Proving if k-algebras are finitely generated and noetherian
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Given two $k$-algebras $R:=koplus x^2 , k[x] subseteq k[x]$ and $S:= koplus xy ,, k[x,y] subseteq k[x,y]$ I need to show if they are finitely generated and if they are noetherian.
For $R$ to be finitely generated k-Algebra, we need to show that there exist finitely many elements, $a_1, a_2,....a_n$ such that $R=k[a_1,a_2,...a_n].$ But we know that it is contained in $k[x]$. For the second Question I Need to show if there is an ascending chain of ideals.
Can somebody give me a hint as how to proceed ? Many thanks.
abstract-algebra
asked Nov 25 at 14:15
user249018
199117
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up vote
1
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favorite
Given two $k$-algebras $R:=koplus x^2 , k[x] subseteq k[x]$ and $S:= koplus xy ,, k[x,y] subseteq k[x,y]$ I need to show if they are finitely generated and if they are noetherian.
For $R$ to be finitely generated k-Algebra, we need to show that there exist finitely many elements, $a_1, a_2,....a_n$ such that $R=k[a_1,a_2,...a_n].$ But we know that it is contained in $k[x]$. For the second Question I Need to show if there is an ascending chain of ideals.
Can somebody give me a hint as how to proceed ? Many thanks.
abstract-algebra
asked Nov 25 at 14:15
user249018
199117
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given two $k$-algebras $R:=koplus x^2 , k[x] subseteq k[x]$ and $S:= koplus xy ,, k[x,y] subseteq k[x,y]$ I need to show if they are finitely generated and if they are noetherian.
For $R$ to be finitely generated k-Algebra, we need to show that there exist finitely many elements, $a_1, a_2,....a_n$ such that $R=k[a_1,a_2,...a_n].$ But we know that it is contained in $k[x]$. For the second Question I Need to show if there is an ascending chain of ideals.
Can somebody give me a hint as how to proceed ? Many thanks.
abstract-algebra
asked Nov 25 at 14:15
user249018
199117
Given two $k$-algebras $R:=koplus x^2 , k[x] subseteq k[x]$ and $S:= koplus xy ,, k[x,y] subseteq k[x,y]$ I need to show if they are finitely generated and if they are noetherian.
For $R$ to be finitely generated k-Algebra, we need to show that there exist finitely many elements, $a_1, a_2,....a_n$ such that $R=k[a_1,a_2,...a_n].$ But we know that it is contained in $k[x]$. For the second Question I Need to show if there is an ascending chain of ideals.
Can somebody give me a hint as how to proceed ? Many thanks.
abstract-algebra
abstract-algebra
asked Nov 25 at 14:15
user249018
199117
asked Nov 25 at 14:15
user249018
199117
asked Nov 25 at 14:15
user249018
199117
asked Nov 25 at 14:15
user249018
199117
asked Nov 25 at 14:15
user249018
199117
199117
add a comment |
add a comment |
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$R = k oplus x^2 k[x] = k [ x^2, x^3 ]$
Also, $S$ is isomorphic to the semi-group algebra $k [ mathbb N times mathbb N ]$, which is not a finitely generated semi-group.
None of the $(n, 1)$ can be written in terms of others. So $S$ is not Noetherian.
answered Nov 25 at 15:39
R.C.Cowsik
31513
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
$R = k oplus x^2 k[x] = k [ x^2, x^3 ]$
Also, $S$ is isomorphic to the semi-group algebra $k [ mathbb N times mathbb N ]$, which is not a finitely generated semi-group.
None of the $(n, 1)$ can be written in terms of others. So $S$ is not Noetherian.
answered Nov 25 at 15:39
R.C.Cowsik
31513
add a comment |
up vote
0
down vote
$R = k oplus x^2 k[x] = k [ x^2, x^3 ]$
Also, $S$ is isomorphic to the semi-group algebra $k [ mathbb N times mathbb N ]$, which is not a finitely generated semi-group.
None of the $(n, 1)$ can be written in terms of others. So $S$ is not Noetherian.
answered Nov 25 at 15:39
R.C.Cowsik
31513
add a comment |
up vote
0
down vote
up vote
0
down vote
$R = k oplus x^2 k[x] = k [ x^2, x^3 ]$
Also, $S$ is isomorphic to the semi-group algebra $k [ mathbb N times mathbb N ]$, which is not a finitely generated semi-group.
None of the $(n, 1)$ can be written in terms of others. So $S$ is not Noetherian.
answered Nov 25 at 15:39
R.C.Cowsik
31513
$R = k oplus x^2 k[x] = k [ x^2, x^3 ]$
Also, $S$ is isomorphic to the semi-group algebra $k [ mathbb N times mathbb N ]$, which is not a finitely generated semi-group.
None of the $(n, 1)$ can be written in terms of others. So $S$ is not Noetherian.
answered Nov 25 at 15:39
R.C.Cowsik
31513
answered Nov 25 at 15:39
R.C.Cowsik
31513
answered Nov 25 at 15:39
R.C.Cowsik
31513
answered Nov 25 at 15:39
R.C.Cowsik
31513
31513
add a comment |
add a comment |
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