Question

Let $A \subset \mathbb{C}[x_1,x_2,\ldots,x_n]$ - be finitely generated graded algebra and $B$ be its subalgebra. How to prove that $A=B.?$

Unfortunalelly I know only one method to do it - to compare theirs Poincare series.

Anybody know more?

Answer 1

This question is posed way too generically in order to obtain an answer that is useful to you by more than mere coincidence, but here are three things that I found of use:

(1) Your algebra is graded, thus in particular filtered. Try induction. Generally, if $A=\bigcup\limits_{n\geq 0} A_n$ is a filtered ring and $B$ is a subring of $A$, and if we know that $A_n\subseteq A_{n-1}+B$ for every $n\geq 0$ (where $A_{-1}$ means $0$), then induction shows that $A=B$.

(2) If $A$ is a $k$-algebra (with $k$ a commutative ring), and $B$ is a subalgebra of $k$, then the inclusion $B\to A$ induces a canonical map $\mathrm{Alg}\left(A,C\right)\to \mathrm{Alg}\left(B,C\right)$ for every $k$-algebra $C$ (where $\mathrm{Alg}\left(U,V\right)$ denotes the set of all $k$-algebra maps from $U$ to $V$). If this map is a bijection for every $k$-algebra $C$, then $A=B$. This follows from the Yoneda lemma (and is pretty easy to see) and doesn't look at all like a simplification (if we don't know $A$ and $B$ well enough to see directly that $A=B$, how can we tell anything about $\mathrm{Alg}\left(A,C\right)\to \mathrm{Alg}\left(B,C\right)$ ?), but if your algebras $A$ and $B$ are defined as coordinate rings of some varieties, then $\mathrm{Alg}\left(A,C\right)$ and $\mathrm{Alg}\left(B,C\right)$ are simpler objects than $A$ and $B$ themselves, so this trick can work well.

(3) If $A$ is a $k$-algebra (with $k$ a field), and $B$ is a subalgebra of $k$, and you wish to show that $A=B$, then it is enough to find a nonzero $k$-algebra $C$ and prove that the inclusion $B\otimes_k C\to A\otimes_k C$ is a bijection. This sounds obvious and stupid again (how is $B\otimes_k C\to A\otimes_k C$ supposed to be more accessible than $B\to A$ ?), but sometimes tensoring with a strategically chosen algebra $C$ helps to "straighten out" the structure of $B$ and $A$ and get some more canonical basis. The most often used particular case is when $k$ is not algebraically closed and $C$ is some field extension of $k$. Now that is not your case, as you have $k=\mathbb C$, but I don't see a real reason why this tactic shouldn't work with more general $C$.