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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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    @Parsa Please consider converting your comment into an answer, so that this question gets removed from the [unanswered tab](http://meta.math.stackexchange.com/q/3138). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868).2013-06-21

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