Looking for alternative argument $(x^2 - y^3, y^2 - z^3)subset k[x,y,z]$ is prime ideal.
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Consider $I=(x^2 - y^3, y^2 - z^3)subset k[x,y,z]$ as an ideal with $k$ a field.
$textbf{Q:}$ I am Looking for alternative argument to conclude $I$ is prime ideal. It is clear that I can use parametrization $k[x,y,z]to k[t]$ by parametrization $(x,y,z)to (t^9, t^6, t^4)$ and argue this descends to a monomorphism under quotient $k[x,y,z]/I$. Note this map is clearly not surjection. Hence this is not isomorphic to $A^1_k$ and this is already indicated by singularity at $(0,0,0)$. Can I conclude primeness of ideal $I$ by some other better argument? Say intersection,grobner basis,...
abstract-algebra algebraic-geometry
asked Nov 19 at 18:11
user45765
2,4352720
add a comment |
up vote
3
down vote
favorite
Consider $I=(x^2 - y^3, y^2 - z^3)subset k[x,y,z]$ as an ideal with $k$ a field.
$textbf{Q:}$ I am Looking for alternative argument to conclude $I$ is prime ideal. It is clear that I can use parametrization $k[x,y,z]to k[t]$ by parametrization $(x,y,z)to (t^9, t^6, t^4)$ and argue this descends to a monomorphism under quotient $k[x,y,z]/I$. Note this map is clearly not surjection. Hence this is not isomorphic to $A^1_k$ and this is already indicated by singularity at $(0,0,0)$. Can I conclude primeness of ideal $I$ by some other better argument? Say intersection,grobner basis,...
abstract-algebra algebraic-geometry
asked Nov 19 at 18:11
user45765
2,4352720
$(x,x^{-1},y,z) mapsto (t^9,t^{-9},t^4,t^6)$ is an isomorphism $k[x,x^{-1},y,z]/I to k(t)$ and $V(I) setminus (0,0,0) simeq A^1_k setminus (0)$ ?
– reuns
Nov 19 at 18:39
@reuns Then you are using embedding implicitly which is essentially showing $k[x,y,z]to k[t]$ inducing an embedding and this will induce $k-$algebra level isomorphism for function fields.
– user45765
Nov 19 at 21:25
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Consider $I=(x^2 - y^3, y^2 - z^3)subset k[x,y,z]$ as an ideal with $k$ a field.
$textbf{Q:}$ I am Looking for alternative argument to conclude $I$ is prime ideal. It is clear that I can use parametrization $k[x,y,z]to k[t]$ by parametrization $(x,y,z)to (t^9, t^6, t^4)$ and argue this descends to a monomorphism under quotient $k[x,y,z]/I$. Note this map is clearly not surjection. Hence this is not isomorphic to $A^1_k$ and this is already indicated by singularity at $(0,0,0)$. Can I conclude primeness of ideal $I$ by some other better argument? Say intersection,grobner basis,...
abstract-algebra algebraic-geometry
asked Nov 19 at 18:11
user45765
2,4352720
Consider $I=(x^2 - y^3, y^2 - z^3)subset k[x,y,z]$ as an ideal with $k$ a field.
$textbf{Q:}$ I am Looking for alternative argument to conclude $I$ is prime ideal. It is clear that I can use parametrization $k[x,y,z]to k[t]$ by parametrization $(x,y,z)to (t^9, t^6, t^4)$ and argue this descends to a monomorphism under quotient $k[x,y,z]/I$. Note this map is clearly not surjection. Hence this is not isomorphic to $A^1_k$ and this is already indicated by singularity at $(0,0,0)$. Can I conclude primeness of ideal $I$ by some other better argument? Say intersection,grobner basis,...
abstract-algebra algebraic-geometry
abstract-algebra algebraic-geometry
asked Nov 19 at 18:11
user45765
2,4352720
asked Nov 19 at 18:11
user45765
2,4352720
asked Nov 19 at 18:11
user45765
2,4352720
asked Nov 19 at 18:11
user45765
2,4352720
asked Nov 19 at 18:11
user45765
2,4352720
2,4352720
$(x,x^{-1},y,z) mapsto (t^9,t^{-9},t^4,t^6)$ is an isomorphism $k[x,x^{-1},y,z]/I to k(t)$ and $V(I) setminus (0,0,0) simeq A^1_k setminus (0)$ ?
– reuns
Nov 19 at 18:39
@reuns Then you are using embedding implicitly which is essentially showing $k[x,y,z]to k[t]$ inducing an embedding and this will induce $k-$algebra level isomorphism for function fields.
– user45765
Nov 19 at 21:25
add a comment |
$(x,x^{-1},y,z) mapsto (t^9,t^{-9},t^4,t^6)$ is an isomorphism $k[x,x^{-1},y,z]/I to k(t)$ and $V(I) setminus (0,0,0) simeq A^1_k setminus (0)$ ?
– reuns
Nov 19 at 18:39
@reuns Then you are using embedding implicitly which is essentially showing $k[x,y,z]to k[t]$ inducing an embedding and this will induce $k-$algebra level isomorphism for function fields.
– user45765
Nov 19 at 21:25
$(x,x^{-1},y,z) mapsto (t^9,t^{-9},t^4,t^6)$ is an isomorphism $k[x,x^{-1},y,z]/I to k(t)$ and $V(I) setminus (0,0,0) simeq A^1_k setminus (0)$ ?
– reuns
Nov 19 at 18:39
$(x,x^{-1},y,z) mapsto (t^9,t^{-9},t^4,t^6)$ is an isomorphism $k[x,x^{-1},y,z]/I to k(t)$ and $V(I) setminus (0,0,0) simeq A^1_k setminus (0)$ ?
– reuns
Nov 19 at 18:39
@reuns Then you are using embedding implicitly which is essentially showing $k[x,y,z]to k[t]$ inducing an embedding and this will induce $k-$algebra level isomorphism for function fields.
– user45765
Nov 19 at 21:25
@reuns Then you are using embedding implicitly which is essentially showing $k[x,y,z]to k[t]$ inducing an embedding and this will induce $k-$algebra level isomorphism for function fields.
– user45765
Nov 19 at 21:25
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Lemma: Let $f:Xto Y$ be a continuous map of topological spaces. If $X$ is irreducible, then $f(X)$ is irreducible.
Proof: If $f(X)$ had a decomposition in to two nontrivial proper closed subsets $Acup B$, then $f^{-1}(A)cup f^{-1}(B)$ would be a decomposition of $X$ in to two nontrivial proper closed subsets, contradicting irreducibility of $X$.
We apply this to the situation at hand in the following fashion: $V(I)$ is the image of the morphism $Bbb A^1toBbb A^3$ given by $tmapsto (t^9,t^6,t^4)$, so $V(I)$ is irreducible. Thus $I$ must be prime.
answered Nov 19 at 21:48
KReiser
9,08711335
Then in this setting, I really should think irreducibility as "connectedness" which transcends down to the map?
– user45765
Nov 19 at 21:56
2
Sort of - one can see that irreducibilty is a little finer than connectedness (it's possible for different irreducible components to meet), but going forwards along the map is exactly the same.
– KReiser
Nov 19 at 22:11
add a comment |
up vote
1
down vote
KReiser has given a really nice answer, but I just wanted to add that you're already essentially there with what you've written in your question. You've observed that the map $(x,y,z)mapsto (t^9,t^6,t^4)$ induces an injection $k[x,y,z]/I hookrightarrow k[t]$. The latter ring is a domain, and any subring of a domain is a domain, thus $I$ must be prime.
answered Nov 20 at 0:34
jgon
10.3k11638
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
Lemma: Let $f:Xto Y$ be a continuous map of topological spaces. If $X$ is irreducible, then $f(X)$ is irreducible.
Proof: If $f(X)$ had a decomposition in to two nontrivial proper closed subsets $Acup B$, then $f^{-1}(A)cup f^{-1}(B)$ would be a decomposition of $X$ in to two nontrivial proper closed subsets, contradicting irreducibility of $X$.
We apply this to the situation at hand in the following fashion: $V(I)$ is the image of the morphism $Bbb A^1toBbb A^3$ given by $tmapsto (t^9,t^6,t^4)$, so $V(I)$ is irreducible. Thus $I$ must be prime.
answered Nov 19 at 21:48
KReiser
9,08711335
Then in this setting, I really should think irreducibility as "connectedness" which transcends down to the map?
– user45765
Nov 19 at 21:56
2
Sort of - one can see that irreducibilty is a little finer than connectedness (it's possible for different irreducible components to meet), but going forwards along the map is exactly the same.
– KReiser
Nov 19 at 22:11
add a comment |
up vote
3
down vote
accepted
Lemma: Let $f:Xto Y$ be a continuous map of topological spaces. If $X$ is irreducible, then $f(X)$ is irreducible.
Proof: If $f(X)$ had a decomposition in to two nontrivial proper closed subsets $Acup B$, then $f^{-1}(A)cup f^{-1}(B)$ would be a decomposition of $X$ in to two nontrivial proper closed subsets, contradicting irreducibility of $X$.
We apply this to the situation at hand in the following fashion: $V(I)$ is the image of the morphism $Bbb A^1toBbb A^3$ given by $tmapsto (t^9,t^6,t^4)$, so $V(I)$ is irreducible. Thus $I$ must be prime.
answered Nov 19 at 21:48
KReiser
9,08711335
Then in this setting, I really should think irreducibility as "connectedness" which transcends down to the map?
– user45765
Nov 19 at 21:56
2
Sort of - one can see that irreducibilty is a little finer than connectedness (it's possible for different irreducible components to meet), but going forwards along the map is exactly the same.
– KReiser
Nov 19 at 22:11
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Lemma: Let $f:Xto Y$ be a continuous map of topological spaces. If $X$ is irreducible, then $f(X)$ is irreducible.
Proof: If $f(X)$ had a decomposition in to two nontrivial proper closed subsets $Acup B$, then $f^{-1}(A)cup f^{-1}(B)$ would be a decomposition of $X$ in to two nontrivial proper closed subsets, contradicting irreducibility of $X$.
We apply this to the situation at hand in the following fashion: $V(I)$ is the image of the morphism $Bbb A^1toBbb A^3$ given by $tmapsto (t^9,t^6,t^4)$, so $V(I)$ is irreducible. Thus $I$ must be prime.
answered Nov 19 at 21:48
KReiser
9,08711335
Lemma: Let $f:Xto Y$ be a continuous map of topological spaces. If $X$ is irreducible, then $f(X)$ is irreducible.
Proof: If $f(X)$ had a decomposition in to two nontrivial proper closed subsets $Acup B$, then $f^{-1}(A)cup f^{-1}(B)$ would be a decomposition of $X$ in to two nontrivial proper closed subsets, contradicting irreducibility of $X$.
We apply this to the situation at hand in the following fashion: $V(I)$ is the image of the morphism $Bbb A^1toBbb A^3$ given by $tmapsto (t^9,t^6,t^4)$, so $V(I)$ is irreducible. Thus $I$ must be prime.
answered Nov 19 at 21:48
KReiser
9,08711335
answered Nov 19 at 21:48
KReiser
9,08711335
answered Nov 19 at 21:48
KReiser
9,08711335
answered Nov 19 at 21:48
KReiser
9,08711335
9,08711335
Then in this setting, I really should think irreducibility as "connectedness" which transcends down to the map?
– user45765
Nov 19 at 21:56
2
Sort of - one can see that irreducibilty is a little finer than connectedness (it's possible for different irreducible components to meet), but going forwards along the map is exactly the same.
– KReiser
Nov 19 at 22:11
add a comment |
Then in this setting, I really should think irreducibility as "connectedness" which transcends down to the map?
– user45765
Nov 19 at 21:56
2
Sort of - one can see that irreducibilty is a little finer than connectedness (it's possible for different irreducible components to meet), but going forwards along the map is exactly the same.
– KReiser
Nov 19 at 22:11
Then in this setting, I really should think irreducibility as "connectedness" which transcends down to the map?
– user45765
Nov 19 at 21:56
Then in this setting, I really should think irreducibility as "connectedness" which transcends down to the map?
– user45765
Nov 19 at 21:56
2
2
Sort of - one can see that irreducibilty is a little finer than connectedness (it's possible for different irreducible components to meet), but going forwards along the map is exactly the same.
– KReiser
Nov 19 at 22:11
Sort of - one can see that irreducibilty is a little finer than connectedness (it's possible for different irreducible components to meet), but going forwards along the map is exactly the same.
– KReiser
Nov 19 at 22:11
add a comment |
up vote
1
down vote
KReiser has given a really nice answer, but I just wanted to add that you're already essentially there with what you've written in your question. You've observed that the map $(x,y,z)mapsto (t^9,t^6,t^4)$ induces an injection $k[x,y,z]/I hookrightarrow k[t]$. The latter ring is a domain, and any subring of a domain is a domain, thus $I$ must be prime.
answered Nov 20 at 0:34
jgon
10.3k11638
add a comment |
up vote
1
down vote
KReiser has given a really nice answer, but I just wanted to add that you're already essentially there with what you've written in your question. You've observed that the map $(x,y,z)mapsto (t^9,t^6,t^4)$ induces an injection $k[x,y,z]/I hookrightarrow k[t]$. The latter ring is a domain, and any subring of a domain is a domain, thus $I$ must be prime.
answered Nov 20 at 0:34
jgon
10.3k11638
add a comment |
up vote
1
down vote
up vote
1
down vote
KReiser has given a really nice answer, but I just wanted to add that you're already essentially there with what you've written in your question. You've observed that the map $(x,y,z)mapsto (t^9,t^6,t^4)$ induces an injection $k[x,y,z]/I hookrightarrow k[t]$. The latter ring is a domain, and any subring of a domain is a domain, thus $I$ must be prime.
answered Nov 20 at 0:34
jgon
10.3k11638
KReiser has given a really nice answer, but I just wanted to add that you're already essentially there with what you've written in your question. You've observed that the map $(x,y,z)mapsto (t^9,t^6,t^4)$ induces an injection $k[x,y,z]/I hookrightarrow k[t]$. The latter ring is a domain, and any subring of a domain is a domain, thus $I$ must be prime.
answered Nov 20 at 0:34
jgon
10.3k11638
answered Nov 20 at 0:34
jgon
10.3k11638
answered Nov 20 at 0:34
jgon
10.3k11638
answered Nov 20 at 0:34
jgon
10.3k11638
10.3k11638
add a comment |
add a comment |
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$(x,x^{-1},y,z) mapsto (t^9,t^{-9},t^4,t^6)$ is an isomorphism $k[x,x^{-1},y,z]/I to k(t)$ and $V(I) setminus (0,0,0) simeq A^1_k setminus (0)$ ?
– reuns
Nov 19 at 18:39
@reuns Then you are using embedding implicitly which is essentially showing $k[x,y,z]to k[t]$ inducing an embedding and this will induce $k-$algebra level isomorphism for function fields.
– user45765
Nov 19 at 21:25