$(X^m-Y^n)$ is a prime ideal in $A[X,Y]$ iff $m, n$ are coprime

Question

Let $A$ be an integral domain and $A[X,Y]$ the polynomial ring in two variables with coefficients in $A$. Let $m, n ∈ \mathbb Z_{≥1}$ be positive integers. Claim: the ideal $(X^m − Y^n)$ is prime in $A[X, Y ]$ iff $m$ and $n$ are coprime.

I got the following hint but I couldn't figure out how to proceed the proof. Please give me a more details hints.
"For the ”if” direction: Show that the map $ϕ : A[X, Y ] → A[T], f(X, Y ) → f(T^n, T^m)$ is a ring homomorphism, which factors over a ring homomorphism $\bar ϕ : A[X, Y ]/(X^m − Y^n) → A[T]$. Then show that $\bar \phi$ is injective."

Answer 1

If $\gcd(n,m)=d>1$, then it is easy to see that $X^m-Y^n=X^{ds}-Y^{dt}=(X^s)^d-(Y^t)^d$ is reducible and thus $(X^m-Y^n)$ is not prime.

Now, if $\gcd(n,m)=1$, your hint seems fine. If you prove that $\bar{\phi}$ is injective, then $A[X, Y ]/(X^m − Y^n)$ is (isomorphic to) a subring of a domain and is thus a domain, which implies that $(X^m-Y^n)$ is prime. To show that $\bar{\phi}$ is injective, take a polynomial $f(X,Y)=\sum a_{ij} X^iY^j$ such that $f(T^n,T^m)=\sum a_{ij} T^{ni}T^{mj}=0$ and show that $X^m-Y^n \vert f(X,Y)$ using the fact that $n$ and $m$ are coprime.