As Hartshorne goes
Let $Y\subseteq \mathbf{A}^3$ be the curve given parametrically by $x=t^3,y=t^4,z=t^5$. Show that $I(Y)$ is a prime ideal of height $2$ in $k[x,y,z]$ which cannot be generated by 2 elements.
It is algebraic routine to check that $I(Y)$ is prime and of height $2$. But how to show it cannot be generated by 2 elements? I can give an elementary solution (see under). But what's more? Any solutions are welcome no matter how less elementary they may be.
Firstly, it is clear that $f$ has not constant coefficient. I claim that for any $f\in \mathfrak{p}$, $f$ has not terms $X^1, X^2, Y^1, Z^1$. Since, for example, $$f(X,Y,Z)=Y^1+\textrm{(other terms)}=Y^1+Y^2(\ldots)+X(\ldots)+Z(\ldots)$$ then $$0=f(T^3,T^4,T^5)=T^4+\underbrace{T^8(\ldots)+T^3(\ldots)+T^4(\ldots)}_{\textrm{has not term $T^4$}}$$ since $3a+4b+5c\neq 1$ for $a,b,c\in \mathbb{Z}_{\geq 0}$. But, $X^3-YZ, Y^2-XZ, Z^2-X^2Y\in \mathfrak{p}$ have terms $X^3, Y^2,Z^2$ respectively. If $\left<f,g\right>=\mathfrak{p}$, assume $a,b,p,q,n,m\in k[X,Y,Z]$ such that $$af+bg=X^3-YZ\qquad pf+qg=Y^2-XZ\qquad nf+mg=Z^2-X^2Y$$ Assume that $$\left(\begin{matrix}f\\ g\end{matrix}\right) =\left(\begin{matrix}x_f& y_f & z_f\\\ x_g& y_g & z_g\end{matrix}\right) \left(\begin{matrix}X^3\\ Y^2\\ Z^2\end{matrix}\right)+ \left(\begin{matrix}\textrm{other terms}\\ \textrm{other terms}\end{matrix}\right)$$ and that $a_0=a(0,0,0)$ and so on. Then by looking at the term $X^3,Y^2,Z^2$, one has $$\left(\begin{matrix}a_0 & b_0\\ p_0& q_0\\ n_0&m_0\end{matrix}\right) \left(\begin{matrix}x_f& y_f & z_f\\\ x_g& y_g & z_g\end{matrix}\right) =\left(\begin{matrix}1 & 0 &0 \\ 0& 1 & 0\\ 0&0 &1\end{matrix}\right)$$ which is a contradiction by standard linear algebra. $\square$
We have an exact sequence $$ I/I^2 \to \Omega_{\mathbb{A}^3}\otimes \mathcal{O}_Y \to \Omega_Y \to 0, $$ where the first arrow takes any function $f \in I$ to $df\vert_Y$. In particular, we have $$ xz - y^2 \mapsto zdx - 2ydy + xdz = t^5dx - 2t^4dy + t^3dz, $$ $$ x^3 - yz \mapsto 3x^2dx - zdy - ydz = 3t^6dx -t^5dy - t^4dz, $$ $$ x^2y - z^2 \mapsto 2xydx + x^2dy -2zdz = 2t^7dx + t^6dy - 2t^5dz. $$ Under the identification $\Omega_{\mathbb{A}^3}\otimes \mathcal{O}_Y \cong \mathcal{O}_Y^{\oplus 3}$ (with the basis $dx$, $dy$, $dz$) the images of all these functions sit in $\mathfrak{m}^{\oplus 3}$, and their images in $$ (\mathfrak{m}/\mathfrak{m}^2)^{\oplus 3} = \Big(k[t^3,t^4,t^5]/(t^6)\Big)^{\oplus 3} $$ are linearly independent, hence $\dim(I/\mathfrak{m}I) \ge 3$. This proves that $I$ cannot be generated by 2 elements.
