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Let $R$ be an integral domain, I want to know the condition for $n,m$ when $(x^n - y^m)$ is a prime ideal in the ring $R[x,y]$.

The followings are my trial

Since $R$ is an integral domain, $R[t]$ is also integral domain. I can define ring homomorphism $\Psi : R[x,y] \rightarrow R[t]$ such that $x \mapsto t^m, y\mapsto t^n$. And if I can make $\operatorname{Ker}(\Psi) = (x^n -y^m)$ then it is done via 1st isomorphism theorem with $R[t]$ is an integral domain.

So my problem changed to find the condition of $\operatorname{ker}(\Psi) = (x^n -y^m)$.

One direction is easy. $\Psi(x^n - y^m) = (t^m)^n - (t^n)^m =0$. The other directions seems unclear to me. i.e., I want to find the condition $\operatorname{ker}(\Psi) \subset (x^n -y^m)$.

I know $(x^3-y^2)$ is a prime ideal so it seems $(n,m)=1$.. but not sure of proving this in general.

Astyx
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phy_math
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    The condition $(n,m)=1$ is definitely necessary, otherwise we could factor $x^n-y^m=(x^{n/d})^d-(y^{m/d})^d = (x^{n/d}-y^{m/d})(\dots)$ with $d=(n,m)$ – leoli1 Jun 13 '21 at 10:44
  • You can do the following operation: If $m>n$ substitute $x$ by $x$ and $y$ by $xy$., and factor out $x$ as much as possible. If $m<n$, then replace $x$ by $xy$, $y$ by $y$, and factor out $y$ all you can. Observe that if the resulting polynomial factors, then so would factor the original. The result of doing this operation to $x^m-y^n$ is $x^{m-n}-y^n$ when $m>n$ or $x^m-y^{n-m}$ when $m<n$. As you can see these steps are doing Euclidean algorithm on the exponents. So, eventually we reach $x^{d}-y^{d}$, where $d=\gcd(m,n)$. If $d>1$, then $x^d-y^d=(x-y)(x^{d-1}+x^{d-2}y+...+y^{d-1})$. – plop Jun 13 '21 at 10:54
  • Therefore, if $x^m-y^n$ is irreducible, then $d=1$. – plop Jun 13 '21 at 10:58
  • Detecting the presence of several irreducible components doesn't require keeping the the variety as it is. If we do an operation that might lose some components and still find many, then can say that there were many to begin with. The operation used above is a blow-up at the origin, which could move apart irreducible components. What we are doing is resolution of singularities in disguise, but we stop one step before the end. When $d>1$ we find at least two irreducible components. Therefore, they were also at the beginning. – plop Jun 13 '21 at 13:19
  • The argument in my first comment is just doing this without all the unnecessary language, ending up with an argument that a high school student can understand. – plop Jun 13 '21 at 13:20
  • @user26857 Step by step. Regardless of whether $x^m-y^n$ is irreducible or not, we do the steps that I mentioned. Eventually we reach $x^d-y^d$, where $d=\gcd(m,n)$. Now, at each of the steps, if the new polynomial factors, then the previous must factor. How to see this elementary? Assume that it factors $f(x,y)g(x,y)$. Multiply by the power of $x$ (or $y$) that was factored out. Its exponent must be the sum of the smallest degree of any $y^k$ term in $f$ and in $g$. And then undoing the substitution. – plop Jun 13 '21 at 17:52
  • Technically, this is just because a blow up at a smooth subvariety doesn't increase the number of irreducible components. The argument finishes, because if $x^m-y^n$ is irreducible, then by the above $x^d-y^d$ must also be irreducible. Hence $d=1$, because we do know that if $d>1$, the latter factors. – plop Jun 13 '21 at 17:53
  • That is also why the argument doesn't need an isomorphism. We do a destructive operation, but one that we know doesn't increase the number of factors. So, if we started we one factor and we didn't end up with zero factors, then we must end up with exactly one factor. Those who know resolution of singularities will recognize that what is being done is monomialization of the ideal, except for the last step, which would really eliminate the last factor and would end up with zero components near the origin. – plop Jun 13 '21 at 18:07
  • @plop Can we please avoid lengthy discussion in the comments if they can be made into an answer? – Pedro Jun 14 '21 at 10:01

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