Why doesn't this definition hold up when we square numbers?












1














So, I've been given the actual definition of a prime for the first time, as opposed to the definition of an irreducible which I was previously taught (as is customary), it goes as follows:



An element $p$ of a ring $R$ is called "prime" if $a,bin R$ and $p|abrightarrow p|a$ or $p|b$



$4^2 = 16$. So inversely $4|16$. Now let's look at all the combinations of two integers $a,bin R$ where $ab = 16$ and whether $p|a$ or $p|b$:



$1 * 16$ ($4|16$)



$2 * 8$ ($4|8$)



$4 * 4$ ($4|4$)



By my logic, $4$ is prime.



However, as we all know $4 = 2^2$. But $2$ is not a unit so $4$ is an irreducible. Since the ring of integers is an integral domain $4$ must, therefore, be prime aswell. Why does this happen?






abstract-algebra group-theory ring-theory prime-numbers







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  • 6




    For an element to be prime this must hold for all a,b, not just one example.
    – Erik Parkinson
    1 hour ago










  • You have shown that there exists $c$ such that $4|c$ and $forall a,b$ such that $ab=c, 4|a$ or $4|b$. Your definition demands this to be true for every $c$, not just 16.
    – RcnSc
    6 mins ago
















1














So, I've been given the actual definition of a prime for the first time, as opposed to the definition of an irreducible which I was previously taught (as is customary), it goes as follows:



An element $p$ of a ring $R$ is called "prime" if $a,bin R$ and $p|abrightarrow p|a$ or $p|b$



$4^2 = 16$. So inversely $4|16$. Now let's look at all the combinations of two integers $a,bin R$ where $ab = 16$ and whether $p|a$ or $p|b$:



$1 * 16$ ($4|16$)



$2 * 8$ ($4|8$)



$4 * 4$ ($4|4$)



By my logic, $4$ is prime.



However, as we all know $4 = 2^2$. But $2$ is not a unit so $4$ is an irreducible. Since the ring of integers is an integral domain $4$ must, therefore, be prime aswell. Why does this happen?






abstract-algebra group-theory ring-theory prime-numbers







share|cite|improve this question




















  • 6




    For an element to be prime this must hold for all a,b, not just one example.
    – Erik Parkinson
    1 hour ago










  • You have shown that there exists $c$ such that $4|c$ and $forall a,b$ such that $ab=c, 4|a$ or $4|b$. Your definition demands this to be true for every $c$, not just 16.
    – RcnSc
    6 mins ago














1












1








1







So, I've been given the actual definition of a prime for the first time, as opposed to the definition of an irreducible which I was previously taught (as is customary), it goes as follows:



An element $p$ of a ring $R$ is called "prime" if $a,bin R$ and $p|abrightarrow p|a$ or $p|b$



$4^2 = 16$. So inversely $4|16$. Now let's look at all the combinations of two integers $a,bin R$ where $ab = 16$ and whether $p|a$ or $p|b$:



$1 * 16$ ($4|16$)



$2 * 8$ ($4|8$)



$4 * 4$ ($4|4$)



By my logic, $4$ is prime.



However, as we all know $4 = 2^2$. But $2$ is not a unit so $4$ is an irreducible. Since the ring of integers is an integral domain $4$ must, therefore, be prime aswell. Why does this happen?






abstract-algebra group-theory ring-theory prime-numbers







share|cite|improve this question















So, I've been given the actual definition of a prime for the first time, as opposed to the definition of an irreducible which I was previously taught (as is customary), it goes as follows:



An element $p$ of a ring $R$ is called "prime" if $a,bin R$ and $p|abrightarrow p|a$ or $p|b$



$4^2 = 16$. So inversely $4|16$. Now let's look at all the combinations of two integers $a,bin R$ where $ab = 16$ and whether $p|a$ or $p|b$:



$1 * 16$ ($4|16$)



$2 * 8$ ($4|8$)



$4 * 4$ ($4|4$)



By my logic, $4$ is prime.



However, as we all know $4 = 2^2$. But $2$ is not a unit so $4$ is an irreducible. Since the ring of integers is an integral domain $4$ must, therefore, be prime aswell. Why does this happen?






abstract-algebra group-theory ring-theory prime-numbers




abstract-algebra group-theory ring-theory prime-numbers






share|cite|improve this question















share|cite|improve this question













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edited 1 hour ago









Asaf Karagila♦

301k32425755




301k32425755










asked 1 hour ago









PolymorphismPrince

1185




1185








  • 6




    For an element to be prime this must hold for all a,b, not just one example.
    – Erik Parkinson
    1 hour ago










  • You have shown that there exists $c$ such that $4|c$ and $forall a,b$ such that $ab=c, 4|a$ or $4|b$. Your definition demands this to be true for every $c$, not just 16.
    – RcnSc
    6 mins ago














  • 6




    For an element to be prime this must hold for all a,b, not just one example.
    – Erik Parkinson
    1 hour ago










  • You have shown that there exists $c$ such that $4|c$ and $forall a,b$ such that $ab=c, 4|a$ or $4|b$. Your definition demands this to be true for every $c$, not just 16.
    – RcnSc
    6 mins ago








6




6




For an element to be prime this must hold for all a,b, not just one example.
– Erik Parkinson
1 hour ago




For an element to be prime this must hold for all a,b, not just one example.
– Erik Parkinson
1 hour ago












You have shown that there exists $c$ such that $4|c$ and $forall a,b$ such that $ab=c, 4|a$ or $4|b$. Your definition demands this to be true for every $c$, not just 16.
– RcnSc
6 mins ago




You have shown that there exists $c$ such that $4|c$ and $forall a,b$ such that $ab=c, 4|a$ or $4|b$. Your definition demands this to be true for every $c$, not just 16.
– RcnSc
6 mins ago










1 Answer
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6














Take $12$.



$12=6cdot2$, but $6$ is not divided by $4$ and $2$ is not divided by $4$, which says that $4$ is not prime.






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    6














    Take $12$.



    $12=6cdot2$, but $6$ is not divided by $4$ and $2$ is not divided by $4$, which says that $4$ is not prime.






    share|cite|improve this answer


























      6














      Take $12$.



      $12=6cdot2$, but $6$ is not divided by $4$ and $2$ is not divided by $4$, which says that $4$ is not prime.






      share|cite|improve this answer
























        6












        6








        6






        Take $12$.



        $12=6cdot2$, but $6$ is not divided by $4$ and $2$ is not divided by $4$, which says that $4$ is not prime.






        share|cite|improve this answer












        Take $12$.



        $12=6cdot2$, but $6$ is not divided by $4$ and $2$ is not divided by $4$, which says that $4$ is not prime.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        Michael Rozenberg

        96.4k1588186




        96.4k1588186






























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