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Does factorisation always exist?
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In the definition of UFD, there are two parts: existence and uniqueness. It is well-understood that the uniqueness property is useful e.g. $mathbbZ[sqrt-5]$. But isn't it true that every integral domain must satisfy the existence property?
Existence: Each element in $R$ which is neither zero nor unit can be written as product of irreducibles.
But this is always true, if an element can not be decomposed as product then it is irreducible.
abstract-algebra ring-theory
asked Jul 22 at 15:22
Xavier Yang
440314
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up vote
4
down vote
favorite
In the definition of UFD, there are two parts: existence and uniqueness. It is well-understood that the uniqueness property is useful e.g. $mathbbZ[sqrt-5]$. But isn't it true that every integral domain must satisfy the existence property?
Existence: Each element in $R$ which is neither zero nor unit can be written as product of irreducibles.
But this is always true, if an element can not be decomposed as product then it is irreducible.
abstract-algebra ring-theory
asked Jul 22 at 15:22
Xavier Yang
440314
6
This is not true, e.g. in $mathscrO(mathbbC)$ the irreducible elements are entire functions with exactly one zero. An entire function with infinitely many zeros, like $sin$, can therefore not be written as a (finite) product of irreducibles.
– Daniel Fischer♦
Jul 22 at 15:26
$O(mathbbC)$ is the set of all entire functions?
– Xavier Yang
Jul 22 at 15:31
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
In the definition of UFD, there are two parts: existence and uniqueness. It is well-understood that the uniqueness property is useful e.g. $mathbbZ[sqrt-5]$. But isn't it true that every integral domain must satisfy the existence property?
Existence: Each element in $R$ which is neither zero nor unit can be written as product of irreducibles.
But this is always true, if an element can not be decomposed as product then it is irreducible.
abstract-algebra ring-theory
asked Jul 22 at 15:22
Xavier Yang
440314
In the definition of UFD, there are two parts: existence and uniqueness. It is well-understood that the uniqueness property is useful e.g. $mathbbZ[sqrt-5]$. But isn't it true that every integral domain must satisfy the existence property?
Existence: Each element in $R$ which is neither zero nor unit can be written as product of irreducibles.
But this is always true, if an element can not be decomposed as product then it is irreducible.
abstract-algebra ring-theory
asked Jul 22 at 15:22
Xavier Yang
440314
asked Jul 22 at 15:22
Xavier Yang
440314
asked Jul 22 at 15:22
Xavier Yang
440314
asked Jul 22 at 15:22
Xavier Yang
440314
440314
6
This is not true, e.g. in $mathscrO(mathbbC)$ the irreducible elements are entire functions with exactly one zero. An entire function with infinitely many zeros, like $sin$, can therefore not be written as a (finite) product of irreducibles.
– Daniel Fischer♦
Jul 22 at 15:26
$O(mathbbC)$ is the set of all entire functions?
– Xavier Yang
Jul 22 at 15:31
add a comment |Â
6
This is not true, e.g. in $mathscrO(mathbbC)$ the irreducible elements are entire functions with exactly one zero. An entire function with infinitely many zeros, like $sin$, can therefore not be written as a (finite) product of irreducibles.
– Daniel Fischer♦
Jul 22 at 15:26
$O(mathbbC)$ is the set of all entire functions?
– Xavier Yang
Jul 22 at 15:31
6
6
This is not true, e.g. in $mathscrO(mathbbC)$ the irreducible elements are entire functions with exactly one zero. An entire function with infinitely many zeros, like $sin$, can therefore not be written as a (finite) product of irreducibles.
– Daniel Fischer♦
Jul 22 at 15:26
This is not true, e.g. in $mathscrO(mathbbC)$ the irreducible elements are entire functions with exactly one zero. An entire function with infinitely many zeros, like $sin$, can therefore not be written as a (finite) product of irreducibles.
– Daniel Fischer♦
Jul 22 at 15:26
$O(mathbbC)$ is the set of all entire functions?
– Xavier Yang
Jul 22 at 15:31
$O(mathbbC)$ is the set of all entire functions?
– Xavier Yang
Jul 22 at 15:31
add a comment |Â
1 Answer
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For another example, consider the ring of algebraic integers. If you try to factor $2$ in this ring, for instance, you'll see that $2 = (sqrt2)^2 = (sqrt[3]2)^3 = cdots$ since $sqrt[n]2$ is integral for each $n$.
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
For another example, consider the ring of algebraic integers. If you try to factor $2$ in this ring, for instance, you'll see that $2 = (sqrt2)^2 = (sqrt[3]2)^3 = cdots$ since $sqrt[n]2$ is integral for each $n$.
add a comment |Â
up vote
7
down vote
accepted
For another example, consider the ring of algebraic integers. If you try to factor $2$ in this ring, for instance, you'll see that $2 = (sqrt2)^2 = (sqrt[3]2)^3 = cdots$ since $sqrt[n]2$ is integral for each $n$.
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
For another example, consider the ring of algebraic integers. If you try to factor $2$ in this ring, for instance, you'll see that $2 = (sqrt2)^2 = (sqrt[3]2)^3 = cdots$ since $sqrt[n]2$ is integral for each $n$.
For another example, consider the ring of algebraic integers. If you try to factor $2$ in this ring, for instance, you'll see that $2 = (sqrt2)^2 = (sqrt[3]2)^3 = cdots$ since $sqrt[n]2$ is integral for each $n$.
answered Jul 22 at 15:49
community wiki
Quasicoherent
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6
This is not true, e.g. in $mathscrO(mathbbC)$ the irreducible elements are entire functions with exactly one zero. An entire function with infinitely many zeros, like $sin$, can therefore not be written as a (finite) product of irreducibles.
– Daniel Fischer♦
Jul 22 at 15:26
$O(mathbbC)$ is the set of all entire functions?
– Xavier Yang
Jul 22 at 15:31