Is a scheme Noetherian if its topological space and its stalks are?











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Is a scheme being Noetherian equivalent to the underlying topological space being Noetherian and all its stalks being Noetherian?










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    Probably should be asked on MSE.
    – Bernie
    2 days ago






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    I think this is a good question for this site : it's an interesting question (and answer) and I at least have not seen the counter example to the seemingly plausible result before.
    – Asvin
    22 hours ago

















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Is a scheme being Noetherian equivalent to the underlying topological space being Noetherian and all its stalks being Noetherian?










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  • 1




    Probably should be asked on MSE.
    – Bernie
    2 days ago






  • 1




    I think this is a good question for this site : it's an interesting question (and answer) and I at least have not seen the counter example to the seemingly plausible result before.
    – Asvin
    22 hours ago















up vote
11
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Is a scheme being Noetherian equivalent to the underlying topological space being Noetherian and all its stalks being Noetherian?










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Is a scheme being Noetherian equivalent to the underlying topological space being Noetherian and all its stalks being Noetherian?







ag.algebraic-geometry ac.commutative-algebra schemes counterexamples






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edited yesterday









R. van Dobben de Bruyn

10.3k23261




10.3k23261










asked 2 days ago









G.-S. Zhou

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1337








  • 1




    Probably should be asked on MSE.
    – Bernie
    2 days ago






  • 1




    I think this is a good question for this site : it's an interesting question (and answer) and I at least have not seen the counter example to the seemingly plausible result before.
    – Asvin
    22 hours ago
















  • 1




    Probably should be asked on MSE.
    – Bernie
    2 days ago






  • 1




    I think this is a good question for this site : it's an interesting question (and answer) and I at least have not seen the counter example to the seemingly plausible result before.
    – Asvin
    22 hours ago










1




1




Probably should be asked on MSE.
– Bernie
2 days ago




Probably should be asked on MSE.
– Bernie
2 days ago




1




1




I think this is a good question for this site : it's an interesting question (and answer) and I at least have not seen the counter example to the seemingly plausible result before.
– Asvin
22 hours ago






I think this is a good question for this site : it's an interesting question (and answer) and I at least have not seen the counter example to the seemingly plausible result before.
– Asvin
22 hours ago












2 Answers
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This is false. The easiest counterexample I could come up with is the following "affine line with embedded points at every closed [rational] point":



Example. Let $k$ be an infinite field, let $R = k[x]$, and for each $alpha in k$ let $R_alpha = R[y_alpha]/((x-alpha)y_alpha,y_alpha^2)$. Then $R_alpha$ is an affine line with an embedded prime $mathfrak p_alpha = (x-alpha,y_alpha)$ at $x = alpha$, sticking out in the $y_alpha$-direction. Finally, let
$$R_infty = bigotimes_{alpha in k} R_alpha = operatorname*{colim}_{substack{longrightarrow\I subseteq k\text{finite}}} bigotimes_{alpha in I} R_alpha$$
be their tensor product over $R$ (not over $k$); that is
$$R_infty = frac{k[x]left[{y_alpha}_{alpha in k}right]}{sum_{alpha in k}((x-alpha)y_alpha, y_alpha^2)}.$$
This is not a Noetherian ring, because the radical $mathfrak r = ({y_alpha}_{alpha in k})$ is not finitely generated. But $operatorname{Spec} R_infty$ agrees as a topological space with $operatorname{Spec} R_infty^{operatorname{red}} = mathbb A^1_k$, hence $|!operatorname{Spec} R_infty|$ is a Noetherian topological space.



On the other hand, the map $R to R_alpha$ is an isomorphism away from $alpha$, and similarly $R_alpha to R_infty$ induces isomorphisms on the stalks at $alpha$. Thus, the stalk $(R_infty)_{mathfrak q_alpha} = (R_alpha)_{mathfrak p_alpha}$ at $mathfrak q_alpha = mathfrak p_alpha R_infty + mathfrak r$ is Noetherian. Similarly, the stalk at the generic point $mathfrak r$ is just $R_{(0)} = k(x)$. Thus, we conclude that all the stalks of $R_infty$ are Noetherian. $square$



Remark. As requested in the comments, my motivation to come up with this example is the following: I was trying to prove by hand that the answer was positive. After an immediate reduction to the affine case, one needs to consider a chain $I_0 subseteq I_1 ldots$ of ideals. Because $|X|$ is Noetherian, we may assume they [eventually] define the same closed set $V$. Intuitively, to check that they agree as ideals, it should suffice to check it at each component of $V$. Then you can try to use the Noetherian rings $mathcal O_{X,x}$ for the generic points $x$ of $V$.



However this is not quite true, because of embedded points. The precise statement [Tags 0311 and 02M3] is that the inclusion $I_{i-1} subseteq I_i$ is an equality if and only if the same holds for the localisation at every associated point of $I_i$. Now you run into trouble if the $I_i$ differ at an embedded point that is not seen by the topology of $V$. Moreover, if this embedded point varies as we move through the chain, there is a chance that the localisations $mathcal O_{X,x}$ at these embedded points are still Noetherian. Once you see this, coming up with the example is not so hard.






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  • 1




    Can you give some motivation (when you are free) to think of this example..
    – Praphulla Koushik
    yesterday










  • @PraphullaKoushik I edited my answer to include some motivation behind the construction.
    – R. van Dobben de Bruyn
    yesterday










  • Thanks for the positive response.. This makes so much sense... Thanks..
    – Praphulla Koushik
    yesterday


















up vote
6
down vote













A further counterexample is Example 2.3 in W. Heinzer, J. Ohm, Locally noetherian commutative rings, Trans. Amer. Math. Soc. 158 (1971), 273-284.






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    2 Answers
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    2 Answers
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    up vote
    23
    down vote



    accepted










    This is false. The easiest counterexample I could come up with is the following "affine line with embedded points at every closed [rational] point":



    Example. Let $k$ be an infinite field, let $R = k[x]$, and for each $alpha in k$ let $R_alpha = R[y_alpha]/((x-alpha)y_alpha,y_alpha^2)$. Then $R_alpha$ is an affine line with an embedded prime $mathfrak p_alpha = (x-alpha,y_alpha)$ at $x = alpha$, sticking out in the $y_alpha$-direction. Finally, let
    $$R_infty = bigotimes_{alpha in k} R_alpha = operatorname*{colim}_{substack{longrightarrow\I subseteq k\text{finite}}} bigotimes_{alpha in I} R_alpha$$
    be their tensor product over $R$ (not over $k$); that is
    $$R_infty = frac{k[x]left[{y_alpha}_{alpha in k}right]}{sum_{alpha in k}((x-alpha)y_alpha, y_alpha^2)}.$$
    This is not a Noetherian ring, because the radical $mathfrak r = ({y_alpha}_{alpha in k})$ is not finitely generated. But $operatorname{Spec} R_infty$ agrees as a topological space with $operatorname{Spec} R_infty^{operatorname{red}} = mathbb A^1_k$, hence $|!operatorname{Spec} R_infty|$ is a Noetherian topological space.



    On the other hand, the map $R to R_alpha$ is an isomorphism away from $alpha$, and similarly $R_alpha to R_infty$ induces isomorphisms on the stalks at $alpha$. Thus, the stalk $(R_infty)_{mathfrak q_alpha} = (R_alpha)_{mathfrak p_alpha}$ at $mathfrak q_alpha = mathfrak p_alpha R_infty + mathfrak r$ is Noetherian. Similarly, the stalk at the generic point $mathfrak r$ is just $R_{(0)} = k(x)$. Thus, we conclude that all the stalks of $R_infty$ are Noetherian. $square$



    Remark. As requested in the comments, my motivation to come up with this example is the following: I was trying to prove by hand that the answer was positive. After an immediate reduction to the affine case, one needs to consider a chain $I_0 subseteq I_1 ldots$ of ideals. Because $|X|$ is Noetherian, we may assume they [eventually] define the same closed set $V$. Intuitively, to check that they agree as ideals, it should suffice to check it at each component of $V$. Then you can try to use the Noetherian rings $mathcal O_{X,x}$ for the generic points $x$ of $V$.



    However this is not quite true, because of embedded points. The precise statement [Tags 0311 and 02M3] is that the inclusion $I_{i-1} subseteq I_i$ is an equality if and only if the same holds for the localisation at every associated point of $I_i$. Now you run into trouble if the $I_i$ differ at an embedded point that is not seen by the topology of $V$. Moreover, if this embedded point varies as we move through the chain, there is a chance that the localisations $mathcal O_{X,x}$ at these embedded points are still Noetherian. Once you see this, coming up with the example is not so hard.






    share|cite|improve this answer



















    • 1




      Can you give some motivation (when you are free) to think of this example..
      – Praphulla Koushik
      yesterday










    • @PraphullaKoushik I edited my answer to include some motivation behind the construction.
      – R. van Dobben de Bruyn
      yesterday










    • Thanks for the positive response.. This makes so much sense... Thanks..
      – Praphulla Koushik
      yesterday















    up vote
    23
    down vote



    accepted










    This is false. The easiest counterexample I could come up with is the following "affine line with embedded points at every closed [rational] point":



    Example. Let $k$ be an infinite field, let $R = k[x]$, and for each $alpha in k$ let $R_alpha = R[y_alpha]/((x-alpha)y_alpha,y_alpha^2)$. Then $R_alpha$ is an affine line with an embedded prime $mathfrak p_alpha = (x-alpha,y_alpha)$ at $x = alpha$, sticking out in the $y_alpha$-direction. Finally, let
    $$R_infty = bigotimes_{alpha in k} R_alpha = operatorname*{colim}_{substack{longrightarrow\I subseteq k\text{finite}}} bigotimes_{alpha in I} R_alpha$$
    be their tensor product over $R$ (not over $k$); that is
    $$R_infty = frac{k[x]left[{y_alpha}_{alpha in k}right]}{sum_{alpha in k}((x-alpha)y_alpha, y_alpha^2)}.$$
    This is not a Noetherian ring, because the radical $mathfrak r = ({y_alpha}_{alpha in k})$ is not finitely generated. But $operatorname{Spec} R_infty$ agrees as a topological space with $operatorname{Spec} R_infty^{operatorname{red}} = mathbb A^1_k$, hence $|!operatorname{Spec} R_infty|$ is a Noetherian topological space.



    On the other hand, the map $R to R_alpha$ is an isomorphism away from $alpha$, and similarly $R_alpha to R_infty$ induces isomorphisms on the stalks at $alpha$. Thus, the stalk $(R_infty)_{mathfrak q_alpha} = (R_alpha)_{mathfrak p_alpha}$ at $mathfrak q_alpha = mathfrak p_alpha R_infty + mathfrak r$ is Noetherian. Similarly, the stalk at the generic point $mathfrak r$ is just $R_{(0)} = k(x)$. Thus, we conclude that all the stalks of $R_infty$ are Noetherian. $square$



    Remark. As requested in the comments, my motivation to come up with this example is the following: I was trying to prove by hand that the answer was positive. After an immediate reduction to the affine case, one needs to consider a chain $I_0 subseteq I_1 ldots$ of ideals. Because $|X|$ is Noetherian, we may assume they [eventually] define the same closed set $V$. Intuitively, to check that they agree as ideals, it should suffice to check it at each component of $V$. Then you can try to use the Noetherian rings $mathcal O_{X,x}$ for the generic points $x$ of $V$.



    However this is not quite true, because of embedded points. The precise statement [Tags 0311 and 02M3] is that the inclusion $I_{i-1} subseteq I_i$ is an equality if and only if the same holds for the localisation at every associated point of $I_i$. Now you run into trouble if the $I_i$ differ at an embedded point that is not seen by the topology of $V$. Moreover, if this embedded point varies as we move through the chain, there is a chance that the localisations $mathcal O_{X,x}$ at these embedded points are still Noetherian. Once you see this, coming up with the example is not so hard.






    share|cite|improve this answer



















    • 1




      Can you give some motivation (when you are free) to think of this example..
      – Praphulla Koushik
      yesterday










    • @PraphullaKoushik I edited my answer to include some motivation behind the construction.
      – R. van Dobben de Bruyn
      yesterday










    • Thanks for the positive response.. This makes so much sense... Thanks..
      – Praphulla Koushik
      yesterday













    up vote
    23
    down vote



    accepted







    up vote
    23
    down vote



    accepted






    This is false. The easiest counterexample I could come up with is the following "affine line with embedded points at every closed [rational] point":



    Example. Let $k$ be an infinite field, let $R = k[x]$, and for each $alpha in k$ let $R_alpha = R[y_alpha]/((x-alpha)y_alpha,y_alpha^2)$. Then $R_alpha$ is an affine line with an embedded prime $mathfrak p_alpha = (x-alpha,y_alpha)$ at $x = alpha$, sticking out in the $y_alpha$-direction. Finally, let
    $$R_infty = bigotimes_{alpha in k} R_alpha = operatorname*{colim}_{substack{longrightarrow\I subseteq k\text{finite}}} bigotimes_{alpha in I} R_alpha$$
    be their tensor product over $R$ (not over $k$); that is
    $$R_infty = frac{k[x]left[{y_alpha}_{alpha in k}right]}{sum_{alpha in k}((x-alpha)y_alpha, y_alpha^2)}.$$
    This is not a Noetherian ring, because the radical $mathfrak r = ({y_alpha}_{alpha in k})$ is not finitely generated. But $operatorname{Spec} R_infty$ agrees as a topological space with $operatorname{Spec} R_infty^{operatorname{red}} = mathbb A^1_k$, hence $|!operatorname{Spec} R_infty|$ is a Noetherian topological space.



    On the other hand, the map $R to R_alpha$ is an isomorphism away from $alpha$, and similarly $R_alpha to R_infty$ induces isomorphisms on the stalks at $alpha$. Thus, the stalk $(R_infty)_{mathfrak q_alpha} = (R_alpha)_{mathfrak p_alpha}$ at $mathfrak q_alpha = mathfrak p_alpha R_infty + mathfrak r$ is Noetherian. Similarly, the stalk at the generic point $mathfrak r$ is just $R_{(0)} = k(x)$. Thus, we conclude that all the stalks of $R_infty$ are Noetherian. $square$



    Remark. As requested in the comments, my motivation to come up with this example is the following: I was trying to prove by hand that the answer was positive. After an immediate reduction to the affine case, one needs to consider a chain $I_0 subseteq I_1 ldots$ of ideals. Because $|X|$ is Noetherian, we may assume they [eventually] define the same closed set $V$. Intuitively, to check that they agree as ideals, it should suffice to check it at each component of $V$. Then you can try to use the Noetherian rings $mathcal O_{X,x}$ for the generic points $x$ of $V$.



    However this is not quite true, because of embedded points. The precise statement [Tags 0311 and 02M3] is that the inclusion $I_{i-1} subseteq I_i$ is an equality if and only if the same holds for the localisation at every associated point of $I_i$. Now you run into trouble if the $I_i$ differ at an embedded point that is not seen by the topology of $V$. Moreover, if this embedded point varies as we move through the chain, there is a chance that the localisations $mathcal O_{X,x}$ at these embedded points are still Noetherian. Once you see this, coming up with the example is not so hard.






    share|cite|improve this answer














    This is false. The easiest counterexample I could come up with is the following "affine line with embedded points at every closed [rational] point":



    Example. Let $k$ be an infinite field, let $R = k[x]$, and for each $alpha in k$ let $R_alpha = R[y_alpha]/((x-alpha)y_alpha,y_alpha^2)$. Then $R_alpha$ is an affine line with an embedded prime $mathfrak p_alpha = (x-alpha,y_alpha)$ at $x = alpha$, sticking out in the $y_alpha$-direction. Finally, let
    $$R_infty = bigotimes_{alpha in k} R_alpha = operatorname*{colim}_{substack{longrightarrow\I subseteq k\text{finite}}} bigotimes_{alpha in I} R_alpha$$
    be their tensor product over $R$ (not over $k$); that is
    $$R_infty = frac{k[x]left[{y_alpha}_{alpha in k}right]}{sum_{alpha in k}((x-alpha)y_alpha, y_alpha^2)}.$$
    This is not a Noetherian ring, because the radical $mathfrak r = ({y_alpha}_{alpha in k})$ is not finitely generated. But $operatorname{Spec} R_infty$ agrees as a topological space with $operatorname{Spec} R_infty^{operatorname{red}} = mathbb A^1_k$, hence $|!operatorname{Spec} R_infty|$ is a Noetherian topological space.



    On the other hand, the map $R to R_alpha$ is an isomorphism away from $alpha$, and similarly $R_alpha to R_infty$ induces isomorphisms on the stalks at $alpha$. Thus, the stalk $(R_infty)_{mathfrak q_alpha} = (R_alpha)_{mathfrak p_alpha}$ at $mathfrak q_alpha = mathfrak p_alpha R_infty + mathfrak r$ is Noetherian. Similarly, the stalk at the generic point $mathfrak r$ is just $R_{(0)} = k(x)$. Thus, we conclude that all the stalks of $R_infty$ are Noetherian. $square$



    Remark. As requested in the comments, my motivation to come up with this example is the following: I was trying to prove by hand that the answer was positive. After an immediate reduction to the affine case, one needs to consider a chain $I_0 subseteq I_1 ldots$ of ideals. Because $|X|$ is Noetherian, we may assume they [eventually] define the same closed set $V$. Intuitively, to check that they agree as ideals, it should suffice to check it at each component of $V$. Then you can try to use the Noetherian rings $mathcal O_{X,x}$ for the generic points $x$ of $V$.



    However this is not quite true, because of embedded points. The precise statement [Tags 0311 and 02M3] is that the inclusion $I_{i-1} subseteq I_i$ is an equality if and only if the same holds for the localisation at every associated point of $I_i$. Now you run into trouble if the $I_i$ differ at an embedded point that is not seen by the topology of $V$. Moreover, if this embedded point varies as we move through the chain, there is a chance that the localisations $mathcal O_{X,x}$ at these embedded points are still Noetherian. Once you see this, coming up with the example is not so hard.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited yesterday

























    answered 2 days ago









    R. van Dobben de Bruyn

    10.3k23261




    10.3k23261








    • 1




      Can you give some motivation (when you are free) to think of this example..
      – Praphulla Koushik
      yesterday










    • @PraphullaKoushik I edited my answer to include some motivation behind the construction.
      – R. van Dobben de Bruyn
      yesterday










    • Thanks for the positive response.. This makes so much sense... Thanks..
      – Praphulla Koushik
      yesterday














    • 1




      Can you give some motivation (when you are free) to think of this example..
      – Praphulla Koushik
      yesterday










    • @PraphullaKoushik I edited my answer to include some motivation behind the construction.
      – R. van Dobben de Bruyn
      yesterday










    • Thanks for the positive response.. This makes so much sense... Thanks..
      – Praphulla Koushik
      yesterday








    1




    1




    Can you give some motivation (when you are free) to think of this example..
    – Praphulla Koushik
    yesterday




    Can you give some motivation (when you are free) to think of this example..
    – Praphulla Koushik
    yesterday












    @PraphullaKoushik I edited my answer to include some motivation behind the construction.
    – R. van Dobben de Bruyn
    yesterday




    @PraphullaKoushik I edited my answer to include some motivation behind the construction.
    – R. van Dobben de Bruyn
    yesterday












    Thanks for the positive response.. This makes so much sense... Thanks..
    – Praphulla Koushik
    yesterday




    Thanks for the positive response.. This makes so much sense... Thanks..
    – Praphulla Koushik
    yesterday










    up vote
    6
    down vote













    A further counterexample is Example 2.3 in W. Heinzer, J. Ohm, Locally noetherian commutative rings, Trans. Amer. Math. Soc. 158 (1971), 273-284.






    share|cite|improve this answer

























      up vote
      6
      down vote













      A further counterexample is Example 2.3 in W. Heinzer, J. Ohm, Locally noetherian commutative rings, Trans. Amer. Math. Soc. 158 (1971), 273-284.






      share|cite|improve this answer























        up vote
        6
        down vote










        up vote
        6
        down vote









        A further counterexample is Example 2.3 in W. Heinzer, J. Ohm, Locally noetherian commutative rings, Trans. Amer. Math. Soc. 158 (1971), 273-284.






        share|cite|improve this answer












        A further counterexample is Example 2.3 in W. Heinzer, J. Ohm, Locally noetherian commutative rings, Trans. Amer. Math. Soc. 158 (1971), 273-284.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        Fred Rohrer

        4,35111634




        4,35111634






























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