Closed subset of an affine variety... is it affine?

Question

Preliminaries

So, first of all let me give you the definitions I'm dealing with. Let $k$ be an algebraically closed field, and $\mathbb{A}^n = k^n$.

An affine variety is a closed and irreducible subset of $\mathbb{A}^n$.

Here we endow $\mathbb{A}^n$ with the Zariski topology, so a closed subset is the zero set of an ideal in the polynomial ring $k[x_1,\dots,x_n]$. The irreducibility condition in the definition of affine variety requires that such an ideal is prime.

A k-space is a topological space $X$ together with a sheaf of $k$-algebras $\mathscr{O}_X$. We require that $\mathscr{O}_X$ is a subsheaf of the sheaf of $k$-valued functions on $X$.

An algebraic variety is an irreducible k-space $X$ such that $\exists$ an open cover $X=\bigcup_{i=1}^n U_i$ where each $U_i$ is an affine variety. Moreover we ask the diagonal $\Delta(X)$ to be closed in $X\times X$.

Now we use the term affine variety also for an algebraic variety which is isomorphic as a $k$-space to an affine variety as defined above.

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The question

Given a closed subset of an affine variety, is this subset affine?

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Remarks

My professor claimed this to be true, and I noticed some people take this fact as granted on this website as well, so I guess this should be true! Nevertheless I don't see why such a closed subset should be irreducible. But the point is: do we really care? My naive understanding of this affine/not affine business is that for an affine variety we can write down something like a global coordinate system (the coordinate ring), while for a general algebraic variety we have just local coordinates, ie we have many affines patched together in a way that cannot be described globally with the usual tools. Hence if a variety is reducible this is not a real obstruction to being affine: we can still find a global coordinate system. The important thing is that all the components lie in the same $\mathbb{A}^n$

As you can see I'm a bit confused... some clarification about these ideas would be really appreciated!

Answer 1

This is just a matter of terminology. In both books I have to hand (Hartshorne, and Eisenbud's "Commutative algebra..."), the authors define an 'affine algebraic set' to be any subset of $\mathbb{A}^n$ given by the vanishing of polynomials, and an 'affine algebraic variety' to be an irreducible such set.

What is perfectly clear (and is possibly what 'The question' really asks, given the absence of the word 'variety' at the end) is that any closed subset of an affine algebraic set is again an affine algebraic set.

In practice though, the word 'variety' often seems to be used more generally, without the irreducibility requirement.

Answer 2

It is not true that any closed subset of an affine variety will be an affine variety. In fact, you may have seen that any affine variety $V\subset \mathbb{A}^n$ has a unique reduction into irreducible components; that is, we can write $V = \cup_{i=1}^n V_i$ such that each $V_i$ is irreducible and $V_i \subsetneq V_j $ for $i\neq j$ and this is unique up to reordering of the $V_i.$ If $V$ has 3 irreducible components, then the union of the first two is a closed subset that is not irreducible (it is already reduced into two pieces!).

Answer 3

The union of two points $\{1\} \cup \{2\}$ is a closed subset of $\Bbb{A}^1$ in the Zariski topology but it is obviously not irreducible.

To answer your querrey about the link in your question above, notice that Georges only claims that

$$X_1 \times \{a_1\} \subseteq X_1 \times X_2$$

is closed. However the fact that $X_1 \times\{ a_1\}$ is irreducible comes from it not being a closed subset, but rather because it is homeomorphic to $X_1$ that is irreducible. Since irreducibility is preserved under homeomorphism thus $X_1$ irreducible implies that the slice $X_1 \times \{a_1\}$ is irreducible and thus affine.