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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."

user26857
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Thomas Edison
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  • Which part do you want more details on? To show that $\overline{phi}$ is injective, you need to show that $f(T^n,T^m)=0$ exactly when $(X^m-Y^n)$ divides $f(X,Y)$. This is where you use that $(m,n)=1$. – Andrew Dudzik Oct 25 '15 at 19:30
  • Hi, I want some hints for both direction. Thanks – Thomas Edison Oct 25 '15 at 20:21
  • Hi @user26857 can you give me the reference to show that (X^m-Y^n) is prime iff $X^n−Y^m$ is irreducible – Thomas Edison Oct 25 '15 at 20:30
  • Have you checked the answers in the linked thread? – user26857 Oct 25 '15 at 20:31
  • @user26857 Yes, I know it before I asked my question. But I don't know is it true or not?? "(X^m-Y^n) is prime iff Xn−Ym is irreducible" Can you show me? – Thomas Edison Oct 25 '15 at 20:33
  • To begin with, if $(m,n)=d>1$, then the ideal $(X^m-Y^n)$ is not prime (why?). Conversely, if $(m,n)=1$ then the kernel of the map from your hint equals $(X^m-Y^n)$ and this follows from the answer of Hagen von Eitzen, so there is no need to reopen this question. – user26857 Oct 25 '15 at 20:50
  • Hi @Slade I still can not figure out how to show that $^m-Y^n | f(X,Y)$. Could you please give me a detail on this part? I am really sorry for my stupidity. – Thomas Edison Oct 26 '15 at 18:17

1 Answers1

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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.

user26857
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Nitrogen
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