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

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 ?

• Do you know Krull's Hauptidealsatz? Sep 13, 2015 at 11:53
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 $(x^p-y^q)=\ker\phi$.
• sorry.. why ker$\phi$ is of height one? Sep 13, 2015 at 15:37
• 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.) Sep 13, 2015 at 16:19