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

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
Jan 6 at 12:55 
1
@JasonStarr But normal means integrally closed in its field of fractions, which isn’t being asked.
– Will Sawin
Jan 6 at 13:04 
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
Jan 6 at 13:16

Thank you for the answer. I am sorry that the title was misleading.
– P. Grape
Jan 6 at 13:24 
2
@JasonStarr Fair enough!
– Will Sawin
Jan 6 at 13:24
add a comment 
Let $k$ be a field and $s$ and $t$ be variables.
Is the ring $k[s][[t]]$ integrally closed in $k[s,s^{1}][[t]]$?
Let $k$ be a field and $s$ and $t$ be variables.
Is the ring $k[s][[t]]$ integrally closed in $k[s,s^{1}][[t]]$?

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
Jan 6 at 12:55 
1
@JasonStarr But normal means integrally closed in its field of fractions, which isn’t being asked.
– Will Sawin
Jan 6 at 13:04 
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
Jan 6 at 13:16

Thank you for the answer. I am sorry that the title was misleading.
– P. Grape
Jan 6 at 13:24 
2
@JasonStarr Fair enough!
– Will Sawin
Jan 6 at 13:24
add a comment 

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
Jan 6 at 12:55 
1
@JasonStarr But normal means integrally closed in its field of fractions, which isn’t being asked.
– Will Sawin
Jan 6 at 13:04 
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
Jan 6 at 13:16

Thank you for the answer. I am sorry that the title was misleading.
– P. Grape
Jan 6 at 13:24 
2
@JasonStarr Fair enough!
– Will Sawin
Jan 6 at 13:24
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
Jan 6 at 12:55
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
Jan 6 at 12:55
@JasonStarr But normal means integrally closed in its field of fractions, which isn’t being asked.
– Will Sawin
Jan 6 at 13:04
@JasonStarr But normal means integrally closed in its field of fractions, which isn’t being asked.
– Will Sawin
Jan 6 at 13:04
@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
Jan 6 at 13:16
@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
Jan 6 at 13:16
Thank you for the answer. I am sorry that the title was misleading.
– P. Grape
Jan 6 at 13:24
Thank you for the answer. I am sorry that the title was misleading.
– P. Grape
Jan 6 at 13:24
@JasonStarr Fair enough!
– Will Sawin
Jan 6 at 13:24
@JasonStarr Fair enough!
– Will Sawin
Jan 6 at 13:24
add a comment 
1 Answer
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No. Let $ell$ be a prime invertible in $k$ and consider
$$x= s (1+ t/s)^{1/ell} = s + frac{t }{ell} – frac{(ell1) t^2}{ 2s ell^2} + frac{ (ell1) (2ell1) t^3}{ 6 s^2 ell^3} + dots in k[s,s^{1}][[t]] $$
Clearly it does not lie in $k[s][[t]]$. But we have $$x^ell = s^{ell} (1+t/s) = s^ell + s^{ell1} t$$ so it satisfies a monic polynomial equation over $k[s][[t]]$ (and even $k[s,t]$).
add a comment 
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1 Answer
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1 Answer
1
active
oldest
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oldest
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active
oldest
votes
No. Let $ell$ be a prime invertible in $k$ and consider
$$x= s (1+ t/s)^{1/ell} = s + frac{t }{ell} – frac{(ell1) t^2}{ 2s ell^2} + frac{ (ell1) (2ell1) t^3}{ 6 s^2 ell^3} + dots in k[s,s^{1}][[t]] $$
Clearly it does not lie in $k[s][[t]]$. But we have $$x^ell = s^{ell} (1+t/s) = s^ell + s^{ell1} t$$ so it satisfies a monic polynomial equation over $k[s][[t]]$ (and even $k[s,t]$).
add a comment 
No. Let $ell$ be a prime invertible in $k$ and consider
$$x= s (1+ t/s)^{1/ell} = s + frac{t }{ell} – frac{(ell1) t^2}{ 2s ell^2} + frac{ (ell1) (2ell1) t^3}{ 6 s^2 ell^3} + dots in k[s,s^{1}][[t]] $$
Clearly it does not lie in $k[s][[t]]$. But we have $$x^ell = s^{ell} (1+t/s) = s^ell + s^{ell1} t$$ so it satisfies a monic polynomial equation over $k[s][[t]]$ (and even $k[s,t]$).
add a comment 
No. Let $ell$ be a prime invertible in $k$ and consider
$$x= s (1+ t/s)^{1/ell} = s + frac{t }{ell} – frac{(ell1) t^2}{ 2s ell^2} + frac{ (ell1) (2ell1) t^3}{ 6 s^2 ell^3} + dots in k[s,s^{1}][[t]] $$
Clearly it does not lie in $k[s][[t]]$. But we have $$x^ell = s^{ell} (1+t/s) = s^ell + s^{ell1} t$$ so it satisfies a monic polynomial equation over $k[s][[t]]$ (and even $k[s,t]$).
No. Let $ell$ be a prime invertible in $k$ and consider
$$x= s (1+ t/s)^{1/ell} = s + frac{t }{ell} – frac{(ell1) t^2}{ 2s ell^2} + frac{ (ell1) (2ell1) t^3}{ 6 s^2 ell^3} + dots in k[s,s^{1}][[t]] $$
Clearly it does not lie in $k[s][[t]]$. But we have $$x^ell = s^{ell} (1+t/s) = s^ell + s^{ell1} t$$ so it satisfies a monic polynomial equation over $k[s][[t]]$ (and even $k[s,t]$).
add a comment 
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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
Jan 6 at 12:55
@JasonStarr But normal means integrally closed in its field of fractions, which isn’t being asked.
– Will Sawin
Jan 6 at 13:04
@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
Jan 6 at 13:16
Thank you for the answer. I am sorry that the title was misleading.
– P. Grape
Jan 6 at 13:24
@JasonStarr Fair enough!
– Will Sawin
Jan 6 at 13:24