal problem
Clash Royale CLAN TAG#URR8PPP
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Let $A$ be a commutative ring and $M$ be a proper maximal ideal in $A$. Show the following properties:
(a) If each $a in A setminus M$ is a unit element in $A$, then $M$ is the only maximal ideal in $A$.
(b) If each element of the form $1+m in A$ with $m in M$ is a unit in $A$, then $A$ has a unique maximal ideal.
I think I could show (a), I have problems with (b), I'll write what I've done so far:
(a) Suppose there exists $M' neq M$ a maximal ideal in $A$. Then there is some $b in M'$ such that $b not in M$. Then $b in A setminus M$, so there $b$ is a unit. Since $M'$ is ideal (bilateral ideal since $A$ is commutative), we have $1=b^-1b in M'$. Now $M$ is a maximal ideal, this means there exists $m in M$ such that $m not in M'$. But $m=m1 in M'$, which is absurd. The absurd comes from the assumption that there is another maximal ideal different from $M$, so $M$ is the only maximal ideal in $A$.
I would appreciate hints for (b).
abstract-algebra ring-theory ideals
asked Sep 18 '14 at 17:57
user16924
1,031618
add a comment |Â
up vote
1
down vote
favorite
Let $A$ be a commutative ring and $M$ be a proper maximal ideal in $A$. Show the following properties:
(a) If each $a in A setminus M$ is a unit element in $A$, then $M$ is the only maximal ideal in $A$.
(b) If each element of the form $1+m in A$ with $m in M$ is a unit in $A$, then $A$ has a unique maximal ideal.
I think I could show (a), I have problems with (b), I'll write what I've done so far:
(a) Suppose there exists $M' neq M$ a maximal ideal in $A$. Then there is some $b in M'$ such that $b not in M$. Then $b in A setminus M$, so there $b$ is a unit. Since $M'$ is ideal (bilateral ideal since $A$ is commutative), we have $1=b^-1b in M'$. Now $M$ is a maximal ideal, this means there exists $m in M$ such that $m not in M'$. But $m=m1 in M'$, which is absurd. The absurd comes from the assumption that there is another maximal ideal different from $M$, so $M$ is the only maximal ideal in $A$.
I would appreciate hints for (b).
abstract-algebra ring-theory ideals
asked Sep 18 '14 at 17:57
user16924
1,031618
3
I assume that in part (b), you mean $1 + m in A$, not $1 + m in M$.
– Bungo
Sep 18 '14 at 18:14
Thanks for the correction.
– user16924
Sep 18 '14 at 19:18
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $A$ be a commutative ring and $M$ be a proper maximal ideal in $A$. Show the following properties:
(a) If each $a in A setminus M$ is a unit element in $A$, then $M$ is the only maximal ideal in $A$.
(b) If each element of the form $1+m in A$ with $m in M$ is a unit in $A$, then $A$ has a unique maximal ideal.
I think I could show (a), I have problems with (b), I'll write what I've done so far:
(a) Suppose there exists $M' neq M$ a maximal ideal in $A$. Then there is some $b in M'$ such that $b not in M$. Then $b in A setminus M$, so there $b$ is a unit. Since $M'$ is ideal (bilateral ideal since $A$ is commutative), we have $1=b^-1b in M'$. Now $M$ is a maximal ideal, this means there exists $m in M$ such that $m not in M'$. But $m=m1 in M'$, which is absurd. The absurd comes from the assumption that there is another maximal ideal different from $M$, so $M$ is the only maximal ideal in $A$.
I would appreciate hints for (b).
abstract-algebra ring-theory ideals
asked Sep 18 '14 at 17:57
user16924
1,031618
Let $A$ be a commutative ring and $M$ be a proper maximal ideal in $A$. Show the following properties:
(a) If each $a in A setminus M$ is a unit element in $A$, then $M$ is the only maximal ideal in $A$.
(b) If each element of the form $1+m in A$ with $m in M$ is a unit in $A$, then $A$ has a unique maximal ideal.
I think I could show (a), I have problems with (b), I'll write what I've done so far:
(a) Suppose there exists $M' neq M$ a maximal ideal in $A$. Then there is some $b in M'$ such that $b not in M$. Then $b in A setminus M$, so there $b$ is a unit. Since $M'$ is ideal (bilateral ideal since $A$ is commutative), we have $1=b^-1b in M'$. Now $M$ is a maximal ideal, this means there exists $m in M$ such that $m not in M'$. But $m=m1 in M'$, which is absurd. The absurd comes from the assumption that there is another maximal ideal different from $M$, so $M$ is the only maximal ideal in $A$.
I would appreciate hints for (b).
abstract-algebra ring-theory ideals
abstract-algebra ring-theory ideals
asked Sep 18 '14 at 17:57
user16924
1,031618
asked Sep 18 '14 at 17:57
user16924
1,031618
edited Sep 18 '14 at 19:18
asked Sep 18 '14 at 17:57
user16924
1,031618
asked Sep 18 '14 at 17:57
user16924
1,031618
asked Sep 18 '14 at 17:57
user16924
1,031618
1,031618
3
I assume that in part (b), you mean $1 + m in A$, not $1 + m in M$.
– Bungo
Sep 18 '14 at 18:14
Thanks for the correction.
– user16924
Sep 18 '14 at 19:18
add a comment |Â
3
I assume that in part (b), you mean $1 + m in A$, not $1 + m in M$.
– Bungo
Sep 18 '14 at 18:14
Thanks for the correction.
– user16924
Sep 18 '14 at 19:18
3
3
I assume that in part (b), you mean $1 + m in A$, not $1 + m in M$.
– Bungo
Sep 18 '14 at 18:14
I assume that in part (b), you mean $1 + m in A$, not $1 + m in M$.
– Bungo
Sep 18 '14 at 18:14
Thanks for the correction.
– user16924
Sep 18 '14 at 19:18
Thanks for the correction.
– user16924
Sep 18 '14 at 19:18
add a comment |Â
2 Answers
2
active
oldest
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up vote
3
down vote
accepted
You can use part (a) to prove the second part.
Let $xin Asetminus M.$ Consider $N=M+xA=,min M, ain A$. Clearly $N$ is an ideal of $A$ and contains $M.$ Thus it has be the whole ring $A.$ In particular $m+xa=1$ for $min M,, ain A$. thus $xa=xa=1-m.$ Now you can apply part (a);) it follows that $a$ is a unit. Done!
answered Sep 18 '14 at 19:33
BigM
2,52411530
"...thus $xa=ax=1-m$...", how one deduces from here that $x$ is a unit? From there I would only conclude that $xa$ is a unit.
– user16924
Sep 18 '14 at 20:02
1
$1=axu=x(au)=(au)x$
– BigM
Sep 18 '14 at 20:10
1
I've just realized the same thing, I got it, thanks!
– user16924
Sep 18 '14 at 20:10
add a comment |Â
up vote
2
down vote
After "$1=b^-1b in M'$" you go a bit off the rails, because you could simply conclude at this point "Thus $M'=A$, contradicting our hypothesis that $M'neq A$.
Earlier I misread something about the statement of (b) and tried an irrelevant example. Then BigM beat me to the most reasonable proof. I can't give a better one: so I recommedn you check that one.
answered Sep 18 '14 at 19:36
rschwieb
101k1198234
Thanks for the remark in (a).
– user16924
Sep 18 '14 at 20:03
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
You can use part (a) to prove the second part.
Let $xin Asetminus M.$ Consider $N=M+xA=,min M, ain A$. Clearly $N$ is an ideal of $A$ and contains $M.$ Thus it has be the whole ring $A.$ In particular $m+xa=1$ for $min M,, ain A$. thus $xa=xa=1-m.$ Now you can apply part (a);) it follows that $a$ is a unit. Done!
answered Sep 18 '14 at 19:33
BigM
2,52411530
"...thus $xa=ax=1-m$...", how one deduces from here that $x$ is a unit? From there I would only conclude that $xa$ is a unit.
– user16924
Sep 18 '14 at 20:02
1
$1=axu=x(au)=(au)x$
– BigM
Sep 18 '14 at 20:10
1
I've just realized the same thing, I got it, thanks!
– user16924
Sep 18 '14 at 20:10
add a comment |Â
up vote
3
down vote
accepted
You can use part (a) to prove the second part.
Let $xin Asetminus M.$ Consider $N=M+xA=,min M, ain A$. Clearly $N$ is an ideal of $A$ and contains $M.$ Thus it has be the whole ring $A.$ In particular $m+xa=1$ for $min M,, ain A$. thus $xa=xa=1-m.$ Now you can apply part (a);) it follows that $a$ is a unit. Done!
answered Sep 18 '14 at 19:33
BigM
2,52411530
"...thus $xa=ax=1-m$...", how one deduces from here that $x$ is a unit? From there I would only conclude that $xa$ is a unit.
– user16924
Sep 18 '14 at 20:02
1
$1=axu=x(au)=(au)x$
– BigM
Sep 18 '14 at 20:10
1
I've just realized the same thing, I got it, thanks!
– user16924
Sep 18 '14 at 20:10
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
You can use part (a) to prove the second part.
Let $xin Asetminus M.$ Consider $N=M+xA=,min M, ain A$. Clearly $N$ is an ideal of $A$ and contains $M.$ Thus it has be the whole ring $A.$ In particular $m+xa=1$ for $min M,, ain A$. thus $xa=xa=1-m.$ Now you can apply part (a);) it follows that $a$ is a unit. Done!
answered Sep 18 '14 at 19:33
BigM
2,52411530
You can use part (a) to prove the second part.
Let $xin Asetminus M.$ Consider $N=M+xA=,min M, ain A$. Clearly $N$ is an ideal of $A$ and contains $M.$ Thus it has be the whole ring $A.$ In particular $m+xa=1$ for $min M,, ain A$. thus $xa=xa=1-m.$ Now you can apply part (a);) it follows that $a$ is a unit. Done!
answered Sep 18 '14 at 19:33
BigM
2,52411530
answered Sep 18 '14 at 19:33
BigM
2,52411530
answered Sep 18 '14 at 19:33
BigM
2,52411530
answered Sep 18 '14 at 19:33
BigM
2,52411530
2,52411530
"...thus $xa=ax=1-m$...", how one deduces from here that $x$ is a unit? From there I would only conclude that $xa$ is a unit.
– user16924
Sep 18 '14 at 20:02
1
$1=axu=x(au)=(au)x$
– BigM
Sep 18 '14 at 20:10
1
I've just realized the same thing, I got it, thanks!
– user16924
Sep 18 '14 at 20:10
add a comment |Â
"...thus $xa=ax=1-m$...", how one deduces from here that $x$ is a unit? From there I would only conclude that $xa$ is a unit.
– user16924
Sep 18 '14 at 20:02
1
$1=axu=x(au)=(au)x$
– BigM
Sep 18 '14 at 20:10
1
I've just realized the same thing, I got it, thanks!
– user16924
Sep 18 '14 at 20:10
"...thus $xa=ax=1-m$...", how one deduces from here that $x$ is a unit? From there I would only conclude that $xa$ is a unit.
– user16924
Sep 18 '14 at 20:02
"...thus $xa=ax=1-m$...", how one deduces from here that $x$ is a unit? From there I would only conclude that $xa$ is a unit.
– user16924
Sep 18 '14 at 20:02
1
1
$1=axu=x(au)=(au)x$
– BigM
Sep 18 '14 at 20:10
$1=axu=x(au)=(au)x$
– BigM
Sep 18 '14 at 20:10
1
1
I've just realized the same thing, I got it, thanks!
– user16924
Sep 18 '14 at 20:10
I've just realized the same thing, I got it, thanks!
– user16924
Sep 18 '14 at 20:10
add a comment |Â
up vote
2
down vote
After "$1=b^-1b in M'$" you go a bit off the rails, because you could simply conclude at this point "Thus $M'=A$, contradicting our hypothesis that $M'neq A$.
Earlier I misread something about the statement of (b) and tried an irrelevant example. Then BigM beat me to the most reasonable proof. I can't give a better one: so I recommedn you check that one.
answered Sep 18 '14 at 19:36
rschwieb
101k1198234
Thanks for the remark in (a).
– user16924
Sep 18 '14 at 20:03
add a comment |Â
up vote
2
down vote
After "$1=b^-1b in M'$" you go a bit off the rails, because you could simply conclude at this point "Thus $M'=A$, contradicting our hypothesis that $M'neq A$.
Earlier I misread something about the statement of (b) and tried an irrelevant example. Then BigM beat me to the most reasonable proof. I can't give a better one: so I recommedn you check that one.
answered Sep 18 '14 at 19:36
rschwieb
101k1198234
Thanks for the remark in (a).
– user16924
Sep 18 '14 at 20:03
add a comment |Â
up vote
2
down vote
up vote
2
down vote
After "$1=b^-1b in M'$" you go a bit off the rails, because you could simply conclude at this point "Thus $M'=A$, contradicting our hypothesis that $M'neq A$.
Earlier I misread something about the statement of (b) and tried an irrelevant example. Then BigM beat me to the most reasonable proof. I can't give a better one: so I recommedn you check that one.
answered Sep 18 '14 at 19:36
rschwieb
101k1198234
After "$1=b^-1b in M'$" you go a bit off the rails, because you could simply conclude at this point "Thus $M'=A$, contradicting our hypothesis that $M'neq A$.
Earlier I misread something about the statement of (b) and tried an irrelevant example. Then BigM beat me to the most reasonable proof. I can't give a better one: so I recommedn you check that one.
answered Sep 18 '14 at 19:36
rschwieb
101k1198234
edited Sep 18 '14 at 19:48
answered Sep 18 '14 at 19:36
rschwieb
101k1198234
answered Sep 18 '14 at 19:36
rschwieb
101k1198234
answered Sep 18 '14 at 19:36
rschwieb
101k1198234
101k1198234
Thanks for the remark in (a).
– user16924
Sep 18 '14 at 20:03
add a comment |Â
Thanks for the remark in (a).
– user16924
Sep 18 '14 at 20:03
Thanks for the remark in (a).
– user16924
Sep 18 '14 at 20:03
Thanks for the remark in (a).
– user16924
Sep 18 '14 at 20:03
add a comment |Â
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3
I assume that in part (b), you mean $1 + m in A$, not $1 + m in M$.
– Bungo
Sep 18 '14 at 18:14
Thanks for the correction.
– user16924
Sep 18 '14 at 19:18