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?
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 $1$.
