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I am trying to solve the following exercise.

Assume that $m$ and $n$ are co-prime positive integers, and that $k$ is a field. Prove that the ideal generated by $x^{m} - {y^n}$ is a prime ideal in $k[x,y]$.

I tried the following approach : Consider the ring homomorphism ${\varphi}:k[x,y]\rightarrow{k[t]}$ defined by $\varphi(x) = t^n$, and $\varphi(y) = t^m$, and extend $\varphi$ to $k[x,y]$ to ensure that it is a ring homomorphism. Then the ideal $(x^m - y^n)$ is clearly contained in the kernel of $\varphi$. It remains to prove the reverse set inclusion. i.e. to prove that this ideal is exactly equal to the kernel of $\varphi$. I don't know how to go about this!


marked as duplicate by user26857 abstract-algebra May 23 '17 at 9:12

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  • $\begingroup$ Do you know about the First Ring Isomorphism Theorem? $\endgroup$ – mathfan27543 May 21 '17 at 20:56
  • $\begingroup$ Let $\varphi(f(x, y)) = 0 \implies f(\varphi(x), \varphi(y)) = 0 \implies f(t^m, t^n) = 0$. This means each term of the polynomial is zero for all coefficients in $K$. But $f$ is polynomial in $t^m$ and $t^n$ with $(m,n)=1$, this means each term in $f$ is multiple of $t^{mn} - t^{mn}$. P.S. not sure if it works. $\endgroup$ – Santosh Linkha May 21 '17 at 21:00
  • $\begingroup$ @SantoshLinkha A monomial $x^iy^j$ from $f$ becomes $t^{im+jn}$ after substitution. This must be cancelled by $t^{i'm+j'n}$ (which comes from a monomial $x^{i'}y^{j'}$) with $i'm+j'n=im+jn$. But since $(m,n)=1$ we get that $m\mid j-j'$ and $n\mid i-i'$. Now try to figure out how looks $x^iy^j-x^{i'}y^{j'}$. $\endgroup$ – user26857 May 21 '17 at 23:20

Since $k[x,y]$ is a Unique Factorization Domain, the ideal $(x^m-y^n)$ is prime there if and only if $P=x^m-y^n$ is irreducible in $k[x,y]$.

I offer two proofs of irreducibility of $x^m-y^n$ when $\gcd(m,n)=1$, both showing that it’s irreducible as an element of $k(x)[y]$. The first is very advanced, the second is elementary, but my argument there is not as simple as it ought to be.

The first argument uses the theory of the Newton Polygon. Consider the discrete valuation $v_x$ on $k(x)$, which makes $v_x(x)=1$ and $v_x(f)=0$ if $x$ does not divide the polynomial $f\in k[x]$. The polygon of $P=x^m-y^n$ has one vertex for each monomial here, one at $(0,m)$ for $x^m$, the other at $(n,0)$ for $y^n$. There is only the one segment, between these two vertices, and it passes through no other integral points. But a factorization $P=P_1P_2$ that’s nontrivial, i.e. with both $y$-degrees positive, will give two segments, of lesser width than $n$ and with the one slope $-m/n$, and with both coordinates of the endpoints integral. Not possible, since the original segment passed through no integral points.

The other proof is elementary but much longer than it should be, and proceeds through several steps.
First, note that if $f(t)\in F[t]$ has degree $d$, where $F$ is a field, and if $\lambda\ne0$ in $F$, then $f(t)$ is irreducible if and only if $f(\lambda t)$ (and similarly $f(\lambda t)/\lambda^d\,$) is irreducible in $k[t]$.
Second, consider the polynomial $x^r-y^s\in k[x,y]$, where $r<s$. Say that $s=qr+d$ with $d<r$, by Euclidean division. Now consider the equivalences: \begin{align} x^r-y^s\text{ irred in }k[x,y]&\Longleftrightarrow x^r-y^s\text{ irred in }k(y)[x]\\ &\Longleftrightarrow\left[(xy^q)^r-y^{rq+d}\right]\big/y^{qr}\text{ irred in }k(y)[x]\\ &\Longleftrightarrow x^r-y^d\text{ irred in }k(y)[x]\\ &\Longleftrightarrow x^r-y^d\text{ irred in }k[x,y] \end{align} You see the strategy now: perform Euclidean division repeatedly on the exponents, just as in the Euclidean algorithm for greatest common divisor, and finally get to a point where one of the exponents is $1$. But you know that $x^\ell-y$ and $y^\ell-x$ are irreducible in $k[x,y]$, which gives the result.


See this answer and adapt it (when $\gcd(n,m)=1$) to show $$\phi : k[x,y]/(y^n-x^m) \to k[t^n,t^m], \qquad \phi(f(x,y))= f(t^n,t^m)$$ is an isomorphism, with inverse homomorphism $\phi^{-1}(f(t^n,t^m)) = f(x,y)+(y^n-x^m)$ which is well-defined because $t^{an+bm} = t^{a'n+bm'} \implies n(a-a') = cnm = m(b'-b)$ so that $ x^{a}y^{b}-x^{a'}y^{b'} = x^{a'}y^{b}(x^{cm}-y^{cn}) \in (y^n-x^m)$ .

Since $k[t^n,t^m]$ is an integral domain, $k[x,y]/(y^n-x^m)$ is an integral domain and $(y^n-x^m)$ is a prime ideal of $k[x,y]$

Otherwise I have this solution using the splitting field.

Let $S = k(\zeta_n)(x)$. Factorizing $y^n-x^m \in S[y]$ in its splitting field : $$y^n-x^m = \prod_{k=1}^n (y-\zeta_n^k \sqrt[n]{x^m}).$$ Thus the minimal polynomial of $\sqrt[n]{x^m}$ over $S$ is of the form $P(y)=\prod_{k \in E} (y-\zeta_n^k \sqrt[n]{x^m})$ for some $E \in \{1 \ldots n\}$.

$P(0) =(-1)^{|E|} \zeta_n^l (\sqrt[n]{x^m})^{|E| } \in S$. Since $\gcd(n,m)=1$, $n$ is the least natural number such that $(\sqrt[n]{x^m})^r \in S$, therefore $|E| = n$, $y^n-x^m$ is irreducible and $(y^n-x^m)$ is a prime (maximal) ideal of $S[y]$, and hence $(y^n-x^m) \cap k[x,y]$ is a prime ideal of $k[x,y] \subset S[y]$.


There might be a detail missing in this ``proof".

With the notation from the original post, $\varphi$ induces a ring homomorphism $\overline\varphi:k[x, y]/(x^m-y^n) \to k[t]$ sending $\bar x \mapsto t^n$ and $\bar y \mapsto t^m$, where the bar means modulo $(x^m-y^n)$. Note $\overline\varphi$ is well defined since $(x^m -y^n) \subset \text{ker}~\varphi$. Note that $\text{im}~\overline\varphi = \text{im}~\varphi = k[t^n,t^m]$.

Now define a ring homomorphism $\psi:k[t^n, t^m] \to k[x,y]/(x^m-y^n)$ as follows: $$\psi(t^n) = \bar x,~~~ \psi(t^m) = \bar y$$ and extend linearly/multiplicatively.

Claim that $\overline\varphi$ and $\psi$ are mutual inverses. We just have to check $(\overline\varphi\circ\psi)(t^n) = t^n$, $(\overline\varphi\circ\psi)(t^m) = t^m$ and $(\psi\circ\overline\varphi)(\bar x) = \bar x$, $(\psi\circ\overline\varphi)(\bar y) = \bar y$ and this is clear. Thus $k[x,y]/(x^m-y^n) \cong k[t^n, t^m]$ and the latter ring is a subring of the integral domain $k[t]$; so $(x^m-y^n)$ is a prime ideal.


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