Talk:Degree of an algebraic variety

What about giving references? —Preceding unsigned comment added by 132.77.4.43 (talk) 14:54, 4 February 2008 (UTC)[reply]

We needn't assume V irreducible - this should be mentioned. Charles Matthews (talk) 08:48, 15 February 2008 (UTC)[reply]

Merger

I have proposed a merger with projective variety because the article only considers the degree of a projective variety. If there is a general notion of the degree of an arbitrary variety, then the best place for it is algebraic variety. -- Taku (talk) 06:12, 11 December 2016 (UTC)[reply]

There is no general notion of the degree of an algebraic variety, as the degree is related to the embedding of the variety in a projective or an affine space. On the other hand, the degree is defined for every affine or projective algebraic set. It is immediate that, if one extends the definition given in the article to affine varieties, the degree of an affine variety equals the degree of its projective completion. It is rather important to define the degree for algebraic sets, as the standard algorithm for computing the degree (via the Hilbert series, deduced from a Gröbner basis) does not allows testing if the algebraic set is irreducible.

I oppose to the merge: If a merger would be done, the best target article is unclear, between algebraic variety, projective variety, dimension of an algebraic variety, ... This article is a stub. If reasonably expanded, merging it in, says, projective variety would break the balance between the various properties that must be considered in this article.

I have asserted that this article is a stub. This is clear as most of the lead is not expanded in the article. Moreover, the main property of the degree is lacking: If V is a projective algebraic set, and H is a hypersurface that does not contain any component of V, then the degree of V ∩ H (counted with multiplicities, if the intersection has multiple components) is the product of the degrees. See Hilbert series and Hilbert polynomial for a proof. This is a generalization of Bézout's theorem, which gives a proof of Bézout's theorem, which is probably the simplest known proof.

Thus, it is worth to expand the article before discussing any merger.

I think we don't really disagree; my merger proposal was because of the current state. For example, it lacks a reference. But, ok, if we're aiming for the expansion, there is no need for the merger. I'm withdrawing the proposal. -- Taku (talk) 19:28, 11 December 2016 (UTC)[reply]