# When to read of the degree of a variety from its defining polynomials

The question concerns algebraic varieties.

I just read the question The degree of an algebraic curve in higher dimensions and great answer by user M P. One of the thing he says is that if a curve in $\mathbb{P}^n$ is given by $n-1$ equations (which is often not the case of course), then its degree is in fact the product of the degree of the polynomials defining it.

I expect this not to hold if there are more than $n-1$ polynomials necessary to define the curve. Could someone tell me if this is true, and why?

I do expect it to hold for general varieties. If a variety in $\mathbb{P}^n$ is of codimension $r$, and also given by exactly $n-r$ polynomials, is its degree in fact the product of the degrees of the polynomials defining it? Could you tell me if this is the case, and more importantly: why?

By the way, degree as in generic number of intersection points with a variety of complementary dimension..

Thanks a lot in advance, Joachim

P.S. Georges E. i owe you one for your effort on my question on étale morphisms. I know this is not the place to say such things, but i'm doing it anyway.

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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 $X^3 - ZW^2 = 0$.

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Thanks! Both answers are great, i'd accept them both if possible. 2 small questions, if that's ok. I quote: "Then you can deform the equation $f_{i+1}=0$ to an equation of the form $l_1\ldots l_{d′}$", could you make this mathematically precise? Or this this just an intuitive argument? Second question, could you confirm that the same reasoning would give us that if $V,W$ are varieties of degree $d,e$, and suppose their intersection is in fact irreducible, then their intersection has degree $de$? You basically applied induction to the case where $W$ is a hypersurface, right? Thanks in advance! – Joachim Aug 9 '12 at 21:59
I just read that you only get a notification if i refer to you, so @Matt E please read my comment before this one.. =) – Joachim Aug 10 '12 at 1:35
@Joachim: Dear Joachim, First, a small point: since this is my answer, I notified about any comment here, whether or not it has "@Matt E". Second: regarding deformation, one can consider the one parameter family $g_t:= t l_1 \cdots l_{d'} + (1-t) f_{i+1}$; for all but possibly finitely many values of $t$, the intersesection of $V(g_t)$ with $V(f_1,\ldots,f_i)$ will be proper, and the degree will be constant in this family. (The technical concept I am using is linear equivalence, or algebraic equivalence if you prefer.) Also, yes, if $V$ and $W$ are of degree $d$ and $e$ and meet properly – Matt E Aug 10 '12 at 3:38
... i.e. their intersection is of the expected dimension, then its degree is equal to $d e$. This is a special part of general intersection theory. Regards, – Matt E Aug 10 '12 at 3:39
Thats clear, thanks. As you noticed i know very little of intersection theory, however i am reading a book that uses its concepts every now and then. Reading your answer i realized it might be helpful to get you know more about the subject. Could you recommend a good reference thats treats basic intersection theory of varieties? – Joachim Aug 10 '12 at 9:35

A variety $X$ in $\mathbb P^n_k$ of codimension $r$ generated by $r$ forms is called a complete intersection. Equivalently, $X$ is defined by a regular sequence $f_0,\ldots,f_r$ of (homogeneous) elements of $k[x_0,\ldots,x_n].$ It is well-known that the Koszul complex associated to a regular sequence $(f_0,\ldots,f_r)$ is a free resolution of the coordinate ring $k[x_0,\ldots,x_n]/(f_0,\ldots,f_r)$ and so enables us to compute the Hilbert polynomial of $X.$ Moreover, it can be shown that the Hilbert polynomial depends only on the degrees of $f_0,\ldots,f_r$, and its leading coefficient is $d_0\cdots d_r/(n-r)!,$ where $d_i$ is the degree of $f_i,$ so in this case the degree of $X$ is $d_0\cdots d_r.$

If $X$ is not a complete intersection, then the Koszul complex of defining equations will not be exact, and thus does not determine the Hilbert polynomial, though I'm not sure how badly behaved the degree can be in this case.

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