Why doesn't this definition hold up when we square numbers?
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
edited 1 hour ago
Asaf Karagila♦
301k32425755
asked 1 hour ago
PolymorphismPrince
1185
add a comment |
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
edited 1 hour ago
Asaf Karagila♦
301k32425755
asked 1 hour ago
PolymorphismPrince
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
add a comment |
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
edited 1 hour ago
Asaf Karagila♦
301k32425755
asked 1 hour ago
PolymorphismPrince
1185
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
edited 1 hour ago
Asaf Karagila♦
301k32425755
asked 1 hour ago
PolymorphismPrince
1185
edited 1 hour ago
Asaf Karagila♦
301k32425755
asked 1 hour ago
PolymorphismPrince
1185
edited 1 hour ago
Asaf Karagila♦
301k32425755
edited 1 hour ago
Asaf Karagila♦
301k32425755
edited 1 hour ago
Asaf Karagila♦
301k32425755
301k32425755
asked 1 hour ago
PolymorphismPrince
1185
asked 1 hour ago
PolymorphismPrince
1185
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
add a comment |
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
add a comment |
1 Answer
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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.
answered 1 hour ago
Michael Rozenberg
96.4k1588186
add a comment |
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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.
answered 1 hour ago
Michael Rozenberg
96.4k1588186
add a comment |
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.
answered 1 hour ago
Michael Rozenberg
96.4k1588186
add a comment |
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.
answered 1 hour ago
Michael Rozenberg
96.4k1588186
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.
answered 1 hour ago
Michael Rozenberg
96.4k1588186
answered 1 hour ago
Michael Rozenberg
96.4k1588186
answered 1 hour ago
Michael Rozenberg
96.4k1588186
answered 1 hour ago
Michael Rozenberg
96.4k1588186
96.4k1588186
add a comment |
add a comment |
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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