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?







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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?







    share|cite|improve this question





















      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?







      share|cite|improve this question












      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?









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      share|cite|improve this question




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      asked Aug 6 at 18:22









      lEm

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          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 $V(g)cap V=g$.






          share|cite|improve this answer





















          • 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










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          up vote
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          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 $V(g)cap V=g$.






          share|cite|improve this answer





















          • 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














          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 $V(g)cap V=g$.






          share|cite|improve this answer





















          • 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












          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 $V(g)cap V=g$.






          share|cite|improve this answer













          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 $V(g)cap V=g$.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          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
















          • 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












           

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