# Homogenization of an irreducible polynomial is irreducible

This is the last detail in an exercise that I'm working on in Hartshorne and I can't seem to figure it out.

If $f$ is an irreducible polynomial in $k[x_{0},\dots,x_{n}]$ (where $x_{i}$ does not appear) and $\beta(f)$ is the homogenization of $f$ with respect to $x_{i}$, i.e., $\beta(f)=x_{i}^{d}f(\tfrac{x_{0}}{x_{i}},\dots ,\tfrac{x_{n}}{x_{i}})$, where $d$ is the maximal total degree of $f$, why is $\beta(f)$ irreducible?

$$\text{My plan}$$

Assume otherwise then $\beta(f)=gh$ where $g$ and $h$ are homogeneous. (A proof of this fact would be nice or a link. I wasn't able to straighten this out.) Then $f=\alpha(g)\alpha(h)$ where $\alpha$ is defined by $\alpha(f)=f(x_{0},\dots ,1,\dots,x_{n})$ where the $1$ appears in the $i^{th}$ spot. $f$ is irreducible so either $\alpha(g)$ or $\alpha(h)$ is a unit. But I don't know how to relate this back to $g$ or $h$.

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Let us assume that $\alpha(g)$ is a unit. But then must be $g = c \cdot x^\alpha_i$. But $g$ cannot be a pure $x_i$-power (This contradicts to the fact that $f$ is irreducible and the construction of the homogenization).
Why the factorization of $\beta(f) = g \cdot h$ ($\deg(f) = d = \deg(g) + \deg(f)$) must be homogeneous. Assume $g$ is not homogeneous. Then $g$ must have a term $t$ which has a smaller then $\deg(g)$. But then $\deg(t \cdot f)$ is smaller then $d = \deg(g) + \deg(f)$.
I don't see how this contradicts the irreducibly of $f$. – TheNumber23 Oct 2 '13 at 18:16
I still don't understand why $g$ can't by $c\cdot x_{i}^{\alpha}$, but the factorization argument was clear than you. – TheNumber23 Oct 2 '13 at 18:26
Yes, I see your counterexample but I this case we homogenize with an indeterminant that is not in $k[x_{1},\cdots,x_{n}]$. I was unclear of the setup in the problem I changed it. – TheNumber23 Oct 2 '13 at 18:34
The homogenization does not change the degree of $f$ e.g $\deg(f) = \deg(\beta(f))$, e.g. the $x_i$ does not appear in the homogenized terms $\beta(t)$ where $t$ is a term of $f$ with $\deg(t) = \deg(f)$. But then $x_i$ can't divide $t = \beta(t)$, hence can't divide $\beta(f)$. – Huehnermelker Oct 2 '13 at 18:43