If spectrum of a commutative ring is empty
I have found this proposition: "If the spectrum of a commutative ring A is empty then A is the zero ring".
By absurdum, if A is not the zero ring, there exists in A an element $xne0$. By Zorn's lemma, there exists a maximal ideal M such that $xnotin M$. If A has not a unit, M is not necessarily a prime ideal... So I have no idea to complete the proof. May you help me with some hint?
Thank You.
commutative-algebra maximal-and-prime-ideals
edited 6 hours ago
Eric Wofsey
179k12204331
asked 8 hours ago
Fabrixady
113
add a comment |
I have found this proposition: "If the spectrum of a commutative ring A is empty then A is the zero ring".
By absurdum, if A is not the zero ring, there exists in A an element $xne0$. By Zorn's lemma, there exists a maximal ideal M such that $xnotin M$. If A has not a unit, M is not necessarily a prime ideal... So I have no idea to complete the proof. May you help me with some hint?
Thank You.
commutative-algebra maximal-and-prime-ideals
edited 6 hours ago
Eric Wofsey
179k12204331
asked 8 hours ago
Fabrixady
113
add a comment |
I have found this proposition: "If the spectrum of a commutative ring A is empty then A is the zero ring".
By absurdum, if A is not the zero ring, there exists in A an element $xne0$. By Zorn's lemma, there exists a maximal ideal M such that $xnotin M$. If A has not a unit, M is not necessarily a prime ideal... So I have no idea to complete the proof. May you help me with some hint?
Thank You.
commutative-algebra maximal-and-prime-ideals
edited 6 hours ago
Eric Wofsey
179k12204331
asked 8 hours ago
Fabrixady
113
I have found this proposition: "If the spectrum of a commutative ring A is empty then A is the zero ring".
By absurdum, if A is not the zero ring, there exists in A an element $xne0$. By Zorn's lemma, there exists a maximal ideal M such that $xnotin M$. If A has not a unit, M is not necessarily a prime ideal... So I have no idea to complete the proof. May you help me with some hint?
Thank You.
commutative-algebra maximal-and-prime-ideals
commutative-algebra maximal-and-prime-ideals
edited 6 hours ago
Eric Wofsey
179k12204331
asked 8 hours ago
Fabrixady
113
edited 6 hours ago
Eric Wofsey
179k12204331
asked 8 hours ago
Fabrixady
113
edited 6 hours ago
Eric Wofsey
179k12204331
edited 6 hours ago
Eric Wofsey
179k12204331
edited 6 hours ago
Eric Wofsey
179k12204331
179k12204331
asked 8 hours ago
Fabrixady
113
asked 8 hours ago
Fabrixady
113
asked 8 hours ago
Fabrixady
113
113
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
I very much doubt you are working with rings lacking identity.
Indeed, the statement is false for rings without identity: $2mathbb Z/4mathbb Z$ has no prime ideals.
answered 8 hours ago
rschwieb
105k1299243
add a comment |
$A$ has a unit in this context. This is necessary both to claim the existence of a maximal ideal using Zorn's lemma and to claim that a maximal ideal is prime.
The example Wikipedia provides, is the ring whose underlying additive group is $mathbf Q$ with the usual addition and whose multiplication is $a cdot b = 0$ for all $a, b$. A subset $A subseteq mathbf Q$ is an ideal if and only if it is an additive subgroup of $mathbf{Q}$.
It is clear that no proper ideal can be prime since if $x notin A$ then $x^2 = 0 in A$.
Also, if $A$ is proper then $(mathbf{Q}/A,+)$ is a non-zero divisible group, which means it has a non-zero proper subgroup, which by correspondence means there is a larger proper ideal than $A$. So no proper ideals of $mathbf{Q}$ are maximal.
answered 8 hours ago
Trevor Gunn
14.1k32046
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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I very much doubt you are working with rings lacking identity.
Indeed, the statement is false for rings without identity: $2mathbb Z/4mathbb Z$ has no prime ideals.
answered 8 hours ago
rschwieb
105k1299243
add a comment |
I very much doubt you are working with rings lacking identity.
Indeed, the statement is false for rings without identity: $2mathbb Z/4mathbb Z$ has no prime ideals.
answered 8 hours ago
rschwieb
105k1299243
add a comment |
I very much doubt you are working with rings lacking identity.
Indeed, the statement is false for rings without identity: $2mathbb Z/4mathbb Z$ has no prime ideals.
answered 8 hours ago
rschwieb
105k1299243
I very much doubt you are working with rings lacking identity.
Indeed, the statement is false for rings without identity: $2mathbb Z/4mathbb Z$ has no prime ideals.
answered 8 hours ago
rschwieb
105k1299243
answered 8 hours ago
rschwieb
105k1299243
answered 8 hours ago
rschwieb
105k1299243
answered 8 hours ago
rschwieb
105k1299243
105k1299243
add a comment |
add a comment |
$A$ has a unit in this context. This is necessary both to claim the existence of a maximal ideal using Zorn's lemma and to claim that a maximal ideal is prime.
The example Wikipedia provides, is the ring whose underlying additive group is $mathbf Q$ with the usual addition and whose multiplication is $a cdot b = 0$ for all $a, b$. A subset $A subseteq mathbf Q$ is an ideal if and only if it is an additive subgroup of $mathbf{Q}$.
It is clear that no proper ideal can be prime since if $x notin A$ then $x^2 = 0 in A$.
Also, if $A$ is proper then $(mathbf{Q}/A,+)$ is a non-zero divisible group, which means it has a non-zero proper subgroup, which by correspondence means there is a larger proper ideal than $A$. So no proper ideals of $mathbf{Q}$ are maximal.
answered 8 hours ago
Trevor Gunn
14.1k32046
add a comment |
$A$ has a unit in this context. This is necessary both to claim the existence of a maximal ideal using Zorn's lemma and to claim that a maximal ideal is prime.
The example Wikipedia provides, is the ring whose underlying additive group is $mathbf Q$ with the usual addition and whose multiplication is $a cdot b = 0$ for all $a, b$. A subset $A subseteq mathbf Q$ is an ideal if and only if it is an additive subgroup of $mathbf{Q}$.
It is clear that no proper ideal can be prime since if $x notin A$ then $x^2 = 0 in A$.
Also, if $A$ is proper then $(mathbf{Q}/A,+)$ is a non-zero divisible group, which means it has a non-zero proper subgroup, which by correspondence means there is a larger proper ideal than $A$. So no proper ideals of $mathbf{Q}$ are maximal.
answered 8 hours ago
Trevor Gunn
14.1k32046
add a comment |
$A$ has a unit in this context. This is necessary both to claim the existence of a maximal ideal using Zorn's lemma and to claim that a maximal ideal is prime.
The example Wikipedia provides, is the ring whose underlying additive group is $mathbf Q$ with the usual addition and whose multiplication is $a cdot b = 0$ for all $a, b$. A subset $A subseteq mathbf Q$ is an ideal if and only if it is an additive subgroup of $mathbf{Q}$.
It is clear that no proper ideal can be prime since if $x notin A$ then $x^2 = 0 in A$.
Also, if $A$ is proper then $(mathbf{Q}/A,+)$ is a non-zero divisible group, which means it has a non-zero proper subgroup, which by correspondence means there is a larger proper ideal than $A$. So no proper ideals of $mathbf{Q}$ are maximal.
answered 8 hours ago
Trevor Gunn
14.1k32046
$A$ has a unit in this context. This is necessary both to claim the existence of a maximal ideal using Zorn's lemma and to claim that a maximal ideal is prime.
The example Wikipedia provides, is the ring whose underlying additive group is $mathbf Q$ with the usual addition and whose multiplication is $a cdot b = 0$ for all $a, b$. A subset $A subseteq mathbf Q$ is an ideal if and only if it is an additive subgroup of $mathbf{Q}$.
It is clear that no proper ideal can be prime since if $x notin A$ then $x^2 = 0 in A$.
Also, if $A$ is proper then $(mathbf{Q}/A,+)$ is a non-zero divisible group, which means it has a non-zero proper subgroup, which by correspondence means there is a larger proper ideal than $A$. So no proper ideals of $mathbf{Q}$ are maximal.
answered 8 hours ago
Trevor Gunn
14.1k32046
answered 8 hours ago
Trevor Gunn
14.1k32046
answered 8 hours ago
Trevor Gunn
14.1k32046
answered 8 hours ago
Trevor Gunn
14.1k32046
14.1k32046
add a comment |
add a comment |
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