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Let $\gcd(p, q) = 1$ and $Y=\{(t^p, t^q) \in \mathbb C^2 \}$. Determine the ideal $I(Y)$.

Definition. The ideal of $X$ is defined as

$$I(X)=\{f\in \Bbb C[x,y]:f(x,y)=0, \forall (x,y)\in X\}.$$

It is clear that $I(Y)$ is the kernel of the homomorphism $ \phi :\mathbb C[x,y] \to \mathbb C[t]$ defined by $ x \to t^p $ and $ y \to t^q$. Further it is clear that $(x^q-y^p)\subseteq\ker\phi$. But I'm unable in showing that $\ker(\phi)$ is contained in $(x^q-y^p)$. Any hints/ideas ?

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  • $\begingroup$ Do you know Krull's Hauptidealsatz? $\endgroup$ – Fredrik Meyer Sep 13 '15 at 11:53
  • $\begingroup$ @FredrikMeyer Sorry.. i don't know about this result..Regard, $\endgroup$ – user270331 Sep 13 '15 at 13:10
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We have $\mathbb C[x,y]/\ker\phi\simeq\mathbb C[t^p,t^q]$, so $\dim \mathbb C[x,y]/\ker\phi=\dim\mathbb C[t^p,t^q]=1$. Moreover, since $\mathbb C[t^p,t^q]$ is an integral domain $\ker\phi$ is a prime ideal (of height one). On the other side, $x^p-y^q$ is an irreducible polynomial contained in $\ker\phi$, so $(x^p-y^q)$ is also a prime ideal. Now it follows $(x^p-y^q)=\ker\phi$.

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  • $\begingroup$ sorry.. why ker$ \phi $ is of height one? $\endgroup$ – user270331 Sep 13 '15 at 15:37
  • $\begingroup$ If $A$ is a finitely generated $k$-algebra, and $I$ an ideal of $A$, then $\dim A=\dim A/I+\mathrm{height}(I)$. (This is proved in any classical textbook in commutative algebra.) $\endgroup$ – user26857 Sep 13 '15 at 16:19

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