$\DeclareMathOperator{\codim}{codim} \newcommand{\dprime}{^{\prime\prime}}\newcommand{\kk}{\mathbb{K}} \newcommand{\local}[2]{\mathcal{O}_{#2, #1}} \newcommand{\pp}{\mathbb{P}} \newcommand{\qq}{\mathbb{Q}} \newcommand{\rr}{\mathbb{R}} \DeclareMathOperator{\res}{Res} \newcommand{\scrL}{\mathcal{L}} \DeclareMathOperator{\sing}{Sing} \newcommand{\zz}{\mathbb{Z}}$

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:

- 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 $\kk(X)$;
- 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

- 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 $|\kk| = \infty$).
- 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. 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 $\kk^{n^2}$.*resultant*