Proposition 5.5 in Undergraduate Algebraic Geometry by Reid says (I only write down a brief idea since the proposition is long and involves some other notations to define):

The affine variety $U$ is irreducible if and only if its closure $\bar{U}$ in a projective space is irreducible.

I can prove it using the correspondence between $U$ and $I(U)$, and between $\bar{U}$ and $I^h(\bar{U})$ where $I^h$ is the homogeneous ideal defined by the projective variety $\bar{U}$. I believe it was correct.

I read this question: Irreducibility of an Affine Variety and its Projective Closure and found a much shorter proof provided by the OP. But I am confused about it.

He assumed $U=Z_1\cup Z_2$ where $Z_1, Z_2$ are closed sets. So $Z_1\cup Z_2 \supset \bar{U}$.

Maybe because I have little background in topology, I am very confused here. The irreducibility of an affine variety $U$ is defined by its not being a union of two nonempty varieties. If $Z_1,Z_2$ are defined this way, then in the sense of Zariski topology in projective space, $Z_1, Z_2$ are open sets.

From another direction, if we define irreducibility of an affine variety in terms of union of closed sets, is it in the sense of the Zariski topology in the affine space or projective space? If it is in affine space, then $Z_1, Z_2$ are not closed in the projective space.

Isn't there an ambiguity in this definition and the above proof?

Thank you for your help.

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    $\begingroup$ Two notes: (1) affine space has the subspace topology inside of projective space (2) irreducibility is an "absolute" notion in the sense that compactness is. $\endgroup$ – Hoot Feb 1 '16 at 15:10
  • $\begingroup$ @Hoot: I think I get it. Thank you for the comment. $\endgroup$ – KittyL Feb 1 '16 at 16:21

What definition of irreducible are you using? The one I have in mind, which is also the definition given in Hartshorne, is:

Definition: A nonempty subset of a topological space $Y \subseteq X$ is irreducible if it cannot be expressed as a union $Y = Y_1 \cup Y_2$ of two proper subsets each of which is closed in $Y$.

Thus, we say that a reducible space is a not irreducible space, hence why we assume $Y = Y_1 \cup Y_2$, where $Y_i$ are proper spaces closed in the subspace topology of $Y$.

The irreducibility of an affine variety is defined by its not being a union of two nonempty varieties.

I'm a bit confused by this. Irreducibility is a notion that comes from general topology. Therefore, when constructing algebraic geometry, we should be careful and ensure that our terminologies coincide with notions from topology. The above statement could be fine as a theorem, but not as a definition.

Also, it should be noted that authors frequently distinguish between algebraic sets and algebraic varieties. For me, algebraic variety means an irreducible algebraic set.

  • $\begingroup$ Thank you for your reply. My definition comes from Ideals, Varieties and Algorithms by Cox. I guess the definition is not rigorous then. But even with your definition, when defining the irreducibility of an affine algebraic set, do you consider it in the Zariski topology in the affine space or in the projective space? I am asking this because a closed set in the affine topology is an open set in the projective topology. $\endgroup$ – KittyL Feb 1 '16 at 14:57
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    $\begingroup$ It's all defined in a relative topology, namely the subspace topology. A closed set in the affine space can be identified via the homeomorphism of hyperplanes with as a closed subspace of an open subspace in projective space, but it doesn't matter because we are still looking at the closed subspace. $\endgroup$ – Future Feb 1 '16 at 17:48
  • $\begingroup$ Thank you again. I got it after I read Hoot's comment and checked up the concept subspace topology in wiki. $\endgroup$ – KittyL Feb 1 '16 at 18:02

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