# Degree of a real algebraic variety and regular morphisms

I'm reading Fulton's "Intersection theory", which i need for some applied needs. And i have two questions on general definition of degree used in Fulton.

1)Let us we have a real algebraic variety defined by a set of equations $f_1=0, f_2=0,\ldots ,f_n=0$ of degrees $d_1,\ldots, d_n$ respectively. Using a well-known real-algebraic-geometry trick we can think about this variety as a variety defined by one equation $\sum_if_i^2=0$ of degree $2\max_{i}d_i.$ Then, we can take the smallest degree of all single polynomials representing fixed real algebraic variety as a degree of a variety.

Will this definition of degree coincide with given in Fulton "Intersection theory" $\S$ 8.4?

2)Let X be a real affine algebraic variety in $R^n$ of degree $p$ and let $f\colon R^n \to R^{n-1}$ be a projection. Will $deg \overline{f(X)}\leq \deg X$ in the sense of Fulton's definition? In sense of my definition?

$\overline{f(X)}$ here is a closure of $f(X)$ in Zarissky topology, not a semialgebraic one.

UPDATE

Being more exact, I have intersection of two hypersurfaces of degrees $d$ and $e$. And i want to project that intersection onto $R^{n−1}$. That projection will be(if everything is nice) a hypersurface in $R^{n−1}$. Can i say that this hypersurface could be represented by a polynomial of degree at most $2max(d,e)$?

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Re 1: a point in $\mathbb{R}^3$ has degree 1 (with any reasonable definition) but it can't be given as the zero locus of a linear polynomial. –  algori Nov 5 '11 at 17:05
Off the top of my head, your definition does not look very nice. Let's say you have a variety which is intersection of two hypersurfaces of degrees $d$ and $e$. It would be natural to say that the degree of the intersection is $de$, which would be the degree of the complexification of your variety (at least when everything is nice). By your definition would have a degree of at most $2\max(d,e)$, so much smaller. I guess it depends what you want to use your degree for, but I don't see anything good coming out of this big discrepancy. –  Thierry Zell Nov 5 '11 at 17:07
I have intersection of two hypersurfaces of degrees d and e. And i want to project that intersection onto $R^{n-1}$. That projection will be(if everything is nice) a hypersurface in $R^{n-1}$. Can i say that this hypersurface could be represented by a polynomial of degree at most $2max(d,e)$? –  Grey Violet Nov 5 '11 at 17:26
@probably: absolutely not. First of all: your projection is a semi-algebraic set, not a variety in general. And the degrees you need will be higher in general. Projection is a pretty nasty operation. –  Thierry Zell Nov 5 '11 at 17:31
The Zariski closure of your semi-algebraic set (the projection) is indeed a real algebraic variety, which gives you also the complexfication. But again, this operation can be complicated. The reference for that is The complexification and degree of a semi-algebraic set by Marie-Françoise Roy and Nicolai Vorobjov. springerlink.com/content/g5u4rpfy0jp1dujm –  Thierry Zell Nov 5 '11 at 21:54

Your definition dramatically underestimates the value of the degree. Here is an example derived from Fulton's book. Take $$f(x,y)= \prod_{i=1}^d (x-i)^2+\prod_{j=1}^d(y-j)^2.$$ Then, $V(f)$ has degree $\leq 2d$ by your definition, but it is made of $d^2$ isolated points. From a geometrc point of view, this is something whose degree should probably be at least $d^2$. The same example works in more variables, of course.