Is the formal power series ring integrally closed?












10














Let $k$ be a field and $s$ and $t$ be variables.
Is the ring $k[s][[t]]$ integrally closed in $k[s,s^{-1}][[t]]$?






ac.commutative-algebra







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  • 1




    Welcome new contributor. Every power series ring over a regular Noetherian ring is a regular Noetherian ring. Every regular Noetherian ring is normal.
    – Jason Starr
    2 days ago






  • 1




    @JasonStarr But normal means integrally closed in its field of fractions, which isn't being asked.
    – Will Sawin
    2 days ago






  • 1




    @WillSawin. I read the question in the title of the post. I see that the OP asks a different question in his post than is in the title of his post.
    – Jason Starr
    2 days ago












  • Thank you for the answer. I am sorry that the title was misleading.
    – P. Grape
    2 days ago






  • 2




    @JasonStarr Fair enough!
    – Will Sawin
    2 days ago
















10














Let $k$ be a field and $s$ and $t$ be variables.
Is the ring $k[s][[t]]$ integrally closed in $k[s,s^{-1}][[t]]$?






ac.commutative-algebra







share|cite|improve this question







New contributor




P. Grape is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 1




    Welcome new contributor. Every power series ring over a regular Noetherian ring is a regular Noetherian ring. Every regular Noetherian ring is normal.
    – Jason Starr
    2 days ago






  • 1




    @JasonStarr But normal means integrally closed in its field of fractions, which isn't being asked.
    – Will Sawin
    2 days ago






  • 1




    @WillSawin. I read the question in the title of the post. I see that the OP asks a different question in his post than is in the title of his post.
    – Jason Starr
    2 days ago












  • Thank you for the answer. I am sorry that the title was misleading.
    – P. Grape
    2 days ago






  • 2




    @JasonStarr Fair enough!
    – Will Sawin
    2 days ago














10












10








10


1





Let $k$ be a field and $s$ and $t$ be variables.
Is the ring $k[s][[t]]$ integrally closed in $k[s,s^{-1}][[t]]$?






ac.commutative-algebra







share|cite|improve this question







New contributor




P. Grape is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











Let $k$ be a field and $s$ and $t$ be variables.
Is the ring $k[s][[t]]$ integrally closed in $k[s,s^{-1}][[t]]$?






ac.commutative-algebra




ac.commutative-algebra






share|cite|improve this question







New contributor




P. Grape is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




P. Grape is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






New contributor




P. Grape is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 2 days ago









P. GrapeP. Grape

814




814




New contributor




P. Grape is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





P. Grape is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






P. Grape is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    Welcome new contributor. Every power series ring over a regular Noetherian ring is a regular Noetherian ring. Every regular Noetherian ring is normal.
    – Jason Starr
    2 days ago






  • 1




    @JasonStarr But normal means integrally closed in its field of fractions, which isn't being asked.
    – Will Sawin
    2 days ago






  • 1




    @WillSawin. I read the question in the title of the post. I see that the OP asks a different question in his post than is in the title of his post.
    – Jason Starr
    2 days ago












  • Thank you for the answer. I am sorry that the title was misleading.
    – P. Grape
    2 days ago






  • 2




    @JasonStarr Fair enough!
    – Will Sawin
    2 days ago














  • 1




    Welcome new contributor. Every power series ring over a regular Noetherian ring is a regular Noetherian ring. Every regular Noetherian ring is normal.
    – Jason Starr
    2 days ago






  • 1




    @JasonStarr But normal means integrally closed in its field of fractions, which isn't being asked.
    – Will Sawin
    2 days ago






  • 1




    @WillSawin. I read the question in the title of the post. I see that the OP asks a different question in his post than is in the title of his post.
    – Jason Starr
    2 days ago












  • Thank you for the answer. I am sorry that the title was misleading.
    – P. Grape
    2 days ago






  • 2




    @JasonStarr Fair enough!
    – Will Sawin
    2 days ago








1




1




Welcome new contributor. Every power series ring over a regular Noetherian ring is a regular Noetherian ring. Every regular Noetherian ring is normal.
– Jason Starr
2 days ago




Welcome new contributor. Every power series ring over a regular Noetherian ring is a regular Noetherian ring. Every regular Noetherian ring is normal.
– Jason Starr
2 days ago




1




1




@JasonStarr But normal means integrally closed in its field of fractions, which isn't being asked.
– Will Sawin
2 days ago




@JasonStarr But normal means integrally closed in its field of fractions, which isn't being asked.
– Will Sawin
2 days ago




1




1




@WillSawin. I read the question in the title of the post. I see that the OP asks a different question in his post than is in the title of his post.
– Jason Starr
2 days ago






@WillSawin. I read the question in the title of the post. I see that the OP asks a different question in his post than is in the title of his post.
– Jason Starr
2 days ago














Thank you for the answer. I am sorry that the title was misleading.
– P. Grape
2 days ago




Thank you for the answer. I am sorry that the title was misleading.
– P. Grape
2 days ago




2




2




@JasonStarr Fair enough!
– Will Sawin
2 days ago




@JasonStarr Fair enough!
– Will Sawin
2 days ago










1 Answer
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15














No. Let $ell$ be a prime invertible in $k$ and consider



$$x= s (1+ t/s)^{1/ell} = s + frac{t }{ell} - frac{(ell-1) t^2}{ 2s ell^2} + frac{ (ell-1) (2ell-1) t^3}{ 6 s^2 ell^3} + dots in k[s,s^{-1}][[t]] $$



Clearly it does not lie in $k[s][[t]]$. But we have $$x^ell = s^{ell} (1+t/s) = s^ell + s^{ell-1} t$$ so it satisfies a monic polynomial equation over $k[s][[t]]$ (and even $k[s,t]$).






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    15














    No. Let $ell$ be a prime invertible in $k$ and consider



    $$x= s (1+ t/s)^{1/ell} = s + frac{t }{ell} - frac{(ell-1) t^2}{ 2s ell^2} + frac{ (ell-1) (2ell-1) t^3}{ 6 s^2 ell^3} + dots in k[s,s^{-1}][[t]] $$



    Clearly it does not lie in $k[s][[t]]$. But we have $$x^ell = s^{ell} (1+t/s) = s^ell + s^{ell-1} t$$ so it satisfies a monic polynomial equation over $k[s][[t]]$ (and even $k[s,t]$).






    share|cite|improve this answer


























      15














      No. Let $ell$ be a prime invertible in $k$ and consider



      $$x= s (1+ t/s)^{1/ell} = s + frac{t }{ell} - frac{(ell-1) t^2}{ 2s ell^2} + frac{ (ell-1) (2ell-1) t^3}{ 6 s^2 ell^3} + dots in k[s,s^{-1}][[t]] $$



      Clearly it does not lie in $k[s][[t]]$. But we have $$x^ell = s^{ell} (1+t/s) = s^ell + s^{ell-1} t$$ so it satisfies a monic polynomial equation over $k[s][[t]]$ (and even $k[s,t]$).






      share|cite|improve this answer
























        15












        15








        15






        No. Let $ell$ be a prime invertible in $k$ and consider



        $$x= s (1+ t/s)^{1/ell} = s + frac{t }{ell} - frac{(ell-1) t^2}{ 2s ell^2} + frac{ (ell-1) (2ell-1) t^3}{ 6 s^2 ell^3} + dots in k[s,s^{-1}][[t]] $$



        Clearly it does not lie in $k[s][[t]]$. But we have $$x^ell = s^{ell} (1+t/s) = s^ell + s^{ell-1} t$$ so it satisfies a monic polynomial equation over $k[s][[t]]$ (and even $k[s,t]$).






        share|cite|improve this answer












        No. Let $ell$ be a prime invertible in $k$ and consider



        $$x= s (1+ t/s)^{1/ell} = s + frac{t }{ell} - frac{(ell-1) t^2}{ 2s ell^2} + frac{ (ell-1) (2ell-1) t^3}{ 6 s^2 ell^3} + dots in k[s,s^{-1}][[t]] $$



        Clearly it does not lie in $k[s][[t]]$. But we have $$x^ell = s^{ell} (1+t/s) = s^ell + s^{ell-1} t$$ so it satisfies a monic polynomial equation over $k[s][[t]]$ (and even $k[s,t]$).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        Will SawinWill Sawin

        67.5k7135280




        67.5k7135280






















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