Is the formal power series ring integrally closed?
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
asked yesterday
P. GrapeP. Grape
814
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P. Grape is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Let $k$ be a field and $s$ and $t$ be variables.
Is the ring $k[s][[t]]$ integrally closed in $k[s,s^{-1}][[t]]$?
ac.commutative-algebra
asked yesterday
P. GrapeP. Grape
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.
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
yesterday
1
@JasonStarr But normal means integrally closed in its field of fractions, which isn't being asked.
– Will Sawin
yesterday
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
yesterday
Thank you for the answer. I am sorry that the title was misleading.
– P. Grape
yesterday
2
@JasonStarr Fair enough!
– Will Sawin
yesterday
add a comment |
Let $k$ be a field and $s$ and $t$ be variables.
Is the ring $k[s][[t]]$ integrally closed in $k[s,s^{-1}][[t]]$?
ac.commutative-algebra
asked yesterday
P. GrapeP. Grape
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.
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
asked yesterday
P. GrapeP. Grape
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.
asked yesterday
P. GrapeP. Grape
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.
asked yesterday
P. GrapeP. Grape
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.
asked yesterday
P. GrapeP. Grape
814
asked yesterday
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
yesterday
1
@JasonStarr But normal means integrally closed in its field of fractions, which isn't being asked.
– Will Sawin
yesterday
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
yesterday
Thank you for the answer. I am sorry that the title was misleading.
– P. Grape
yesterday
2
@JasonStarr Fair enough!
– Will Sawin
yesterday
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
yesterday
1
@JasonStarr But normal means integrally closed in its field of fractions, which isn't being asked.
– Will Sawin
yesterday
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
yesterday
Thank you for the answer. I am sorry that the title was misleading.
– P. Grape
yesterday
2
@JasonStarr Fair enough!
– Will Sawin
yesterday
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
yesterday
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
yesterday
1
1
@JasonStarr But normal means integrally closed in its field of fractions, which isn't being asked.
– Will Sawin
yesterday
@JasonStarr But normal means integrally closed in its field of fractions, which isn't being asked.
– Will Sawin
yesterday
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
yesterday
@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
yesterday
Thank you for the answer. I am sorry that the title was misleading.
– P. Grape
yesterday
Thank you for the answer. I am sorry that the title was misleading.
– P. Grape
yesterday
2
2
@JasonStarr Fair enough!
– Will Sawin
yesterday
@JasonStarr Fair enough!
– Will Sawin
yesterday
add a comment |
1 Answer
1
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No. Let $ell$ be a prime invertible in $k$ and consider
$$x= s (1+ t/s)^{1/ell} = s + frac{t }{ell} - frac{(ell-1) t^2}{ 2s ell^2} + frac{ (ell-1) (2ell-1) t^3}{ 6 s^2 ell^3} + dots in k[s,s^{-1}][[t]] $$
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 $k[s,t]$).
answered yesterday
Will SawinWill Sawin
67.4k7135280
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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votes
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 $k[s,t]$).
answered yesterday
Will SawinWill Sawin
67.4k7135280
add a comment |
No. Let $ell$ be a prime invertible in $k$ and consider
$$x= s (1+ t/s)^{1/ell} = s + frac{t }{ell} - frac{(ell-1) t^2}{ 2s ell^2} + frac{ (ell-1) (2ell-1) t^3}{ 6 s^2 ell^3} + dots in k[s,s^{-1}][[t]] $$
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 $k[s,t]$).
answered yesterday
Will SawinWill Sawin
67.4k7135280
add a comment |
No. Let $ell$ be a prime invertible in $k$ and consider
$$x= s (1+ t/s)^{1/ell} = s + frac{t }{ell} - frac{(ell-1) t^2}{ 2s ell^2} + frac{ (ell-1) (2ell-1) t^3}{ 6 s^2 ell^3} + dots in k[s,s^{-1}][[t]] $$
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 $k[s,t]$).
answered yesterday
Will SawinWill Sawin
67.4k7135280
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 $k[s,t]$).
answered yesterday
Will SawinWill Sawin
67.4k7135280
answered yesterday
Will SawinWill Sawin
67.4k7135280
answered yesterday
Will SawinWill Sawin
67.4k7135280
answered yesterday
Will SawinWill Sawin
67.4k7135280
67.4k7135280
add a comment |
add a comment |
P. Grape is a new contributor. Be nice, and check out our Code of Conduct.
P. Grape is a new contributor. Be nice, and check out our Code of Conduct.
P. Grape is a new contributor. Be nice, and check out our Code of Conduct.
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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
yesterday
1
@JasonStarr But normal means integrally closed in its field of fractions, which isn't being asked.
– Will Sawin
yesterday
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
yesterday
Thank you for the answer. I am sorry that the title was misleading.
– P. Grape
yesterday
2
@JasonStarr Fair enough!
– Will Sawin
yesterday