Let $A \subset k[x\_1, \dots, x\_n]$ be a subalgebra, which is also a graded subspace $A = \oplus\_{i \ge 0} A\_i$. One can write $A = A\_0 \oplus A\_{> 0}$ where we have $A\_0 = k^0[x\_1, \dots, x\_n] = k$ and $A\_{> 0} := \oplus\_{i \ge 1} A\_i$ is a graded ideal of $A$, known as the augmentation ideal.


My question is, is it necessary the case that the ideal $A\_{> 0}$ is finitely generated (as an $A$-module) if and only if $A$ is finitely generated as a $k$-algebra?



Augmentation ideal is finitely generated if and only if $A$ is finitely generated as a $k$-algebra?
This amounts to an exercise, applying what are perhaps the first few theorems you encounter in the context of noetherian rings. Moreover, the assumption that $A$ is a graded subring of $k[x\_1,\ldots,x\_n]$ is stronger than necessary. It is enough to assume that $k$ is a commutative noetherian ring and that $A$ is a commutative augmented $k$-algebra (but even these assumptions are stronger than necessary).


I will write $A\_+$ for the augmentation ideal of $k$. For one direction of the equivalence, if you know that $A\_+$ is a finitely-generated $A$-module, then it follows more or less immediately that $A$ is generated as a $k$-algebra by your set of $A$-module generators for $A\_+$ together with $1$.


For the other direction, assume that $A$ is finitely-generated as a $k$-algebra, so that $A$ is a quotient of a polynomial algebra of the form $k[x\_1,\ldots,x\_n]$. Then it is a consequence of Hilbert's Basis Theorem that $A$ is a noetherian ring. Now $A\_+$ is an $A$-submodule of the left $A$-module $A$, so $A\_+$ must be a noetherian $A$-module. Then by one of the first theorems you encounter for noetherian rings, $A\_+$ must be finitely-generated as a left $A$-module.


**Edit:** As Eric pointed out in the comments, I made a careless error in the second paragraph of my answer. As he says, if you take $A$ to be the power series ring $k[[x]]$, then $A\_+$ is generated as a left $A$-module by $\{x\}$, but the set $\{x,1\}$ does not generate $k[[x]]$ as an algebra (it generates only the subring $k[x]$). On the other hand, if you assume that $A$ is a positively graded algebra, then the inductive argument indicated by Todd in his comment shows that $A$ is generated as a $k$-algebra by the generating set for $A\_+$ together with $1$.


