Do we have $\mathbb{C}[T] = \mathbb{C}[\Lambda] = \oplus_{\lambda \in \Lambda} \mathbb{C}[\lambda]$? Let $\mathbb{C}[T]$ be the coordinate ring of a torus $T$. Suppose that $T$ acts on some variety $X$. Then $T$ acts on $\mathbb{C}[X]$: $t(f) = \lambda(t)f$ ($f$ is a homogenous function on $\mathbb{C}[X]$), where $\lambda$ is the weight of $f$. Do we have 
$$
\mathbb{C}[T] = \mathbb{C}[\Lambda] = \oplus_{\lambda \in \Lambda} \mathbb{C}[\lambda]?
$$
Here $\Lambda$ is the set of weights. Thank you very much.
 A: Yes, that's true. More generally, consider the following: Suppose $D$ is an abelian group, ${\mathbb k}$ is a field and $\text{diag}_D$ is the associated algebraic group over ${\mathbb k}$ given by $\text{diag}_D(A) := \text{Hom}_{\text{Grp}}(D,A^{\times})$ for a commutative ${\mathbb k}$-algebra $A$. Then, given a ${\mathbb k}$-vector space $V$, putting the structure of a $\text{diag}_D$-representation on $V$ is equialent to equipping $V$ with a $D$-grading.
If $D={\mathbb Z}$, then $\text{diag}_D = \text{diag}_{\mathbb Z} = \text{Spec}({\mathbb k}[t,t^{-1}])$ is the torus, so you see that endowing a ${\mathbb k}$-vector space with an action of the torus is equivalent to giving it a ${\mathbb Z}$-grading.
For the proof, note that $\text{diag}_D$ is represented by the ${\mathbb k}$ group ring ${\mathbb k}[D]$, so giving $V$ the structure of a $\text{diag}_D$-representation is the same as providing it with the structure of a comodule over ${\mathbb k}[D]$. Now, you can check that given a comodule structure $\Delta: V\to V\otimes_{\mathbb k} {\mathbb k}[D]$, $V$ decomposes as $V = \bigoplus_{d\in D} V_d$ with $V_d := \{v\in V\ |\ \Delta(v) = v\otimes d\}$; conversely, any decomposition $V=\bigoplus_{d\in D} V_d$ defines a coaction by $\Delta(v) := v\otimes d$ for $v\in V_d$ and linear extension.
