This is related to my question [here](https://mathoverflow.net/questions/250489/m-is-finitely-generated-as-a-module-iff-m-a-0m-is-finitely-generated-as). My question is as follows. How do I see that a nonnegatively graded algebra $A$ is finitely generated as a $k$-algebra if and only if $A\_0$ is finitely generated as a $k$-algebra and $A\_{>0}$ is finitely generated as an $A$-module (i.e. as a left ideal of $A$)?


Our algebras here are associative, unital, and not necessarily commutative.



Nonnegatively graded algebra $A$ finitely generated as $k$-algebra iff $A_0$ finitely generated, $A_{>0}$ finitely generated as $A$-module?
Seems like the "greedy algorithm" works...


Suppose $A\_0$ is finitely generated as a $k$-algebra, say by $a\_1, \ldots, a\_m$ and $A\_{>0}$ is finitely generated as an $A$-module, say by $b\_1, \ldots, b\_n$. I claim that $\{a\_1, \ldots, a\_m, b\_1, \ldots, b\_n\}$ generate $A$ as a $k$-algebra. 


Indeed, given any $x \in A$, if $x \in A\_0$ then this is obvious. Otherwise, $x \in A\_{>0}$ so we can write 


$$ x = b\_1 x\_1' + b\_2 x\_2' + \ldots $$


with the $x\_i' \in A$ having strictly lower degrees. Then we are done by induction. 


Now for the other direction. If $A$ is finitely generated as a $k$-algebra, write a finite list of generators $\{a\_1, \ldots, a\_m, b\_1, \ldots, b\_n\}$ with $\deg a\_i = 0$ and $\deg b\_i >0$. Then $A\_0$ is generated as a $k$-algebra by $\{a\_1, \ldots, a\_m\}$, since by definition everything in $A$ is a polynomial expression in these generators, but by degree considerations no such expression involving a $b$ can be in $A\_0$. 


Next I claim that $\{b\_1, \ldots, b\_n\}$ generates $A\_{>0}$ as an $A$-module. Indeed, any $x \in A\_{>0}$ is a polynomial expression in the generators, and again by degree considerations every monomial must involve at least one of the $b\_i$. 


