# Why degree of a reducible projective variety is the sum of the degree of its irreducible components

Could anyone show me how to prove that

The degree of a reducible projective variety is the sum of the degree of its irreducible components?

The definition of the degree I know is quite vague, saying that the degree of a projective variety $X$ is the maximal number of possible finite intersections of $X$ with a linear hyperspace in general position.

I'm not comfortable with the concept in general position. The note I'm reading explains it by an example. I guess that in general position just means there isn't multiple intersection.

• You may find this interesting: math.stackexchange.com/questions/461361/… – Brenin Nov 24 '13 at 22:57
• With the correct (and precise) definition of degree (for example using the Hilbert polynomial) it's quite easy. – Martin Brandenburg Nov 24 '13 at 23:12
• @MartinBrandenburg, could you provide some more details? I know the degree is determined by the leading coefficient of the Hilbert polynomial, but also involves the dimension. I cannot figure out the relation between the polynomial of $X$ and its irreducible components. – hxhxhx88 Nov 25 '13 at 2:14
• @Brenin, thanks. It is helpful, but I am still not sure how to prove the question in my post.. – hxhxhx88 Nov 25 '13 at 8:20

As mentioned in the comments, the best way to prove statements like this is to use the Hilbert polynomial definition of degree. For this point of view, see section I.7 of Algebraic Geometry by Hartshorne.

Let $R = k[x_0,\ldots, x_n]$ where $k$ is an algebraically closed field viewed as a graded ring. For any graded $R$-module $M$, the Hilbert function is the function

$$h(l) = \dim_k M_l$$

giving the dimension as a $k$-vector space of the graded pieces of $M$. The idea is that for large enough $l$ this agrees with a polynomial $P_M(l)$ which is the Hilbert polynomial of $M$. Then for a projective variety $X \hookrightarrow \mathbb{P}^n$, the Hilbert polynomial $P_X(l)$ is just the Hilbert polynomial of the homogeneous coordinate ring of $X$ as a graded module over $R$. You can show that the degree of $P_X(l)$ is $d = \dim X$ and then we define the degree of $X$ to be $d!$ times the leading coefficient of $P_X(l)$

The nice thing about the Hilbert polynomial is that it behaves well with exact sequences and this gives it the geometric properties we want and expect. This is because $\dim_k$ is additive on exact sequences.

In particular, if we have that $X = Y_1 \cup Y_2$ with $Y_1$ and $Y_2$ the same dimension and intersecting in a lower dimension, then we can write the exact sequence

$$0 \to R/I \to R/I_1 \oplus R/I_2 \to R/(I_1 + I_2) \to 0$$

where $I_i$ is the homogeneous ideal of $Y_i$ and $I$ is the homogeneous ideal of $X$. Then by additivity of the Hilbert polynomial,

$$P_{R/I_1 \oplus R/I_2} = P_{R/I} + P_{R/(I_1 + I_2)}.$$

Applying additivity of the Hilbert polynomial again, we see that the left hand side of this equation is in fact $P_{R/I_1} + P_{R/I_2}$. Rephrasing this geometrically, we see that

$$P_{Y_1} + P_{Y_2} = P_{X} + P_{Y_1 \cap Y_2}.$$

Since $Y_i$ were assumed to be the same dimension, the leading coefficient of the left hand side is the sum $\deg{Y_1}/d! + \deg{Y_2}/d!$. Similarly, on the right hand side, since we assumed $Y_1 \cap Y_2$ is lower dimensional than all of $X$, we have that the leading coefficient of the right hand side is just that of $P_{X}$, that is, $\deg{X}/d!$, giving us the equality $\deg{X} = \deg{Y_1} + \deg{Y_2}$.

Now you can deduce the more general case with a little more work by applying this to the irreducible components.

The proof that this gives the same definition of degree as the one you gave is a little involved but it uses exactly the same technique. Write down an exact sequence whose terms correspond to the varieties we are intersecting and compare the two sides of the equation we get for the Hilbert polynomials. However, it requires some commutative algebra.

Edit: I wanted to add a bit about the dimension considerations since you brought that up as something you had issue with. If you notice, my argument above implies something a little different from your statement. It says that the degree of a variety is the sum of degrees of the highest dimensional irreducible components. This is because the contribution of the lower dimensional components to the Hilbert polynomial will not affect the leading coefficient which is the same degree as the dimension.

How does this reconcile with the classical notion of degree? The idea is that the lower dimensional components won't affect the intersection with your general plane. The reason for this is that if we have a $k$ dimensional subvariety $Y$ of $\mathbb{P}^n$, the classical degree is the number of points in the intersection with an $n - k$ plane. "Most" $n-k$ planes will certainly miss any components of dimension less than $k$ (think for example a point and a line in $\mathbb{P}^3$) and thus shouldn't contribute to the classical notion of degree, and indeed with the Hilbert polynomial argument, we see that they don't. Hopefully this fixes part of your confusion about how the dimension affects things.

• Is there a way of doing this without using the Hilbert polynomial definition, but instead the number of intersections with a general linear hyperspace? – Tim Aug 24 '15 at 18:21
• @Tim You can make the argument that the condition that a linear subspace intersects $Y_1 \cap Y_2$ is a closed condition and so a general linear subspace $H$ of dimension equal to the codimension of $Y_1$ and $Y_2$ will miss the intersection. Therefore it will intersect $Y_i$ in $\deg Y_i$ points and none of these will overlap so it will intersect $Y_1 \cup Y_2$ in $\deg Y_1 + \deg Y_2$ points. – Dori Bejleri Aug 24 '15 at 19:48