This is an exercise from the book Introduction to algebra by А. И. КОСТРИКИН.
The irreducible factors of the homogeneous polynomial function $$ f(X, Y)=a_{0} X^{n}+a_{1} X^{n-1} Y+\cdots+a_{n-1} X Y^{n-1}+a_{n} Y^{n} \in \mathbb{Q}[X, Y] $$ are also homogeneous, and $f(X,Y)$ is irreducible if and only if $f(X,1)=a_0X^n+a_1X^{n-1}+\cdots+a_{n-1}X+a_n\in \mathbb Q[X]$ is irreducible.
My question is, how to prove these two statements? I can intuitively know the correctness of the statements, but I cannot prove them strictly.
Let's look at the first claim first. Say first to set the stage that we're dealing with polynomials with coefficients in one variable. Then we would want to show that the factors of a monomial must also be monomials. If they weren't, then we'd have some factorization $f=gh$, where $f$ is a monomial but one of the factors, say $g$, is not. To show that this means that $gh$ is not a monomial, what we can do is take the largest-degree and smallest-degree terms from $g$ and $h$ and multiply them. Since $\mathbb{Q}$ is a domain, the products of these terms will be nonzero. The product of the largest-degree terms will have a power different from the other pairwise products of monomials in $gh$, as will the product of the smallest-degree terms. That means that these terms can't be cancelled out, so they are both in $f$, so $f$ is not a monomial.
When we look at $f$ in 2 variables, we kind of have to extend our definition of "largest-degree." We'll call our extended definition the extended degree of a monomial. Say that the extended degree of a larger-degree monomial is larger than that of a smaller-degree polynomial, as you would expect, but say also that given two monomials of the same degree, the one with a higher power of $X$ has a higher extended degree. Then we do the same thing as before. Consider some product $gh$. Take their terms of largest and smallest extended degrees and multiply them. Again, nothing can cancel those two terms out, so that they are both present in $f$. And if $g,h$ are not homogenous, then these terms will have different degrees so $f$ is not homogenous.
I'm not so sure about the second claim. The function $XY$ is definitely not irreducible, but the function $X$ definitely is. But I think the intention of this problem is that if we have a factorization in one, then we have a factorization in the other. If we have a factorization of $f(X,1)$, then we can definitely multiply powers of $Y$ by the various terms so that each factor is homogenous, and then multiplying by some power of $Y$ at the end we will have a factorization of $f(X,Y)$. The contrapositive of this is that if $f(X,Y)$ is irreducible then $f(X,1)$ is irreducible.
Now like I said the first direction isn't totally correct as-is. But it's right in spirit: if we have a factorization of $f(X,Y)$, then we can plug in 1 for $Y$ and we will have a factorization of $f(X,1)$. The only problem is that it's possible that some of these factors become trivial: in particular, any factor which is a power of $Y$ will simply get mapped to 1. Thus, if we add the stipulation that $f(X,Y)$ has a factorization where at least two factors are not powers of $Y$, then this is correct. In particular, if we only consider $f(X,Y)$ where $a_0\neq0$ so that $Y$ is not a factor, then $f(X,1)$ being irreducible will imply that $f(X,Y)$ is irreducible.