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I am stuck at the following exercise:

Let $K$ be an algebraically closed field and let $f \in K[X,Y]$ be a homogenuous polynomial such that $f(0,0) = 0$. Show that $f$ can be split into a product of linear polynomials.

First we observe that the condition $f(0,0) = 0$ entails that $f$ has no constant term, i.e. if $$f := \sum_{\substack{i = 0 \\ i_1+i_2 = d}}^n a_iX^{i_1}Y^{i_2},$$

then $a_0 = 0$. I guess the best way to prove the claim is to use induction. For $n = 1$ the claim is clear, so suppose that all homogenuous polynomials with $(0,0)$ as a root split into linear polynomials. Now let $f$ have $deg(f) = n+1$. We need to show that $f$ splits into linear factors. However, I do not see how to proceed from here. Could you please give me a hint?

3nondatur
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    You can dehomogeneize the polynomial taking $Y=1$. Use the property that $K$ is algebraically closed, and then homogeneize again. – Marcos Dec 16 '21 at 17:43
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    Because $f$ is homogeneous, all monomials have the same degree, so $i_1+i_2$ is constant. On the other hand their coefficients may vary, whereas your expression for $f$ has $a_i$ constant for different $i_1$ and $i_2$. Perhaps you should revise what it means for a polynomial to be homogeneous? – Servaes Dec 16 '21 at 20:01
  • @Servaes: You are right, I corrected the mistake. – 3nondatur Dec 16 '21 at 21:44
  • @Marcos: Thanks for your help, I got it now. – 3nondatur Dec 16 '21 at 21:44

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