Let $k$ be an algebraically closed field. Show that $\{(x,y)\in k^2\mid xy=0\}$ is closed and connected but not irreducible in $k^2$.

algebraic-geometrygeneral-topologyzariski-topology

This is Exercise 1.2.8(3) of Springer's, "Linear Algebraic Groups (Second Edition)". According to this search on Approach0, it is new to MSE. The exercise comprises of three parts, whose relationship with each other is, I believe, strong enough to include into one post here.

The Details:

Since definitions vary:

A topological space $(X,\tau)$ is a set $\tau$ of subsets of $X$, called closed subsets, such that

  • $\varnothing, X\in\tau$,
  • The intersection $$\bigcap_{i\in I}X_i$$ of any closed subsets $(X_i)_{i\in I}$ is closed, where $I$ is arbitrary, and
  • The union of finitely many closed sets is closed.

Note that $\tau$ is omitted sometimes when the context is clear.

Let $k$ be an algebraically closed field.

We denote by $\mathcal{V}(I)$ the set of all zeros of $I$.

The topology on $V:=k^n$ whose closed subsets are $\mathcal{V}(I)$ for ideals $I$ of $S:=k[T_1,\dots, T_n]$ is known as the Zariski topology.

On page 2 of Springer's book, 1.2.1 reads,

A topological space $X\neq\varnothing $ is reducible if it is the union of two proper closed subsets. Otherwise $X$ is irreducible. A subset $A\subset X$ is irreducible if it is irreducible for the induced topology. Notice that $X$ is irreducible if and only if any two nonempty open subsets of $X$ have a nonempty intersection.

On page 4, 1.2.7 reads,

[A] topological space is connected if it is not the union of two disjoint proper closed subsets.

The Question:

Let $$X=\{(x,y)\in k^2\mid xy=0\}.$$ Show that $X$ is a closed subset of $k^2$ which is connected but not irreducible, all with respect to the Zariski topology.

Thoughts:

Since $xy$ is a polynomial $f$ in $x,y$ with coefficients in $k$, we have $X=\mathcal{V}((f))$. Hence $X$ is closed under the Zariski topology.

For connectedness, I can start as follows. If $k=\Bbb R$, then $X$ is the union of the $x$-axis with the $y$-axis; this is clearly connected under the usual topology on $\Bbb R^2$. I'm not sure where to go from here.${}^\dagger$

A similar consideration strikes me for reducibility: We can write

$$X=\mathcal V((g))\cup\mathcal V((h))$$

for $g=x$ and $h=y$.

Please help 🙂

$\dagger$: I am aware that $\Bbb R$ is not algebraically closed.

Best Answer

First of all, $X$ is the union of the closed sets $A=\{(x,y)\in k^2: x=0\}$ and $B=\{(x,y)\in k^2: y=0\}$, and so it is reducible.

Since $A$ and $B$ have a common point, it is sufficient to show they are both connected, and this will imply that $X$ is connected as well. So let's show $A=\{(0,y): y\in k\}$ is connected. Clearly it is sufficient to prove $A$ is irreducible, which is equivalent to the ideal $I(A)\subseteq k[x,y]$ being a prime ideal. And indeed, it can be shown that $I(A)=(x)$, the ideal generated by $x$. The right hand side is clearly contained in the left hand side. Conversely, suppose that $f(x,y)\in I(A)$, i.e $f(0,y)=0$ for all $y\in k$. Write:

$f(x,y)=\sum\limits_{i=0}^n f_i(x)y^i$

So the polynomial $f(0,y)=\sum\limits_{i=0}^n f_i(0)y^i\in k[y]$ vanishes at all points of $k$. Since $k$ is an infinite field (it is algebraically closed) it follows that $f(0,y)$ must be the zero polynomial, i.e $f_i(0)=0$ for all $i$. But this exactly means that all the polynomials $f_i(x)$ are divisible by $x$, and so $f$ is divisible by $x$ as well. So indeed $f\in (x)$.

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