Chapter 1, Exercise *1.11: A variety that is not a local complete intersection

Let $Y\subseteq\mathbb A^3$ 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. We say $Y$ is not a local complete intersection-- cf. (Ex. 2.17).

This is a starred exercise, however we will provide a solution here since it is quite fun and yields some enlightening computations in commutative algebra. The difficulty in this problem is that it is not immediately obvious as to how we should parameterize this curve - indeed, the obvious method would no longer be based on a polynomial system and would use radicals which are disallowed in algebraic geometry. Thus we must turn to a more subtle way of dealing with this problem. Note that we have $\operatorname{height}\mathfrak p = \operatorname{codim}Y$ where $Y$ is the associated variety, and thus we need to show that $Y$ has dimension one.

To show $\dim Y=1$ (and that $I(Y)$ is prime in the process), we will construct a topological homeomorphism $\mathbb A^1\to Y$. Dimension and irreducibility are both preserved under homeomorphism and our desired conclusion would follow. The homeomorphism, as with most things in algebraic geometry, is canonical: by definition $Y$ is the image of $\varphi:\mathbb A^1\to\mathbb A^3$ taking $t\mapsto(t^3,t^4,t^5)$. To see that $\varphi$ is injective, suppose $\varphi(s)=\varphi(t)$ which simply means $$s^3=t^3\qquad s^4=t^4\qquad s^5=t^5.$$ If either variable is zero we are done (since we are working in an integral domain). If they are both nonzero, divide equation two by equation one to find $s=t$ again. Hence $\varphi$ is a bijection and it remains to show it is continuous. This means that if $W$ is any algebraic set in $\mathbb A^3$, then $\varphi^{-1}(W)$ is algebraic in $\mathbb A^1$. In other words, it is either finite or all of $\mathbb A^1$. Suppose that $\varphi^{-1}(W)$ is infinite, that is, there are infinitely many $t\in\mathbb A^1$ such that $(t^3,t^4,t^5)\in W$. If $W=Z(f_1,\dots,f_r)$ then each $f_i$ must vanish on infinitely many points, thus $f_i\equiv 0$ and $W=\mathbb A^3$. Consequently, $\varphi^{-1}(W)=\mathbb A^1$ which is indeed closed. Hence $\varphi$ is a homeomorphism and the first half of the exercise follows.

To speak about generators of $I(Y)$, we should give a concrete description of this ideal. Suppose $f\in A=k[x,y,z]$ such that $f(t^3,t^4,t^5)=0$ for all $t\in k$. In general we can write $$f(x,y,z)=\sum_{a_1,a_2,a_3}c_{a_1a_2a_3}x^{a_1}y^{a_2}z^{a_3}.$$ Plugging in $x=t^3,y=t^4,z=t^5$, we find $$f(t^3,t^4,t^5)=\sum_{a_1,a_2,a_3}c_{a_1,a_2,a_3}t^{3a_1+4a_2+5a_3}\stackrel{?}{=}0\quad\forall t\in k.$$ For the last condition to be true we must have that for any given value of $3a_1+4a_2+5a_3$ all valid $c_{a_1a_2a_3}$ sum to zero, in other words $$\tag{*}\sum_{3a_1+4a_2+5a_3=s}c_{a_1a_2a_3}=0$$ for any $s$. These polynomials make up $I(Y)$.

The next step is an extraordinarily tedious calculation, in which we work out the meaning of condition $(*)$ for $s\in\{0,1,2,\dots,10\}$. These eleven calculations are all extremely trivial: we will work out $s=5$ and $s=10$. For $s=5$, the only valid set of $a_i$ is $(0,0,1)$, and the sum collapses to a single term giving $c_{001}=0$. In other words, $f$ must have no term proportional to $z$. For $s=10$, the two valid combinations are $(2,1,0)$ and $(0,0,2)$ and thus $c_{210}+c_{002}=0$. In other words, the $x^2y$ and $z^2$ terms must have equal and opposite coefficients. In the end, we rule out the following terms: $$\text{constant},x,y,z,x^2,xy.$$ The following pairs must have equal and opposite coefficients: $$xz,y^2\qquad x^2,yz\qquad x^2y,z^2.$$ Thus, the polynomials $y^2-xz,x^2-yz,x^2y-z^2$ lie in $I(Y)$, and further generate it. They are minimal - if we were to remove one polynomial then the associated condition would no longer be assured. Thus $I(Y)$ cannot have fewer than three generators.