Cohen-Macaulay ring without non-trivial idempotent is homomorphic image of Noetherian domain?
Let $R$ be a Cohen-Macaulay ring with no non-trivial idempotent element. Then is it true that there is a Noetherian domain $S$ such that $Rcong S/I$ for some ideal $I$ of $S$ ?
If this is not true in general, what if we also assume $R$ has finite Krull-dimension ?
MOTIVATION: A theorem of Hungerford https://msp.org/pjm/1968/25-3/pjm-v25-n3-p11-p.pdf says that every PIR is a homomorphic image of a direct product of finitely many PID s. So I was thinking about the same question for special kind of Noetherian rings
algebraic-geometry commutative-algebra homological-algebra krull-dimension cohen-macaulay
asked Dec 3 '18 at 23:31
user521337user521337
9961315
add a comment |
Let $R$ be a Cohen-Macaulay ring with no non-trivial idempotent element. Then is it true that there is a Noetherian domain $S$ such that $Rcong S/I$ for some ideal $I$ of $S$ ?
If this is not true in general, what if we also assume $R$ has finite Krull-dimension ?
MOTIVATION: A theorem of Hungerford https://msp.org/pjm/1968/25-3/pjm-v25-n3-p11-p.pdf says that every PIR is a homomorphic image of a direct product of finitely many PID s. So I was thinking about the same question for special kind of Noetherian rings
algebraic-geometry commutative-algebra homological-algebra krull-dimension cohen-macaulay
asked Dec 3 '18 at 23:31
user521337user521337
9961315
1
Can you motivate the question a little bit?
– Badam Baplan
Dec 4 '18 at 0:30
add a comment |
Let $R$ be a Cohen-Macaulay ring with no non-trivial idempotent element. Then is it true that there is a Noetherian domain $S$ such that $Rcong S/I$ for some ideal $I$ of $S$ ?
If this is not true in general, what if we also assume $R$ has finite Krull-dimension ?
MOTIVATION: A theorem of Hungerford https://msp.org/pjm/1968/25-3/pjm-v25-n3-p11-p.pdf says that every PIR is a homomorphic image of a direct product of finitely many PID s. So I was thinking about the same question for special kind of Noetherian rings
algebraic-geometry commutative-algebra homological-algebra krull-dimension cohen-macaulay
asked Dec 3 '18 at 23:31
user521337user521337
9961315
Let $R$ be a Cohen-Macaulay ring with no non-trivial idempotent element. Then is it true that there is a Noetherian domain $S$ such that $Rcong S/I$ for some ideal $I$ of $S$ ?
If this is not true in general, what if we also assume $R$ has finite Krull-dimension ?
MOTIVATION: A theorem of Hungerford https://msp.org/pjm/1968/25-3/pjm-v25-n3-p11-p.pdf says that every PIR is a homomorphic image of a direct product of finitely many PID s. So I was thinking about the same question for special kind of Noetherian rings
algebraic-geometry commutative-algebra homological-algebra krull-dimension cohen-macaulay
algebraic-geometry commutative-algebra homological-algebra krull-dimension cohen-macaulay
asked Dec 3 '18 at 23:31
user521337user521337
9961315
asked Dec 3 '18 at 23:31
user521337user521337
9961315
edited Jan 4 at 13:40
user521337
asked Dec 3 '18 at 23:31
user521337user521337
9961315
asked Dec 3 '18 at 23:31
user521337user521337
9961315
asked Dec 3 '18 at 23:31
user521337user521337
9961315
9961315
1
Can you motivate the question a little bit?
– Badam Baplan
Dec 4 '18 at 0:30
add a comment |
1
Can you motivate the question a little bit?
– Badam Baplan
Dec 4 '18 at 0:30
1
1
Can you motivate the question a little bit?
– Badam Baplan
Dec 4 '18 at 0:30
Can you motivate the question a little bit?
– Badam Baplan
Dec 4 '18 at 0:30
add a comment |
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Can you motivate the question a little bit?
– Badam Baplan
Dec 4 '18 at 0:30