Exercise 3.30 from Görtz-Wedhorn: local affine algebras
Here's the Exercise 3.30 from the textbook of Görtz and Wedhorn:
Let $k$ be a field, and let $A$ be a local $k$-algebra of finite type.
Prove that $operatorname{Spec} A$ consists of a single point, and
that $A$ is finite-dimensional as a $k$-vector space. In particular
$A$ is a local Artin ring (why?), and $kappa (A)/k$ is a finite field
extension.
I see how this follows from some well-known results:
Every finitely generated algebra over a Jacobson ring is Jacobson, so that $A$ is Jacobson. As $A$ is local, this amounts to saying that its maximal ideal is the only prime ideal.
In general, for every maximal ideal $mathfrak{m}$ in a finitely generated $k$-algebra $A$, the extension $kappa (mathfrak{m})/k$ is finite (which follows again from the general results about Jacobson rings).
A ring is Artinian iff it is Noetherian and every prime ideal is maximal, which is the case here.
A finitely generated $k$-algebra is Artinian iff it is finite (Atiyah-Macdonald, Exercise 8.3).
However, all this seems to be an overkill, and I think the authors had in mind some direct argument for the very special case when $A$ is local. Do you see one?
commutative-algebra
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Here's the Exercise 3.30 from the textbook of Görtz and Wedhorn:
Let $k$ be a field, and let $A$ be a local $k$-algebra of finite type.
Prove that $operatorname{Spec} A$ consists of a single point, and
that $A$ is finite-dimensional as a $k$-vector space. In particular
$A$ is a local Artin ring (why?), and $kappa (A)/k$ is a finite field
extension.
I see how this follows from some well-known results:
Every finitely generated algebra over a Jacobson ring is Jacobson, so that $A$ is Jacobson. As $A$ is local, this amounts to saying that its maximal ideal is the only prime ideal.
In general, for every maximal ideal $mathfrak{m}$ in a finitely generated $k$-algebra $A$, the extension $kappa (mathfrak{m})/k$ is finite (which follows again from the general results about Jacobson rings).
A ring is Artinian iff it is Noetherian and every prime ideal is maximal, which is the case here.
A finitely generated $k$-algebra is Artinian iff it is finite (Atiyah-Macdonald, Exercise 8.3).
However, all this seems to be an overkill, and I think the authors had in mind some direct argument for the very special case when $A$ is local. Do you see one?
commutative-algebra
asked 2 days ago
user632025user632025
61
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user632025 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Just an opinion, but this doesn't seem like overkill to me.
– RghtHndSd
2 days ago
add a comment |
Here's the Exercise 3.30 from the textbook of Görtz and Wedhorn:
Let $k$ be a field, and let $A$ be a local $k$-algebra of finite type.
Prove that $operatorname{Spec} A$ consists of a single point, and
that $A$ is finite-dimensional as a $k$-vector space. In particular
$A$ is a local Artin ring (why?), and $kappa (A)/k$ is a finite field
extension.
I see how this follows from some well-known results:
Every finitely generated algebra over a Jacobson ring is Jacobson, so that $A$ is Jacobson. As $A$ is local, this amounts to saying that its maximal ideal is the only prime ideal.
In general, for every maximal ideal $mathfrak{m}$ in a finitely generated $k$-algebra $A$, the extension $kappa (mathfrak{m})/k$ is finite (which follows again from the general results about Jacobson rings).
A ring is Artinian iff it is Noetherian and every prime ideal is maximal, which is the case here.
A finitely generated $k$-algebra is Artinian iff it is finite (Atiyah-Macdonald, Exercise 8.3).
However, all this seems to be an overkill, and I think the authors had in mind some direct argument for the very special case when $A$ is local. Do you see one?
commutative-algebra
asked 2 days ago
user632025user632025
61
New contributor
user632025 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Here's the Exercise 3.30 from the textbook of Görtz and Wedhorn:
Let $k$ be a field, and let $A$ be a local $k$-algebra of finite type.
Prove that $operatorname{Spec} A$ consists of a single point, and
that $A$ is finite-dimensional as a $k$-vector space. In particular
$A$ is a local Artin ring (why?), and $kappa (A)/k$ is a finite field
extension.
I see how this follows from some well-known results:
Every finitely generated algebra over a Jacobson ring is Jacobson, so that $A$ is Jacobson. As $A$ is local, this amounts to saying that its maximal ideal is the only prime ideal.
In general, for every maximal ideal $mathfrak{m}$ in a finitely generated $k$-algebra $A$, the extension $kappa (mathfrak{m})/k$ is finite (which follows again from the general results about Jacobson rings).
A ring is Artinian iff it is Noetherian and every prime ideal is maximal, which is the case here.
A finitely generated $k$-algebra is Artinian iff it is finite (Atiyah-Macdonald, Exercise 8.3).
However, all this seems to be an overkill, and I think the authors had in mind some direct argument for the very special case when $A$ is local. Do you see one?
commutative-algebra
commutative-algebra
asked 2 days ago
user632025user632025
61
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user632025 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 2 days ago
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asked 2 days ago
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asked 2 days ago
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asked 2 days ago
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61
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Just an opinion, but this doesn't seem like overkill to me.
– RghtHndSd
2 days ago
add a comment |
Just an opinion, but this doesn't seem like overkill to me.
– RghtHndSd
2 days ago
Just an opinion, but this doesn't seem like overkill to me.
– RghtHndSd
2 days ago
Just an opinion, but this doesn't seem like overkill to me.
– RghtHndSd
2 days ago
add a comment |
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Just an opinion, but this doesn't seem like overkill to me.
– RghtHndSd
2 days ago