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?


1 Answer 1


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.

  • $\begingroup$ Thanks, this is great! I had never shown irreducibility in this way before, so I'm very happy to be aware of this method now! $\endgroup$
    – Sha Vuklia
    Dec 13, 2021 at 15:18

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