We say that a polynomial $F$ is irreducible over $\mathbb{C}$ if it cannot be factored into a product of two non-constant polynomials over $\mathbb{C}$. I want to show that
If $F(x,y,z)$ is a homogenous polynomial that is irreducible over $\mathbb{C}$, then $f(y,z):=F(1,y,z)$ is also irreducible over $\mathbb{C}$.
If $x|F(x,y,z)$, then we can write $F(x,y,z)=xh(x,y,z)$, however as $F$ is assumed to be irreducible, $h(x,y,z)$ is forced to be a complex constant, say $h(x,y,z)=C$, and thus $F(x,y,z)=Cx$. Therefore, $f(y,z):=F(1,y,z)=C$ which is irreducible, because it cannot be factored into a product of two non-constant polynomials.
The problem is that I don't know how to deal with the case where $x\nmid F(x,y,z)$. I tried to argue by contradiction:
If $f(y,z)$ is reducible, then we can write $F(1,y,z)=f(y,z)=h(y,z)g(y,z)$ where $h,g$ are non-constants polynomials. Then, I tried to use the property of being homogenous: say $F$ is homogenous of degree $d$, then \begin{align*} &F(1,y,z)=h(y,z)g(y,z)\\ &\implies F\Bigg(1,\dfrac{y}{x},\dfrac{z}{x}\Bigg)=h\Bigg(\dfrac{y}{x},\dfrac{z}{x}\Bigg)g\Bigg(\dfrac{y}{x},\dfrac{z}{x}\Bigg)\ \ \text{for all}\ \ x\in\mathbb{C}^{*}\\ &\implies x^{d}F\Bigg(1,\dfrac{y}{x},\dfrac{z}{x}\Bigg)=x^{d}h\Bigg(\dfrac{y}{x},\dfrac{z}{x}\Bigg)g\Bigg(\dfrac{y}{x},\dfrac{z}{x}\Bigg)\ \ \text{for all}\ \ x\in\mathbb{C}^{*}\\ &\implies F(x,y,z)=x^{d}h\Bigg(\dfrac{y}{x},\dfrac{z}{x}\Bigg)g\Bigg(\dfrac{y}{x},\dfrac{z}{x}\Bigg)\ \ \text{for all}\ \ x\in\mathbb{C}^{*}. \end{align*}
But then I don't know what to do to find a contradiction. Is there anything like, the highest degree of $h$ plus the highest degree of $g$ is less than $d$, so we can factor out at least one $x$ in the RHS?
Thank you!
Edit:
As discussed, this post is a duplication of the link provided in the comment. After reading that post, I think that my proof above is basically done. The only confusion I had before was that if $d=\deg(g)+\deg(h)$. This is actually clear, because $F(1,y,z)=h(y,z)g(y,z)$, so $\deg(F)=d=\deg(hg)=\deg(g)+\deg(h)$, and then the proof can go through.
The following is a complete proof I wrote:
Proof:
Firstly, note that if $x\mid F(x,y,z)$, then we can write $F(x,y,z)=xh(x,y,z)$, but $F$ is irreducible, which forces $h(x,y,z)$ to be a constant, say $h(x,y,z)=C$. Then, $F(x,y,z)=Cx$ and thus $f(y,z):=F(1,y,z)=C$ which is clearly irreducible over $\mathbb{C}$ because it cannot be factored into a product of two non-constant polynomials.
If $x\nmid F(x,y,z)$, then there must be some monomial summand of $F$ which contains no $x$. As all monomials in $F$ have the same degree $d$, the degree of $F(1,y,z)$ is also $d$ (the degree here is not homogenous degree, but is the highest degree among all the monomial summands in $F(1,y,z)$), because monomials in $F(x,y,z)$ that have $x-$term will not have $x-$term anymore in $F(1,y,z)$, so the degree of these monomials will decrease, but the degree of those monomials that do not have $x-$term will stay the same as $d$.
Hence, if $F(1,x,y)$ is not irreducible, i.e. $F(1,y,z)=g(y,z)h(y,z)$ for some non-constant polynomials $g,h\in\mathbb{C}[y,z]$, as $\deg(F)=d$, we must have $\deg(gh)=\deg(g)+\deg(h)=d$, and using the property of $F$ being homogenous, we have that $$F(1,y,z)=g(y,z)h(y,z)\implies F\Bigg(1,\dfrac{y}{x},\dfrac{z}{x}\Bigg)=g\Bigg(\dfrac{y}{x},\dfrac{z}{x}\Bigg)h\Bigg(\dfrac{y}{x},\dfrac{z}{x}\Bigg)\ \ \text{for all}\ \ x\in\mathbb{C}^{*}$$ $$\implies x^{d}F\Bigg(1,\dfrac{y}{x},\dfrac{z}{x}\Bigg)=x^{d}g\Bigg(\dfrac{y}{x},\dfrac{z}{x}\Bigg)h\Bigg(\dfrac{y}{x},\dfrac{z}{x}\Bigg)\ \ \text{for all}\ \ x\in\mathbb{C}^{*}$$ $$\implies F(x,y,z)=x^{\deg(g)+\deg(h)}g\Bigg(\dfrac{y}{x},\dfrac{z}{x}\Bigg)h\Bigg(\dfrac{y}{x},\dfrac{z}{x}\Bigg)\ \ \text{for all}\ \ x\in\mathbb{C}^{*}$$ $$\implies F(x,y,z)=x^{\deg(g)}g\Bigg(\dfrac{y}{x},\dfrac{z}{x}\Bigg)\cdot x^{\deg(h)}h\Bigg(\dfrac{y}{x},\dfrac{z}{x}\Bigg)\ \ \text{for all}\ \ x\in\mathbb{C}^{*}.$$
Note that both $x^{\deg(g)}g(\frac{y}{x},\frac{z}{x})$ and $x^{\deg(h)}(\frac{y}{x},\frac{z}{x})$ are polynomials, because for example, the highest degree of those monomial summands in $g$ is $\deg(g)$, so $x^{\deg(g)}$ will cancel the denominator, and for other monomial summands we have extra factors of $x$. But $F$ is assumed to be irreducible, and thus one of $x^{\deg(g)}g(\frac{y}{x},\frac{z}{x})$ or $x^{\deg(h)}(\frac{y}{x},\frac{z}{x})$ must be constant. This implies that $\deg(g)=0$ or $\deg(h)=0$ must hold, either of which will contradict the hypothesis that none of $g$ or $h$ are constant.
Hence, $F(1,y,z)$ must be irreducible