$K[X^2,X^3]$ is a non-unique factorization ring
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Let $K$ be any field. I would like to prove that any element of $K[X^2,X^3]$ can be written as a product of irreducible elements, in a possibly non-unique fashion. The non-unique part can be easily proven when noticing that
$$X^6 = (X^2)^3=(X^3)^2$$
and given that $X^2$ and $X^3$ are both irreducible in $K[X^2,X^3]$ (writing any of them as a product of two non-invertible elements would lie a factor of degree $1$, which can not be an element of $K[X^2,X^3]$).
However, I can't find how to prove the existence of such a decomposition. Could someone please give a hand for this exercise ?
abstract-algebra polynomials ring-theory
asked Nov 17 at 13:58
Suzet
2,574527
add a comment |
up vote
4
down vote
favorite
Let $K$ be any field. I would like to prove that any element of $K[X^2,X^3]$ can be written as a product of irreducible elements, in a possibly non-unique fashion. The non-unique part can be easily proven when noticing that
$$X^6 = (X^2)^3=(X^3)^2$$
and given that $X^2$ and $X^3$ are both irreducible in $K[X^2,X^3]$ (writing any of them as a product of two non-invertible elements would lie a factor of degree $1$, which can not be an element of $K[X^2,X^3]$).
However, I can't find how to prove the existence of such a decomposition. Could someone please give a hand for this exercise ?
abstract-algebra polynomials ring-theory
asked Nov 17 at 13:58
Suzet
2,574527
1
Isn't that ring Noetherian? Then you can use the fact that every element in a Noetherian ring is product of irreducibles.
– Bias of Priene
Nov 17 at 14:11
I was not aware of this result - or rather, I totally forgot it. Thank you very much !
– Suzet
Nov 17 at 14:25
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $K$ be any field. I would like to prove that any element of $K[X^2,X^3]$ can be written as a product of irreducible elements, in a possibly non-unique fashion. The non-unique part can be easily proven when noticing that
$$X^6 = (X^2)^3=(X^3)^2$$
and given that $X^2$ and $X^3$ are both irreducible in $K[X^2,X^3]$ (writing any of them as a product of two non-invertible elements would lie a factor of degree $1$, which can not be an element of $K[X^2,X^3]$).
However, I can't find how to prove the existence of such a decomposition. Could someone please give a hand for this exercise ?
abstract-algebra polynomials ring-theory
asked Nov 17 at 13:58
Suzet
2,574527
Let $K$ be any field. I would like to prove that any element of $K[X^2,X^3]$ can be written as a product of irreducible elements, in a possibly non-unique fashion. The non-unique part can be easily proven when noticing that
$$X^6 = (X^2)^3=(X^3)^2$$
and given that $X^2$ and $X^3$ are both irreducible in $K[X^2,X^3]$ (writing any of them as a product of two non-invertible elements would lie a factor of degree $1$, which can not be an element of $K[X^2,X^3]$).
However, I can't find how to prove the existence of such a decomposition. Could someone please give a hand for this exercise ?
abstract-algebra polynomials ring-theory
abstract-algebra polynomials ring-theory
asked Nov 17 at 13:58
Suzet
2,574527
asked Nov 17 at 13:58
Suzet
2,574527
asked Nov 17 at 13:58
Suzet
2,574527
asked Nov 17 at 13:58
Suzet
2,574527
asked Nov 17 at 13:58
Suzet
2,574527
2,574527
1
Isn't that ring Noetherian? Then you can use the fact that every element in a Noetherian ring is product of irreducibles.
– Bias of Priene
Nov 17 at 14:11
I was not aware of this result - or rather, I totally forgot it. Thank you very much !
– Suzet
Nov 17 at 14:25
add a comment |
1
Isn't that ring Noetherian? Then you can use the fact that every element in a Noetherian ring is product of irreducibles.
– Bias of Priene
Nov 17 at 14:11
I was not aware of this result - or rather, I totally forgot it. Thank you very much !
– Suzet
Nov 17 at 14:25
1
1
Isn't that ring Noetherian? Then you can use the fact that every element in a Noetherian ring is product of irreducibles.
– Bias of Priene
Nov 17 at 14:11
Isn't that ring Noetherian? Then you can use the fact that every element in a Noetherian ring is product of irreducibles.
– Bias of Priene
Nov 17 at 14:11
I was not aware of this result - or rather, I totally forgot it. Thank you very much !
– Suzet
Nov 17 at 14:25
I was not aware of this result - or rather, I totally forgot it. Thank you very much !
– Suzet
Nov 17 at 14:25
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
The $K$-algebra $K[X^2,X^3]$ is finitely generated over $K$, so by Hilbert's basis theorem, it is noetherian. In particular, it satisfies ACC on principal ideals, hence every element has a factorization into irreducibles.
answered Nov 17 at 14:13
barto
13.6k32682
Thank you very much ! I absolutely forgot about this result.
– Suzet
Nov 17 at 14:26
add a comment |
up vote
1
down vote
Another approach: unique factorization domains are normal, that is, integrally closed in their field of fractions. But the field of fractions of your domain is $k(x)$, and the closure of your domain in this is $k[x]$.
answered Nov 17 at 15:06
Pedro Tamaroff♦
95.7k10150295
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The $K$-algebra $K[X^2,X^3]$ is finitely generated over $K$, so by Hilbert's basis theorem, it is noetherian. In particular, it satisfies ACC on principal ideals, hence every element has a factorization into irreducibles.
answered Nov 17 at 14:13
barto
13.6k32682
Thank you very much ! I absolutely forgot about this result.
– Suzet
Nov 17 at 14:26
add a comment |
up vote
3
down vote
accepted
The $K$-algebra $K[X^2,X^3]$ is finitely generated over $K$, so by Hilbert's basis theorem, it is noetherian. In particular, it satisfies ACC on principal ideals, hence every element has a factorization into irreducibles.
answered Nov 17 at 14:13
barto
13.6k32682
Thank you very much ! I absolutely forgot about this result.
– Suzet
Nov 17 at 14:26
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The $K$-algebra $K[X^2,X^3]$ is finitely generated over $K$, so by Hilbert's basis theorem, it is noetherian. In particular, it satisfies ACC on principal ideals, hence every element has a factorization into irreducibles.
answered Nov 17 at 14:13
barto
13.6k32682
The $K$-algebra $K[X^2,X^3]$ is finitely generated over $K$, so by Hilbert's basis theorem, it is noetherian. In particular, it satisfies ACC on principal ideals, hence every element has a factorization into irreducibles.
answered Nov 17 at 14:13
barto
13.6k32682
answered Nov 17 at 14:13
barto
13.6k32682
answered Nov 17 at 14:13
barto
13.6k32682
answered Nov 17 at 14:13
barto
13.6k32682
13.6k32682
Thank you very much ! I absolutely forgot about this result.
– Suzet
Nov 17 at 14:26
add a comment |
Thank you very much ! I absolutely forgot about this result.
– Suzet
Nov 17 at 14:26
Thank you very much ! I absolutely forgot about this result.
– Suzet
Nov 17 at 14:26
Thank you very much ! I absolutely forgot about this result.
– Suzet
Nov 17 at 14:26
add a comment |
up vote
1
down vote
Another approach: unique factorization domains are normal, that is, integrally closed in their field of fractions. But the field of fractions of your domain is $k(x)$, and the closure of your domain in this is $k[x]$.
answered Nov 17 at 15:06
Pedro Tamaroff♦
95.7k10150295
add a comment |
up vote
1
down vote
Another approach: unique factorization domains are normal, that is, integrally closed in their field of fractions. But the field of fractions of your domain is $k(x)$, and the closure of your domain in this is $k[x]$.
answered Nov 17 at 15:06
Pedro Tamaroff♦
95.7k10150295
add a comment |
up vote
1
down vote
up vote
1
down vote
Another approach: unique factorization domains are normal, that is, integrally closed in their field of fractions. But the field of fractions of your domain is $k(x)$, and the closure of your domain in this is $k[x]$.
answered Nov 17 at 15:06
Pedro Tamaroff♦
95.7k10150295
Another approach: unique factorization domains are normal, that is, integrally closed in their field of fractions. But the field of fractions of your domain is $k(x)$, and the closure of your domain in this is $k[x]$.
answered Nov 17 at 15:06
Pedro Tamaroff♦
95.7k10150295
answered Nov 17 at 15:06
Pedro Tamaroff♦
95.7k10150295
answered Nov 17 at 15:06
Pedro Tamaroff♦
95.7k10150295
answered Nov 17 at 15:06
Pedro Tamaroff♦
95.7k10150295
95.7k10150295
add a comment |
add a comment |
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Isn't that ring Noetherian? Then you can use the fact that every element in a Noetherian ring is product of irreducibles.
– Bias of Priene
Nov 17 at 14:11
I was not aware of this result - or rather, I totally forgot it. Thank you very much !
– Suzet
Nov 17 at 14:25