Are there rings with uncountably many irreducible elements (prime elements, if in a PID)?

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I'm playing around and trying to construct rings with different numbers of irreducible elements, hence the above question.






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    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














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







share|cite|improve this question









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












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







share|cite|improve this question









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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






share|cite|improve this question









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Federica Zanni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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edited 3 hours ago









Andrés E. Caicedo

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Federica Zanni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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












  • 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










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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.






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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.






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      up vote
      5
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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.






      share|cite|improve this answer






















        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












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        share|cite|improve this answer










        answered 3 hours ago









        P Vanchinathan

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