Let $G$ be an affine, connected, reductive group and $X$ a $G$-module. Choose a maximal torus $T\subseteq G$, a Borel $B\subseteq G$ containing $T$ and let $U$ be the unipotent radical of $B$. Denote by the character group of $T$. Let $\Lambda\subseteq\mathbb{X}$ be the set of dominant weights of $G$ with respect to these choices. We can decompose the graded coordinate ring $\mathbb{C}[X]=\bigoplus_{\lambda\in\Lambda} V_{(\lambda)}$ into its isotypic components $V_{(\lambda)}$ of weight $\lambda$. Let
\[ \Lambda_X=\{ \lambda\in\Lambda \mid V_{(\lambda)}\ne\{0\}\} \]
be the set of weights that occur in $\mathbb{C}[X]$.
Let $V_{(\lambda)}\cong V_\lambda^{\oplus n_\lambda}$, where $V_\lambda$ is the irreducible module of highest weight $\lambda$. Each $V_\lambda$ has a highest weight vector which is unique up to scaling - let $f_{\lambda 1}, \ldots, f_{\lambda n_\lambda}\in V_{(\lambda)}$ be linearly independent highest weight vectors.
Claim. *For each $1\le k\le n_\lambda$, if the function $f:=f_{\lambda k}\in\mathbb{C}[X]$ is reducible, then there exist weights $\lambda_1,\ldots,\lambda_r\in\Lambda_X$ such that $\lambda$ is an $\mathbb N$-linear combination of the $\lambda_i$.*
Indeed, about a year ago this statement was completely unclear to me. However, it's actually not that hard to see and I felt like sharing my proof.
Proof. Let $Z=\mathrm{Z}(f)$ be the vanishing set of $f$ (with the reduced subvariety structure). Note that $Z$ is $G$-invariant, so we have an action $\alpha:G\times Z\to Z$ induced by the action of $G$ on $X$. Since $\mathbb{C}[X]$ is factorial, we can write $f=f_1^{e_1}\cdots f_r^{e_r}$ as a product of powers of irreducible elements where any two of the $f_i$ are not associated. This corresponds to the decomposition of $Z=Z_1\cup\cdots\cup Z_r$ into its irreducible components $Z_i=\mathrm{Z}(f_i^{e_i})=\mathrm{Z}(f_i)$. Note that all these components have the same dimension. Let $\alpha_i:G\times Z_i\to Z$ be the restriction of the action morphism $\alpha$. Then, $Z_i\subseteq \mathrm{im}(\alpha_i)$ and $\mathrm{im}(\alpha_i)$ is irreducible of maximal dimension, hence $\mathrm{im}(\alpha_i)=Z_i$. This implies that $Z_i$ is $G$-invariant. In particular, $g.f_i=uf_i$ for some $u\in\mathbb{C}[X]$ and since the action of $G$ preserves the degree on $\mathbb{C}[X]$, we also have $\deg(u)=0$. Hence, $u\in\mathbb{C}[X]_0=\mathbb{C}$. This means that $G.f_i = \mathbb{C} f_i$, so $G$ acts by a character on $f_i$. This means that $f_i$ is a highest weight vector of some weight $\lambda_i$. Consequently, we must have $\lambda = e_1\lambda_1 + \cdots + e_r\lambda_r$ with $\lambda_i\in\Lambda$ corresponding to an irreducible highest weight vector. QED
For $\lambda\in\Lambda_X$, denote by $q(\lambda)$ the number of irreducible highest weight vectors in $V_{(\lambda)}$. We define $M(\lambda):=\{ \mu\in\Lambda_X \mid \lambda-\mu\in\Lambda_X \}$, then
\[ q(\lambda) = n_\lambda - \sum_{\mu\in M(\lambda)} q(\mu). \]
This formula was the reason why the question interested me in the first place: Let $I\subseteq\mathbb{C}[X]$ be a homogeneous, $G$-invariant ideal. If we already know the multiplicities $m_\lambda$ of $\mathbb{C}[X]\cong\bigoplus_{\lambda\in\Lambda} V_\lambda^{\oplus m_\lambda}$ and we want to know candidates for generators of $I$, there is a randomized procedure that works as follows:
+ Take $m=m_\lambda$ highest weight vectors $F_1,\ldots,F_m\in\mathbb{C}[X]$ of weight $\lambda$.
+ Evaluate these vectors at $m$ random points $x_1,\ldots,x_m\in Z(I)$.
+ If the corresponding matrix $(F_i(x_j))_{ij}$ has full rank, then it is a certificate for the fact that the weight $\lambda$ does not occur in $I$.
Now, the above claim tells us that we can start in low degrees and cache evaluations of low-degree highest weight vectors to use them in certificates in higher degrees, saving a lot of work.