algebraic-geometry
Let $X$ be a complete intersection in $\mathbb P^n$, does anyone know how to calculate the degree of a dual variety of $X$?
Some preliminaries can be found in 3264 and all that, chapter 10. For instance, such dual varieties are always codimension one (except when $X$ is linear); and when $X$ is a hypersurface, the answer is well know.
In particular, I'd like to know the answer when $X$ is the curve as a complete intersection by a cubic and quadric in $\mathbb P^3$.
(1) If you ask that $E$ and $H$ both be nonsingular, the answer is no. In $\mathbb{P}^3$, let $Q$ be a degree $2$ hypersurface with a singularity at one point $p$ (the cone on a smooth conic). Let $C$ be a generic smooth cubic; so $C$ does not pass through $p$ and is transverse to $Q$. Then $X:=C \cap Q$ is a degree $6$, genus $4$, smooth curve, embedded by its canonical section.
If we want to write $X$ as $A \cap B$, for hypersurfaces $A$ and $B$, then we must have $(\deg A)(\deg B) = 6$. Since $X$ does not lie in any plane, one of $A$ and $B$ must have degree $2$. There is only one degree two hypersurface containing $X$, namely $Q$, and $Q$ is not smooth.
In fact, I can give a smaller example. In $\mathbb{P}^2$, take a smooth cubic $C$ and intersect it with $Q$, the union of two lines, where the crossing point of the lines is not in $C$. You get $6$ points, and the same logic applies.
(2) If you work in $\mathbb{P}^n$, over a characteristic zero field, and only ask that $E$ be smooth (and not $H$), the answer is yes. Let $X = \{ f_1 = f_2 = \cdots = f_r =0 \}$ with $\deg f_i = d_i$, and reorder the terms such that $d_1 \geq d_2 \geq \cdots \geq d_r$. Choose $h_{ij}$ a polymonial of degree $d_i - d_j$, and otherwise generic. Define $g_i = f_i + \sum_{j>i} h_{ij} f_j$. Clearly, the $g_i$ and $f_i$ generate the same homogeneous ideal. We claim that, for generic values of the $h_{ij}$, every intersection $\{ g_1 \cap g_2 \cap \cdots \cap g_s =0 \}$ is smooth, for $s \leq r$.
Proof: By induction on $s$. When $s=0$, the claim is trivial. By induction, fix $h_{ij}$ for $i<s$ such that $\{ g_1 \cap g_2 \cap \cdots \cap g_{s-1} =0 \}$ is smooth; call this smooth variety $A$. We want to show that for generic $h_{sj}$, the intersection $A \cap \{ f_s + \sum h_{sj} f_j=0 \}$ is smooth.
We now use a variant of Corollary III.10.9 in Hartshorne. (This is where we use the characteristic zero hypothesis.) Look at the linear system on $A$ spanned by $f_s$ and by $f_j \mathcal{O}(d_s-d_j)|_A$. The variant we want is that, over a field of characteristic zero, on a smooth variety, a generic element of a finite dimensional linear system has singularities contained within the basepoints of the linear system. Hartshorne sketches the proof of this in Remark 10.9.2. (Hartshorne fails to mention that the algebraic closure hypothesis can be removed, but I think that's just an oversight on his part.) So, for generic $h_{sj}$, the singularities of $A \cap \{ f_s + \sum h_{sj} f_j=0 \}$ are contained in $X$.
Suppose for the sake of contradiction that $x \in X$ is a singular point of $A \cap \{ f_s + \sum h_{sj} f_j=0 \}$. But then $X = A \cap \{ f_s + \sum h_{sj} f_j=0 \} \cap \{ f_{s+1} = f_{s+2} = \cdots = f_r =0 \}$. Slicing a singular point by a regular sequence leaves a singular point behind, so this shows that $X$ is singular at $x$, contradicting the hypothesis that $X$ is smooth. QED
(3) If you take $X=\mathbb{P}^2 \times \mathbb{P}^2$, the answer is no. Take the intersection of a $(2,1)$ hypersurface and a $(1,2)$ hypersurface, each with an isolated singularity and meeting transversely. The intersection is a smooth surface $S$. The coholomogy ring of $\mathbb{P}^2 \times \mathbb{P}^2$ is $\mathbb{Z}[u,v]/\langle u^3, v^3 \rangle$.
Suppose we could write $S$ as the intersection of a hypersurface of degree $(a_1, a_2)$ and one of degree $(b_1, b_2)$. Then, in $H^2(\mathbb{P}^2 \times \mathbb{P}^2)$, we have $(a_1 u + a_2 v)(b_1 u + b_2 v) = (2 u + v)(u + 2 v)$. By unique factorization of polynomials, we must have $(a_1, a_2) = (2,1)$ and $(b_1, b_2) = (1,2)$, or vice versa. (Note that I had to go up to $\mathbb{P}^2 \times \mathbb{P}^2$, so that the degree $2$ polynomials would inject in $H^*$; this argument doesn't work in $\mathbb{P}^1 \times \mathbb{P}^1$.)
But the only $(1,2)$ form which vanishes on $S$ is the singular $(1,2)$ form we started with, and the same for $(2,1)$ forms. So $S$ cannot be written as a complete intersection with either factor being smooth. QED
I couldn't figure out:
(4) Characteristic $p$, in $\mathbb{P}^n$. I tried to build a counter-example based on Hartshorne Exercise III.10.7 and failed. I don't have an intuition for whether that is because I was not clever enough, or because the result is true.
Andrew's answer is correct, but here is a more geometric way to think about the same answer.
If a codimension $r$ variety is cut out by $r$ equations, then each equation genuinely cuts out a new dimension, i.e. $V(f_1,\ldots,f_{i+1})$ is a divisor in $V(f_1,\ldots,f_i)$ cut out by $f_{i+1} = 0$. Thus the intersection of $V(f_1,\ldots,f_i)$ and $V(f_{i+1})$ is a proper intersection, and so the degrees multiply.
Here is one way to think about this last statement:
We can assume that the degree of $V(f_1,\ldots,f_i)$ is equal to $d,$ the product of the degrees of $f_1, \ldots,f_i$, by induction. Geometrically, this means that a generic linear subspace $L$ of dimension $i$ meets $V(f_1,\ldots,f_i)$ in $d$ points.
To compute the degree of $V(f_1,\ldots,f_{i+1})$ we intersect with a generic linear subspace $L'$ of dimension $i+1$.
Suppose that $d'$ is the degree of $f_{i+1}$. Then you can deform the equation $f_{i+1} = 0$ to an equation of the form $l_1\ldots l_{d'}$, where each $l_j$ is a generically chosen linear equation.
Thus $V(f_{i+1})$ can be deformed to the union of the hyperplanes $V(l_1)\cup \cdots \cup V(l_{d'})$, and so
$$V(f_1,\ldots,f_{i+1}) \cap L' = V(f_1,\ldots,f_i) \cap V(f_{i+1}) \cap L',$$ which can be deformed to
$$V(f_1,\ldots,f_i) \cap \bigl(V(l_1)\cup \cdots \cup V(l_{d'})\bigr) \cap L'
= \bigl(V(f_1,\ldots , f_i) \cap V(l_1) \cap L'\bigr)
\cup \cdots \cup
\bigl(V(f_1,\ldots,f_i)\cap V(l_1)\cap L'\bigr) \bigr).$$
Now each $V(l_j)\cap L'$ is the intersection of a generic hyperplane and the generic $i+1$-dimensional linear subspace $L'$, and so is a generic linear subspace of dimension $i$. Thus each intersection $V(f_1,\ldots,f_i)\cap V(l_j)\cap L'$ consists of $d$ points, and so their union consists of $dd'$ points. QED
If $V$ is an $r$-codimensional variety cut out by more than $r$ equations, then we can write it in the form $V(f_1,\ldots,f_r, f_{r+1},\ldots,f_s)$ for some $s > r$, where $V(f_1,\ldots,f_r)$ is already of codimension $r$ (but not irreducible). The closed subscheme $V(f_1,\ldots,f_r)$ of $\mathbb P^n$ is then of degree equal to the product of the degrees of the $f_1,\ldots,f_r$, but (by assumption) the additional equations $f_{r+1}, \ldots, f_s$ do not cut down the dimension any further. Instead, they cut out some particular irreducible component of $V(f_1,\ldots,f_r)$. In particular, the intersections $V(f_1,\ldots,f_r) \cap V(f_j)$ for $j > r$ are not proper, and the degree of a non-proper intersection is not given as the product of the degrees.
So the basic phenomenon here is as follows: if $V$ is a reducible algebraic set of (equi)codimension $r$, a union of components $V_1,\ldots,V_m$, then $\deg V = \sum_i \deg V_i$, but there is no immediate relationship between the degrees of the additional polynomials needed to cut out the various $V_i$ and the degrees of those $V_i$.
A good example to think about is the case of a twisted cubic curve in $\mathbb P^3$ (with hom. coords. $X,Y,Z,W$). It is obtained by first intersecting the quadrics $V(X^2 - YW)$ and $V(XZ - Y^2)$. This intersection is a degree $4$ curve which is reducible; it is the union of a line $L$ (the line cut out by $X = Y = 0$) and the twisted cubic curve $C$. To cut out $C$ we have to impose the additional equation $X^3 - ZW^2 = 0$.
Best Answer
Let $X$ be a smooth complete intersection of type $(d_1,\dots,d_k)$ in $\mathbb{P}(V)$. Consider the bundle $$ E := \bigoplus \mathcal{O}_X(1 - d_i) \hookrightarrow V^\vee \otimes \mathcal{O}_X, $$ where the morphism is given by the derivatives of the equations of $X$. Then $\mathbb{P}(E)$ is the universal tangent hyperplane to $X$, hence the dual variety is the image of the map $$ \mathbb{P}_X(E) \to \mathbb{P}(V^\vee) $$ induced by the embedding $E \to V^\vee \otimes \mathcal{O}_X$. Therefore, the degree of the dual variety is $$ \deg(X^\vee) = s_{n}(E), $$ where $n = \dim(X)$ and $s_n$ is the $n$-th Segre class (this is true under the assumption that the map $\mathbb{P}_X(E) \to X^\vee$ is birational, otherwise you need to divide by its degree). This class is easy to compute: this is the coefficient of $h^n$ in $$ \left(\prod_{i=1}^k(1 - (d_i - 1)h)^{-1}\right)\prod_{i=1}^k d_i. $$
In the case of a $(2,3)$ intersection in $\mathbb{P}^3$ this gives $3 \cdot 2 \cdot 3 = 18$.