Why are ideals specifically additive subgroups?
$begingroup$
I'm trying to understand the notion of ideals better, and something I can't quite figure out is why and where the additive portion of the ideal definition comes into use. If I have an ideal $I$ of $(R,+,cdot)$, why can't $I$ also be a multiplicative subgroup? Or if it can but is rarely used, why is this the case?
abstract-algebra ring-theory ideals
asked Nov 28 '18 at 5:47
rb612rb612
1,848923
$endgroup$
add a comment |
$begingroup$
I'm trying to understand the notion of ideals better, and something I can't quite figure out is why and where the additive portion of the ideal definition comes into use. If I have an ideal $I$ of $(R,+,cdot)$, why can't $I$ also be a multiplicative subgroup? Or if it can but is rarely used, why is this the case?
abstract-algebra ring-theory ideals
asked Nov 28 '18 at 5:47
rb612rb612
1,848923
$endgroup$
$begingroup$
Note that in a ring, $(R,cdot)$ is not necessarily a multiplicative group, since it may not have inverses (for example, $(mathbb{Z}, cdot)$ is not a multiplicative group). So taking ideals to be additive subgroups makes them more widely applicable, since $(R,+)$ is always a group.
$endgroup$
– platty
Nov 28 '18 at 5:51
add a comment |
$begingroup$
I'm trying to understand the notion of ideals better, and something I can't quite figure out is why and where the additive portion of the ideal definition comes into use. If I have an ideal $I$ of $(R,+,cdot)$, why can't $I$ also be a multiplicative subgroup? Or if it can but is rarely used, why is this the case?
abstract-algebra ring-theory ideals
asked Nov 28 '18 at 5:47
rb612rb612
1,848923
$endgroup$
I'm trying to understand the notion of ideals better, and something I can't quite figure out is why and where the additive portion of the ideal definition comes into use. If I have an ideal $I$ of $(R,+,cdot)$, why can't $I$ also be a multiplicative subgroup? Or if it can but is rarely used, why is this the case?
abstract-algebra ring-theory ideals
abstract-algebra ring-theory ideals
asked Nov 28 '18 at 5:47
rb612rb612
1,848923
asked Nov 28 '18 at 5:47
rb612rb612
1,848923
asked Nov 28 '18 at 5:47
rb612rb612
1,848923
asked Nov 28 '18 at 5:47
rb612rb612
1,848923
asked Nov 28 '18 at 5:47
rb612rb612
1,848923
1,848923
$begingroup$
Note that in a ring, $(R,cdot)$ is not necessarily a multiplicative group, since it may not have inverses (for example, $(mathbb{Z}, cdot)$ is not a multiplicative group). So taking ideals to be additive subgroups makes them more widely applicable, since $(R,+)$ is always a group.
$endgroup$
– platty
Nov 28 '18 at 5:51
add a comment |
$begingroup$
Note that in a ring, $(R,cdot)$ is not necessarily a multiplicative group, since it may not have inverses (for example, $(mathbb{Z}, cdot)$ is not a multiplicative group). So taking ideals to be additive subgroups makes them more widely applicable, since $(R,+)$ is always a group.
$endgroup$
– platty
Nov 28 '18 at 5:51
$begingroup$
Note that in a ring, $(R,cdot)$ is not necessarily a multiplicative group, since it may not have inverses (for example, $(mathbb{Z}, cdot)$ is not a multiplicative group). So taking ideals to be additive subgroups makes them more widely applicable, since $(R,+)$ is always a group.
$endgroup$
– platty
Nov 28 '18 at 5:51
$begingroup$
Note that in a ring, $(R,cdot)$ is not necessarily a multiplicative group, since it may not have inverses (for example, $(mathbb{Z}, cdot)$ is not a multiplicative group). So taking ideals to be additive subgroups makes them more widely applicable, since $(R,+)$ is always a group.
$endgroup$
– platty
Nov 28 '18 at 5:51
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
One way to think about ideals is as "things you can quotient by to still get a ring", or "things that can be kernels of ring homomorphisms $R to S$". You can check that these two "definitions" are equivalent to each other, and those are what ideals are. But sometimes it's convenient to write down a more symbolic definition of an ideal in terms of the ring operations in $R$, and that's the "standard" definition you're likely referencing.
answered Nov 28 '18 at 5:49
user263190user263190
803410
$endgroup$
add a comment |
$begingroup$
If it were a multiplicative subgroup it would have $1$ in it. But an ideal having $1$ as a member is the whole ring. That's sometimes called a "trivial" ideal, not much use in working with them.
answered Nov 28 '18 at 5:50
coffeemathcoffeemath
2,7501415
$endgroup$
add a comment |
$begingroup$
Keep in mind that $R$ is not a group under multiplication - for one thing, $0$ is never invertible, and even if you exclude 0 it is not always the case that $R setminus {0}$ is a group (such a ring is called a division ring). When $R$ has a multiplicative identity, the set of all elements of $R$ that are invertible is called the group of units, as an invertible element in a ring is called a unit. Now you can't expect an ideal to be a subgroup of the group of units, since an ideal that contains any unit at all must actually contain the entire ring! So you can't expect the ideal to be a multiplicative "subgroup" in any sense. However, it is true that an ideal is always a subring, in the sense that is an additive subgroup that is closed under multiplication.
answered Nov 28 '18 at 5:58
Monstrous MoonshinerMonstrous Moonshiner
2,25011337
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
One way to think about ideals is as "things you can quotient by to still get a ring", or "things that can be kernels of ring homomorphisms $R to S$". You can check that these two "definitions" are equivalent to each other, and those are what ideals are. But sometimes it's convenient to write down a more symbolic definition of an ideal in terms of the ring operations in $R$, and that's the "standard" definition you're likely referencing.
answered Nov 28 '18 at 5:49
user263190user263190
803410
$endgroup$
add a comment |
$begingroup$
One way to think about ideals is as "things you can quotient by to still get a ring", or "things that can be kernels of ring homomorphisms $R to S$". You can check that these two "definitions" are equivalent to each other, and those are what ideals are. But sometimes it's convenient to write down a more symbolic definition of an ideal in terms of the ring operations in $R$, and that's the "standard" definition you're likely referencing.
answered Nov 28 '18 at 5:49
user263190user263190
803410
$endgroup$
add a comment |
$begingroup$
One way to think about ideals is as "things you can quotient by to still get a ring", or "things that can be kernels of ring homomorphisms $R to S$". You can check that these two "definitions" are equivalent to each other, and those are what ideals are. But sometimes it's convenient to write down a more symbolic definition of an ideal in terms of the ring operations in $R$, and that's the "standard" definition you're likely referencing.
answered Nov 28 '18 at 5:49
user263190user263190
803410
$endgroup$
One way to think about ideals is as "things you can quotient by to still get a ring", or "things that can be kernels of ring homomorphisms $R to S$". You can check that these two "definitions" are equivalent to each other, and those are what ideals are. But sometimes it's convenient to write down a more symbolic definition of an ideal in terms of the ring operations in $R$, and that's the "standard" definition you're likely referencing.
answered Nov 28 '18 at 5:49
user263190user263190
803410
answered Nov 28 '18 at 5:49
user263190user263190
803410
answered Nov 28 '18 at 5:49
user263190user263190
803410
answered Nov 28 '18 at 5:49
user263190user263190
803410
803410
add a comment |
add a comment |
$begingroup$
If it were a multiplicative subgroup it would have $1$ in it. But an ideal having $1$ as a member is the whole ring. That's sometimes called a "trivial" ideal, not much use in working with them.
answered Nov 28 '18 at 5:50
coffeemathcoffeemath
2,7501415
$endgroup$
add a comment |
$begingroup$
If it were a multiplicative subgroup it would have $1$ in it. But an ideal having $1$ as a member is the whole ring. That's sometimes called a "trivial" ideal, not much use in working with them.
answered Nov 28 '18 at 5:50
coffeemathcoffeemath
2,7501415
$endgroup$
add a comment |
$begingroup$
If it were a multiplicative subgroup it would have $1$ in it. But an ideal having $1$ as a member is the whole ring. That's sometimes called a "trivial" ideal, not much use in working with them.
answered Nov 28 '18 at 5:50
coffeemathcoffeemath
2,7501415
$endgroup$
If it were a multiplicative subgroup it would have $1$ in it. But an ideal having $1$ as a member is the whole ring. That's sometimes called a "trivial" ideal, not much use in working with them.
answered Nov 28 '18 at 5:50
coffeemathcoffeemath
2,7501415
answered Nov 28 '18 at 5:50
coffeemathcoffeemath
2,7501415
answered Nov 28 '18 at 5:50
coffeemathcoffeemath
2,7501415
answered Nov 28 '18 at 5:50
coffeemathcoffeemath
2,7501415
2,7501415
add a comment |
add a comment |
$begingroup$
Keep in mind that $R$ is not a group under multiplication - for one thing, $0$ is never invertible, and even if you exclude 0 it is not always the case that $R setminus {0}$ is a group (such a ring is called a division ring). When $R$ has a multiplicative identity, the set of all elements of $R$ that are invertible is called the group of units, as an invertible element in a ring is called a unit. Now you can't expect an ideal to be a subgroup of the group of units, since an ideal that contains any unit at all must actually contain the entire ring! So you can't expect the ideal to be a multiplicative "subgroup" in any sense. However, it is true that an ideal is always a subring, in the sense that is an additive subgroup that is closed under multiplication.
answered Nov 28 '18 at 5:58
Monstrous MoonshinerMonstrous Moonshiner
2,25011337
$endgroup$
add a comment |
$begingroup$
Keep in mind that $R$ is not a group under multiplication - for one thing, $0$ is never invertible, and even if you exclude 0 it is not always the case that $R setminus {0}$ is a group (such a ring is called a division ring). When $R$ has a multiplicative identity, the set of all elements of $R$ that are invertible is called the group of units, as an invertible element in a ring is called a unit. Now you can't expect an ideal to be a subgroup of the group of units, since an ideal that contains any unit at all must actually contain the entire ring! So you can't expect the ideal to be a multiplicative "subgroup" in any sense. However, it is true that an ideal is always a subring, in the sense that is an additive subgroup that is closed under multiplication.
answered Nov 28 '18 at 5:58
Monstrous MoonshinerMonstrous Moonshiner
2,25011337
$endgroup$
add a comment |
$begingroup$
Keep in mind that $R$ is not a group under multiplication - for one thing, $0$ is never invertible, and even if you exclude 0 it is not always the case that $R setminus {0}$ is a group (such a ring is called a division ring). When $R$ has a multiplicative identity, the set of all elements of $R$ that are invertible is called the group of units, as an invertible element in a ring is called a unit. Now you can't expect an ideal to be a subgroup of the group of units, since an ideal that contains any unit at all must actually contain the entire ring! So you can't expect the ideal to be a multiplicative "subgroup" in any sense. However, it is true that an ideal is always a subring, in the sense that is an additive subgroup that is closed under multiplication.
answered Nov 28 '18 at 5:58
Monstrous MoonshinerMonstrous Moonshiner
2,25011337
$endgroup$
Keep in mind that $R$ is not a group under multiplication - for one thing, $0$ is never invertible, and even if you exclude 0 it is not always the case that $R setminus {0}$ is a group (such a ring is called a division ring). When $R$ has a multiplicative identity, the set of all elements of $R$ that are invertible is called the group of units, as an invertible element in a ring is called a unit. Now you can't expect an ideal to be a subgroup of the group of units, since an ideal that contains any unit at all must actually contain the entire ring! So you can't expect the ideal to be a multiplicative "subgroup" in any sense. However, it is true that an ideal is always a subring, in the sense that is an additive subgroup that is closed under multiplication.
answered Nov 28 '18 at 5:58
Monstrous MoonshinerMonstrous Moonshiner
2,25011337
answered Nov 28 '18 at 5:58
Monstrous MoonshinerMonstrous Moonshiner
2,25011337
answered Nov 28 '18 at 5:58
Monstrous MoonshinerMonstrous Moonshiner
2,25011337
answered Nov 28 '18 at 5:58
Monstrous MoonshinerMonstrous Moonshiner
2,25011337
2,25011337
add a comment |
add a comment |
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$begingroup$
Note that in a ring, $(R,cdot)$ is not necessarily a multiplicative group, since it may not have inverses (for example, $(mathbb{Z}, cdot)$ is not a multiplicative group). So taking ideals to be additive subgroups makes them more widely applicable, since $(R,+)$ is always a group.
$endgroup$
– platty
Nov 28 '18 at 5:51