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Let $k$ be a field. Suppose I have a homogeneous polynomial $f$ in $k[x,y,z]$. If $f$ is reducible, does it always decompose as a product of homogeneous polynomials? Thanks!

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Yes. More generally:

Let $A = \bigoplus_{i\in\mathbb{Z}} A_i$ be a graded domain und $f\in A\setminus \{0\}$ homogeneous. If $f$ factors in $A$ as $f=gh$ then $g,h$ are homogeneous.

Proof: Since we are in a domain and $f\neq 0$, the factors $g,h$ are non-zero as well. Let $g$ have non-zero component of lowest degree $d_{\min}$ and of highest degree $d_{\max}$. Similarly for $h$ (with degree symbol $e$ in place of $d$). Then $f=gh$ has non-zero component of lowest degree $d_{\min}+ e_{\min}$ and of highest degree $d_{\max} + e_{\max}$ (here we use that $A$ is a domain!). But since $f$ is homogeneous, lowest and highest degree of $f$ agree, i.e. $d_{\min}+ e_{\min} = d_{\max}+ e_{\max}$. From $$0 = (\;d_{\max} - d_{\min}\;) + (\;e_{\max} - e_{\min}\;)$$ and the non-negativity of the summands we conclude $d_{\min} = d_{\max}$ and $e_{\min} = e_{\max}$, i.e., $g$ and $h$ are homogeneous. qed

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    $\begingroup$ One way to say this is as follows: the width of a non-zero element $a$ of a graded domain is the difference $d_{\max}(a)-d_{\min}(a)$ of the degrees of the topmost nonzero component of $a$ and that of the bottommost (?) component. Then $w$ is multiplicative: $w(a)+w(b)=w(ab)$. The result follows formally from this, just as for one-variable polynomials and the usual degree. $\endgroup$ Jul 30 '15 at 22:56
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    $\begingroup$ This has the following nice generalization: If we had a multigraded domain $A=\bigoplus_{\alpha\in\mathbb Z^n}A_\alpha$, we could let $w(a)$ be the convex hull in $\mathbb R^n$ of the set of multiudegrees of its nonzerocomponents. This "width" is also multiplicative, with the sum being the Minkowski sum of convex sets. $\endgroup$ Jul 30 '15 at 23:09
  • $\begingroup$ @Mariano Suárez-Alvarez: Yes, that's a really nice, compact formulation. $\endgroup$
    – tj_
    Jul 30 '15 at 23:11

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