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Are there rings with uncountably many irreducible elements (prime elements, if in a PID)?
Clash Royale CLAN TAG#URR8PPP
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I'm playing around and trying to construct rings with different numbers of irreducible elements, hence the above question.
elementary-set-theory ring-theory
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Andrés E. Caicedo
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Federica Zanni
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up vote
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I'm playing around and trying to construct rings with different numbers of irreducible elements, hence the above question.
elementary-set-theory ring-theory
edited 3 hours ago
Andrés E. Caicedo
64k7156239
asked 3 hours ago
Federica Zanni
112
New contributor
Federica Zanni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Massive overkill: since being irreducible/prime is a first-order condition, take a nontrivial ultrapower of any ring with infinitely many irreducible elements.
– Noah Schweber
3 hours ago
@NoahSchweber: The term "first-order" is very common but seems to be one used in different branches of mathematics with different meanings. And ultrafilter is also unfamiliar to me. But if a a result in algebra can be proved from other branches I'd be happy to learn more about that branch. Is that logic?
– P Vanchinathan
2 hours ago
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up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm playing around and trying to construct rings with different numbers of irreducible elements, hence the above question.
elementary-set-theory ring-theory
edited 3 hours ago
Andrés E. Caicedo
64k7156239
asked 3 hours ago
Federica Zanni
112
New contributor
Federica Zanni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm playing around and trying to construct rings with different numbers of irreducible elements, hence the above question.
elementary-set-theory ring-theory
elementary-set-theory ring-theory
edited 3 hours ago
Andrés E. Caicedo
64k7156239
asked 3 hours ago
Federica Zanni
112
New contributor
Federica Zanni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
Andrés E. Caicedo
64k7156239
asked 3 hours ago
Federica Zanni
112
New contributor
Federica Zanni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
Andrés E. Caicedo
64k7156239
edited 3 hours ago
Andrés E. Caicedo
64k7156239
edited 3 hours ago
Andrés E. Caicedo
64k7156239
64k7156239
asked 3 hours ago
Federica Zanni
112
New contributor
Federica Zanni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Federica Zanni
112
asked 3 hours ago
Federica Zanni
112
112
New contributor
Federica Zanni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Federica Zanni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Federica Zanni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Massive overkill: since being irreducible/prime is a first-order condition, take a nontrivial ultrapower of any ring with infinitely many irreducible elements.
– Noah Schweber
3 hours ago
@NoahSchweber: The term "first-order" is very common but seems to be one used in different branches of mathematics with different meanings. And ultrafilter is also unfamiliar to me. But if a a result in algebra can be proved from other branches I'd be happy to learn more about that branch. Is that logic?
– P Vanchinathan
2 hours ago
add a comment |Â
1
Massive overkill: since being irreducible/prime is a first-order condition, take a nontrivial ultrapower of any ring with infinitely many irreducible elements.
– Noah Schweber
3 hours ago
@NoahSchweber: The term "first-order" is very common but seems to be one used in different branches of mathematics with different meanings. And ultrafilter is also unfamiliar to me. But if a a result in algebra can be proved from other branches I'd be happy to learn more about that branch. Is that logic?
– P Vanchinathan
2 hours ago
1
1
Massive overkill: since being irreducible/prime is a first-order condition, take a nontrivial ultrapower of any ring with infinitely many irreducible elements.
– Noah Schweber
3 hours ago
Massive overkill: since being irreducible/prime is a first-order condition, take a nontrivial ultrapower of any ring with infinitely many irreducible elements.
– Noah Schweber
3 hours ago
@NoahSchweber: The term "first-order" is very common but seems to be one used in different branches of mathematics with different meanings. And ultrafilter is also unfamiliar to me. But if a a result in algebra can be proved from other branches I'd be happy to learn more about that branch. Is that logic?
– P Vanchinathan
2 hours ago
@NoahSchweber: The term "first-order" is very common but seems to be one used in different branches of mathematics with different meanings. And ultrafilter is also unfamiliar to me. But if a a result in algebra can be proved from other branches I'd be happy to learn more about that branch. Is that logic?
– P Vanchinathan
2 hours ago
add a comment |Â
1 Answer
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Take a polynomial ring in a single variable $x$ with coefficients in real numbers (or complex numbers, what is important is the uncountablity). Then all degree one polynomials $(x-alpha)$ are irreducible. Being degree 1 they cannot be products of lower degree polynomials!
This being a PID irreducible, prime are one and the same.
answered 3 hours ago
P Vanchinathan
14.4k12036
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
Take a polynomial ring in a single variable $x$ with coefficients in real numbers (or complex numbers, what is important is the uncountablity). Then all degree one polynomials $(x-alpha)$ are irreducible. Being degree 1 they cannot be products of lower degree polynomials!
This being a PID irreducible, prime are one and the same.
answered 3 hours ago
P Vanchinathan
14.4k12036
add a comment |Â
up vote
5
down vote
Take a polynomial ring in a single variable $x$ with coefficients in real numbers (or complex numbers, what is important is the uncountablity). Then all degree one polynomials $(x-alpha)$ are irreducible. Being degree 1 they cannot be products of lower degree polynomials!
This being a PID irreducible, prime are one and the same.
answered 3 hours ago
P Vanchinathan
14.4k12036
add a comment |Â
up vote
5
down vote
up vote
5
down vote
Take a polynomial ring in a single variable $x$ with coefficients in real numbers (or complex numbers, what is important is the uncountablity). Then all degree one polynomials $(x-alpha)$ are irreducible. Being degree 1 they cannot be products of lower degree polynomials!
This being a PID irreducible, prime are one and the same.
answered 3 hours ago
P Vanchinathan
14.4k12036
Take a polynomial ring in a single variable $x$ with coefficients in real numbers (or complex numbers, what is important is the uncountablity). Then all degree one polynomials $(x-alpha)$ are irreducible. Being degree 1 they cannot be products of lower degree polynomials!
This being a PID irreducible, prime are one and the same.
answered 3 hours ago
P Vanchinathan
14.4k12036
answered 3 hours ago
P Vanchinathan
14.4k12036
answered 3 hours ago
P Vanchinathan
14.4k12036
answered 3 hours ago
P Vanchinathan
14.4k12036
14.4k12036
add a comment |Â
add a comment |Â
Federica Zanni is a new contributor. Be nice, and check out our Code of Conduct.
Federica Zanni is a new contributor. Be nice, and check out our Code of Conduct.
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1
Massive overkill: since being irreducible/prime is a first-order condition, take a nontrivial ultrapower of any ring with infinitely many irreducible elements.
– Noah Schweber
3 hours ago
@NoahSchweber: The term "first-order" is very common but seems to be one used in different branches of mathematics with different meanings. And ultrafilter is also unfamiliar to me. But if a a result in algebra can be proved from other branches I'd be happy to learn more about that branch. Is that logic?
– P Vanchinathan
2 hours ago