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Let F be algebraically closed field. Let C be a curve in F^n defined as zeroes of polynomials $p_1(x_1,\ldots,x_n),..,p_{n-1}(x_1,\ldots x_n)$. Let us define degree of the curve as $\max_S \{ S\cap C \}$ were $S$ $n-1$ dimensional linear subspace such that $ \{ S\cap C \}$ is finite.

How does possible to calculate this degree of the curve?

In general degree of curve should be product of degrees of $p_1,\ldots p_{n-1}$ but for example if $C=(x-f_1(z),y-f_2(z))$ were $f_1,f_2$ are of degree d then degree of $C$ is $d$ and not $d^2$.

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1 Answer 1

What's happening is that indeed the degree of your curve (when it is a curve) is the product of the degrees of the $p_i$, counted with multiplicity! In other words, the intersection scheme has that degree.

In your example in the last paragraph, the intersection is actually $d\cdot C$, so getting degree $d$ for $C$ is the correct answer. The issue is that in this example the intersection multiplicity of the defining equations is $d$ everywhere along the intersection.

Here is a specific example to see what's happening: Let's say that $d=2$, $f_1(z)=z^2+l_1(z)$ and $f_2(z)=z^2+l_2(z)$ where the $l_i$ are linear polynomials in $z$. Then the ideal generated by $x-f_1(z)$ and $y-f_2(z)$ contains $x-y+l_1(z)-l_2(z)$, a linear polynomial and you get the same ideal with the generators $x-f_1(z)$ and $x-y+l_1(z)-l_2(z)$. If you take the intersection of these two hypersurfaces, then you get the correct degree $d=2$. You can easily see that if you take a local ring of the ambient space at a point of the intersection curve, then the original defining equations are both in the square of the maximal ideal, so their intersection multiplicity has to be (at least) $2$.

By the way, intersection theory should be really done in the projective space.

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Thanks for you answer. How can I calculate this degree for a given set of polynomials. One specific example is what will be degree if I have two random polynomials $p(x,y)$ of degree $d_1$ and $g(x_1,x_2,x_3,x_4)$ of degree $d_2$ what will be typical degree of $C=<p(x_1,x_2),p(x_3,x_4), g(x_1,x_2 x_3,x_4)>$ –  Klim Efremenko Jan 13 '11 at 20:31
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I think in this case you should expect a degree $d_1^2 d_2$ curve. Basically, that's your degree, divided by the multiplicity of the curve. This is of course assuming that the curve is irreducible. If $C$ is reducible, you can get different multiplicities along different components. In any case, what you need to determine is the intersection multiplicities along each irreducible component of the intersection. In the case of three polynomials as in your comment, you would probably be best off by doing it first for the intersection of just two of them and then intersect with the third. –  Sándor Kovács Jan 13 '11 at 21:01
    
ps: the reason I would expect degree $d_1^2d_2$ in the particular case you are asking about is because the two occurrences of $p$ happens in separate variables, so their zero loci are transversal. Then if $g$ is general, then it should be transversal to the intersection. –  Sándor Kovács Jan 13 '11 at 21:04
    
But, this still does not answer the question: How can I calculate degree of the curve? Say given two polynomials $p(x,y,z), g(x,y,z)$ Does there exist algorithm for calculating degree of intersection? Or what is the multiplicity of the intersection? –  Klim Efremenko Jan 16 '11 at 8:45

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