Let $A$ be any commutative ring and consider $operatorname{Spec}(A)$, then for any $E subseteq A$ we have...
Let $A$ be any commutative ring and consider $operatorname{Spec}(A)$, then for any $E subseteq A$ we have $V(E) = emptyset iff E = A$
Recall that $V(E) = {text{prime ideals of $A$ containing $E$}}$
Proof: That Since $V(A) = V(1) = emptyset$ the reverse direction is trivial. To prove the forward direction suppose that $V(E) = emptyset$, then no prime ideal of $A$ contains $E$. Let $(E)$ be the ideal generated by $E$, then we have $V(E) = V(operatorname{rad}(E)) = emptyset$ and so no prime ideal contains the intersection of all prime ideals containing $E$.
Now if $(E) neq (1)$ then $(E) subsetneq M$ for some maximal ideal of $A$ containing $(E)$. But then $M$ is a prime ideal containing $(E)$ and so must contain the intersection of all prime ideals containing $E$, so $M in V(E)$ a contradiction. Thus $(E) = (1) = A$. $square$
I just wanted to check if my proof was correct and my reasoning was sound and efficient.
ring-theory commutative-algebra
asked Nov 27 at 0:26
Perturbative
4,07011449
add a comment |
Let $A$ be any commutative ring and consider $operatorname{Spec}(A)$, then for any $E subseteq A$ we have $V(E) = emptyset iff E = A$
Recall that $V(E) = {text{prime ideals of $A$ containing $E$}}$
Proof: That Since $V(A) = V(1) = emptyset$ the reverse direction is trivial. To prove the forward direction suppose that $V(E) = emptyset$, then no prime ideal of $A$ contains $E$. Let $(E)$ be the ideal generated by $E$, then we have $V(E) = V(operatorname{rad}(E)) = emptyset$ and so no prime ideal contains the intersection of all prime ideals containing $E$.
Now if $(E) neq (1)$ then $(E) subsetneq M$ for some maximal ideal of $A$ containing $(E)$. But then $M$ is a prime ideal containing $(E)$ and so must contain the intersection of all prime ideals containing $E$, so $M in V(E)$ a contradiction. Thus $(E) = (1) = A$. $square$
I just wanted to check if my proof was correct and my reasoning was sound and efficient.
ring-theory commutative-algebra
asked Nov 27 at 0:26
Perturbative
4,07011449
add a comment |
Let $A$ be any commutative ring and consider $operatorname{Spec}(A)$, then for any $E subseteq A$ we have $V(E) = emptyset iff E = A$
Recall that $V(E) = {text{prime ideals of $A$ containing $E$}}$
Proof: That Since $V(A) = V(1) = emptyset$ the reverse direction is trivial. To prove the forward direction suppose that $V(E) = emptyset$, then no prime ideal of $A$ contains $E$. Let $(E)$ be the ideal generated by $E$, then we have $V(E) = V(operatorname{rad}(E)) = emptyset$ and so no prime ideal contains the intersection of all prime ideals containing $E$.
Now if $(E) neq (1)$ then $(E) subsetneq M$ for some maximal ideal of $A$ containing $(E)$. But then $M$ is a prime ideal containing $(E)$ and so must contain the intersection of all prime ideals containing $E$, so $M in V(E)$ a contradiction. Thus $(E) = (1) = A$. $square$
I just wanted to check if my proof was correct and my reasoning was sound and efficient.
ring-theory commutative-algebra
asked Nov 27 at 0:26
Perturbative
4,07011449
Let $A$ be any commutative ring and consider $operatorname{Spec}(A)$, then for any $E subseteq A$ we have $V(E) = emptyset iff E = A$
Recall that $V(E) = {text{prime ideals of $A$ containing $E$}}$
Proof: That Since $V(A) = V(1) = emptyset$ the reverse direction is trivial. To prove the forward direction suppose that $V(E) = emptyset$, then no prime ideal of $A$ contains $E$. Let $(E)$ be the ideal generated by $E$, then we have $V(E) = V(operatorname{rad}(E)) = emptyset$ and so no prime ideal contains the intersection of all prime ideals containing $E$.
Now if $(E) neq (1)$ then $(E) subsetneq M$ for some maximal ideal of $A$ containing $(E)$. But then $M$ is a prime ideal containing $(E)$ and so must contain the intersection of all prime ideals containing $E$, so $M in V(E)$ a contradiction. Thus $(E) = (1) = A$. $square$
I just wanted to check if my proof was correct and my reasoning was sound and efficient.
ring-theory commutative-algebra
ring-theory commutative-algebra
asked Nov 27 at 0:26
Perturbative
4,07011449
asked Nov 27 at 0:26
Perturbative
4,07011449
asked Nov 27 at 0:26
Perturbative
4,07011449
asked Nov 27 at 0:26
Perturbative
4,07011449
asked Nov 27 at 0:26
Perturbative
4,07011449
4,07011449
add a comment |
add a comment |
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For starters, surely $E$ itself is supposed to be an ideal and not just a subset (or even subring). Otherwise consider $E = {1 }$ for which (generally) $E not= A$ but $V(E) = emptyset$. Also, when you say $E subsetneq M$, that should be $E subseteq M$, since $E$ might itself be a prime ideal.
Beyond that, your proof is correct but needlessly complicated. Your reasoning is perfectly fine without mentioning radicals and intersections of primes. You also are structuring your reasoning like a proof by contradiction but proceeding by contraposition, which is confusing.
Indeed, what you are really saying is: $$E not= A implies 1 notin E implies E subseteq M text{ for some maximal ideal $M$ by Zorn's lemma} implies V(E) not= emptyset text{ by definition}$$
Contrapositively, $V(E) = emptyset implies E = A$
answered Nov 27 at 1:48
Badam Baplan
4,361722
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1 Answer
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1 Answer
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For starters, surely $E$ itself is supposed to be an ideal and not just a subset (or even subring). Otherwise consider $E = {1 }$ for which (generally) $E not= A$ but $V(E) = emptyset$. Also, when you say $E subsetneq M$, that should be $E subseteq M$, since $E$ might itself be a prime ideal.
Beyond that, your proof is correct but needlessly complicated. Your reasoning is perfectly fine without mentioning radicals and intersections of primes. You also are structuring your reasoning like a proof by contradiction but proceeding by contraposition, which is confusing.
Indeed, what you are really saying is: $$E not= A implies 1 notin E implies E subseteq M text{ for some maximal ideal $M$ by Zorn's lemma} implies V(E) not= emptyset text{ by definition}$$
Contrapositively, $V(E) = emptyset implies E = A$
answered Nov 27 at 1:48
Badam Baplan
4,361722
add a comment |
For starters, surely $E$ itself is supposed to be an ideal and not just a subset (or even subring). Otherwise consider $E = {1 }$ for which (generally) $E not= A$ but $V(E) = emptyset$. Also, when you say $E subsetneq M$, that should be $E subseteq M$, since $E$ might itself be a prime ideal.
Beyond that, your proof is correct but needlessly complicated. Your reasoning is perfectly fine without mentioning radicals and intersections of primes. You also are structuring your reasoning like a proof by contradiction but proceeding by contraposition, which is confusing.
Indeed, what you are really saying is: $$E not= A implies 1 notin E implies E subseteq M text{ for some maximal ideal $M$ by Zorn's lemma} implies V(E) not= emptyset text{ by definition}$$
Contrapositively, $V(E) = emptyset implies E = A$
answered Nov 27 at 1:48
Badam Baplan
4,361722
add a comment |
For starters, surely $E$ itself is supposed to be an ideal and not just a subset (or even subring). Otherwise consider $E = {1 }$ for which (generally) $E not= A$ but $V(E) = emptyset$. Also, when you say $E subsetneq M$, that should be $E subseteq M$, since $E$ might itself be a prime ideal.
Beyond that, your proof is correct but needlessly complicated. Your reasoning is perfectly fine without mentioning radicals and intersections of primes. You also are structuring your reasoning like a proof by contradiction but proceeding by contraposition, which is confusing.
Indeed, what you are really saying is: $$E not= A implies 1 notin E implies E subseteq M text{ for some maximal ideal $M$ by Zorn's lemma} implies V(E) not= emptyset text{ by definition}$$
Contrapositively, $V(E) = emptyset implies E = A$
answered Nov 27 at 1:48
Badam Baplan
4,361722
For starters, surely $E$ itself is supposed to be an ideal and not just a subset (or even subring). Otherwise consider $E = {1 }$ for which (generally) $E not= A$ but $V(E) = emptyset$. Also, when you say $E subsetneq M$, that should be $E subseteq M$, since $E$ might itself be a prime ideal.
Beyond that, your proof is correct but needlessly complicated. Your reasoning is perfectly fine without mentioning radicals and intersections of primes. You also are structuring your reasoning like a proof by contradiction but proceeding by contraposition, which is confusing.
Indeed, what you are really saying is: $$E not= A implies 1 notin E implies E subseteq M text{ for some maximal ideal $M$ by Zorn's lemma} implies V(E) not= emptyset text{ by definition}$$
Contrapositively, $V(E) = emptyset implies E = A$
answered Nov 27 at 1:48
Badam Baplan
4,361722
answered Nov 27 at 1:48
Badam Baplan
4,361722
answered Nov 27 at 1:48
Badam Baplan
4,361722
answered Nov 27 at 1:48
Badam Baplan
4,361722
4,361722
add a comment |
add a comment |
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