$K[X^2,X^3]$ is a non-unique factorization ring
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
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♦
95.6k10149295
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♦
95.6k10149295
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♦
95.6k10149295
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♦
95.6k10149295
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♦
95.6k10149295
answered Nov 17 at 15:06
Pedro Tamaroff♦
95.6k10149295
answered Nov 17 at 15:06
Pedro Tamaroff♦
95.6k10149295
answered Nov 17 at 15:06
Pedro Tamaroff♦
95.6k10149295
95.6k10149295
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3002390%2fkx2-x3-is-a-non-unique-factorization-ring%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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