$(x^a-y^b)$ is a prime ideal in $R[x,y]$ where $R$ is a domain and $\gcd(a,b)=1$. 
Let $R$ be an integral domain and $a,b$ be positive integers with $\gcd(a,b)=1.$ Then I want to show that the ideal $(x^a-y^b)$ is a prime ideal in $R[x,y].$ 

I approached in the following way.
Consider an algebraically independent element $t$ over $R$ and consider the ring homomorphism $\phi : R[x,y] \to R[t]$ which maps $f(x,y) \in R[x,y]$ to $f(t^b,t^a)$ and very clearly $(x^a-y^b) \in \ker(\phi).$ Now to show the other containment pick $g(x,y) \in \ker(\phi)$ and since $x^a-y^b$ is a monic polynomial in the indeterminate $x$ one can divide $g$ by the monic polynomial $x^a-y^b$ over the domain $R[y],$ so that we have $g(x,y)=(x^a-y^b)q(x,y)+ s_0(y)+s_1(y)x+ \cdots +s_{(a-1)}(y)x^{a-1}.$ Now I tried to divide each $s_i(y)$ by $x^a-y^b$ so that we have a final remainder with degree in $x$ and $y$ are less than $a$ and $b$ respectively so that $\gcd(a,b)=1$ comes into play. But it is not happening because in the last division loosing control in the degree of $x$. Can I manage this way to conclude that $g$ is in the ideal. I need some help to complete it in the way I am trying to prove it. Many thanks.

With your notation, we know that $r(x, y) = s_{0}(y)+s_{1}(y)x+\cdots+s_{(a-1)}(y)x^{a-1}$ satisfies $r(t^{b}, t^{a}) = 0$, since $g(t^{b}, t^{a}) = 0$. We want to show that $r = 0$, i.e. that $s_{i}(y) = 0$ for each $i = 0, \ldots, a-1$. It is clear that if some $s_{i}(y)$ is nonzero, then some $s_{j}(y)$ must also be nonzero for $j \neq i$. I claim that 
$$\deg_{t}(s_{i}(t^{a})(t^{b})^{i}) \neq \deg_{t}(s_{j}(t^{a})(t^{b})^{j})$$ if $s_{i}(y), s_{j}(y)$ are nonzero and $i \neq j$. 
Note that $s_{i}(t^{a})(t^{b})^{i}$ has terms of the form $t^{ka+bi}$ for $k$ a non-negative integer, so it suffices to show that we can never have $ka+bi = la+bj$ for any non-negative integers $k, l$ and $0 \leqslant i \neq j \leqslant a-1$. Indeed, if $ka+bi = la+bj$ with $k, l, i, j$ as specified, then $(k-l)a = b(j-i)$; since $a$ and $b$ are coprime, $a$ must divide $j-i$, which is a contradiction since $0 < j-i < a$.
