# Irreducible algebraic variety of dimension $d$ cannot necessarily be given by $n-d$ equations

Let $$k$$ be an algebraically closed field. For an algebraic set $$Y\subset k^n$$ it is true that $$Y$$ is irreducible and of dimension $$n-1$$ iff $$Y=Z(f)$$ for some irreducible $$f\in k[x_1,\dots,x_n]$$. My notes give a counter example for the wrong statement that for every closed irreducible $$Y\subset k^n$$ of dimension $$d$$ there are $$f_1,\dots,f_{n-d}\in k[x_1,\dots,x_n]$$ s.t. $$Y=Z((f_1,\dots,f_{n-d}))$$. They claim that the following algebraic set is a counter example: $$Y=\{(s^3,s^2t,st^2,t^3):s,t\in k\}\subset k^4.$$ For this we have to verify that $$Y$$

1. is closed

2. is irreducible

3. is of dimension $$2$$

4. cannot be defined using only two equations.

I verified that $$Y=Z(x_0x_3-x_1x_2,x_1^2-x_0x_2,x_2^2-x_1x_3)$$, so $$Y$$ is indeed closed. Since $$Y\neq k^4$$ and $$Y$$ contains $$(1,t,t^2,t^3)$$, we know that if $$Y$$ is irreducible, then it is of dimension $$1$$, $$2$$, or $$3$$. Assuming $$Y$$ can't be given by only two equations (and hence in particular not by one), the result stated above tells us that $$Y$$ is not of dimension $$3$$. What's left to argue in that case is that $$Y$$ is not of dimension $$1$$. If I can prove that $$\{(1,t,t^2,t^3):t\in K\}=Z(x_0-1,x_2-x_1^2,x_3-x_1^3)$$ is irreducible, then we have the chain $$Y\supset \{(1,t,t^2,t^3):t\in K\}\supset \{(1,0,0,0)\},$$ so $$Y$$ must then be of dimension $$2$$. So I can't prove the following:

a. $$Y$$ is irreducible

b. $$Y$$ can't be given by two equations

c. $$\{(1,t,t^2,t^3):t\in K\}$$ is irreducible.

Could someone help me out with any of those?

For (a) and (c), you can try to show the sets $$Y$$ and $$Z=\{(1, t, t^2, t^3) \mid t \in K\}$$ as images of $$\mathbb{A}^1$$ and $$\mathbb{A}^2$$. For instance, note that the morphism $$\phi\colon \mathbb{A}^2 \to \mathbb{A}^4, \qquad (s, t) \mapsto (s^3, s^2t, st^2, t^3)$$ surejcts onto $$Y$$. This is induced by the ring homomorphism $$\alpha\colon k[x, y, z, w] \to k[s, t]$$, $$x \mapsto s^3$$, $$y \mapsto s^2t$$, $$z \mapsto st^2$$, $$w \mapsto t^3$$.
Now, if $$I=I(Y)$$ is the ideal of functions that vanish on $$Y$$, then $$I = \ker \alpha$$ (why?). This shows that $$k[x, y, z, w]/I$$ is a subring of $$k[s, t]$$, and in particular it is an integral domain. Therefore, $$I$$ is a prime ideal, i.e. $$Y$$ is irreducible. A similar argument applies for (c).
For (b), assume that $$I(Y)=(f)$$ for some equation $$f \in k[x, y, z, w]$$. Note that $$y^2-xz$$ and $$z^2-yw$$ are both in $$I(Y)$$, and so $$f$$ must divide them. The problem is that these are "different" irreducible elements (why?), and so $$f$$ must be constant, a contradiction.