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The following situation occurs in a proof that I would like to understand: we have polynomials $F_1,\ldots, F_N$ in $k[X_1,\ldots,X_M]$, where $k$ is of characteristic zero and algebraically closed. The polynomials $F_i$ are homogeneous of positive degree. And we have $M>N$.

Now I would like to conclude that these polynomials have a common zero $\neq (0,\ldots,0)$.

(the following seems to be wrong) By Hilbert's Nullstellensatz it would be enough to show that the ideal $(F_1,\ldots,F_N)$ is proper. This should be easy (probably by looking at dimensions and using $M>N$?), but somehow I don't see a nice argument for that.

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If it weren't proper it would contain units (constant polynomials) which are of grade zero. This is impossible here, since elements of the ideal must have degree larger than that of the generators. –  rschwieb Jun 9 '12 at 12:28

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up vote 1 down vote accepted

Since the $F_i$ are homogeneous of positive degree $(F_1, \ldots, F_N) \subset (X_1, \ldots, X_M) \neq k[X_1,\ldots,X_M]$. So do we need $M > N$?

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Ok you're right about that. It might be not enough to show that the ideal is proper, since for example $x+y$ and $x^2+y^2$ have only one common zero: $(0,0)$. So, how can I check if there are non-trivial common zeros, if not by checking properness of the ideal? –  Fabmor Jun 9 '12 at 12:32
I have found a proof now, it's a proposition in Chapter 1, §6.2 in Safarevich's Basic Algebraic Geometry. –  Fabmor Jun 9 '12 at 13:36

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