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?
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?
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 $C$.