Is the homogenization of an irreducible polynomial irreducible?

This statement is something I have thought to be true for a long time. However, lately I realized that I did not have any reference for it, so I was a bit unsure. I could find an MSE question, whose answer matched what I thought, and a couple of friends came up with the same arguments, which made me believe the statement. I also think that this fact is available in a book by Cox, Little, and O'Shea (Exercise 7 on p.392), although many places in this book assumes that the ambient field $k$ is algebraically closed. Thus, I wanted to record this here for later purpose.

Why do I care? I care about this statement mostly because I want to think of the notion of hypersurfaces of $\mathbb{P}^{n}$ over a field $k$ somewhat intrinsically. That is, if I have an integral closed subscheme $H \hookrightarrow \mathbb{P}^{n}$ with codimension $1$ and degree $d,$ then I want to say there is a homogenous polynomial $f \in k[x_{0}, \dots, x_{n}]$ of degree $d$ such that $$H \simeq \mathrm{Proj}(k[x_{0}, \dots, x_{n}]/(f))$$ as $k$-schemes. Take any standard affine open subset $$U_{i} := D_{+}(x_{i}) \simeq \mathrm{Spec}(k[x_{0}/x_{i}, \dots, x_{n}/x_{i}])$$ such that $H \cap U_{i} \neq \emptyset.$ Since $H \cap U_{i}$ is of codimension $1$ in $U_{i},$ there must be a (monic) nonzero $f_{i} \in k[x_{0}/x_{i}, \dots, x_{n}/x_{i}]$ such that $$H \cap U_{i} = \mathrm{Spec}(k[x_{0}/x_{i}, \dots, x_{n}/x_{i}]/(f_{i})).$$ Since $H$ is the closure of $H \cap U_{i}$ in $\mathbb{P}^{n},$ it follows (e.g., from a previous posting) that topologically, we have $$V_{+}(f) = H,$$ where $f(x_{0}, \dots, x_{n}) = x_{i}^{\deg(f_{i})}f_{i}(x_{0}/x_{i}, \dots, x_{n}/x_{i}),$ where $\deg(f_{i})$ denotes the maximum among the degrees of the monomials of $f_{i}$ in $x_{0}/x_{i}, \dots, x_{n}/x_{i},$ except $x_{i}/x_{i} = 1.$ However, to show that $$H \simeq \mathrm{Proj}(k[x_{0}, \dots, x_{n}]/(f)),$$ we also need to check that $$x_{i}^{\deg(f_{i})}f_{i}(x_{0}/x_{i}, \dots, x_{n}/x_{i}) = x_{j}^{\deg(f_{j})}f_{j}(x_{0}/x_{j}, \dots, x_{n}/x_{j})$$ for all $i, j$ such that $U_{i}$ and $U_{j}$ both intersect $H,$ which holds when $f$ is irreducible.

Geometric argument. If $f$ is not irreducible, then $f = gh$ for some homogeneous polynomials $g, h$ with positive degrees. (To see this, decompose $g$ and $h$ into homogeneous pieces and argue by degree.) Thus, we would have $$V_{+}(f) = V_{+}(g) \cup V_{+}(h).$$ Since $V(f_{i})$ is irreducible, we know $V_{+}(f) = \overline{V(f_{i})}$ is an irreducible topological space, at least one of $V_{+}(g)$ and $V_{+}(h)$ is empty. This implies that one of $\sqrt{(g)}$ and $\sqrt{(h)}$ is equal to $(x_{0}, \dots, x_{n}).$ Without loss of generality, say $\sqrt{(g)} = (x_{0}, \dots, x_{n}),$ which is a maximal ideal. Then we have $\{(x_{0}, \dots, x_{n})\} = V(x_{0}, \dots, x_{n}) = V(g)$ in $\mathbb{A}^{n+1},$ although $V(g)$ has dimension $\geq n + 1 - 1 = n \geq 1,$ by Krull's principal ideal theorem. This is a contradiction as the singleton of a closed point has dimension $0.$

Combinatorial argument. If $f$ is not irreducible, then $f = gh$ for some homogeneous polynomials $g, h$ with positive degrees. We may assume that $g, h$ are monic since $k$ is a field. Then $$\begin{align*}f_{i}(x_{0}/x_{i}, \dots, x_{n}/x_{i}) &= x_{i}^{-\deg(f_{i})}f(x_{0}, \dots, x_{n}) \\ &= g(x_{0}/x_{i}, \dots, x_{n}/x_{i}) h(x_{0}/x_{i}, \dots, x_{n}/x_{i}).\end{align*}$$ Since $f_{i}$ is irreducible, at least one of the polynomials on the right-hand side must be constant in $k[x_{0}/x_{i}, \dots, x_{n}/x_{i}],$ which necessarily mean that it is $1$ (due to the monic assumption). Without loss of generality, say $g(x_{0}/x_{i}, \dots, x_{n}/x_{i}) = 1.$ Then we have $g(x_{0}, \dots, x_{n}) = x_{i}^{\deg(g)},$ so $$f(x_{0}, \dots, x_{n}) = x_{i}^{\deg(g)}h(x_{0}, \dots, x_{n}).$$ Now, note from the definition of $f$ that it cannot vanish at the homogenizing variable $x_{i}.$ Thus, we must have $\deg(g) = 0,$ so $f$ must be irreducible.