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Let $K[x_1, x_2,\dots, x_n] $ be a polynomial ring. If it is a graded ring, then under certain conditions, its subalgebras may be finitely generated.

Isn't a subalgebra of a finitely generated k-algebra always finitely generated?

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marked as duplicate by user26857 commutative-algebra Jan 16 '16 at 17:21

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    $\begingroup$ No. Consider the free $k$-algebra $k\langle x,y\rangle$, say. $\endgroup$ – Pedro Tamaroff Jan 16 '16 at 16:14
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    $\begingroup$ A more familiar example might be that subgroups of finitely generated free groups need not be free. One can convert this to an example of algebras by taking group rings. $\endgroup$ – Cheerful Parsnip Jan 16 '16 at 16:17
  • $\begingroup$ @GrumpyParsnip Yes, that also came to mind. $\endgroup$ – Pedro Tamaroff Jan 16 '16 at 16:18
  • $\begingroup$ @Grumpy: do you mean that subgroups of f.g. free groups need not be f.g.? Of course they are always free. I also don't understand Pedro's remark; he gives an example of a f.g. algebra and then does not give an example of a non-f.g. subalgebra of it. $\endgroup$ – Qiaochu Yuan Jan 16 '16 at 16:50
  • $\begingroup$ @QiaochuYuan: oops, that was a typo. Yes I meant a subgroup need not be f.g. $\endgroup$ – Cheerful Parsnip Jan 16 '16 at 22:30
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This is false even in the commutative case. For example, $k[x, y]$ is finitely generated, but it has a subalgebra $k[x, xy, xy^2, xy^3, \dots ]$ which is not (this is a nice exercise).

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  • $\begingroup$ I'm sorry I'm having trouble understanding this. Isn't $k [x, xy, xy^2, xy^3, \dots]=k [x] $? $\endgroup$ – fierydemon Jan 16 '16 at 16:56
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    $\begingroup$ @Ayush: no. $k[x]$ doesn't contain $xy$. This is a subalgebra, not an ideal. $\endgroup$ – Qiaochu Yuan Jan 16 '16 at 16:58

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