$\DeclareMathOperator{\codim}{codim} \newcommand{\dprime}{^{\prime\prime}}\newcommand{\kk}{\mathbb{K}} \newcommand{\local}[2]{\mathcal{O}_{#2, #1}} \newcommand{\pp}{\mathbb{P}} \newcommand{\qq}{\mathbb{Q}} \newcommand{\rr}{\mathbb{R}} \DeclareMathOperator{\res}{Res} \newcommand{\scrL}{\mathcal{L}} \DeclareMathOperator{\sing}{Sing} \newcommand{\zz}{\mathbb{Z}}$
The degree of a subvariety $X$ of a projective space $\pp^n$ defined over an algebraically closed field $\kk$ is the number of points of intersection of $X$ and a “generic” linear subspace of $\pp^n$ of “complementary dimension”, i.e. of dimension equal to $n – \dim(X)$. The goal of this post is to show that this definition is well defined – this is the statement of Theorem 3 below. We also show (in Theorem 4 below) that the degree also equals the number of points of such a generic intersection if every point is counted with an appropriate “multiplicity” (in particular, all points of intersection of $X$ and a generic complementary dimensional linear space has intersection multiplicity one). First we clarify what “generic” means in this context.

Generic Linear Subspaces. Let $\scrL_m$ be the set of $m+1$ points in “general position” in $\pp^n$, i.e. the set of all $(P_0, \ldots, P_m) \in ((\pp^n)^*)^{m+1}$ such that every $(m+1) \times (m+1)$ minor of the $(n+1) \times (m+1)$ matrix whose columns are (the homogeneous coordinates of) $P_0, \ldots, P_m$ is nonzero. Then $\scrL_m$ is a nonempty Zariski open subset of $((\pp^n)^*)^{m+1}$ and every $(P_0, \ldots, P_m) \in \scrL_m$ determines a unique $m$-dimensional linear subspace of $\pp^n$, which is the span of the $P_i$. We say that a property holds for a “generic linear subspace of dimension $m$” if it holds for the subspaces corresponding to all elements of a nonempty Zariski open subset of $\scrL_m$.

It is clear that if $X$ is a linear subspace of $\pp^n$, then the degree of $X$ is well defined and equals $1$. The first non-trivial case is the following:

Proposition 1 (Degree of a hypersurface). Let $X = V(f)$ be a hypersurface in $\pp^n$ defined by an irreducible homogeneous polynomial $f$ of degree $d$. Then a generic line intersects $X$ at $d$ points. In particular, $\deg(X)$ is well defined and equals $\deg(f)$.

Proof. Consider a line $L := \{t_0P_0 + t_1P_1: (t_0,t_1) \in \kk^2\setminus\{(0,0)\}\}$. For generic $P_0, P_1$, $f|_L$ is a degree $d$ homogeneous polynomial, and if we count the points in $L \cap V(f)$ with the corresponding multiplicity as a root of $f|_L$, then the number of points in $L \cap V(f)$ is $d$ (since $\kk$ is algebraically closed). However, we are going to show that $f|_L$ has no repeated root for generic choices of $P_0, P_1$. Indeed, choose homogeneous coordinates $[x’_0: \cdots: x’_n]$ on $\pp^n$ such that $L$ is parallel to $x’_n$-axis. Then in affine coordinates $u_i := x’_i/x_0$, $i = 1, \ldots, n$, $L$ has a parametrization of the form
\[\{(a’_1, \ldots, a’_n) + t(0, \ldots, 0, 1): t \in \kk\}\]
Consider the dehomogenization $\tilde f := f/x_0^d$ of $f$. For a generic choice of $L$ one can ensure that

• $\deg(\tilde f|_L) = \deg(\tilde f) = \deg(f) = d$,
• $\partial \tilde f/\partial u_n \not\equiv 0$

Since $\tilde f$ is irreducible, $\dim(X \cap V(\partial \tilde f/\partial u_n)) < \dim(X) = n -1$. Consequently, the complement in $\kk^{n-1}$ of the image of the projection of $X \cap V(\partial \tilde f/\partial u_n)$ onto the first $n-1$ coordinates contains a nonempty Zariski open subset of $\kk^{n-1}$. Therefore, for a generic choice of $L$ one can also ensure that

• $(\partial \tilde f/\partial u_n)(a’_1, \ldots, a’_{n-1}, t) \neq 0$ for each $t \in \kk$ such that $f(a’_1, \ldots, a’_{n-1}, t) = 0$.

It follows that for generic $L$ there are precisely $d$ elements in $L \cap X$, as required to complete the proof of Proposition 1.

We reduce the general case to hypersurfaces using generic linear projections: Given a hyperplane $H$ in $\pp^n$, every point $P \in \pp^n \setminus H$ defines a projection $\pp^n \setminus \{P\} \to H$ which maps a point $x$ to the (unique) point of intersection of $H$ and the line joining $x$ and $P$. Consequently, we can identify the set of “projections onto a hyperplane of $\pp^n$” with the set of pairs $(P, H)$ such that $H$ is a hyperplane of $\pp^n$ and $P \in \pp^n \setminus H$. In general, every pair $(H’,H)$ of linear subspaces of $\pp^n$ such that

• $H’ \cap H = \emptyset$, and
• $\dim(H’) + \dim(H) = n-1$

defines a projection map $\pi_{H’,H}: \pp^n \setminus H’ \to H$ which maps $x \in \pp^n \setminus H’$ to the unique point where $H$ intersects with the complementary dimensional linear subspace of $\pp^n$ spanned by $x$ and $H’$. Note that we can choose coordinates $[x_0: \cdots: x_n]$ on $\pp^n$ such that $H’ = \{x_0 = \cdots = x_k=0\}$ and $H$ is the coordinate subspace spanned by $x_0, \ldots, x_k$; in that case
\[\pi_{H’,H}: [x_0: \cdots :x_n] \in \pp^n \setminus H’ \mapsto [x_0: \cdots : x_k]\]
We say that a property holds for a “generic linear projection onto a $k$-dimensional subspace” if it holds for the projection map corresponding to pairs of generic $n-k-1$ and $k$ dimensional subspaces in the sense defined above.

Linear projections can be used to make the standard result that every variety is birational to a hypersurface a bit more precise:

Lemma 2. Let $X$ be an irreducible subvariety of $\pp^n$ of dimension $m \leq n – 2$. Then a generic $m$ dimensional hyperplane $H_0$ satisfies the following properties:

1. If $u_0, \ldots, u_m$ are linearly independent linear forms over $H_0$, then $u_1/u_0, \ldots, u_m/u_0$ restrict to algebraically independent elements in $\kk(X)$;
2. if $H$ is a generic $m+1$ dimensional linear subspace of $\pp^n$ containing $H_0$ and $H’$ is a generic $n-m-2$ dimensional linear subspace $\pp^n$ such that $H’ \cap X = H’ \cap H = \emptyset$, then
   • the image of $X$ under the linear projection $\pi_{H’,H}: \pp^n \setminus V(H’) \to H$ is birational to $X$, and
   • the degree $d$ of the polynomial defining $\pi_{H’,H}(X)$ as a hypersurface of $H$ equals the degree of the field extension $[\kk(X): \kk(u_1/u_0, \ldots, u_m/u_0)]$. In particular, $d$ does not depend on $H$ or $H’$.

We sketch a proof of Lemma 2 at the end of this post (note that the arguments from the sketch show that for Lemma 2 it is sufficient to have $|\kk| = \infty$ in place of being algebraically closed). Now we use the lemma to show that degree is well defined.

Indeed, fix generic $H_0 \subseteq H$ as in Lemma 2. For generic $H’$ as in Lemma 2, let $U_{H’, H}$ be the Zariski open subset of $X$ such that

• $\pi = \pi_{H’, H}$ induces an isomorphism between $U_{H’, H}$ and $\pi(U_{H’, H})$, and
• $U_{H’,H} = \pi^{-1}(\pi(U_{H’,H})) \cap X$.

Observation. Since $\pi(U_{H’,H})$ is open in $\pi(X)$ and $\pi(X)$ is irreducible of dimension $m$, the complement $\pi(X) \setminus \pi(U_{H’,H})$ has dimension less than $m$, and consequently, if $L$ is a generic $n-m$ dimensional linear subspace of $\pp^n$ containing $H’$, then $\pi(L\setminus H’)$ is a line which does not intersect $\pi(X) \setminus \pi(U_{H’,H})$.

We now reformulate this observation in a more elaborate way: let $S$ be the subset of
\[\scrL := \scrL_{n-m} \times \scrL_{n-m-2} \times \scrL_{m+1}\]
consisting of all $(L, H’,H)$ such that $L \supseteq H’$. Note that $S$ is a subvariety of $\scrL$, since the condition $L \subseteq H’$ can be defined by vanishing of certain polynomials in coordinates of $\scrL$. Let $S’$ be the subset of $S \times X$ consisting of all $(L,H’,H,x)$ such that

• $H’ \cap X = H’ \cap H = \emptyset$,
• $x \in U_{H’,H}$
• $\pi_{H’,H}(L\setminus H’) \cap \pi_{H’,H}(X) \setminus \pi(U_{H’,H}) = \emptyset$

(Here is a bit of hand waving:) $S’$ can be defined as a subset of $S \times X$ by certain polynomial equalities and inequalities, and consequently, $S’$ is a constructible subset of $S \times X$.

Therefore, by Chevalley’s theorem on images of morphisms, the projection $S\dprime$ of $S’$ to $S$ is also constructible. Consider the projection
\[\psi: S \subseteq \scrL_{n-m} \times \scrL_{n-m-2} \times \scrL_{m+1}
\to \scrL_{n-m-2} \times \scrL_{m+1}\]
The “observation” above says that for all $(H’, H)$ in a nonempty Zariski open subset of $\scrL_{n-m-2} \times \scrL_{m+1}$, a nonempty Zariski open subset of $\psi^{-1}(H’,H)$ is included in $S\dprime$. Since $S\dprime$ is constructible, this can only be possible if $S\dprime$ contains a nonempty Zariski open subset of $S$. Consequently, for a generic $n-m$ dimensional linear subspace $L$ of $\pp^n$ one can pick an $n-m-2$ dimensional linear subspace $H’$ of $L$ which satisfies the second assertion of Lemma 2, and in addition, $L \cap (X \setminus U_{H’, H}) = \emptyset$. Then Proposition 1 and Lemma 2 together imply that the number of points in the intersection of $L \cap X$ does not depend on $L$. Consequently, we have proved the following result in the case that $X$ is irreducible:

Theorem 3. The degree $\deg(X)$ of a “pure dimensional” subvariety $X$ of $\pp^n$ is well defined (recall that a variety is “pure dimensional” if each of its irreducible components has the same dimension). In addition, $\deg(X) = \deg(\pi(X))$ for a generic linear projection $\pi$ onto a linear subspace of dimension $\dim(X) + 1$ in $\pp^n$.

Proof. As noted above, we already proved the theorem in the case that $X$ is irreducible. In general, if $\dim(X) = m$, we can take an $n-m$ dimensional linear subspace of $\pp^n$ which is generic for each irreducible component of $X$ (since “generic” properties hold for nonempty Zariski open subsets of $\scrL_{n-m}$ and since $\scrL_{n-m}$ itself is irreducible, the intersection of finitely many generic conditions is also generic). This completes the proof of the theorem.

Recall that the intersection multiplicity of $n$ hypersurfaces $H_i := \{f_i = 0\}$, $i = 1, \ldots, n$, on an $n$ dimensional variety $X$ (where $f_i$ are regular functions on $X$) at a nonsingular point $x \in X$ is the dimension (as a vector space over $\kk$) of
\[\local{X}{x}/\langle f_1, \ldots, f_n \rangle\]

Theorem 4. Let $X$ be a subvariety of $\pp^n$ of pure dimension $m$ and degree $d$. Then for generic linear homogeneous polynomials $l_1, \ldots, l_m$, the hypersurfaces of $X$ defined by $l_i = 0$ intersect at precisely $d$ points, each of which is a nonsingular point of $X$, and the intersection multiplicity of $\{l_1|_X = 0\}, \ldots, \{l_m|_X = 0\}$ at each point of intersection is one.

Proof. If $l_1, \ldots, l_m$ are generic, we already know that $L := \{l_1 = \cdots = l_m = 0\}$ intersects $X$ in precisely $d$ points (Theorem 3), and since the set $\sing(X)$ of singular points of $X$ has dimension $< m -1$, in the generic case $L$ does not intersect $\sing(X)$. It remains to show that at each point of intersection, the intersection multiplicity is one in the generic case. However, the arguments from the proof of Proposition 1 show that this is true when $X$ is an hypersurface, and then the general case follows from choosing $(L,H’,H)$ from the constructible subset $S\dprime$ defined above, since in that case the projection $X \to \pi_{H’,H}(X)$ is an isomorphism near every point of $L \cap X$. This completes the proof of the theorem.

Sketch of a Proof of Lemma 2. Since $\dim(X) = m$, if $u_0, \ldots, u_m$ are generic (homogeneous) linear forms in $(x_0, \ldots, x_n)$, then $u_i/u_0$, $i = 1, \ldots, m$, are algebraically independent over $\kk$. Consequently, a generic $m$-dimensional linear subspace $H_0$ of $\pp^n$ satisfies assertion 1 of Lemma 2. For the second assertion, choose homogeneous coordinates $[x_0: \cdots :x_n]$ on $\pp^n$ such that $X \not\subseteq V(x_0)$. Write $x’_i := x_i/x_0$, $i = 1, \ldots, n$. Arguments from standard proofs of the primitive element theorem and Schmidt’s “separable extension theorem” (see e.g. How Many Zeroes?, Theorem B.35 and Corollary B.37) show that

1. There are linear combinations $v_i := \sum_j \lambda_{i,j} x’_i$ with $\lambda_{i,j} \in \kk$ such that $v_1|_X, \ldots, v_m|_X$ are algebraically independent over $\kk$ and $\kk(X) = \kk(v_1, \ldots, v_m)(v_{m+1})$ (here one needs to follow the arguments of the proofs of Theorem B.35 and Corollary B.37 in How Many Zeroes? and observe that one can choose the $\lambda_{i,j}$ from $\kk$ since $|\kk| = \infty$).
2. To see that the above can be achieved with generic $\lambda_1, \ldots, \lambda_m$, observe the following from the proof of Theorem B.35 in How Many Zeroes?: the requirement on $\lambda_1, \ldots, \lambda_m$ boils down to a finite sequence of conditions of the following form: given a certain field $F$ containing $\kk$ and $\alpha_1, \alpha_2 \in F$ and $f_1, f_2 \in F(t)$, where $t$ is an indeterminate, such that $f_1(\alpha_1) = f_2(\alpha_2) = 0$, there is $\lambda \in \kk$ such that
   \[\gcd(f_1, f_2(\lambda \alpha_1 + \alpha_2 – \lambda t)) = t – \alpha_1\]
   (where $\gcd$ is computed in $F[t]$). Since $t – \alpha_1$ divides both of these polynomials, the above condition is equivalent to requiring that
   \[\res(f’_1, f’_2) \neq 0\]
   where
   \[
   \begin{align*}
   f’_1 &:= \frac{f_1}{t-\alpha_1} \\
   f’_2 &:= \frac{f_2(\lambda \alpha_1 + \alpha_2 – \lambda t)}{t – \alpha_1}
   \end{align*}
   \]
   and $\res$ denotes the resultant. Since the resultant is a polynomial in the coefficients, its non-vanishing is a Zariski open condition on the coefficients. Since the condition is satisfied for at least one $\lambda \in \kk$ (that’s the first observation above), it is satisfied for all $\lambda$ in a non-empty Zariski open subset of $\kk$. It follows that the full set of conditions hold for all $\lambda_{i,j}$ in a non-empty Zariski open subset of $\kk^{n^2}$.