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Suppose $f=f(x)=f(x_1,\cdots,x_n)$ is a homogeneous polynomial of degree $n$ in the ring $k[x_1,\cdots,x_n]$, with $k$ a field. Denote by $H_f$ the hypersurface given by $f=0$. Suppose that for every smooth point $z$ of $H_f$, $f(x)$ is divisible by the linear form $f_x(x,z) = \sum_{i=1}^n x_i \frac{\partial f(\xi)}{\partial \xi_i}|_{\xi = z}$.

Question 1: Why is it true that $H_f$ coincides with the tangent hyperplane in the neighborhood of $z$? (i can see that $\sum_{i=1}^n x_i \frac{\partial f(\xi)}{\partial \xi_i}|_{\xi = z}$ is the equation of the tangent hyperplane at $z$).

Question 2: Why is $f$ a union of hyperplanes?

PS: I can see the validity of 1 and 2 intuitively, i am looking for a rigorous algebraic geometric argument, requiring as little background as possible.

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  • $\begingroup$ The basic point is that if f=gh, then the associated hypersurface is a union: $H_f = H_g \cup H_h$, and a smooth point of H_f can only belong to one of these components, so locally near such a point $H_f$ coincides with either H_g or H_h. Apply this in your case with g (say) being one of your linear forms. $\endgroup$ – user64687 Mar 14 '13 at 21:34
  • $\begingroup$ @user64687: Ok, this answers Question 1. $\endgroup$ – Manos Mar 14 '13 at 21:45
  • $\begingroup$ It's not hard to deduce Question 2 now. You need to show that f factorises completely into linear forms: to do this, suppose it didn't, and see if your hypotheses could really hold. (Say f has an irreducible factor g which is of degree $\geq 2$, and think about a point z on $H_g$.) $\endgroup$ – user64687 Mar 14 '13 at 21:54
  • $\begingroup$ @user64687: Got it. If you make this into an answer i will gladly accept it. $\endgroup$ – Manos Mar 14 '13 at 22:08
  • $\begingroup$ ok, will do.... $\endgroup$ – user64687 Mar 14 '13 at 22:11
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(answerifying my comments)

The basic point is that if $f=gh$, then the associated hypersurface is a union: $H_f=H_g \cup H_h$, and a smooth point of $H_f$ can only belong to one of these components, so locally near such a point $H_f$ coincides with either $H_g$ or $H_h$. Apply this in your case with $g$ (say) being one of your linear forms.

It's not hard to deduce Question 2 now. You need to show that $f$ factorises completely into linear forms: to do this, suppose it didn't, and see if your hypotheses could really hold. (Say $f$ has an irreducible factor $g$ which is of degree $\geq 2$, and think about a point $z$ on $H_g$.)

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