Mixing
How do you square an ideal?
Clash Royale CLAN TAG#URR8PPP
up vote
5
down vote
favorite
For some of you, this question is going to seem extremely basic.
I think I understand what an ideal is. Such as $langle 2, 1 + sqrt-5 rangle$, it consists of all numbers in this ring of the form $2a + (1 + sqrt-5)b$.
But then what is $langle 2, 1 + sqrt-5 rangle^2$? My first thought was $langle 4, -4 + 2 sqrt-5 rangle$, but that seems wrong somehow. I also had something in the back of my mind saying $langle 2 rangle$, but I'm not sure about that one either.
Then I thought about trying to figure out $langle 2, 1 + sqrt-5 rangle langle 2, 1 + sqrt-5 rangle$ when I realized I don't actually understand how to multiply ideals to begin with.
I'm only using $langle 2, 1 + sqrt-5 rangle$ as an example (though that does draw in one question identified as similar that looks much more relevant than all the questions identified as "Questions that may already have your answer"). In a principal ideal domain, would it be correct to think that $langle a rangle langle b rangle = langle ab rangle$?
Any help would be much appreciated.
abstract-algebra ring-theory ideals
edited Oct 13 '16 at 21:00
Xam
4,45451445
asked Nov 9 '15 at 22:16
Bob Happ
3651222
add a comment |Â
up vote
5
down vote
favorite
For some of you, this question is going to seem extremely basic.
I think I understand what an ideal is. Such as $langle 2, 1 + sqrt-5 rangle$, it consists of all numbers in this ring of the form $2a + (1 + sqrt-5)b$.
But then what is $langle 2, 1 + sqrt-5 rangle^2$? My first thought was $langle 4, -4 + 2 sqrt-5 rangle$, but that seems wrong somehow. I also had something in the back of my mind saying $langle 2 rangle$, but I'm not sure about that one either.
Then I thought about trying to figure out $langle 2, 1 + sqrt-5 rangle langle 2, 1 + sqrt-5 rangle$ when I realized I don't actually understand how to multiply ideals to begin with.
I'm only using $langle 2, 1 + sqrt-5 rangle$ as an example (though that does draw in one question identified as similar that looks much more relevant than all the questions identified as "Questions that may already have your answer"). In a principal ideal domain, would it be correct to think that $langle a rangle langle b rangle = langle ab rangle$?
Any help would be much appreciated.
abstract-algebra ring-theory ideals
edited Oct 13 '16 at 21:00
Xam
4,45451445
asked Nov 9 '15 at 22:16
Bob Happ
3651222
3
The product of two ideals $IJ$ is (in a commutative ring) the collection of all sums of products $ij$ where $i in I, j in J$.
– Qiaochu Yuan
Nov 9 '15 at 22:26
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
For some of you, this question is going to seem extremely basic.
I think I understand what an ideal is. Such as $langle 2, 1 + sqrt-5 rangle$, it consists of all numbers in this ring of the form $2a + (1 + sqrt-5)b$.
But then what is $langle 2, 1 + sqrt-5 rangle^2$? My first thought was $langle 4, -4 + 2 sqrt-5 rangle$, but that seems wrong somehow. I also had something in the back of my mind saying $langle 2 rangle$, but I'm not sure about that one either.
Then I thought about trying to figure out $langle 2, 1 + sqrt-5 rangle langle 2, 1 + sqrt-5 rangle$ when I realized I don't actually understand how to multiply ideals to begin with.
I'm only using $langle 2, 1 + sqrt-5 rangle$ as an example (though that does draw in one question identified as similar that looks much more relevant than all the questions identified as "Questions that may already have your answer"). In a principal ideal domain, would it be correct to think that $langle a rangle langle b rangle = langle ab rangle$?
Any help would be much appreciated.
abstract-algebra ring-theory ideals
edited Oct 13 '16 at 21:00
Xam
4,45451445
asked Nov 9 '15 at 22:16
Bob Happ
3651222
For some of you, this question is going to seem extremely basic.
I think I understand what an ideal is. Such as $langle 2, 1 + sqrt-5 rangle$, it consists of all numbers in this ring of the form $2a + (1 + sqrt-5)b$.
But then what is $langle 2, 1 + sqrt-5 rangle^2$? My first thought was $langle 4, -4 + 2 sqrt-5 rangle$, but that seems wrong somehow. I also had something in the back of my mind saying $langle 2 rangle$, but I'm not sure about that one either.
Then I thought about trying to figure out $langle 2, 1 + sqrt-5 rangle langle 2, 1 + sqrt-5 rangle$ when I realized I don't actually understand how to multiply ideals to begin with.
I'm only using $langle 2, 1 + sqrt-5 rangle$ as an example (though that does draw in one question identified as similar that looks much more relevant than all the questions identified as "Questions that may already have your answer"). In a principal ideal domain, would it be correct to think that $langle a rangle langle b rangle = langle ab rangle$?
Any help would be much appreciated.
abstract-algebra ring-theory ideals
edited Oct 13 '16 at 21:00
Xam
4,45451445
asked Nov 9 '15 at 22:16
Bob Happ
3651222
edited Oct 13 '16 at 21:00
Xam
4,45451445
edited Oct 13 '16 at 21:00
Xam
4,45451445
edited Oct 13 '16 at 21:00
Xam
4,45451445
4,45451445
asked Nov 9 '15 at 22:16
Bob Happ
3651222
asked Nov 9 '15 at 22:16
Bob Happ
3651222
asked Nov 9 '15 at 22:16
Bob Happ
3651222
3651222
3
The product of two ideals $IJ$ is (in a commutative ring) the collection of all sums of products $ij$ where $i in I, j in J$.
– Qiaochu Yuan
Nov 9 '15 at 22:26
add a comment |Â
3
The product of two ideals $IJ$ is (in a commutative ring) the collection of all sums of products $ij$ where $i in I, j in J$.
– Qiaochu Yuan
Nov 9 '15 at 22:26
3
3
The product of two ideals $IJ$ is (in a commutative ring) the collection of all sums of products $ij$ where $i in I, j in J$.
– Qiaochu Yuan
Nov 9 '15 at 22:26
The product of two ideals $IJ$ is (in a commutative ring) the collection of all sums of products $ij$ where $i in I, j in J$.
– Qiaochu Yuan
Nov 9 '15 at 22:26
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
7
down vote
accepted
Both other answers are right but fail to give the answer in its simplest form. Note that
$$-4+(2 + 2 sqrt-5) - (-4+2 sqrt-5) = 2.$$
So $2$ is in the ideal $langle 4, (2 + 2 sqrt-5), (-4+2 sqrt-5) rangle$ and, conversely, it is easy to see that each of $4$, $2 + 2 sqrt-5$ and $-4+2 sqrt-5$ is divisible by $2$. So
$$langle 2, 1 + sqrt-5 rangle^2 = langle 4, (2 + 2 sqrt-5), (-4+2 sqrt-5) rangle = langle 2 rangle.$$
answered Oct 11 '16 at 14:25
David E Speyer
44.2k4120197
I think it should be $-4+2sqrtcolorred-5$ everywhere.
– Bernard
Aug 5 at 19:13
@Bernard Thanks, fixed!
– David E Speyer
Aug 6 at 21:17
add a comment |Â
up vote
4
down vote
For a P. I. D., yes. For the general case, the product of two ideals $I=(a_1,dots, a_n)$ and $J=(b_1,dots, b_p)$ is the ideal generated by all possible products $a_ib_j$.
Hence the square of $I=(2, 1+sqrt-5)$ is the ideal $I^2=(4,2+2sqrt-5,-4+2sqrt-5)$.
answered Nov 9 '15 at 22:45
Bernard
110k635103
I think the third generator of $I^2$ should be $-4+2sqrt-5$? And I don't think $I=mathbbZ[1+sqrt-5]$.
– Sam Weatherhog
Nov 9 '15 at 22:46
I don't think we can get $-6+2sqrt-5$ from these generators either. I don't think $2in langle 2, 1+sqrt-5 rangle^2$.
– Sam Weatherhog
Nov 9 '15 at 22:50
Yes I've just checked, as it seemed unlikely. It's corrected
– Bernard
Nov 9 '15 at 22:51
Telescoping errors. Thanks for pointing it.
– Bernard
Nov 9 '15 at 22:52
No worries. I had similar issues trying to find relations between generators and also forgot to square 2 :)
– Sam Weatherhog
Nov 9 '15 at 22:53
add a comment |Â
up vote
2
down vote
When you square an ideal you're really looking at products of two things coming from your ideal. In you case $langle 2, 1+sqrt-5rangle^2$ means take something in $langle 2, 1+sqrt-5rangle$ and multiply it by something else in the ideal; the set of all the things you get by doing this is the ideal $langle 2, 1+sqrt-5rangle^2$.
In order to work with this you can just multiply your generators together, so $langle 2, 1+sqrt-5rangle^2=langle 4,2+2sqrt-5,-4+2sqrt-5rangle$. Then you might look for relations between these generators to reduce this set. In this case it does not appear that there are any relations between them.
You are correct that in a P.I.D., $langle a rangle langle brangle=langle abrangle$ provided $a,b$ aren't units (in which case $langle a rangle$ or $langle b rangle$ would be the entire ring).
answered Nov 9 '15 at 22:26
Sam Weatherhog
1,222515
As long as neither is a unit ideal, right?
– Bob Happ
Nov 9 '15 at 22:33
@BobHapp yes sorry, let me sort out the other glaring errors first..
– Sam Weatherhog
Nov 9 '15 at 22:33
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
Both other answers are right but fail to give the answer in its simplest form. Note that
$$-4+(2 + 2 sqrt-5) - (-4+2 sqrt-5) = 2.$$
So $2$ is in the ideal $langle 4, (2 + 2 sqrt-5), (-4+2 sqrt-5) rangle$ and, conversely, it is easy to see that each of $4$, $2 + 2 sqrt-5$ and $-4+2 sqrt-5$ is divisible by $2$. So
$$langle 2, 1 + sqrt-5 rangle^2 = langle 4, (2 + 2 sqrt-5), (-4+2 sqrt-5) rangle = langle 2 rangle.$$
answered Oct 11 '16 at 14:25
David E Speyer
44.2k4120197
I think it should be $-4+2sqrtcolorred-5$ everywhere.
– Bernard
Aug 5 at 19:13
@Bernard Thanks, fixed!
– David E Speyer
Aug 6 at 21:17
add a comment |Â
up vote
7
down vote
accepted
Both other answers are right but fail to give the answer in its simplest form. Note that
$$-4+(2 + 2 sqrt-5) - (-4+2 sqrt-5) = 2.$$
So $2$ is in the ideal $langle 4, (2 + 2 sqrt-5), (-4+2 sqrt-5) rangle$ and, conversely, it is easy to see that each of $4$, $2 + 2 sqrt-5$ and $-4+2 sqrt-5$ is divisible by $2$. So
$$langle 2, 1 + sqrt-5 rangle^2 = langle 4, (2 + 2 sqrt-5), (-4+2 sqrt-5) rangle = langle 2 rangle.$$
answered Oct 11 '16 at 14:25
David E Speyer
44.2k4120197
I think it should be $-4+2sqrtcolorred-5$ everywhere.
– Bernard
Aug 5 at 19:13
@Bernard Thanks, fixed!
– David E Speyer
Aug 6 at 21:17
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
Both other answers are right but fail to give the answer in its simplest form. Note that
$$-4+(2 + 2 sqrt-5) - (-4+2 sqrt-5) = 2.$$
So $2$ is in the ideal $langle 4, (2 + 2 sqrt-5), (-4+2 sqrt-5) rangle$ and, conversely, it is easy to see that each of $4$, $2 + 2 sqrt-5$ and $-4+2 sqrt-5$ is divisible by $2$. So
$$langle 2, 1 + sqrt-5 rangle^2 = langle 4, (2 + 2 sqrt-5), (-4+2 sqrt-5) rangle = langle 2 rangle.$$
answered Oct 11 '16 at 14:25
David E Speyer
44.2k4120197
Both other answers are right but fail to give the answer in its simplest form. Note that
$$-4+(2 + 2 sqrt-5) - (-4+2 sqrt-5) = 2.$$
So $2$ is in the ideal $langle 4, (2 + 2 sqrt-5), (-4+2 sqrt-5) rangle$ and, conversely, it is easy to see that each of $4$, $2 + 2 sqrt-5$ and $-4+2 sqrt-5$ is divisible by $2$. So
$$langle 2, 1 + sqrt-5 rangle^2 = langle 4, (2 + 2 sqrt-5), (-4+2 sqrt-5) rangle = langle 2 rangle.$$
answered Oct 11 '16 at 14:25
David E Speyer
44.2k4120197
edited Aug 6 at 21:17
answered Oct 11 '16 at 14:25
David E Speyer
44.2k4120197
answered Oct 11 '16 at 14:25
David E Speyer
44.2k4120197
answered Oct 11 '16 at 14:25
David E Speyer
44.2k4120197
44.2k4120197
I think it should be $-4+2sqrtcolorred-5$ everywhere.
– Bernard
Aug 5 at 19:13
@Bernard Thanks, fixed!
– David E Speyer
Aug 6 at 21:17
add a comment |Â
I think it should be $-4+2sqrtcolorred-5$ everywhere.
– Bernard
Aug 5 at 19:13
@Bernard Thanks, fixed!
– David E Speyer
Aug 6 at 21:17
I think it should be $-4+2sqrtcolorred-5$ everywhere.
– Bernard
Aug 5 at 19:13
I think it should be $-4+2sqrtcolorred-5$ everywhere.
– Bernard
Aug 5 at 19:13
@Bernard Thanks, fixed!
– David E Speyer
Aug 6 at 21:17
@Bernard Thanks, fixed!
– David E Speyer
Aug 6 at 21:17
add a comment |Â
up vote
4
down vote
For a P. I. D., yes. For the general case, the product of two ideals $I=(a_1,dots, a_n)$ and $J=(b_1,dots, b_p)$ is the ideal generated by all possible products $a_ib_j$.
Hence the square of $I=(2, 1+sqrt-5)$ is the ideal $I^2=(4,2+2sqrt-5,-4+2sqrt-5)$.
answered Nov 9 '15 at 22:45
Bernard
110k635103
I think the third generator of $I^2$ should be $-4+2sqrt-5$? And I don't think $I=mathbbZ[1+sqrt-5]$.
– Sam Weatherhog
Nov 9 '15 at 22:46
I don't think we can get $-6+2sqrt-5$ from these generators either. I don't think $2in langle 2, 1+sqrt-5 rangle^2$.
– Sam Weatherhog
Nov 9 '15 at 22:50
Yes I've just checked, as it seemed unlikely. It's corrected
– Bernard
Nov 9 '15 at 22:51
Telescoping errors. Thanks for pointing it.
– Bernard
Nov 9 '15 at 22:52
No worries. I had similar issues trying to find relations between generators and also forgot to square 2 :)
– Sam Weatherhog
Nov 9 '15 at 22:53
add a comment |Â
up vote
4
down vote
For a P. I. D., yes. For the general case, the product of two ideals $I=(a_1,dots, a_n)$ and $J=(b_1,dots, b_p)$ is the ideal generated by all possible products $a_ib_j$.
Hence the square of $I=(2, 1+sqrt-5)$ is the ideal $I^2=(4,2+2sqrt-5,-4+2sqrt-5)$.
answered Nov 9 '15 at 22:45
Bernard
110k635103
I think the third generator of $I^2$ should be $-4+2sqrt-5$? And I don't think $I=mathbbZ[1+sqrt-5]$.
– Sam Weatherhog
Nov 9 '15 at 22:46
I don't think we can get $-6+2sqrt-5$ from these generators either. I don't think $2in langle 2, 1+sqrt-5 rangle^2$.
– Sam Weatherhog
Nov 9 '15 at 22:50
Yes I've just checked, as it seemed unlikely. It's corrected
– Bernard
Nov 9 '15 at 22:51
Telescoping errors. Thanks for pointing it.
– Bernard
Nov 9 '15 at 22:52
No worries. I had similar issues trying to find relations between generators and also forgot to square 2 :)
– Sam Weatherhog
Nov 9 '15 at 22:53
add a comment |Â
up vote
4
down vote
up vote
4
down vote
For a P. I. D., yes. For the general case, the product of two ideals $I=(a_1,dots, a_n)$ and $J=(b_1,dots, b_p)$ is the ideal generated by all possible products $a_ib_j$.
Hence the square of $I=(2, 1+sqrt-5)$ is the ideal $I^2=(4,2+2sqrt-5,-4+2sqrt-5)$.
answered Nov 9 '15 at 22:45
Bernard
110k635103
For a P. I. D., yes. For the general case, the product of two ideals $I=(a_1,dots, a_n)$ and $J=(b_1,dots, b_p)$ is the ideal generated by all possible products $a_ib_j$.
Hence the square of $I=(2, 1+sqrt-5)$ is the ideal $I^2=(4,2+2sqrt-5,-4+2sqrt-5)$.
answered Nov 9 '15 at 22:45
Bernard
110k635103
edited Nov 9 '15 at 22:52
answered Nov 9 '15 at 22:45
Bernard
110k635103
answered Nov 9 '15 at 22:45
Bernard
110k635103
answered Nov 9 '15 at 22:45
Bernard
110k635103
110k635103
I think the third generator of $I^2$ should be $-4+2sqrt-5$? And I don't think $I=mathbbZ[1+sqrt-5]$.
– Sam Weatherhog
Nov 9 '15 at 22:46
I don't think we can get $-6+2sqrt-5$ from these generators either. I don't think $2in langle 2, 1+sqrt-5 rangle^2$.
– Sam Weatherhog
Nov 9 '15 at 22:50
Yes I've just checked, as it seemed unlikely. It's corrected
– Bernard
Nov 9 '15 at 22:51
Telescoping errors. Thanks for pointing it.
– Bernard
Nov 9 '15 at 22:52
No worries. I had similar issues trying to find relations between generators and also forgot to square 2 :)
– Sam Weatherhog
Nov 9 '15 at 22:53
add a comment |Â
I think the third generator of $I^2$ should be $-4+2sqrt-5$? And I don't think $I=mathbbZ[1+sqrt-5]$.
– Sam Weatherhog
Nov 9 '15 at 22:46
I don't think we can get $-6+2sqrt-5$ from these generators either. I don't think $2in langle 2, 1+sqrt-5 rangle^2$.
– Sam Weatherhog
Nov 9 '15 at 22:50
Yes I've just checked, as it seemed unlikely. It's corrected
– Bernard
Nov 9 '15 at 22:51
Telescoping errors. Thanks for pointing it.
– Bernard
Nov 9 '15 at 22:52
No worries. I had similar issues trying to find relations between generators and also forgot to square 2 :)
– Sam Weatherhog
Nov 9 '15 at 22:53
I think the third generator of $I^2$ should be $-4+2sqrt-5$? And I don't think $I=mathbbZ[1+sqrt-5]$.
– Sam Weatherhog
Nov 9 '15 at 22:46
I think the third generator of $I^2$ should be $-4+2sqrt-5$? And I don't think $I=mathbbZ[1+sqrt-5]$.
– Sam Weatherhog
Nov 9 '15 at 22:46
I don't think we can get $-6+2sqrt-5$ from these generators either. I don't think $2in langle 2, 1+sqrt-5 rangle^2$.
– Sam Weatherhog
Nov 9 '15 at 22:50
I don't think we can get $-6+2sqrt-5$ from these generators either. I don't think $2in langle 2, 1+sqrt-5 rangle^2$.
– Sam Weatherhog
Nov 9 '15 at 22:50
Yes I've just checked, as it seemed unlikely. It's corrected
– Bernard
Nov 9 '15 at 22:51
Yes I've just checked, as it seemed unlikely. It's corrected
– Bernard
Nov 9 '15 at 22:51
Telescoping errors. Thanks for pointing it.
– Bernard
Nov 9 '15 at 22:52
Telescoping errors. Thanks for pointing it.
– Bernard
Nov 9 '15 at 22:52
No worries. I had similar issues trying to find relations between generators and also forgot to square 2 :)
– Sam Weatherhog
Nov 9 '15 at 22:53
No worries. I had similar issues trying to find relations between generators and also forgot to square 2 :)
– Sam Weatherhog
Nov 9 '15 at 22:53
add a comment |Â
up vote
2
down vote
When you square an ideal you're really looking at products of two things coming from your ideal. In you case $langle 2, 1+sqrt-5rangle^2$ means take something in $langle 2, 1+sqrt-5rangle$ and multiply it by something else in the ideal; the set of all the things you get by doing this is the ideal $langle 2, 1+sqrt-5rangle^2$.
In order to work with this you can just multiply your generators together, so $langle 2, 1+sqrt-5rangle^2=langle 4,2+2sqrt-5,-4+2sqrt-5rangle$. Then you might look for relations between these generators to reduce this set. In this case it does not appear that there are any relations between them.
You are correct that in a P.I.D., $langle a rangle langle brangle=langle abrangle$ provided $a,b$ aren't units (in which case $langle a rangle$ or $langle b rangle$ would be the entire ring).
answered Nov 9 '15 at 22:26
Sam Weatherhog
1,222515
As long as neither is a unit ideal, right?
– Bob Happ
Nov 9 '15 at 22:33
@BobHapp yes sorry, let me sort out the other glaring errors first..
– Sam Weatherhog
Nov 9 '15 at 22:33
add a comment |Â
up vote
2
down vote
When you square an ideal you're really looking at products of two things coming from your ideal. In you case $langle 2, 1+sqrt-5rangle^2$ means take something in $langle 2, 1+sqrt-5rangle$ and multiply it by something else in the ideal; the set of all the things you get by doing this is the ideal $langle 2, 1+sqrt-5rangle^2$.
In order to work with this you can just multiply your generators together, so $langle 2, 1+sqrt-5rangle^2=langle 4,2+2sqrt-5,-4+2sqrt-5rangle$. Then you might look for relations between these generators to reduce this set. In this case it does not appear that there are any relations between them.
You are correct that in a P.I.D., $langle a rangle langle brangle=langle abrangle$ provided $a,b$ aren't units (in which case $langle a rangle$ or $langle b rangle$ would be the entire ring).
answered Nov 9 '15 at 22:26
Sam Weatherhog
1,222515
As long as neither is a unit ideal, right?
– Bob Happ
Nov 9 '15 at 22:33
@BobHapp yes sorry, let me sort out the other glaring errors first..
– Sam Weatherhog
Nov 9 '15 at 22:33
add a comment |Â
up vote
2
down vote
up vote
2
down vote
When you square an ideal you're really looking at products of two things coming from your ideal. In you case $langle 2, 1+sqrt-5rangle^2$ means take something in $langle 2, 1+sqrt-5rangle$ and multiply it by something else in the ideal; the set of all the things you get by doing this is the ideal $langle 2, 1+sqrt-5rangle^2$.
In order to work with this you can just multiply your generators together, so $langle 2, 1+sqrt-5rangle^2=langle 4,2+2sqrt-5,-4+2sqrt-5rangle$. Then you might look for relations between these generators to reduce this set. In this case it does not appear that there are any relations between them.
You are correct that in a P.I.D., $langle a rangle langle brangle=langle abrangle$ provided $a,b$ aren't units (in which case $langle a rangle$ or $langle b rangle$ would be the entire ring).
answered Nov 9 '15 at 22:26
Sam Weatherhog
1,222515
When you square an ideal you're really looking at products of two things coming from your ideal. In you case $langle 2, 1+sqrt-5rangle^2$ means take something in $langle 2, 1+sqrt-5rangle$ and multiply it by something else in the ideal; the set of all the things you get by doing this is the ideal $langle 2, 1+sqrt-5rangle^2$.
In order to work with this you can just multiply your generators together, so $langle 2, 1+sqrt-5rangle^2=langle 4,2+2sqrt-5,-4+2sqrt-5rangle$. Then you might look for relations between these generators to reduce this set. In this case it does not appear that there are any relations between them.
You are correct that in a P.I.D., $langle a rangle langle brangle=langle abrangle$ provided $a,b$ aren't units (in which case $langle a rangle$ or $langle b rangle$ would be the entire ring).
answered Nov 9 '15 at 22:26
Sam Weatherhog
1,222515
edited Nov 9 '15 at 22:45
answered Nov 9 '15 at 22:26
Sam Weatherhog
1,222515
answered Nov 9 '15 at 22:26
Sam Weatherhog
1,222515
answered Nov 9 '15 at 22:26
Sam Weatherhog
1,222515
1,222515
As long as neither is a unit ideal, right?
– Bob Happ
Nov 9 '15 at 22:33
@BobHapp yes sorry, let me sort out the other glaring errors first..
– Sam Weatherhog
Nov 9 '15 at 22:33
add a comment |Â
As long as neither is a unit ideal, right?
– Bob Happ
Nov 9 '15 at 22:33
@BobHapp yes sorry, let me sort out the other glaring errors first..
– Sam Weatherhog
Nov 9 '15 at 22:33
As long as neither is a unit ideal, right?
– Bob Happ
Nov 9 '15 at 22:33
As long as neither is a unit ideal, right?
– Bob Happ
Nov 9 '15 at 22:33
@BobHapp yes sorry, let me sort out the other glaring errors first..
– Sam Weatherhog
Nov 9 '15 at 22:33
@BobHapp yes sorry, let me sort out the other glaring errors first..
– Sam Weatherhog
Nov 9 '15 at 22:33
add a comment |Â
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3
The product of two ideals $IJ$ is (in a commutative ring) the collection of all sums of products $ij$ where $i in I, j in J$.
– Qiaochu Yuan
Nov 9 '15 at 22:26