If I erase one of the variables in an irreducible homogeneous polynomial, it's still irreducible If $f$ is an irreducible homogeneous polynomial in $k[x_0,...,x_n]$ then $f$ is supposed to cut out an irreducible hypersurface in $\mathbb{P}_k^n$.  So if I look in an affine chart, I should see an irreducible hypersurface. I.e. algebraically, if I set $x_0$ equal to $1$ in $f$ I should get an irreducible polynomial in $k[x_1,...,x_n]$. How to easily see this simple algebraic fact?
I tried taking a factorization of $f(1,x_1,...,x_n)$ and "homogenizing" it, but it gave me a headache trying to show that homogenization is multiplicative and i'm not sure it's the best way to proceed.
 A: Just some mildly tedious bookkeeping. Here's my arrangement.
Suppose we have

*

*$F(x_0, x_1, \ldots, x_n)$ homogeneous of degree $D$ and irreducible

*$f(x_1, \ldots, x_n) = F(1, x_1, \ldots, x_n)$ of total degree $d \leq D$ (not necessarily homogeneous)

*$g(x_1, \ldots, x_n)$ of total degree $a$ and $h(x_1, \ldots, x_n)$ of total degree $b$ where $f = gh$, so $a+b=d$.

Now let

*

*$G(x_0, x_1, \ldots, x_n) = x_0^a g(x_1/x_0, \ldots, x_n/x_0)$

*$H(x_0, x_1, \ldots, x_n) = x_0^b h(x_1/x_0, \ldots, x_n/x_0)$
Note that $G, H$ are homogeneous of degrees $a, b$, respectively. Hence $x_0^{D-d} GH$ is homogeneous of degree $D-d+a+b=D$. Further,
\begin{align*}
  x_0^{D-d} G(x_0, x_1, \ldots, x_n) H(x_0, x_1, \ldots, x_n)
  &= x_0^D g(x_1/x_0, \ldots, x_n/x_0) h(x_1/x_0, \ldots, x_n/x_0) \\
  &= x_0^D f(x_1/x_0, \ldots, x_n/x_0) \\
  &= x_0^D F(1, x_1/x_0, \ldots, x_n/x_0) \\
  &= F(x_0, x_1, \ldots, x_n).
\end{align*}
Since $F$ is irreducible, $d=D$ and either $G$ or $H$ is constant. Setting $x_0=1$ shows that $g$ or $h$ is constant. Hence $f$ is irreducible.
A: First of all, let's denote the affine chart by $U = \mathbb P^n - Z(x_0)$. Now, we run into a problem when $Z(f) \cap U = \emptyset$. The empty space is not considered irreducible, so this is case is worth considering for the algebraic result you're looking for. Indeed, $Z(f) \cap U = \emptyset$ iff $Z(f) \subseteq Z(x_0)$ iff $x_0 \mid f$. Of course, $f$ is irreducible so that would mean that $f = x_0$ up to a multiplicative constant. The point is that the dehomogenization $f(1, x_1, \dots, x_n) = 1$, so the degree decreased. Thus, in this bad case, dehomogenization loses information.
All this is to say that once we stipulate that we are not in this bad case, the result is clearer. We assume that $x_0 \nmid f$. Let me now denote $\alpha(f) = f(1, x_1, \dots, x_n)$ the dehomogenization in the variable $x_0$. I will also denote, for $g \in k[x_1, \dots, x_n]$, $\beta(g) = x_0^{\deg g} g(x_1/x_0, \dots, x_n/x_0)$ the honogenization in the variable $x_0$.
Let's say $\deg f = d$. Now, because $x_0 \nmid f$, there must be some monomial summand of $f$ which contains no $x_0$. Hence, as all monomials in $f$ have degree $d$, $\alpha(f)$ will also have degree $d$. This furthermore tells us that $\beta(\alpha(f)) = x_0^d f(1, x_1/x_0, \dots, x_n/x_0)$. Because $f$ is homogeneous of degree $d$, this equals $f$ itself. That is, dehomogenizing didn't lose any information.
Now, you also mentioned having trouble showing multiplicativity of $\beta$ the homogenization map, so let me include this here. Let $g, h \in k[x_1, \dots, x_n]$. Then $\beta(gh) = x_0^{\deg(gh)} (gh)(x_1/x_0, \dots, x_n/x_0)$. We have $\deg(gh) = \deg(g) + \deg(h)$ so $x_0^{\deg(gh)} = x_0^{\deg(g)} x_0^{\deg(h)}$. Furthermore, $(gh)(x_1/x_0, \dots, x_n/x_0) = g(x_1/x_0, \dots, x_n/x_0) h(x_1/x_0, \dots, x_n/x_0)$. This arises from the evaluation map $x_i \mapsto x_i/x_0$, and evaluation is a ring homomorphism. Putting this together, we therefore have that $\beta(gh) = \beta(g) \beta(h)$.
Now, we are ready to prove irreducibility of $\alpha(f)$, again assuming that $x_0 \nmid f$. Say we had a factorization $\alpha(f) = gh$ in $k[x_1, \dots, x_n]$. Then by the results above, we'd have $f = \beta(\alpha(f)) = \beta(g) \beta(h)$. As $f$ is irreducible, one of $\beta(g)$ or $\beta(h)$ must be a nonzero constant. That is, one of these must be degree $0$. As $\beta$ takes degree $e$ polynomials to degree $e$ homogeneous polynomials, one of $g$ or $h$ must be degree $0$, hence a nonzero constant. In conclusion, $\alpha(f)$ is irreducible.
