If $I$ is a maximal ideal and $ain R -I$, then the assumption that $I + (a) = R$ gives a contradiction
$begingroup$
While I'm trying to prove that
Let $S$ be a multiplicative set in the commutative ring $R$ with
identity s.t $0 not in S$.Let $I$ be a maximal ideal in $S^c = R -
S$. Then show that $I$ is a prime ideal.
.I stuck at showing that
For $ab in I$ and $a,b not in I$, if $a in S^c - I$, then $$I +
(a) = R$$ gives a contradiction.
My first (and in fact only) attempt was observing that $exists (j in I, r in R)$ s.t
$$j + ra = 1_R.$$
But I couldn't find anything wrong by with such a claim. Of course, this might have throw any contradiction, but I'm not sure, so my question is that do we have any contradiction in here, if so, how to show it ?
abstract-algebra ring-theory ideals maximal-and-prime-ideals
asked Jan 8 at 8:12
onurcanbektasonurcanbektas
3,40711036
$endgroup$
add a comment |
$begingroup$
While I'm trying to prove that
Let $S$ be a multiplicative set in the commutative ring $R$ with
identity s.t $0 not in S$.Let $I$ be a maximal ideal in $S^c = R -
S$. Then show that $I$ is a prime ideal.
.I stuck at showing that
For $ab in I$ and $a,b not in I$, if $a in S^c - I$, then $$I +
(a) = R$$ gives a contradiction.
My first (and in fact only) attempt was observing that $exists (j in I, r in R)$ s.t
$$j + ra = 1_R.$$
But I couldn't find anything wrong by with such a claim. Of course, this might have throw any contradiction, but I'm not sure, so my question is that do we have any contradiction in here, if so, how to show it ?
abstract-algebra ring-theory ideals maximal-and-prime-ideals
asked Jan 8 at 8:12
onurcanbektasonurcanbektas
3,40711036
$endgroup$
$begingroup$
Sounds like you're just leaving something out about what you're proving, or else misunderstood what the intended contradiction is.
$endgroup$
– rschwieb
Jan 8 at 14:50
1
$begingroup$
@rschwieb Yes, I left out some part intentionally because I didn't want someone to give the answer to the main thing that I was trying to prove; I just wanted help at a particular point; see my edit.
$endgroup$
– onurcanbektas
Jan 8 at 16:08
$begingroup$
Yeah: that completely illuminates where your misunderstanding was.
$endgroup$
– rschwieb
Jan 8 at 16:24
add a comment |
$begingroup$
While I'm trying to prove that
Let $S$ be a multiplicative set in the commutative ring $R$ with
identity s.t $0 not in S$.Let $I$ be a maximal ideal in $S^c = R -
S$. Then show that $I$ is a prime ideal.
.I stuck at showing that
For $ab in I$ and $a,b not in I$, if $a in S^c - I$, then $$I +
(a) = R$$ gives a contradiction.
My first (and in fact only) attempt was observing that $exists (j in I, r in R)$ s.t
$$j + ra = 1_R.$$
But I couldn't find anything wrong by with such a claim. Of course, this might have throw any contradiction, but I'm not sure, so my question is that do we have any contradiction in here, if so, how to show it ?
abstract-algebra ring-theory ideals maximal-and-prime-ideals
asked Jan 8 at 8:12
onurcanbektasonurcanbektas
3,40711036
$endgroup$
While I'm trying to prove that
Let $S$ be a multiplicative set in the commutative ring $R$ with
identity s.t $0 not in S$.Let $I$ be a maximal ideal in $S^c = R -
S$. Then show that $I$ is a prime ideal.
.I stuck at showing that
For $ab in I$ and $a,b not in I$, if $a in S^c - I$, then $$I +
(a) = R$$ gives a contradiction.
My first (and in fact only) attempt was observing that $exists (j in I, r in R)$ s.t
$$j + ra = 1_R.$$
But I couldn't find anything wrong by with such a claim. Of course, this might have throw any contradiction, but I'm not sure, so my question is that do we have any contradiction in here, if so, how to show it ?
abstract-algebra ring-theory ideals maximal-and-prime-ideals
abstract-algebra ring-theory ideals maximal-and-prime-ideals
asked Jan 8 at 8:12
onurcanbektasonurcanbektas
3,40711036
asked Jan 8 at 8:12
onurcanbektasonurcanbektas
3,40711036
edited Jan 8 at 16:07
onurcanbektas
asked Jan 8 at 8:12
onurcanbektasonurcanbektas
3,40711036
asked Jan 8 at 8:12
onurcanbektasonurcanbektas
3,40711036
asked Jan 8 at 8:12
onurcanbektasonurcanbektas
3,40711036
3,40711036
$begingroup$
Sounds like you're just leaving something out about what you're proving, or else misunderstood what the intended contradiction is.
$endgroup$
– rschwieb
Jan 8 at 14:50
1
$begingroup$
@rschwieb Yes, I left out some part intentionally because I didn't want someone to give the answer to the main thing that I was trying to prove; I just wanted help at a particular point; see my edit.
$endgroup$
– onurcanbektas
Jan 8 at 16:08
$begingroup$
Yeah: that completely illuminates where your misunderstanding was.
$endgroup$
– rschwieb
Jan 8 at 16:24
add a comment |
$begingroup$
Sounds like you're just leaving something out about what you're proving, or else misunderstood what the intended contradiction is.
$endgroup$
– rschwieb
Jan 8 at 14:50
1
$begingroup$
@rschwieb Yes, I left out some part intentionally because I didn't want someone to give the answer to the main thing that I was trying to prove; I just wanted help at a particular point; see my edit.
$endgroup$
– onurcanbektas
Jan 8 at 16:08
$begingroup$
Yeah: that completely illuminates where your misunderstanding was.
$endgroup$
– rschwieb
Jan 8 at 16:24
$begingroup$
Sounds like you're just leaving something out about what you're proving, or else misunderstood what the intended contradiction is.
$endgroup$
– rschwieb
Jan 8 at 14:50
$begingroup$
Sounds like you're just leaving something out about what you're proving, or else misunderstood what the intended contradiction is.
$endgroup$
– rschwieb
Jan 8 at 14:50
1
1
$begingroup$
@rschwieb Yes, I left out some part intentionally because I didn't want someone to give the answer to the main thing that I was trying to prove; I just wanted help at a particular point; see my edit.
$endgroup$
– onurcanbektas
Jan 8 at 16:08
$begingroup$
@rschwieb Yes, I left out some part intentionally because I didn't want someone to give the answer to the main thing that I was trying to prove; I just wanted help at a particular point; see my edit.
$endgroup$
– onurcanbektas
Jan 8 at 16:08
$begingroup$
Yeah: that completely illuminates where your misunderstanding was.
$endgroup$
– rschwieb
Jan 8 at 16:24
$begingroup$
Yeah: that completely illuminates where your misunderstanding was.
$endgroup$
– rschwieb
Jan 8 at 16:24
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
With the added context, we can now see where your problem lies:
While I'm trying to prove that
Let $S$ be a multiplicative set in the commutative ring R with identity s.t $0∉S$. Let I be a maximal ideal in $S^c=R−S$. Then show that $I$ is a prime ideal.
The goal is to show that if $I$ is an ideal maximal with respect to containment in $S^c$, then $I$ is prime. In fact, $I$ is not necessarily a maximal ideal, it's just maximal with respect to the property I just stated.
That seems to be your misunderstanding. You need to proceed instead with $(a,I)cap Sneqemptyset$, rather than $(a,I)=R$. Also consider further assuming $(b,I)cap Sneqemptyset$ and see what you can do with the two statements.
answered Jan 8 at 16:19
rschwiebrschwieb
107k12102250
$endgroup$
add a comment |
$begingroup$
A contradiction? No.
Observe that $I+(a)$ is an ideal and is larger than maximal ideal $I$ because it contains $I$ as a subset and contains $anotin I$ as an element.
Then $I+(a)=R$ is the only possibility, because $I+(a)neq R$ would violate the maximality of $I$.
It is not a contradiction but a correct conclusion.
P.S.
This answer was posted before the illuminating edit of the OP.
answered Jan 8 at 8:19
drhabdrhab
102k545136
$endgroup$
$begingroup$
I hate doing controversial edits to the question after I've asked, but I think I missed a lot of point in the question s.t a reader cannot post a helpful answer, so see my edit please.
$endgroup$
– onurcanbektas
Jan 8 at 16:13
add a comment |
$begingroup$
This is not a contradiction. Show that $(2)$ is maximal (the quotient is a field!) in $mathbb Z$ and that $(2)+(3)=mathbb Z$ (show that $1$ is in the sum.)
In fact, if $I subset A$ is a maximal ideal and $I+(a) neq A$ for $a notin I$, we have that $I subset I+(a) subsetneq A$ so how could $I$ have been maximal?
answered Jan 8 at 8:18
Andres MejiaAndres Mejia
16.2k21548
$endgroup$
add a comment |
$begingroup$
As mentioned above, this is not a contradiction.
If you're trying to prove that $I$ is NOT a maximal ideal, I think what you need to say is that there exists some $a in R-I$ such that $I+(a) subsetneq R$, which probably can be achieved by yielding a contradiction from $j + ra = 1_R$.
answered Jan 8 at 8:38
Yanger MaYanger Ma
1014
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
With the added context, we can now see where your problem lies:
While I'm trying to prove that
Let $S$ be a multiplicative set in the commutative ring R with identity s.t $0∉S$. Let I be a maximal ideal in $S^c=R−S$. Then show that $I$ is a prime ideal.
The goal is to show that if $I$ is an ideal maximal with respect to containment in $S^c$, then $I$ is prime. In fact, $I$ is not necessarily a maximal ideal, it's just maximal with respect to the property I just stated.
That seems to be your misunderstanding. You need to proceed instead with $(a,I)cap Sneqemptyset$, rather than $(a,I)=R$. Also consider further assuming $(b,I)cap Sneqemptyset$ and see what you can do with the two statements.
answered Jan 8 at 16:19
rschwiebrschwieb
107k12102250
$endgroup$
add a comment |
$begingroup$
With the added context, we can now see where your problem lies:
While I'm trying to prove that
Let $S$ be a multiplicative set in the commutative ring R with identity s.t $0∉S$. Let I be a maximal ideal in $S^c=R−S$. Then show that $I$ is a prime ideal.
The goal is to show that if $I$ is an ideal maximal with respect to containment in $S^c$, then $I$ is prime. In fact, $I$ is not necessarily a maximal ideal, it's just maximal with respect to the property I just stated.
That seems to be your misunderstanding. You need to proceed instead with $(a,I)cap Sneqemptyset$, rather than $(a,I)=R$. Also consider further assuming $(b,I)cap Sneqemptyset$ and see what you can do with the two statements.
answered Jan 8 at 16:19
rschwiebrschwieb
107k12102250
$endgroup$
add a comment |
$begingroup$
With the added context, we can now see where your problem lies:
While I'm trying to prove that
Let $S$ be a multiplicative set in the commutative ring R with identity s.t $0∉S$. Let I be a maximal ideal in $S^c=R−S$. Then show that $I$ is a prime ideal.
The goal is to show that if $I$ is an ideal maximal with respect to containment in $S^c$, then $I$ is prime. In fact, $I$ is not necessarily a maximal ideal, it's just maximal with respect to the property I just stated.
That seems to be your misunderstanding. You need to proceed instead with $(a,I)cap Sneqemptyset$, rather than $(a,I)=R$. Also consider further assuming $(b,I)cap Sneqemptyset$ and see what you can do with the two statements.
answered Jan 8 at 16:19
rschwiebrschwieb
107k12102250
$endgroup$
With the added context, we can now see where your problem lies:
While I'm trying to prove that
Let $S$ be a multiplicative set in the commutative ring R with identity s.t $0∉S$. Let I be a maximal ideal in $S^c=R−S$. Then show that $I$ is a prime ideal.
The goal is to show that if $I$ is an ideal maximal with respect to containment in $S^c$, then $I$ is prime. In fact, $I$ is not necessarily a maximal ideal, it's just maximal with respect to the property I just stated.
That seems to be your misunderstanding. You need to proceed instead with $(a,I)cap Sneqemptyset$, rather than $(a,I)=R$. Also consider further assuming $(b,I)cap Sneqemptyset$ and see what you can do with the two statements.
answered Jan 8 at 16:19
rschwiebrschwieb
107k12102250
answered Jan 8 at 16:19
rschwiebrschwieb
107k12102250
answered Jan 8 at 16:19
rschwiebrschwieb
107k12102250
answered Jan 8 at 16:19
rschwiebrschwieb
107k12102250
107k12102250
add a comment |
add a comment |
$begingroup$
A contradiction? No.
Observe that $I+(a)$ is an ideal and is larger than maximal ideal $I$ because it contains $I$ as a subset and contains $anotin I$ as an element.
Then $I+(a)=R$ is the only possibility, because $I+(a)neq R$ would violate the maximality of $I$.
It is not a contradiction but a correct conclusion.
P.S.
This answer was posted before the illuminating edit of the OP.
answered Jan 8 at 8:19
drhabdrhab
102k545136
$endgroup$
$begingroup$
I hate doing controversial edits to the question after I've asked, but I think I missed a lot of point in the question s.t a reader cannot post a helpful answer, so see my edit please.
$endgroup$
– onurcanbektas
Jan 8 at 16:13
add a comment |
$begingroup$
A contradiction? No.
Observe that $I+(a)$ is an ideal and is larger than maximal ideal $I$ because it contains $I$ as a subset and contains $anotin I$ as an element.
Then $I+(a)=R$ is the only possibility, because $I+(a)neq R$ would violate the maximality of $I$.
It is not a contradiction but a correct conclusion.
P.S.
This answer was posted before the illuminating edit of the OP.
answered Jan 8 at 8:19
drhabdrhab
102k545136
$endgroup$
$begingroup$
I hate doing controversial edits to the question after I've asked, but I think I missed a lot of point in the question s.t a reader cannot post a helpful answer, so see my edit please.
$endgroup$
– onurcanbektas
Jan 8 at 16:13
add a comment |
$begingroup$
A contradiction? No.
Observe that $I+(a)$ is an ideal and is larger than maximal ideal $I$ because it contains $I$ as a subset and contains $anotin I$ as an element.
Then $I+(a)=R$ is the only possibility, because $I+(a)neq R$ would violate the maximality of $I$.
It is not a contradiction but a correct conclusion.
P.S.
This answer was posted before the illuminating edit of the OP.
answered Jan 8 at 8:19
drhabdrhab
102k545136
$endgroup$
A contradiction? No.
Observe that $I+(a)$ is an ideal and is larger than maximal ideal $I$ because it contains $I$ as a subset and contains $anotin I$ as an element.
Then $I+(a)=R$ is the only possibility, because $I+(a)neq R$ would violate the maximality of $I$.
It is not a contradiction but a correct conclusion.
P.S.
This answer was posted before the illuminating edit of the OP.
answered Jan 8 at 8:19
drhabdrhab
102k545136
edited Jan 8 at 16:42
answered Jan 8 at 8:19
drhabdrhab
102k545136
answered Jan 8 at 8:19
drhabdrhab
102k545136
answered Jan 8 at 8:19
drhabdrhab
102k545136
102k545136
$begingroup$
I hate doing controversial edits to the question after I've asked, but I think I missed a lot of point in the question s.t a reader cannot post a helpful answer, so see my edit please.
$endgroup$
– onurcanbektas
Jan 8 at 16:13
add a comment |
$begingroup$
I hate doing controversial edits to the question after I've asked, but I think I missed a lot of point in the question s.t a reader cannot post a helpful answer, so see my edit please.
$endgroup$
– onurcanbektas
Jan 8 at 16:13
$begingroup$
I hate doing controversial edits to the question after I've asked, but I think I missed a lot of point in the question s.t a reader cannot post a helpful answer, so see my edit please.
$endgroup$
– onurcanbektas
Jan 8 at 16:13
$begingroup$
I hate doing controversial edits to the question after I've asked, but I think I missed a lot of point in the question s.t a reader cannot post a helpful answer, so see my edit please.
$endgroup$
– onurcanbektas
Jan 8 at 16:13
add a comment |
$begingroup$
This is not a contradiction. Show that $(2)$ is maximal (the quotient is a field!) in $mathbb Z$ and that $(2)+(3)=mathbb Z$ (show that $1$ is in the sum.)
In fact, if $I subset A$ is a maximal ideal and $I+(a) neq A$ for $a notin I$, we have that $I subset I+(a) subsetneq A$ so how could $I$ have been maximal?
answered Jan 8 at 8:18
Andres MejiaAndres Mejia
16.2k21548
$endgroup$
add a comment |
$begingroup$
This is not a contradiction. Show that $(2)$ is maximal (the quotient is a field!) in $mathbb Z$ and that $(2)+(3)=mathbb Z$ (show that $1$ is in the sum.)
In fact, if $I subset A$ is a maximal ideal and $I+(a) neq A$ for $a notin I$, we have that $I subset I+(a) subsetneq A$ so how could $I$ have been maximal?
answered Jan 8 at 8:18
Andres MejiaAndres Mejia
16.2k21548
$endgroup$
add a comment |
$begingroup$
This is not a contradiction. Show that $(2)$ is maximal (the quotient is a field!) in $mathbb Z$ and that $(2)+(3)=mathbb Z$ (show that $1$ is in the sum.)
In fact, if $I subset A$ is a maximal ideal and $I+(a) neq A$ for $a notin I$, we have that $I subset I+(a) subsetneq A$ so how could $I$ have been maximal?
answered Jan 8 at 8:18
Andres MejiaAndres Mejia
16.2k21548
$endgroup$
This is not a contradiction. Show that $(2)$ is maximal (the quotient is a field!) in $mathbb Z$ and that $(2)+(3)=mathbb Z$ (show that $1$ is in the sum.)
In fact, if $I subset A$ is a maximal ideal and $I+(a) neq A$ for $a notin I$, we have that $I subset I+(a) subsetneq A$ so how could $I$ have been maximal?
answered Jan 8 at 8:18
Andres MejiaAndres Mejia
16.2k21548
answered Jan 8 at 8:18
Andres MejiaAndres Mejia
16.2k21548
answered Jan 8 at 8:18
Andres MejiaAndres Mejia
16.2k21548
answered Jan 8 at 8:18
Andres MejiaAndres Mejia
16.2k21548
16.2k21548
add a comment |
add a comment |
$begingroup$
As mentioned above, this is not a contradiction.
If you're trying to prove that $I$ is NOT a maximal ideal, I think what you need to say is that there exists some $a in R-I$ such that $I+(a) subsetneq R$, which probably can be achieved by yielding a contradiction from $j + ra = 1_R$.
answered Jan 8 at 8:38
Yanger MaYanger Ma
1014
$endgroup$
add a comment |
$begingroup$
As mentioned above, this is not a contradiction.
If you're trying to prove that $I$ is NOT a maximal ideal, I think what you need to say is that there exists some $a in R-I$ such that $I+(a) subsetneq R$, which probably can be achieved by yielding a contradiction from $j + ra = 1_R$.
answered Jan 8 at 8:38
Yanger MaYanger Ma
1014
$endgroup$
add a comment |
$begingroup$
As mentioned above, this is not a contradiction.
If you're trying to prove that $I$ is NOT a maximal ideal, I think what you need to say is that there exists some $a in R-I$ such that $I+(a) subsetneq R$, which probably can be achieved by yielding a contradiction from $j + ra = 1_R$.
answered Jan 8 at 8:38
Yanger MaYanger Ma
1014
$endgroup$
As mentioned above, this is not a contradiction.
If you're trying to prove that $I$ is NOT a maximal ideal, I think what you need to say is that there exists some $a in R-I$ such that $I+(a) subsetneq R$, which probably can be achieved by yielding a contradiction from $j + ra = 1_R$.
answered Jan 8 at 8:38
Yanger MaYanger Ma
1014
answered Jan 8 at 8:38
Yanger MaYanger Ma
1014
answered Jan 8 at 8:38
Yanger MaYanger Ma
1014
answered Jan 8 at 8:38
Yanger MaYanger Ma
1014
1014
add a comment |
add a comment |
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$begingroup$
Sounds like you're just leaving something out about what you're proving, or else misunderstood what the intended contradiction is.
$endgroup$
– rschwieb
Jan 8 at 14:50
1
$begingroup$
@rschwieb Yes, I left out some part intentionally because I didn't want someone to give the answer to the main thing that I was trying to prove; I just wanted help at a particular point; see my edit.
$endgroup$
– onurcanbektas
Jan 8 at 16:08
$begingroup$
Yeah: that completely illuminates where your misunderstanding was.
$endgroup$
– rschwieb
Jan 8 at 16:24