Degree of union is the sum of degrees

Question

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.

Answer 1

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.