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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. But that is not enough.

More explicitly, choose $v_1$, $\dots$, $v_m$ in $R$ so that their images together with $1$ form a basis of the vector space $R/I$. Then choose $x_1$, $\dots$, $x_n$ in $I$ so that together with the $v_i$ they generate the $k$ algebra $R$ and so that every product $v_iv_j$ can be rewritten as a $k$ linear combination of the generators $1$, $v_1$, $\dots$, $v_m$, $x_1$, $\dots$, $x_n$. (First achieve the generation, then throw in more elements to achieve the rewriting. Use that for every $r\in R$ there is a $\tilde r\in I$ so that $r$ can be written as a  $k$ linear combination of the generators $1$, $v_1$, $\dots$, $v_m$ and $\tilde r$.) Now the $x_j$ together with the $v_ix_j$ generate the $k$ algebra $A=1+I$. Indeed every element of $R$ can be written as a  $k$ linear combination of the generators $1$, $v_1$, $\dots$, $v_m$ plus a polynomial in the $x_j$, $v_ix_j$. And the elements of $A$ can thus be written as a polynomial in the $x_j$, $v_ix_j$.

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. But that is not enough. It only shows that the hypotheses are reasonable.

Choose $v_1$, $\dots$, $v_m$ in $R$ so that their images together with $1$ form a basis of the vector space $R/I$. Then $R$ is generated as an $A$-module by $1$, $v_1,\dots,v_m$. So $A$ is a finitely generated $k$-algebra by the Artin-Tate lemma in wikipedia.

One may argue more directly. Every $r\in R$ may be written as a $k$ linear combination of $1$, $v_1,\dots,v_m$ plus an element, say $f(r)$, in $I$. Now take a generating set $y_1,\dots,y_m$ of the $k$-algebra $R$ and choose $x_1$, $\dots$, $x_n$ in $I$ so that the $f(y_i)$ and the $f(v_iv_j)$ are amongst the $x_j$.

We claim that every element of $R$ can be written as a  $k$ linear combination of the generators $1$, $v_1$, $\dots$, $v_m$ plus a polynomial in the $x_j$, $v_ix_j$. Indeed it is easy to check that the set of elements that can be written this way is invariant under multiplication by the $v_i$ and the $x_j$, hence also by the $y_j$.

If an element of $A$ is written as a  $k$ linear combination of the generators $1$, $v_1$, $\dots$, $v_m$ plus a polynomial in the $x_j$, $v_ix_j$, then it must be a polynomial in the $x_j$, $v_ix_j$.

So the $x_j$ together with the $v_ix_j$ generate the $k$ algebra $A=1+I$.

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. But that is not enough.

More explicitly, choose $v_1$, $\dots$, $v_m$ in $R$ so that their images together with $1$ form a basis of the vector space $R/I$. Then choose $x_1$, $\dots$, $x_n$ in $I$ so that together with the $v_i$ they generate the $k$ algebra $R$ and so that every product $v_iv_j$ can be rewritten as a $k$ linear combination of the generators $1$, $v_1$, $\dots$, $v_m$, $x_1$, $\dots$, $x_n$. (First achieve the generation, then throw in more elements to achieve the rewriting.) Now the $x_j$ together with the $v_ix_j$ generate the $k$ algebra $A=1+I$. Indeed every element of $R$ can be written as a $k$ linear combination of the generators $1$, $v_1$, $\dots$, $v_m$ plus a polynomial in the $x_j$, $v_ix_j$. And the elements of $A$ can thus be written as a polynomial in the $x_j$, $v_ix_j$.

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. But that is not enough.

More explicitly, choose $v_1$, $\dots$, $v_m$ in $R$ so that their images together with $1$ form a basis of the vector space $R/I$. Then choose $x_1$, $\dots$, $x_n$ in $I$ so that together with the $v_i$ they generate the $k$ algebra $R$ and so that every product $v_iv_j$ can be rewritten as a $k$ linear combination of the generators $1$, $v_1$, $\dots$, $v_m$, $x_1$, $\dots$, $x_n$. (First achieve the generation, then throw in more elements to achieve the rewriting. Use that for every $r\in R$ there is a $\tilde r\in I$ so that $r$ can be written as a $k$ linear combination of the generators $1$, $v_1$, $\dots$, $v_m$ and $\tilde r$.) Now the $x_j$ together with the $v_ix_j$ generate the $k$ algebra $A=1+I$. Indeed every element of $R$ can be written as a $k$ linear combination of the generators $1$, $v_1$, $\dots$, $v_m$ plus a polynomial in the $x_j$, $v_ix_j$. And the elements of $A$ can thus be written as a polynomial in the $x_j$, $v_ix_j$.

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 $A=k+I$?

More explicitly, choose $v_1$, $\dots$, $v_m$ in $R$ so that their images together with $1$ form a basis of the vector space $R/I$. Then choose $x_1$, $\dots$, $x_n$ in $I$ so that together with the $v_i$ they generate the $k$ algebra $R$ and so that every product $v_iv_j$ can be rewritten as a $k$ linear combination of the generators $1$, $v_1$, $\dots$, $v_m$, $x_1$, $\dots$, $x_n$. Now the $x_j$ together with the $v_ix_j$ generate the $k$ algebra $A=1+I$. Indeed every element of $R$ can be written as a $k$ linear combination of the generators $1$, $v_1$, $\dots$, $v_m$ plus a polynomial in the $x_j$, $v_ix_j$. And the elements of $A$ can thus be written as a polynomial in the $x_j$, $v_ix_j$.

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. But that is not enough.

More explicitly, choose $v_1$, $\dots$, $v_m$ in $R$ so that their images together with $1$ form a basis of the vector space $R/I$. Then choose $x_1$, $\dots$, $x_n$ in $I$ so that together with the $v_i$ they generate the $k$ algebra $R$ and so that every product $v_iv_j$ can be rewritten as a $k$ linear combination of the generators $1$, $v_1$, $\dots$, $v_m$, $x_1$, $\dots$, $x_n$. (First achieve the generation, then throw in more elements to achieve the rewriting.) Now the $x_j$ together with the $v_ix_j$ generate the $k$ algebra $A=1+I$. Indeed every element of $R$ can be written as a $k$ linear combination of the generators $1$, $v_1$, $\dots$, $v_m$ plus a polynomial in the $x_j$, $v_ix_j$. And the elements of $A$ can thus be written as a polynomial in the $x_j$, $v_ix_j$.