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Irreducible affine algebraic varieties corresponds to prime ideals
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If $V$ is an irreducible affine variety, then $I(V)$ is prime.
A quick search on the site, I have found questions on the other direction (prime ideals implies irreducibility) but not in this direction.
I have translated the defintion of irreducibility to a condition on ideals: if $V_1$ and $V_2$ are subvarieties of $V$ such that $I(V) =I(V_1)cap I(V_2)$, then $I(V)$ equals to $I(V_1)$ or $I(V_2)$.
Now to show that it is a prime ideal, suppose $fgin I(V)$. My idea is to construct two ideals where each contains one of $f$ and $g$. I have tried putting $I_1 = I(V(f) cap V) $ and so on but it doesn't work so far. Any hint on how I could continue?
algebraic-geometry maximal-and-prime-ideals affine-varieties
asked Aug 6 at 18:22
lEm
3,0581617
add a comment |Â
up vote
1
down vote
favorite
If $V$ is an irreducible affine variety, then $I(V)$ is prime.
A quick search on the site, I have found questions on the other direction (prime ideals implies irreducibility) but not in this direction.
I have translated the defintion of irreducibility to a condition on ideals: if $V_1$ and $V_2$ are subvarieties of $V$ such that $I(V) =I(V_1)cap I(V_2)$, then $I(V)$ equals to $I(V_1)$ or $I(V_2)$.
Now to show that it is a prime ideal, suppose $fgin I(V)$. My idea is to construct two ideals where each contains one of $f$ and $g$. I have tried putting $I_1 = I(V(f) cap V) $ and so on but it doesn't work so far. Any hint on how I could continue?
algebraic-geometry maximal-and-prime-ideals affine-varieties
asked Aug 6 at 18:22
lEm
3,0581617
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If $V$ is an irreducible affine variety, then $I(V)$ is prime.
A quick search on the site, I have found questions on the other direction (prime ideals implies irreducibility) but not in this direction.
I have translated the defintion of irreducibility to a condition on ideals: if $V_1$ and $V_2$ are subvarieties of $V$ such that $I(V) =I(V_1)cap I(V_2)$, then $I(V)$ equals to $I(V_1)$ or $I(V_2)$.
Now to show that it is a prime ideal, suppose $fgin I(V)$. My idea is to construct two ideals where each contains one of $f$ and $g$. I have tried putting $I_1 = I(V(f) cap V) $ and so on but it doesn't work so far. Any hint on how I could continue?
algebraic-geometry maximal-and-prime-ideals affine-varieties
asked Aug 6 at 18:22
lEm
3,0581617
If $V$ is an irreducible affine variety, then $I(V)$ is prime.
A quick search on the site, I have found questions on the other direction (prime ideals implies irreducibility) but not in this direction.
I have translated the defintion of irreducibility to a condition on ideals: if $V_1$ and $V_2$ are subvarieties of $V$ such that $I(V) =I(V_1)cap I(V_2)$, then $I(V)$ equals to $I(V_1)$ or $I(V_2)$.
Now to show that it is a prime ideal, suppose $fgin I(V)$. My idea is to construct two ideals where each contains one of $f$ and $g$. I have tried putting $I_1 = I(V(f) cap V) $ and so on but it doesn't work so far. Any hint on how I could continue?
algebraic-geometry maximal-and-prime-ideals affine-varieties
asked Aug 6 at 18:22
lEm
3,0581617
asked Aug 6 at 18:22
lEm
3,0581617
asked Aug 6 at 18:22
lEm
3,0581617
asked Aug 6 at 18:22
lEm
3,0581617
3,0581617
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
Hint: $V(fg)=V(f)cup V(g)$ and $V=Vcap V(fg)=(V(f)cap V)cup (V(g)cap V)$. This implies that $V(f)cap V=V$ or $V(g)cap V=g$.
answered Aug 6 at 18:39
Tsemo Aristide
51.5k11243
I knew it would have a simple answer. Somehow I didn't think about working with varieties instead of dealing ideals. And since I'm new to it, it's slightly confusing at times. Thanks for the answer :)
– lEm
Aug 6 at 18:45
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Hint: $V(fg)=V(f)cup V(g)$ and $V=Vcap V(fg)=(V(f)cap V)cup (V(g)cap V)$. This implies that $V(f)cap V=V$ or $V(g)cap V=g$.
answered Aug 6 at 18:39
Tsemo Aristide
51.5k11243
I knew it would have a simple answer. Somehow I didn't think about working with varieties instead of dealing ideals. And since I'm new to it, it's slightly confusing at times. Thanks for the answer :)
– lEm
Aug 6 at 18:45
add a comment |Â
up vote
3
down vote
accepted
Hint: $V(fg)=V(f)cup V(g)$ and $V=Vcap V(fg)=(V(f)cap V)cup (V(g)cap V)$. This implies that $V(f)cap V=V$ or $V(g)cap V=g$.
answered Aug 6 at 18:39
Tsemo Aristide
51.5k11243
I knew it would have a simple answer. Somehow I didn't think about working with varieties instead of dealing ideals. And since I'm new to it, it's slightly confusing at times. Thanks for the answer :)
– lEm
Aug 6 at 18:45
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Hint: $V(fg)=V(f)cup V(g)$ and $V=Vcap V(fg)=(V(f)cap V)cup (V(g)cap V)$. This implies that $V(f)cap V=V$ or $V(g)cap V=g$.
answered Aug 6 at 18:39
Tsemo Aristide
51.5k11243
Hint: $V(fg)=V(f)cup V(g)$ and $V=Vcap V(fg)=(V(f)cap V)cup (V(g)cap V)$. This implies that $V(f)cap V=V$ or $V(g)cap V=g$.
answered Aug 6 at 18:39
Tsemo Aristide
51.5k11243
answered Aug 6 at 18:39
Tsemo Aristide
51.5k11243
answered Aug 6 at 18:39
Tsemo Aristide
51.5k11243
answered Aug 6 at 18:39
Tsemo Aristide
51.5k11243
51.5k11243
I knew it would have a simple answer. Somehow I didn't think about working with varieties instead of dealing ideals. And since I'm new to it, it's slightly confusing at times. Thanks for the answer :)
– lEm
Aug 6 at 18:45
add a comment |Â
I knew it would have a simple answer. Somehow I didn't think about working with varieties instead of dealing ideals. And since I'm new to it, it's slightly confusing at times. Thanks for the answer :)
– lEm
Aug 6 at 18:45
I knew it would have a simple answer. Somehow I didn't think about working with varieties instead of dealing ideals. And since I'm new to it, it's slightly confusing at times. Thanks for the answer :)
– lEm
Aug 6 at 18:45
I knew it would have a simple answer. Somehow I didn't think about working with varieties instead of dealing ideals. And since I'm new to it, it's slightly confusing at times. Thanks for the answer :)
– lEm
Aug 6 at 18:45
add a comment |Â
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