must a ring (commutative, with 1), in which every non-zero ideal is prime, be a field?Non-Noetherian rings with an ideal not containing a product of prime idealsExistence of minimal non-zero prime ideals: Counter examples?Where does the proof for commutative rings break down in the non-commutative ring when showing only two ideals implies the ring is a field?Does this characterize rings in which every ideal is principal?Can we usefully characterize those rings whose every non-zero prime ideal is maximal?Is there an adjective for rings whose every non-zero prime ideal is maximal?Does there exist a non-PIR in which every countably generated prime ideal is principal?Commutative rings with unity over which every non-zero module has an associated primeCommutative, infinite, Noetherian ring with no non-trivial nilpotent and no non-trivial idempotentExample of a principal prime ideal containing a proper prime non-zero ideal.
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must a ring (commutative, with 1), in which every non-zero ideal is prime, be a field?
Non-Noetherian rings with an ideal not containing a product of prime idealsExistence of minimal non-zero prime ideals: Counter examples?Where does the proof for commutative rings break down in the non-commutative ring when showing only two ideals implies the ring is a field?Does this characterize rings in which every ideal is principal?Can we usefully characterize those rings whose every non-zero prime ideal is maximal?Is there an adjective for rings whose every non-zero prime ideal is maximal?Does there exist a non-PIR in which every countably generated prime ideal is principal?Commutative rings with unity over which every non-zero module has an associated primeCommutative, infinite, Noetherian ring with no non-trivial nilpotent and no non-trivial idempotentExample of a principal prime ideal containing a proper prime non-zero ideal.
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
An early exercise in Irving Kaplansky's commutative rings asks:
Let R be a ring. Suppose that every ideal in R (other than R) is prime. Prove that R is a field.
This is easy if we assume the zero ideal is prime. But is this assumption necessary?
If every non-zero ideal is prime, then for any non-unit $x in R$ and with $x^n ne 0$ we must have $langle x rangle subseteq langle x^n rangle$, which requires the existence of an element $y$ satisfying:
$$
x(1-x^ny) = 0
$$
The collection of these and similar relations on the elements seems rather restrictive, but i would appreciate a simple and incisive argument to show that the condition that all non-zero ideals are prime can only be met by rings with trivial spectrum, or, if my guess is incorrect and this is untrue, a counter-example.
abstract-algebra ring-theory commutative-algebra
asked 8 hours ago
David HoldenDavid Holden
15.4k2 gold badges13 silver badges27 bronze badges
$endgroup$
add a comment |
$begingroup$
An early exercise in Irving Kaplansky's commutative rings asks:
Let R be a ring. Suppose that every ideal in R (other than R) is prime. Prove that R is a field.
This is easy if we assume the zero ideal is prime. But is this assumption necessary?
If every non-zero ideal is prime, then for any non-unit $x in R$ and with $x^n ne 0$ we must have $langle x rangle subseteq langle x^n rangle$, which requires the existence of an element $y$ satisfying:
$$
x(1-x^ny) = 0
$$
The collection of these and similar relations on the elements seems rather restrictive, but i would appreciate a simple and incisive argument to show that the condition that all non-zero ideals are prime can only be met by rings with trivial spectrum, or, if my guess is incorrect and this is untrue, a counter-example.
abstract-algebra ring-theory commutative-algebra
asked 8 hours ago
David HoldenDavid Holden
15.4k2 gold badges13 silver badges27 bronze badges
$endgroup$
$begingroup$
Note that Kaplansky's original statement is not quite right, since you also have to require that $R$ is nonzero. (If $R$ is the zero ring, then there are no ideals other than $R$ so the condition is vacuously true.)
$endgroup$
– Eric Wofsey
8 hours ago
$begingroup$
@EricWofsey: I think your point is that fields are usually required to have $1 neq 0$, hence (somewhat unusually by comparison with groups and rings and modules and ...) the trivial structure with just one element isn't a field. If Kaplansky allows $1 = 0$ in a field as some authors do, then his statement is right.
$endgroup$
– Rob Arthan
7 hours ago
1
$begingroup$
I have never seen any author that allows $1=0$ in a field.
$endgroup$
– Eric Wofsey
7 hours ago
$begingroup$
I agree that it is usual to require $1 neq 0$, but Weber's original definition allowed it: he says explicitly that $0$ is distinct from $1$ except in the uninteresting case when the field has only one element. See gdz.sub.uni-goettingen.de/id/PPN235181684_0043?tify page 527. (Please forgive me for going back to the 19th century, but I've become a fan of going back to original sources in mathematics recently.)
$endgroup$
– Rob Arthan
7 hours ago
$begingroup$
... and in modern definitions the requirement that $1 neq 0$ is often sneaked in surreptitiously by saying that that then non-zero elements of the field form a group under multiplication (which implies there is at leat one non-zero element, because the traditional definition of a group requires a group to have a non-empty universe).
$endgroup$
– Rob Arthan
6 hours ago
add a comment |
$begingroup$
An early exercise in Irving Kaplansky's commutative rings asks:
Let R be a ring. Suppose that every ideal in R (other than R) is prime. Prove that R is a field.
This is easy if we assume the zero ideal is prime. But is this assumption necessary?
If every non-zero ideal is prime, then for any non-unit $x in R$ and with $x^n ne 0$ we must have $langle x rangle subseteq langle x^n rangle$, which requires the existence of an element $y$ satisfying:
$$
x(1-x^ny) = 0
$$
The collection of these and similar relations on the elements seems rather restrictive, but i would appreciate a simple and incisive argument to show that the condition that all non-zero ideals are prime can only be met by rings with trivial spectrum, or, if my guess is incorrect and this is untrue, a counter-example.
abstract-algebra ring-theory commutative-algebra
asked 8 hours ago
David HoldenDavid Holden
15.4k2 gold badges13 silver badges27 bronze badges
$endgroup$
An early exercise in Irving Kaplansky's commutative rings asks:
Let R be a ring. Suppose that every ideal in R (other than R) is prime. Prove that R is a field.
This is easy if we assume the zero ideal is prime. But is this assumption necessary?
If every non-zero ideal is prime, then for any non-unit $x in R$ and with $x^n ne 0$ we must have $langle x rangle subseteq langle x^n rangle$, which requires the existence of an element $y$ satisfying:
$$
x(1-x^ny) = 0
$$
The collection of these and similar relations on the elements seems rather restrictive, but i would appreciate a simple and incisive argument to show that the condition that all non-zero ideals are prime can only be met by rings with trivial spectrum, or, if my guess is incorrect and this is untrue, a counter-example.
abstract-algebra ring-theory commutative-algebra
abstract-algebra ring-theory commutative-algebra
asked 8 hours ago
David HoldenDavid Holden
15.4k2 gold badges13 silver badges27 bronze badges
asked 8 hours ago
David HoldenDavid Holden
15.4k2 gold badges13 silver badges27 bronze badges
asked 8 hours ago
David HoldenDavid Holden
15.4k2 gold badges13 silver badges27 bronze badges
asked 8 hours ago
David HoldenDavid Holden
15.4k2 gold badges13 silver badges27 bronze badges
asked 8 hours ago
David HoldenDavid Holden
15.4k2 gold badges13 silver badges27 bronze badges
15.4k2 gold badges13 silver badges27 bronze badges
$begingroup$
Note that Kaplansky's original statement is not quite right, since you also have to require that $R$ is nonzero. (If $R$ is the zero ring, then there are no ideals other than $R$ so the condition is vacuously true.)
$endgroup$
– Eric Wofsey
8 hours ago
$begingroup$
@EricWofsey: I think your point is that fields are usually required to have $1 neq 0$, hence (somewhat unusually by comparison with groups and rings and modules and ...) the trivial structure with just one element isn't a field. If Kaplansky allows $1 = 0$ in a field as some authors do, then his statement is right.
$endgroup$
– Rob Arthan
7 hours ago
1
$begingroup$
I have never seen any author that allows $1=0$ in a field.
$endgroup$
– Eric Wofsey
7 hours ago
$begingroup$
I agree that it is usual to require $1 neq 0$, but Weber's original definition allowed it: he says explicitly that $0$ is distinct from $1$ except in the uninteresting case when the field has only one element. See gdz.sub.uni-goettingen.de/id/PPN235181684_0043?tify page 527. (Please forgive me for going back to the 19th century, but I've become a fan of going back to original sources in mathematics recently.)
$endgroup$
– Rob Arthan
7 hours ago
$begingroup$
... and in modern definitions the requirement that $1 neq 0$ is often sneaked in surreptitiously by saying that that then non-zero elements of the field form a group under multiplication (which implies there is at leat one non-zero element, because the traditional definition of a group requires a group to have a non-empty universe).
$endgroup$
– Rob Arthan
6 hours ago
add a comment |
$begingroup$
Note that Kaplansky's original statement is not quite right, since you also have to require that $R$ is nonzero. (If $R$ is the zero ring, then there are no ideals other than $R$ so the condition is vacuously true.)
$endgroup$
– Eric Wofsey
8 hours ago
$begingroup$
@EricWofsey: I think your point is that fields are usually required to have $1 neq 0$, hence (somewhat unusually by comparison with groups and rings and modules and ...) the trivial structure with just one element isn't a field. If Kaplansky allows $1 = 0$ in a field as some authors do, then his statement is right.
$endgroup$
– Rob Arthan
7 hours ago
1
$begingroup$
I have never seen any author that allows $1=0$ in a field.
$endgroup$
– Eric Wofsey
7 hours ago
$begingroup$
I agree that it is usual to require $1 neq 0$, but Weber's original definition allowed it: he says explicitly that $0$ is distinct from $1$ except in the uninteresting case when the field has only one element. See gdz.sub.uni-goettingen.de/id/PPN235181684_0043?tify page 527. (Please forgive me for going back to the 19th century, but I've become a fan of going back to original sources in mathematics recently.)
$endgroup$
– Rob Arthan
7 hours ago
$begingroup$
... and in modern definitions the requirement that $1 neq 0$ is often sneaked in surreptitiously by saying that that then non-zero elements of the field form a group under multiplication (which implies there is at leat one non-zero element, because the traditional definition of a group requires a group to have a non-empty universe).
$endgroup$
– Rob Arthan
6 hours ago
$begingroup$
Note that Kaplansky's original statement is not quite right, since you also have to require that $R$ is nonzero. (If $R$ is the zero ring, then there are no ideals other than $R$ so the condition is vacuously true.)
$endgroup$
– Eric Wofsey
8 hours ago
$begingroup$
Note that Kaplansky's original statement is not quite right, since you also have to require that $R$ is nonzero. (If $R$ is the zero ring, then there are no ideals other than $R$ so the condition is vacuously true.)
$endgroup$
– Eric Wofsey
8 hours ago
$begingroup$
@EricWofsey: I think your point is that fields are usually required to have $1 neq 0$, hence (somewhat unusually by comparison with groups and rings and modules and ...) the trivial structure with just one element isn't a field. If Kaplansky allows $1 = 0$ in a field as some authors do, then his statement is right.
$endgroup$
– Rob Arthan
7 hours ago
$begingroup$
@EricWofsey: I think your point is that fields are usually required to have $1 neq 0$, hence (somewhat unusually by comparison with groups and rings and modules and ...) the trivial structure with just one element isn't a field. If Kaplansky allows $1 = 0$ in a field as some authors do, then his statement is right.
$endgroup$
– Rob Arthan
7 hours ago
1
1
$begingroup$
I have never seen any author that allows $1=0$ in a field.
$endgroup$
– Eric Wofsey
7 hours ago
$begingroup$
I have never seen any author that allows $1=0$ in a field.
$endgroup$
– Eric Wofsey
7 hours ago
$begingroup$
I agree that it is usual to require $1 neq 0$, but Weber's original definition allowed it: he says explicitly that $0$ is distinct from $1$ except in the uninteresting case when the field has only one element. See gdz.sub.uni-goettingen.de/id/PPN235181684_0043?tify page 527. (Please forgive me for going back to the 19th century, but I've become a fan of going back to original sources in mathematics recently.)
$endgroup$
– Rob Arthan
7 hours ago
$begingroup$
I agree that it is usual to require $1 neq 0$, but Weber's original definition allowed it: he says explicitly that $0$ is distinct from $1$ except in the uninteresting case when the field has only one element. See gdz.sub.uni-goettingen.de/id/PPN235181684_0043?tify page 527. (Please forgive me for going back to the 19th century, but I've become a fan of going back to original sources in mathematics recently.)
$endgroup$
– Rob Arthan
7 hours ago
$begingroup$
... and in modern definitions the requirement that $1 neq 0$ is often sneaked in surreptitiously by saying that that then non-zero elements of the field form a group under multiplication (which implies there is at leat one non-zero element, because the traditional definition of a group requires a group to have a non-empty universe).
$endgroup$
– Rob Arthan
6 hours ago
$begingroup$
... and in modern definitions the requirement that $1 neq 0$ is often sneaked in surreptitiously by saying that that then non-zero elements of the field form a group under multiplication (which implies there is at leat one non-zero element, because the traditional definition of a group requires a group to have a non-empty universe).
$endgroup$
– Rob Arthan
6 hours ago
add a comment |
1 Answer
1
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$begingroup$
This is false. For instance, let $R=Ktimes L$ where $K$ and $L$ are fields. Then the only nonzero proper ideals in $R$ are $Ktimes 0$ and $0times L$, which are both prime, but $R$ is not a field.
For another example, consider $R=mathbbZ/(p^2)$ for any prime $p$. The only nonzero proper ideal is $(p)$ which is prime.
Here is a classification of all the examples. Suppose $R$ is a ring in which every nonzero proper ideal is prime. For any prime $Psubseteq R$, then $R/P$ has the same property but is a domain, and so must be a field. Thus in fact every nonzero proper ideal is maximal.
If $R$ has two different nonzero proper ideals $P$ and $Q$, then we must have $Pcap Q=0$ (since the intersection is a non-maximal proper ideal). By the Chinese remainder theorem we then get an isomorphism $Rcong R/Ptimes R/Q$ and so $R$ is a product of two fields.
If $R$ has exactly one nonzero proper ideal $P$, then $P$ is the nilradical of $R$ (since it is the unique prime ideal) and is principal (generated by any of its nonzero elements). This implies $P^2=0$ (otherwise it would be a smaller nonzero proper ideal) and that $Pcong R/P$ as an $R$-module (otherwise $P$ would be an $R/P$-vector space of dimension greater than $1$ and so would have a nontrivial proper subspace). If the quotient map $Rto R/P$ has a section which is a ring-homomorphism, then we can identify $R$ with $K[x]/(x^2)$ where $K$ is the field $R/P$. But such a section may not exist, as shown by the example $R=mathbbZ/(p^2)$ above.
Finally, if $R$ has no nonzero proper ideals, it is either a field or the zero ring.
All of these cases can be joined together into the following equivalent characterization: $R$ is a ring in which every nonzero proper ideal is prime iff $R$ is an artinian ring of length at most $2$ as a module over itself.
answered 8 hours ago
Eric WofseyEric Wofsey
208k14 gold badges243 silver badges375 bronze badges
$endgroup$
add a comment |
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$begingroup$
This is false. For instance, let $R=Ktimes L$ where $K$ and $L$ are fields. Then the only nonzero proper ideals in $R$ are $Ktimes 0$ and $0times L$, which are both prime, but $R$ is not a field.
For another example, consider $R=mathbbZ/(p^2)$ for any prime $p$. The only nonzero proper ideal is $(p)$ which is prime.
Here is a classification of all the examples. Suppose $R$ is a ring in which every nonzero proper ideal is prime. For any prime $Psubseteq R$, then $R/P$ has the same property but is a domain, and so must be a field. Thus in fact every nonzero proper ideal is maximal.
If $R$ has two different nonzero proper ideals $P$ and $Q$, then we must have $Pcap Q=0$ (since the intersection is a non-maximal proper ideal). By the Chinese remainder theorem we then get an isomorphism $Rcong R/Ptimes R/Q$ and so $R$ is a product of two fields.
If $R$ has exactly one nonzero proper ideal $P$, then $P$ is the nilradical of $R$ (since it is the unique prime ideal) and is principal (generated by any of its nonzero elements). This implies $P^2=0$ (otherwise it would be a smaller nonzero proper ideal) and that $Pcong R/P$ as an $R$-module (otherwise $P$ would be an $R/P$-vector space of dimension greater than $1$ and so would have a nontrivial proper subspace). If the quotient map $Rto R/P$ has a section which is a ring-homomorphism, then we can identify $R$ with $K[x]/(x^2)$ where $K$ is the field $R/P$. But such a section may not exist, as shown by the example $R=mathbbZ/(p^2)$ above.
Finally, if $R$ has no nonzero proper ideals, it is either a field or the zero ring.
All of these cases can be joined together into the following equivalent characterization: $R$ is a ring in which every nonzero proper ideal is prime iff $R$ is an artinian ring of length at most $2$ as a module over itself.
answered 8 hours ago
Eric WofseyEric Wofsey
208k14 gold badges243 silver badges375 bronze badges
$endgroup$
add a comment |
$begingroup$
This is false. For instance, let $R=Ktimes L$ where $K$ and $L$ are fields. Then the only nonzero proper ideals in $R$ are $Ktimes 0$ and $0times L$, which are both prime, but $R$ is not a field.
For another example, consider $R=mathbbZ/(p^2)$ for any prime $p$. The only nonzero proper ideal is $(p)$ which is prime.
Here is a classification of all the examples. Suppose $R$ is a ring in which every nonzero proper ideal is prime. For any prime $Psubseteq R$, then $R/P$ has the same property but is a domain, and so must be a field. Thus in fact every nonzero proper ideal is maximal.
If $R$ has two different nonzero proper ideals $P$ and $Q$, then we must have $Pcap Q=0$ (since the intersection is a non-maximal proper ideal). By the Chinese remainder theorem we then get an isomorphism $Rcong R/Ptimes R/Q$ and so $R$ is a product of two fields.
If $R$ has exactly one nonzero proper ideal $P$, then $P$ is the nilradical of $R$ (since it is the unique prime ideal) and is principal (generated by any of its nonzero elements). This implies $P^2=0$ (otherwise it would be a smaller nonzero proper ideal) and that $Pcong R/P$ as an $R$-module (otherwise $P$ would be an $R/P$-vector space of dimension greater than $1$ and so would have a nontrivial proper subspace). If the quotient map $Rto R/P$ has a section which is a ring-homomorphism, then we can identify $R$ with $K[x]/(x^2)$ where $K$ is the field $R/P$. But such a section may not exist, as shown by the example $R=mathbbZ/(p^2)$ above.
Finally, if $R$ has no nonzero proper ideals, it is either a field or the zero ring.
All of these cases can be joined together into the following equivalent characterization: $R$ is a ring in which every nonzero proper ideal is prime iff $R$ is an artinian ring of length at most $2$ as a module over itself.
answered 8 hours ago
Eric WofseyEric Wofsey
208k14 gold badges243 silver badges375 bronze badges
$endgroup$
add a comment |
$begingroup$
This is false. For instance, let $R=Ktimes L$ where $K$ and $L$ are fields. Then the only nonzero proper ideals in $R$ are $Ktimes 0$ and $0times L$, which are both prime, but $R$ is not a field.
For another example, consider $R=mathbbZ/(p^2)$ for any prime $p$. The only nonzero proper ideal is $(p)$ which is prime.
Here is a classification of all the examples. Suppose $R$ is a ring in which every nonzero proper ideal is prime. For any prime $Psubseteq R$, then $R/P$ has the same property but is a domain, and so must be a field. Thus in fact every nonzero proper ideal is maximal.
If $R$ has two different nonzero proper ideals $P$ and $Q$, then we must have $Pcap Q=0$ (since the intersection is a non-maximal proper ideal). By the Chinese remainder theorem we then get an isomorphism $Rcong R/Ptimes R/Q$ and so $R$ is a product of two fields.
If $R$ has exactly one nonzero proper ideal $P$, then $P$ is the nilradical of $R$ (since it is the unique prime ideal) and is principal (generated by any of its nonzero elements). This implies $P^2=0$ (otherwise it would be a smaller nonzero proper ideal) and that $Pcong R/P$ as an $R$-module (otherwise $P$ would be an $R/P$-vector space of dimension greater than $1$ and so would have a nontrivial proper subspace). If the quotient map $Rto R/P$ has a section which is a ring-homomorphism, then we can identify $R$ with $K[x]/(x^2)$ where $K$ is the field $R/P$. But such a section may not exist, as shown by the example $R=mathbbZ/(p^2)$ above.
Finally, if $R$ has no nonzero proper ideals, it is either a field or the zero ring.
All of these cases can be joined together into the following equivalent characterization: $R$ is a ring in which every nonzero proper ideal is prime iff $R$ is an artinian ring of length at most $2$ as a module over itself.
answered 8 hours ago
Eric WofseyEric Wofsey
208k14 gold badges243 silver badges375 bronze badges
$endgroup$
This is false. For instance, let $R=Ktimes L$ where $K$ and $L$ are fields. Then the only nonzero proper ideals in $R$ are $Ktimes 0$ and $0times L$, which are both prime, but $R$ is not a field.
For another example, consider $R=mathbbZ/(p^2)$ for any prime $p$. The only nonzero proper ideal is $(p)$ which is prime.
Here is a classification of all the examples. Suppose $R$ is a ring in which every nonzero proper ideal is prime. For any prime $Psubseteq R$, then $R/P$ has the same property but is a domain, and so must be a field. Thus in fact every nonzero proper ideal is maximal.
If $R$ has two different nonzero proper ideals $P$ and $Q$, then we must have $Pcap Q=0$ (since the intersection is a non-maximal proper ideal). By the Chinese remainder theorem we then get an isomorphism $Rcong R/Ptimes R/Q$ and so $R$ is a product of two fields.
If $R$ has exactly one nonzero proper ideal $P$, then $P$ is the nilradical of $R$ (since it is the unique prime ideal) and is principal (generated by any of its nonzero elements). This implies $P^2=0$ (otherwise it would be a smaller nonzero proper ideal) and that $Pcong R/P$ as an $R$-module (otherwise $P$ would be an $R/P$-vector space of dimension greater than $1$ and so would have a nontrivial proper subspace). If the quotient map $Rto R/P$ has a section which is a ring-homomorphism, then we can identify $R$ with $K[x]/(x^2)$ where $K$ is the field $R/P$. But such a section may not exist, as shown by the example $R=mathbbZ/(p^2)$ above.
Finally, if $R$ has no nonzero proper ideals, it is either a field or the zero ring.
All of these cases can be joined together into the following equivalent characterization: $R$ is a ring in which every nonzero proper ideal is prime iff $R$ is an artinian ring of length at most $2$ as a module over itself.
answered 8 hours ago
Eric WofseyEric Wofsey
208k14 gold badges243 silver badges375 bronze badges
edited 7 hours ago
answered 8 hours ago
Eric WofseyEric Wofsey
208k14 gold badges243 silver badges375 bronze badges
answered 8 hours ago
Eric WofseyEric Wofsey
208k14 gold badges243 silver badges375 bronze badges
answered 8 hours ago
Eric WofseyEric Wofsey
208k14 gold badges243 silver badges375 bronze badges
208k14 gold badges243 silver badges375 bronze badges
add a comment |
add a comment |
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$begingroup$
Note that Kaplansky's original statement is not quite right, since you also have to require that $R$ is nonzero. (If $R$ is the zero ring, then there are no ideals other than $R$ so the condition is vacuously true.)
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– Eric Wofsey
8 hours ago
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@EricWofsey: I think your point is that fields are usually required to have $1 neq 0$, hence (somewhat unusually by comparison with groups and rings and modules and ...) the trivial structure with just one element isn't a field. If Kaplansky allows $1 = 0$ in a field as some authors do, then his statement is right.
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– Rob Arthan
7 hours ago
1
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I have never seen any author that allows $1=0$ in a field.
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– Eric Wofsey
7 hours ago
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I agree that it is usual to require $1 neq 0$, but Weber's original definition allowed it: he says explicitly that $0$ is distinct from $1$ except in the uninteresting case when the field has only one element. See gdz.sub.uni-goettingen.de/id/PPN235181684_0043?tify page 527. (Please forgive me for going back to the 19th century, but I've become a fan of going back to original sources in mathematics recently.)
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– Rob Arthan
7 hours ago
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... and in modern definitions the requirement that $1 neq 0$ is often sneaked in surreptitiously by saying that that then non-zero elements of the field form a group under multiplication (which implies there is at leat one non-zero element, because the traditional definition of a group requires a group to have a non-empty universe).
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– Rob Arthan
6 hours ago