Are ideals also rings?
$begingroup$
I am learning about rings and ideals. But I am confused about something. My book (Gallian) says that an ideal of a ring by definition is a subring. But I have talked to other people who insist that an ideal is not a ring itself. I am confused about this. According to Wikipedia an ideal isn't necessarily a subring. But maybe it follows from the definition in Wikipedia that an ideal is a subring?
So my question is: is an ideal a ring?
abstract-algebra definition ideals
asked Apr 20 '15 at 11:54
John DoeJohn Doe
25621346
$endgroup$
add a comment |
$begingroup$
I am learning about rings and ideals. But I am confused about something. My book (Gallian) says that an ideal of a ring by definition is a subring. But I have talked to other people who insist that an ideal is not a ring itself. I am confused about this. According to Wikipedia an ideal isn't necessarily a subring. But maybe it follows from the definition in Wikipedia that an ideal is a subring?
So my question is: is an ideal a ring?
abstract-algebra definition ideals
asked Apr 20 '15 at 11:54
John DoeJohn Doe
25621346
$endgroup$
1
$begingroup$
As the answers have pointed out, it depends on whether or not your rings are required to have 1. In previous discussions of this (on MO or MSE, but which I can no longer find), two excellent discussions of how one might make this choice were posted, one by Keith Conrad (math.uconn.edu/~kconrad/blurbs/ringtheory/ringdefs.pdf) and one by Bjorn Poonen (math.mit.edu/~poonen/papers/ring.pdf).
$endgroup$
– LSpice
Apr 20 '15 at 21:49
add a comment |
$begingroup$
I am learning about rings and ideals. But I am confused about something. My book (Gallian) says that an ideal of a ring by definition is a subring. But I have talked to other people who insist that an ideal is not a ring itself. I am confused about this. According to Wikipedia an ideal isn't necessarily a subring. But maybe it follows from the definition in Wikipedia that an ideal is a subring?
So my question is: is an ideal a ring?
abstract-algebra definition ideals
asked Apr 20 '15 at 11:54
John DoeJohn Doe
25621346
$endgroup$
I am learning about rings and ideals. But I am confused about something. My book (Gallian) says that an ideal of a ring by definition is a subring. But I have talked to other people who insist that an ideal is not a ring itself. I am confused about this. According to Wikipedia an ideal isn't necessarily a subring. But maybe it follows from the definition in Wikipedia that an ideal is a subring?
So my question is: is an ideal a ring?
abstract-algebra definition ideals
abstract-algebra definition ideals
asked Apr 20 '15 at 11:54
John DoeJohn Doe
25621346
asked Apr 20 '15 at 11:54
John DoeJohn Doe
25621346
asked Apr 20 '15 at 11:54
John DoeJohn Doe
25621346
asked Apr 20 '15 at 11:54
John DoeJohn Doe
25621346
asked Apr 20 '15 at 11:54
John DoeJohn Doe
25621346
25621346
1
$begingroup$
As the answers have pointed out, it depends on whether or not your rings are required to have 1. In previous discussions of this (on MO or MSE, but which I can no longer find), two excellent discussions of how one might make this choice were posted, one by Keith Conrad (math.uconn.edu/~kconrad/blurbs/ringtheory/ringdefs.pdf) and one by Bjorn Poonen (math.mit.edu/~poonen/papers/ring.pdf).
$endgroup$
– LSpice
Apr 20 '15 at 21:49
add a comment |
1
$begingroup$
As the answers have pointed out, it depends on whether or not your rings are required to have 1. In previous discussions of this (on MO or MSE, but which I can no longer find), two excellent discussions of how one might make this choice were posted, one by Keith Conrad (math.uconn.edu/~kconrad/blurbs/ringtheory/ringdefs.pdf) and one by Bjorn Poonen (math.mit.edu/~poonen/papers/ring.pdf).
$endgroup$
– LSpice
Apr 20 '15 at 21:49
1
1
$begingroup$
As the answers have pointed out, it depends on whether or not your rings are required to have 1. In previous discussions of this (on MO or MSE, but which I can no longer find), two excellent discussions of how one might make this choice were posted, one by Keith Conrad (math.uconn.edu/~kconrad/blurbs/ringtheory/ringdefs.pdf) and one by Bjorn Poonen (math.mit.edu/~poonen/papers/ring.pdf).
$endgroup$
– LSpice
Apr 20 '15 at 21:49
$begingroup$
As the answers have pointed out, it depends on whether or not your rings are required to have 1. In previous discussions of this (on MO or MSE, but which I can no longer find), two excellent discussions of how one might make this choice were posted, one by Keith Conrad (math.uconn.edu/~kconrad/blurbs/ringtheory/ringdefs.pdf) and one by Bjorn Poonen (math.mit.edu/~poonen/papers/ring.pdf).
$endgroup$
– LSpice
Apr 20 '15 at 21:49
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The answer is both yes and no, so this will take a bit of elaboration.
There are two ways to define a ring. One of them require the existence of a $1$, the other does not.
Let's start with the one that does. In this case, a subring is required to contain the $1$ from the larger ring, and hence no proper ideal can be a subring (as any ideal containing $1$ will be the entire ring).
If we removed the requirement that the subring contained the original $1$, then the answer would be "sometimes", since the ideal might or might not have a "local" $1$ (I invite you to try some small examples of rings to find examples of either case).
If we instead take the case where a ring need not have a $1$, then it follows straight from the definitions that any ideal will also be a subring.
Here is a general way to get a proper non-trivial ideal which has a "local" $1$: Take any $xin R$ such that $x^2 = x$ (such $x$ are called idempotent) and such that $xneq 0$ and $xneq 1$ (which need not exist, but will for example exist in any ring of the form $R_1times R_2$). Then the ideal generated by $x$ will have $x$ as its "local" $1$ (I am assuming the ring to be commutative, or at least $x$ to be central here).
Conversely, if we have such an $x$ then the ring will be the direct product of the ideals generated by $x$ and $1-x$ (note that $1-x$ is also idempotent).
answered Apr 20 '15 at 12:00
Tobias KildetoftTobias Kildetoft
16.7k14274
$endgroup$
$begingroup$
Any ideal with $1$ coincides with whole ring?
$endgroup$
– Leox
Apr 20 '15 at 12:25
$begingroup$
Yes, since any $ain R$ can be written as $1a$ or $a1$, so if the ideal contains $1$ then it also contains $a$ (whether it is a left-, right- or twosided ideal).
$endgroup$
– Tobias Kildetoft
Apr 20 '15 at 12:28
1
$begingroup$
@Leox But note that this is when the $1$ is the same as for the whole ring. It is possible to have an element behave like a $1$ for a proper ideal but not for the entire ring.
$endgroup$
– Tobias Kildetoft
Apr 20 '15 at 12:29
$begingroup$
yes, but in this case any ideal of a ring be trivial one
$endgroup$
– Leox
Apr 20 '15 at 12:32
2
$begingroup$
@Leox I am not sure what you mean by trivial here. It is certainly possible to have a non-trivial ideal which is a ring with unity and the unit just happens to be different from the one of the original ring.
$endgroup$
– Tobias Kildetoft
Apr 20 '15 at 12:33
|
show 2 more comments
$begingroup$
In general, an ideal is a ring without unity - i.e. without a multiplicative identity - even if the ring it is an ideal of has unity. For example $mathbb{2Z} subset mathbb{Z}$ is an ideal but $mathbb{2Z}$ is not a ring with unity. So if you require your rings to have unity - and a lot of the time one does - then an ideal is in general not a ring. This is where the disagreement comes.
Sometimes the ideal can have a different identity (e.g., as quid points out in the comments, $mathbb{Z}^2$ has $mathbb{Z}times {0}$ as an ideal, which is a ring with unity). But the only way the ideal can have the same multiplicative identity - and so be a sub-ring-with-identity - is if it is the whole ring.
answered Apr 20 '15 at 11:58
ChristopherChristopher
6,46711628
$endgroup$
$begingroup$
Thanks for following up I think it is fine now. (I delete the original comment an will detete that one if I don't forger about it.)
$endgroup$
– quid♦
Apr 20 '15 at 12:21
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer is both yes and no, so this will take a bit of elaboration.
There are two ways to define a ring. One of them require the existence of a $1$, the other does not.
Let's start with the one that does. In this case, a subring is required to contain the $1$ from the larger ring, and hence no proper ideal can be a subring (as any ideal containing $1$ will be the entire ring).
If we removed the requirement that the subring contained the original $1$, then the answer would be "sometimes", since the ideal might or might not have a "local" $1$ (I invite you to try some small examples of rings to find examples of either case).
If we instead take the case where a ring need not have a $1$, then it follows straight from the definitions that any ideal will also be a subring.
Here is a general way to get a proper non-trivial ideal which has a "local" $1$: Take any $xin R$ such that $x^2 = x$ (such $x$ are called idempotent) and such that $xneq 0$ and $xneq 1$ (which need not exist, but will for example exist in any ring of the form $R_1times R_2$). Then the ideal generated by $x$ will have $x$ as its "local" $1$ (I am assuming the ring to be commutative, or at least $x$ to be central here).
Conversely, if we have such an $x$ then the ring will be the direct product of the ideals generated by $x$ and $1-x$ (note that $1-x$ is also idempotent).
answered Apr 20 '15 at 12:00
Tobias KildetoftTobias Kildetoft
16.7k14274
$endgroup$
$begingroup$
Any ideal with $1$ coincides with whole ring?
$endgroup$
– Leox
Apr 20 '15 at 12:25
$begingroup$
Yes, since any $ain R$ can be written as $1a$ or $a1$, so if the ideal contains $1$ then it also contains $a$ (whether it is a left-, right- or twosided ideal).
$endgroup$
– Tobias Kildetoft
Apr 20 '15 at 12:28
1
$begingroup$
@Leox But note that this is when the $1$ is the same as for the whole ring. It is possible to have an element behave like a $1$ for a proper ideal but not for the entire ring.
$endgroup$
– Tobias Kildetoft
Apr 20 '15 at 12:29
$begingroup$
yes, but in this case any ideal of a ring be trivial one
$endgroup$
– Leox
Apr 20 '15 at 12:32
2
$begingroup$
@Leox I am not sure what you mean by trivial here. It is certainly possible to have a non-trivial ideal which is a ring with unity and the unit just happens to be different from the one of the original ring.
$endgroup$
– Tobias Kildetoft
Apr 20 '15 at 12:33
|
show 2 more comments
$begingroup$
The answer is both yes and no, so this will take a bit of elaboration.
There are two ways to define a ring. One of them require the existence of a $1$, the other does not.
Let's start with the one that does. In this case, a subring is required to contain the $1$ from the larger ring, and hence no proper ideal can be a subring (as any ideal containing $1$ will be the entire ring).
If we removed the requirement that the subring contained the original $1$, then the answer would be "sometimes", since the ideal might or might not have a "local" $1$ (I invite you to try some small examples of rings to find examples of either case).
If we instead take the case where a ring need not have a $1$, then it follows straight from the definitions that any ideal will also be a subring.
Here is a general way to get a proper non-trivial ideal which has a "local" $1$: Take any $xin R$ such that $x^2 = x$ (such $x$ are called idempotent) and such that $xneq 0$ and $xneq 1$ (which need not exist, but will for example exist in any ring of the form $R_1times R_2$). Then the ideal generated by $x$ will have $x$ as its "local" $1$ (I am assuming the ring to be commutative, or at least $x$ to be central here).
Conversely, if we have such an $x$ then the ring will be the direct product of the ideals generated by $x$ and $1-x$ (note that $1-x$ is also idempotent).
answered Apr 20 '15 at 12:00
Tobias KildetoftTobias Kildetoft
16.7k14274
$endgroup$
$begingroup$
Any ideal with $1$ coincides with whole ring?
$endgroup$
– Leox
Apr 20 '15 at 12:25
$begingroup$
Yes, since any $ain R$ can be written as $1a$ or $a1$, so if the ideal contains $1$ then it also contains $a$ (whether it is a left-, right- or twosided ideal).
$endgroup$
– Tobias Kildetoft
Apr 20 '15 at 12:28
1
$begingroup$
@Leox But note that this is when the $1$ is the same as for the whole ring. It is possible to have an element behave like a $1$ for a proper ideal but not for the entire ring.
$endgroup$
– Tobias Kildetoft
Apr 20 '15 at 12:29
$begingroup$
yes, but in this case any ideal of a ring be trivial one
$endgroup$
– Leox
Apr 20 '15 at 12:32
2
$begingroup$
@Leox I am not sure what you mean by trivial here. It is certainly possible to have a non-trivial ideal which is a ring with unity and the unit just happens to be different from the one of the original ring.
$endgroup$
– Tobias Kildetoft
Apr 20 '15 at 12:33
|
show 2 more comments
$begingroup$
The answer is both yes and no, so this will take a bit of elaboration.
There are two ways to define a ring. One of them require the existence of a $1$, the other does not.
Let's start with the one that does. In this case, a subring is required to contain the $1$ from the larger ring, and hence no proper ideal can be a subring (as any ideal containing $1$ will be the entire ring).
If we removed the requirement that the subring contained the original $1$, then the answer would be "sometimes", since the ideal might or might not have a "local" $1$ (I invite you to try some small examples of rings to find examples of either case).
If we instead take the case where a ring need not have a $1$, then it follows straight from the definitions that any ideal will also be a subring.
Here is a general way to get a proper non-trivial ideal which has a "local" $1$: Take any $xin R$ such that $x^2 = x$ (such $x$ are called idempotent) and such that $xneq 0$ and $xneq 1$ (which need not exist, but will for example exist in any ring of the form $R_1times R_2$). Then the ideal generated by $x$ will have $x$ as its "local" $1$ (I am assuming the ring to be commutative, or at least $x$ to be central here).
Conversely, if we have such an $x$ then the ring will be the direct product of the ideals generated by $x$ and $1-x$ (note that $1-x$ is also idempotent).
answered Apr 20 '15 at 12:00
Tobias KildetoftTobias Kildetoft
16.7k14274
$endgroup$
The answer is both yes and no, so this will take a bit of elaboration.
There are two ways to define a ring. One of them require the existence of a $1$, the other does not.
Let's start with the one that does. In this case, a subring is required to contain the $1$ from the larger ring, and hence no proper ideal can be a subring (as any ideal containing $1$ will be the entire ring).
If we removed the requirement that the subring contained the original $1$, then the answer would be "sometimes", since the ideal might or might not have a "local" $1$ (I invite you to try some small examples of rings to find examples of either case).
If we instead take the case where a ring need not have a $1$, then it follows straight from the definitions that any ideal will also be a subring.
Here is a general way to get a proper non-trivial ideal which has a "local" $1$: Take any $xin R$ such that $x^2 = x$ (such $x$ are called idempotent) and such that $xneq 0$ and $xneq 1$ (which need not exist, but will for example exist in any ring of the form $R_1times R_2$). Then the ideal generated by $x$ will have $x$ as its "local" $1$ (I am assuming the ring to be commutative, or at least $x$ to be central here).
Conversely, if we have such an $x$ then the ring will be the direct product of the ideals generated by $x$ and $1-x$ (note that $1-x$ is also idempotent).
answered Apr 20 '15 at 12:00
Tobias KildetoftTobias Kildetoft
16.7k14274
edited Apr 20 '15 at 12:56
answered Apr 20 '15 at 12:00
Tobias KildetoftTobias Kildetoft
16.7k14274
answered Apr 20 '15 at 12:00
Tobias KildetoftTobias Kildetoft
16.7k14274
answered Apr 20 '15 at 12:00
Tobias KildetoftTobias Kildetoft
16.7k14274
16.7k14274
$begingroup$
Any ideal with $1$ coincides with whole ring?
$endgroup$
– Leox
Apr 20 '15 at 12:25
$begingroup$
Yes, since any $ain R$ can be written as $1a$ or $a1$, so if the ideal contains $1$ then it also contains $a$ (whether it is a left-, right- or twosided ideal).
$endgroup$
– Tobias Kildetoft
Apr 20 '15 at 12:28
1
$begingroup$
@Leox But note that this is when the $1$ is the same as for the whole ring. It is possible to have an element behave like a $1$ for a proper ideal but not for the entire ring.
$endgroup$
– Tobias Kildetoft
Apr 20 '15 at 12:29
$begingroup$
yes, but in this case any ideal of a ring be trivial one
$endgroup$
– Leox
Apr 20 '15 at 12:32
2
$begingroup$
@Leox I am not sure what you mean by trivial here. It is certainly possible to have a non-trivial ideal which is a ring with unity and the unit just happens to be different from the one of the original ring.
$endgroup$
– Tobias Kildetoft
Apr 20 '15 at 12:33
|
show 2 more comments
$begingroup$
Any ideal with $1$ coincides with whole ring?
$endgroup$
– Leox
Apr 20 '15 at 12:25
$begingroup$
Yes, since any $ain R$ can be written as $1a$ or $a1$, so if the ideal contains $1$ then it also contains $a$ (whether it is a left-, right- or twosided ideal).
$endgroup$
– Tobias Kildetoft
Apr 20 '15 at 12:28
1
$begingroup$
@Leox But note that this is when the $1$ is the same as for the whole ring. It is possible to have an element behave like a $1$ for a proper ideal but not for the entire ring.
$endgroup$
– Tobias Kildetoft
Apr 20 '15 at 12:29
$begingroup$
yes, but in this case any ideal of a ring be trivial one
$endgroup$
– Leox
Apr 20 '15 at 12:32
2
$begingroup$
@Leox I am not sure what you mean by trivial here. It is certainly possible to have a non-trivial ideal which is a ring with unity and the unit just happens to be different from the one of the original ring.
$endgroup$
– Tobias Kildetoft
Apr 20 '15 at 12:33
$begingroup$
Any ideal with $1$ coincides with whole ring?
$endgroup$
– Leox
Apr 20 '15 at 12:25
$begingroup$
Any ideal with $1$ coincides with whole ring?
$endgroup$
– Leox
Apr 20 '15 at 12:25
$begingroup$
Yes, since any $ain R$ can be written as $1a$ or $a1$, so if the ideal contains $1$ then it also contains $a$ (whether it is a left-, right- or twosided ideal).
$endgroup$
– Tobias Kildetoft
Apr 20 '15 at 12:28
$begingroup$
Yes, since any $ain R$ can be written as $1a$ or $a1$, so if the ideal contains $1$ then it also contains $a$ (whether it is a left-, right- or twosided ideal).
$endgroup$
– Tobias Kildetoft
Apr 20 '15 at 12:28
1
1
$begingroup$
@Leox But note that this is when the $1$ is the same as for the whole ring. It is possible to have an element behave like a $1$ for a proper ideal but not for the entire ring.
$endgroup$
– Tobias Kildetoft
Apr 20 '15 at 12:29
$begingroup$
@Leox But note that this is when the $1$ is the same as for the whole ring. It is possible to have an element behave like a $1$ for a proper ideal but not for the entire ring.
$endgroup$
– Tobias Kildetoft
Apr 20 '15 at 12:29
$begingroup$
yes, but in this case any ideal of a ring be trivial one
$endgroup$
– Leox
Apr 20 '15 at 12:32
$begingroup$
yes, but in this case any ideal of a ring be trivial one
$endgroup$
– Leox
Apr 20 '15 at 12:32
2
2
$begingroup$
@Leox I am not sure what you mean by trivial here. It is certainly possible to have a non-trivial ideal which is a ring with unity and the unit just happens to be different from the one of the original ring.
$endgroup$
– Tobias Kildetoft
Apr 20 '15 at 12:33
$begingroup$
@Leox I am not sure what you mean by trivial here. It is certainly possible to have a non-trivial ideal which is a ring with unity and the unit just happens to be different from the one of the original ring.
$endgroup$
– Tobias Kildetoft
Apr 20 '15 at 12:33
|
show 2 more comments
$begingroup$
In general, an ideal is a ring without unity - i.e. without a multiplicative identity - even if the ring it is an ideal of has unity. For example $mathbb{2Z} subset mathbb{Z}$ is an ideal but $mathbb{2Z}$ is not a ring with unity. So if you require your rings to have unity - and a lot of the time one does - then an ideal is in general not a ring. This is where the disagreement comes.
Sometimes the ideal can have a different identity (e.g., as quid points out in the comments, $mathbb{Z}^2$ has $mathbb{Z}times {0}$ as an ideal, which is a ring with unity). But the only way the ideal can have the same multiplicative identity - and so be a sub-ring-with-identity - is if it is the whole ring.
answered Apr 20 '15 at 11:58
ChristopherChristopher
6,46711628
$endgroup$
$begingroup$
Thanks for following up I think it is fine now. (I delete the original comment an will detete that one if I don't forger about it.)
$endgroup$
– quid♦
Apr 20 '15 at 12:21
add a comment |
$begingroup$
In general, an ideal is a ring without unity - i.e. without a multiplicative identity - even if the ring it is an ideal of has unity. For example $mathbb{2Z} subset mathbb{Z}$ is an ideal but $mathbb{2Z}$ is not a ring with unity. So if you require your rings to have unity - and a lot of the time one does - then an ideal is in general not a ring. This is where the disagreement comes.
Sometimes the ideal can have a different identity (e.g., as quid points out in the comments, $mathbb{Z}^2$ has $mathbb{Z}times {0}$ as an ideal, which is a ring with unity). But the only way the ideal can have the same multiplicative identity - and so be a sub-ring-with-identity - is if it is the whole ring.
answered Apr 20 '15 at 11:58
ChristopherChristopher
6,46711628
$endgroup$
$begingroup$
Thanks for following up I think it is fine now. (I delete the original comment an will detete that one if I don't forger about it.)
$endgroup$
– quid♦
Apr 20 '15 at 12:21
add a comment |
$begingroup$
In general, an ideal is a ring without unity - i.e. without a multiplicative identity - even if the ring it is an ideal of has unity. For example $mathbb{2Z} subset mathbb{Z}$ is an ideal but $mathbb{2Z}$ is not a ring with unity. So if you require your rings to have unity - and a lot of the time one does - then an ideal is in general not a ring. This is where the disagreement comes.
Sometimes the ideal can have a different identity (e.g., as quid points out in the comments, $mathbb{Z}^2$ has $mathbb{Z}times {0}$ as an ideal, which is a ring with unity). But the only way the ideal can have the same multiplicative identity - and so be a sub-ring-with-identity - is if it is the whole ring.
answered Apr 20 '15 at 11:58
ChristopherChristopher
6,46711628
$endgroup$
In general, an ideal is a ring without unity - i.e. without a multiplicative identity - even if the ring it is an ideal of has unity. For example $mathbb{2Z} subset mathbb{Z}$ is an ideal but $mathbb{2Z}$ is not a ring with unity. So if you require your rings to have unity - and a lot of the time one does - then an ideal is in general not a ring. This is where the disagreement comes.
Sometimes the ideal can have a different identity (e.g., as quid points out in the comments, $mathbb{Z}^2$ has $mathbb{Z}times {0}$ as an ideal, which is a ring with unity). But the only way the ideal can have the same multiplicative identity - and so be a sub-ring-with-identity - is if it is the whole ring.
answered Apr 20 '15 at 11:58
ChristopherChristopher
6,46711628
edited Apr 20 '15 at 12:15
answered Apr 20 '15 at 11:58
ChristopherChristopher
6,46711628
answered Apr 20 '15 at 11:58
ChristopherChristopher
6,46711628
answered Apr 20 '15 at 11:58
ChristopherChristopher
6,46711628
6,46711628
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Thanks for following up I think it is fine now. (I delete the original comment an will detete that one if I don't forger about it.)
$endgroup$
– quid♦
Apr 20 '15 at 12:21
add a comment |
$begingroup$
Thanks for following up I think it is fine now. (I delete the original comment an will detete that one if I don't forger about it.)
$endgroup$
– quid♦
Apr 20 '15 at 12:21
$begingroup$
Thanks for following up I think it is fine now. (I delete the original comment an will detete that one if I don't forger about it.)
$endgroup$
– quid♦
Apr 20 '15 at 12:21
$begingroup$
Thanks for following up I think it is fine now. (I delete the original comment an will detete that one if I don't forger about it.)
$endgroup$
– quid♦
Apr 20 '15 at 12:21
add a comment |
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As the answers have pointed out, it depends on whether or not your rings are required to have 1. In previous discussions of this (on MO or MSE, but which I can no longer find), two excellent discussions of how one might make this choice were posted, one by Keith Conrad (math.uconn.edu/~kconrad/blurbs/ringtheory/ringdefs.pdf) and one by Bjorn Poonen (math.mit.edu/~poonen/papers/ring.pdf).
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– LSpice
Apr 20 '15 at 21:49