Probably yes. Both $R$ and $A=k+I$ are filtered by powers of $I$ and we may look at the associated graded rings. Now $I/I^2$ is a finitely generated $R/I$ module and $R/I$ is a finitely generated vector space, by Zariski's lemma on finitely generated $k$ algebras that are fields. So $I/I^2$ is a finitely generated vector space and the associated graded of $A=k+I$ is a finitely generated $k$ algebra. Then so is $A=k+I$?
Wilberd van der Kallen
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