Question. Suppose that $\phi: \Bbb Q[X, Y]\rightarrow \Bbb Q[T]$ is the homomorphism between polynomial rings defined by $\phi(X)=T^2$ and $\phi(Y)=T^3$. Is $\phi$ surjective?
Attempt. Now from what I understand, a function is surjective if the image of the function is the entire codomain, $\Bbb Q[T]$ in this case. Now it seems like I can get all of $\Bbb Q[T]$: $\phi(1)=1$, $\phi(Y/X)=T$, $\phi(X)=T^2$ and so on, so I would say that $\phi$ is surjective.
Am I correct with my answer and reasoning? If not, then what would be the right approach?
Thanks in advance!
The rational function $Y/X$ does not belong to the polynomial ring $\mathbb{Q}[X,Y].$
As such, you can't say that $\phi(Y/X)=T.$
In fact, there does not exist a polynomial $f(X,Y) \in \mathbb{Q}[X,Y]$ such that $f(T^2,T^3)=T.$
Therefore $\phi$ is not surjective.