Element which is prime but not irreducible.
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I wanted to find an element which is irreducible but not prime. I found on wiki the example that says that $x^2$ is prime but not irreducible in $mathbb{Q}[x]/(x^2+x)$.
My reasoning why this is not a irreducible element is as following.
Suppose $x^2$ is irreducible. In particular, this means it isn't a unit.
But $x^2=xx=(-x^2) (-x^2)=x^2x^2$ And so this would be a product of two non-units. Does this makes sense?
To show that it is not a prime I have to show that $x^2 vert ab implies x^2 vert a$ or $x^2 vert b$ but I don't know how to show this.
abstract-algebra
asked Nov 17 at 18:09
roi_saumon
31117
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I wanted to find an element which is irreducible but not prime. I found on wiki the example that says that $x^2$ is prime but not irreducible in $mathbb{Q}[x]/(x^2+x)$.
My reasoning why this is not a irreducible element is as following.
Suppose $x^2$ is irreducible. In particular, this means it isn't a unit.
But $x^2=xx=(-x^2) (-x^2)=x^2x^2$ And so this would be a product of two non-units. Does this makes sense?
To show that it is not a prime I have to show that $x^2 vert ab implies x^2 vert a$ or $x^2 vert b$ but I don't know how to show this.
abstract-algebra
asked Nov 17 at 18:09
roi_saumon
31117
1
"Suppose $x^2$ is irreducible. In particular, this means it isn't a prime." You meant to say it isn't a unit, right?
– saulspatz
Nov 17 at 18:18
Beware $ $ There is no standard of "irreducible" in rings with zero-divisors. Various incompatible definitions are in use. e.g. see the paper linked here, where (Corollary 2.7) idempotents are irreducible $iff$ prime.
– Bill Dubuque
Nov 17 at 22:22
@BillDubuque okay, I see your point. So is it true that for all $a$, $b$ such that $x^2=ab$ we have either $x^2 vert a$ and $a vert x^2$ or $x^2 vert b$ and $b vert x^2$? Also, for the definition of prime there is no problem no? $ab in (p) iff p vert ab$
– roi_saumon
Nov 17 at 23:37
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I wanted to find an element which is irreducible but not prime. I found on wiki the example that says that $x^2$ is prime but not irreducible in $mathbb{Q}[x]/(x^2+x)$.
My reasoning why this is not a irreducible element is as following.
Suppose $x^2$ is irreducible. In particular, this means it isn't a unit.
But $x^2=xx=(-x^2) (-x^2)=x^2x^2$ And so this would be a product of two non-units. Does this makes sense?
To show that it is not a prime I have to show that $x^2 vert ab implies x^2 vert a$ or $x^2 vert b$ but I don't know how to show this.
abstract-algebra
asked Nov 17 at 18:09
roi_saumon
31117
I wanted to find an element which is irreducible but not prime. I found on wiki the example that says that $x^2$ is prime but not irreducible in $mathbb{Q}[x]/(x^2+x)$.
My reasoning why this is not a irreducible element is as following.
Suppose $x^2$ is irreducible. In particular, this means it isn't a unit.
But $x^2=xx=(-x^2) (-x^2)=x^2x^2$ And so this would be a product of two non-units. Does this makes sense?
To show that it is not a prime I have to show that $x^2 vert ab implies x^2 vert a$ or $x^2 vert b$ but I don't know how to show this.
abstract-algebra
abstract-algebra
asked Nov 17 at 18:09
roi_saumon
31117
asked Nov 17 at 18:09
roi_saumon
31117
edited Nov 17 at 18:25
asked Nov 17 at 18:09
roi_saumon
31117
asked Nov 17 at 18:09
roi_saumon
31117
asked Nov 17 at 18:09
roi_saumon
31117
31117
1
"Suppose $x^2$ is irreducible. In particular, this means it isn't a prime." You meant to say it isn't a unit, right?
– saulspatz
Nov 17 at 18:18
Beware $ $ There is no standard of "irreducible" in rings with zero-divisors. Various incompatible definitions are in use. e.g. see the paper linked here, where (Corollary 2.7) idempotents are irreducible $iff$ prime.
– Bill Dubuque
Nov 17 at 22:22
@BillDubuque okay, I see your point. So is it true that for all $a$, $b$ such that $x^2=ab$ we have either $x^2 vert a$ and $a vert x^2$ or $x^2 vert b$ and $b vert x^2$? Also, for the definition of prime there is no problem no? $ab in (p) iff p vert ab$
– roi_saumon
Nov 17 at 23:37
add a comment |
1
"Suppose $x^2$ is irreducible. In particular, this means it isn't a prime." You meant to say it isn't a unit, right?
– saulspatz
Nov 17 at 18:18
Beware $ $ There is no standard of "irreducible" in rings with zero-divisors. Various incompatible definitions are in use. e.g. see the paper linked here, where (Corollary 2.7) idempotents are irreducible $iff$ prime.
– Bill Dubuque
Nov 17 at 22:22
@BillDubuque okay, I see your point. So is it true that for all $a$, $b$ such that $x^2=ab$ we have either $x^2 vert a$ and $a vert x^2$ or $x^2 vert b$ and $b vert x^2$? Also, for the definition of prime there is no problem no? $ab in (p) iff p vert ab$
– roi_saumon
Nov 17 at 23:37
1
1
"Suppose $x^2$ is irreducible. In particular, this means it isn't a prime." You meant to say it isn't a unit, right?
– saulspatz
Nov 17 at 18:18
"Suppose $x^2$ is irreducible. In particular, this means it isn't a prime." You meant to say it isn't a unit, right?
– saulspatz
Nov 17 at 18:18
Beware $ $ There is no standard of "irreducible" in rings with zero-divisors. Various incompatible definitions are in use. e.g. see the paper linked here, where (Corollary 2.7) idempotents are irreducible $iff$ prime.
– Bill Dubuque
Nov 17 at 22:22
Beware $ $ There is no standard of "irreducible" in rings with zero-divisors. Various incompatible definitions are in use. e.g. see the paper linked here, where (Corollary 2.7) idempotents are irreducible $iff$ prime.
– Bill Dubuque
Nov 17 at 22:22
@BillDubuque okay, I see your point. So is it true that for all $a$, $b$ such that $x^2=ab$ we have either $x^2 vert a$ and $a vert x^2$ or $x^2 vert b$ and $b vert x^2$? Also, for the definition of prime there is no problem no? $ab in (p) iff p vert ab$
– roi_saumon
Nov 17 at 23:37
@BillDubuque okay, I see your point. So is it true that for all $a$, $b$ such that $x^2=ab$ we have either $x^2 vert a$ and $a vert x^2$ or $x^2 vert b$ and $b vert x^2$? Also, for the definition of prime there is no problem no? $ab in (p) iff p vert ab$
– roi_saumon
Nov 17 at 23:37
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"Suppose $x^2$ is irreducible. In particular, this means it isn't a prime." You meant to say it isn't a unit, right?
– saulspatz
Nov 17 at 18:18
Beware $ $ There is no standard of "irreducible" in rings with zero-divisors. Various incompatible definitions are in use. e.g. see the paper linked here, where (Corollary 2.7) idempotents are irreducible $iff$ prime.
– Bill Dubuque
Nov 17 at 22:22
@BillDubuque okay, I see your point. So is it true that for all $a$, $b$ such that $x^2=ab$ we have either $x^2 vert a$ and $a vert x^2$ or $x^2 vert b$ and $b vert x^2$? Also, for the definition of prime there is no problem no? $ab in (p) iff p vert ab$
– roi_saumon
Nov 17 at 23:37