Let $$A=\left\{ \sum_{i=1}^n a_i X^{i}\in R[X]\mid n\in\mathbb{N}, a_i\in R, a_1=0 \right\}\subseteq R[X]$$ be a subring of a commutative ring $R[X]$.
Now I have to show that $$\varphi:R[X_1, X_2]\rightarrow A\\ f(X_1,X_2)\mapsto f(X^2, X^3)$$
is surjective.
I think it is basically a technicality, but I just wont get it. Can someone please help me?
Hint: The monomial $X_1^{\alpha_1}X_2^{\alpha_2}$ with $\alpha_1,\alpha_2\geq 0$ is mapped to $(X^2)^{\alpha_1}(X^3)^{\alpha_2} = X^{2\alpha_1+3\alpha_2}$. The exponent $2\alpha_1+3\alpha_2$ can be any integer $\geq 0$ but not be equal to $1$. This implies the surjectivity.
Note that for each even integer $n\geq 0$, $n=2\alpha_1$, and for each odd integer $n\geq 3$, $n=2\alpha_1+3$.