Image of an evaluation homomorphism 
I have the following  evaluation-homomorphism:
$$\phi: K[X,Y]  \to K[T]$$
$$X \to T^3$$
$$Y \to T^2$$
I have to prove that $\text{Im}(\phi)=K[T^2,T^3]$.

How can I prove it? I have already seen that $K[T^2,T^3]=\{ \sum_{i\geq0}c_iT^i\in K[T]: c_1=0\}$.
Thank you for your time.
 A: $K[T^2,T^3]$ is the smallest $K$-subalgebra of $K[T]$ containing $T^2$ and $T^3$. If a $K$-subalgebra of $K[T]$ contains $T^2$ and $T^3$, then it also contains all polynomial expressions in $T^2$ and $T^3$, i.e., it contains $\text{Im}(\phi)$. Since $\text{Im}(\phi)$ is a $K$-subalgebra of $K[T]$ containing $T^2$ and $T^3$, it is automatically the smallest one. Therefore $K[T^2, T^3] = \text{Im}(\phi)$.
A: A polynomial $f\in K[X,Y]$ is a sum of monomials and these have the form $X^iY^j$ with $i,j$ non-negative integers. When you consider $\phi(f)=f(T^2,T^3)$ the monomials of $f$ are sent to monomials in $T$ of the form $(T^2)^i(T^3)^j=T^{2i+3j}$. Now let's see what values can take $2i+3j$? For $i=0$ and $j=0$ you get $0$; for $i=1$ and $j=0$ you get $2$; for $i=0$ and $j=1$ you get $3$, and so on. To conclude (well, a little bit more effort is necessary at this point), $2i+3j$ covers the set $\mathbb N-\{1\}$, so $\phi(f)\in K[T^2,T^3]$ (which you already identified with the subring of $K[T]$ consisting by polynomials that don't have $T$ is their representation as a sum of monomials).
