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The question came up while reading a bit more into the Hilbert-Zariski theorem I asked about the other week.

Suppose $V$ is an algebraic variety over arbitrary field $k$. (For this situation, I'll take the definition $\dim\ V=\deg_k(k(x))$, where $(x)\in V$ is a generic point, and by $\deg$ I mean the transcendence degree.) As usual, $V(f_1,\dots,f_s)$ is the set of zeroes of the homogeneous forms $f_1,\dots,f_s)$ in the affine space.

Let $\dim\ V=d$. Suppose I take $d$ generic linear forms $f_1=\sum_{i=1}^n a_{1i}X_i,\dots,f_d=\sum_{i=1}^n a_{di}X_i$, so that the $a_{ji}$ are algebraically independent. In this scenario, is it true that $$ V(f_1,\dots,f_d)\cap V=\{0\}? $$

If so, why is there only the trivial zero? Many thanks.

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If your field is infinite, then by choosing your linear forms sufficiently general, you can assure that your intersection $V \cap V(f_1, \ldots ,f_d)$ is irreducible, so it can only be the origin. But if your field is finite, I don't think this is true anymore. For instance, take $k=\mathbb F_2$ and $V=V(X^2+Y^2+X+Y) = \mathbb A_k^2$. The only homogenous linear forms here are $X,Y,X+Y$ and each one intersected with $V$ gives a reducible algebraic set consisting of the origin and one extra point. –  Parsa Feb 11 '12 at 4:47
    
@Parsa Dear Parsa, thank you for your comment. If you find the time, do you mind expanding the idea for when $k$ is an infinite field into an answer? No hurry of course! –  Vika Feb 11 '12 at 5:22
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You can read Prop 7.1 in Hartshorne (AG), page 48. That way you see that for a general hyperplane not contained in $V$, the dimension drops by $1$ with each intersection, so $V(f_1, \ldots ,f_d) \cap V$ is $0$-dimensional, i.e. a collection of points, and obviously non-empty since $0$ is in there. Then read p. 174 of Griffiths & Harris (PAG) to see that a general hyperplane intersected with an irreducible variety stays irreducible, so your intersection must contain only the trivial zero. –  Parsa Feb 11 '12 at 21:10
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@Parsa Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. –  Julian Kuelshammer Jun 21 '13 at 19:09
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