Rings of formal power series is finitely generated as module.

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The question rises when I was reading Kemper's "A course in Commutative Algebra."
Let $K$ be a field and consider the ring of formal power series $K[[x]]$. In an exercise I proved $K[[x]]$ is not Jacobson, hence not finitely generated as a K-algebra. But since $K$ is Noetherian and $K[[x]]$ is a K-module, $K[[x]]$ is Noetherian is equivalent to $K[[x]]$ is finitely generated as a K-module. There is a proof of $K[[x]]$ being Noetherian. I can't see why $K[[x]]$ is a finitely generated K-module but not a finitely generated K-algebra. Is there a mistake in my understanding? Any help is appreciated.




abstract-algebra commutative-algebra noetherian formal-power-series




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    It cannot be a finitely generated module without being a finitely generated algebra : if $F$ is a finite generating set as a module, then a fortiori $F$ generates it as an algebra
    – Max
    14 hours ago














up vote
0
down vote

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The question rises when I was reading Kemper's "A course in Commutative Algebra."
Let $K$ be a field and consider the ring of formal power series $K[[x]]$. In an exercise I proved $K[[x]]$ is not Jacobson, hence not finitely generated as a K-algebra. But since $K$ is Noetherian and $K[[x]]$ is a K-module, $K[[x]]$ is Noetherian is equivalent to $K[[x]]$ is finitely generated as a K-module. There is a proof of $K[[x]]$ being Noetherian. I can't see why $K[[x]]$ is a finitely generated K-module but not a finitely generated K-algebra. Is there a mistake in my understanding? Any help is appreciated.




abstract-algebra commutative-algebra noetherian formal-power-series




share|cite|improve this question

















  • 1




    It cannot be a finitely generated module without being a finitely generated algebra : if $F$ is a finite generating set as a module, then a fortiori $F$ generates it as an algebra
    – Max
    14 hours ago












up vote
0
down vote

favorite









up vote
0
down vote

favorite











The question rises when I was reading Kemper's "A course in Commutative Algebra."
Let $K$ be a field and consider the ring of formal power series $K[[x]]$. In an exercise I proved $K[[x]]$ is not Jacobson, hence not finitely generated as a K-algebra. But since $K$ is Noetherian and $K[[x]]$ is a K-module, $K[[x]]$ is Noetherian is equivalent to $K[[x]]$ is finitely generated as a K-module. There is a proof of $K[[x]]$ being Noetherian. I can't see why $K[[x]]$ is a finitely generated K-module but not a finitely generated K-algebra. Is there a mistake in my understanding? Any help is appreciated.




abstract-algebra commutative-algebra noetherian formal-power-series




share|cite|improve this question













The question rises when I was reading Kemper's "A course in Commutative Algebra."
Let $K$ be a field and consider the ring of formal power series $K[[x]]$. In an exercise I proved $K[[x]]$ is not Jacobson, hence not finitely generated as a K-algebra. But since $K$ is Noetherian and $K[[x]]$ is a K-module, $K[[x]]$ is Noetherian is equivalent to $K[[x]]$ is finitely generated as a K-module. There is a proof of $K[[x]]$ being Noetherian. I can't see why $K[[x]]$ is a finitely generated K-module but not a finitely generated K-algebra. Is there a mistake in my understanding? Any help is appreciated.





abstract-algebra commutative-algebra noetherian formal-power-series





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edited 13 hours ago









Eric Wofsey

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  • 1




    It cannot be a finitely generated module without being a finitely generated algebra : if $F$ is a finite generating set as a module, then a fortiori $F$ generates it as an algebra
    – Max
    14 hours ago












  • 1




    It cannot be a finitely generated module without being a finitely generated algebra : if $F$ is a finite generating set as a module, then a fortiori $F$ generates it as an algebra
    – Max
    14 hours ago







1




1




It cannot be a finitely generated module without being a finitely generated algebra : if $F$ is a finite generating set as a module, then a fortiori $F$ generates it as an algebra
– Max
14 hours ago




It cannot be a finitely generated module without being a finitely generated algebra : if $F$ is a finite generating set as a module, then a fortiori $F$ generates it as an algebra
– Max
14 hours ago










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$K[[x]]$ is a Noetherian ring, not a Noetherian $K$-module. In other words, every ideal in the ring $K[[x]]$ is finitely generated, rather than every $K$-submodule of $K[[x]]$ being finitely generated.






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    $K[[x]]$ is a Noetherian ring, not a Noetherian $K$-module. In other words, every ideal in the ring $K[[x]]$ is finitely generated, rather than every $K$-submodule of $K[[x]]$ being finitely generated.






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      $K[[x]]$ is a Noetherian ring, not a Noetherian $K$-module. In other words, every ideal in the ring $K[[x]]$ is finitely generated, rather than every $K$-submodule of $K[[x]]$ being finitely generated.






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        $K[[x]]$ is a Noetherian ring, not a Noetherian $K$-module. In other words, every ideal in the ring $K[[x]]$ is finitely generated, rather than every $K$-submodule of $K[[x]]$ being finitely generated.






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        $K[[x]]$ is a Noetherian ring, not a Noetherian $K$-module. In other words, every ideal in the ring $K[[x]]$ is finitely generated, rather than every $K$-submodule of $K[[x]]$ being finitely generated.







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        answered 13 hours ago









        Eric Wofsey

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