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Suppose that $X$ is a reducible projective variety with equidimensional irreducible components $X_i$. Then I am trying to show that $\deg X = \sum_i \deg X_i$.

This has been asked before. See here and here.

Both answers here seem to make use of the fact that, for projective varieties $X, Y$ with homogeneous ideals $I, J$, we have $P_{R/(I + J)} = P_{X\cap Y}$, where $P$ is the Hilbert polynomial. I cannot see why this is true.

We have that $P_{X\cap Y}(m) = \dim_k(k[x_0,\ldots, x_n]_m/\mathcal I(X \cap Y)_m)$, where the subscript denotes the $m^{th}$ sumand of the standard grading. On the other hand, $P_{R/(I+J)} = \dim_k(k[x_0,\ldots, x_n]_m/\mathcal (I+J)_m)$. Now, the problem is that $$ \mathcal{I}(X\cap Y) = \mathcal I(\mathcal V(I)\cap\mathcal{V}(J)) = \mathcal{I}(\mathcal{V}(I + J)) = \text{rad}(I + J), $$ by the projective Nullstellensatz, where $\text{rad}$ denotes the radical of $I + J$. This is a problem, because in general the radical of $I + J$ does not equal $I + J$, so I don't see why the two vector spaces should have the same dimension.

Arctic Char
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Sebastian Monnet
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1 Answers1

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There's two things going on here:

  1. If you take the scheme-theoretic intersection, $I(X\cap Y)= I(X)+I(Y)$ by definition and everything's fine. This is a common perspective to take as you get more experienced in algebraic geometry.
  2. Even if you don't take the scheme-theoretic intersection, the Hilbert polynomials of $R/\sqrt{I+J}$ and $R/(I+J)$ have the same degree, since the degree of the Hilbert polynomial only depends on the dimension and that is unaffected by whether you pick $I+J$ or $\sqrt{I+J}$ as your ideal. Since all we're using is the additivity of the Hilbert polynomial across short exact sequences, we're still fine.
KReiser
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  • My lecture notes only define the Hilbert polynomial in terms of homogeneous ideals generated by varieties, in which case they state that the degree of the polynomial is equal to the dimension of the variety. Should I infer from your answer that for a general ideal $I$, the degree of the Hilbert polynomial $P_{R/I}$ is equal to the Krull dimension of the ring $R/I$? This would make sense as a generalisation of what I know, but it's not a fact I am aware of. – Sebastian Monnet Sep 18 '20 at 22:18
  • @Qwertiops Yes, the Hilbert polynomial can be defined for any finitely-generated graded $k$-algebra, and the same proof that $\deg P_R = \dim R$ can be used in both cases (homogeneous Noether normaliztion). – KReiser Sep 18 '20 at 22:22
  • Fantastic, thank you very much. Regarding your first point, I haven’t learned about schemes yet, but I look forward to the new perspective! – Sebastian Monnet Sep 18 '20 at 22:36