Is the follow proposition
Prop Let $f \in k\left[x_1, x_2, \ldots, x_n\right]$ be a polynomial with the form $l+h$, where $l$ is an irreducible non constant homogeneous polynomial and $h$ is a homogeneous polynomial with $\operatorname{deg}h \neq \operatorname{deg} l$, and $l$ not dividing $h$. Then $f$ is irreducibile.
covered by some more general criteria? Or does it have a more general form? I proved it via argument ragarding to the degree. But I am still wondering if things like Eisenstein’s criteria or other tools can apply to it.
Thank you in advanced.
I made a mistake and thought I could prove it using Eisenstein's criterion via the method in the related link. But In fact, I cannot do it. Supposing $\operatorname{deg}l < \operatorname{deg}h$ then homogenization gives a polynomial like $lY^n+h$ with $n = \operatorname{deg}h - \operatorname{deg}l$. It has the form that $l$ divides all the coefficients except the constant term $h$, and $l^2$ does not divide the highest term's coefficients. This condition is not for Eisenstein's criterion. Hence one cannot apply it.
Nevertheless, I post my original proof for it as an answer.
First of all, one has the following basic proposition:
Proposition 1 Let $f$ be a non-constant polynomial in $k[x_1, x_2, \dots, x_n]$ such that the highest degree term of it is irreducible, then $f$ is also irreducible.
Proof: Say $f = g + h$ with $h$ homogeneous and $\operatorname{deg}g < \operatorname{deg} h$ , and $h$ is irreducible, and $f=pq$ with both $p$ and $q$ non-constant. Then comparing the highest degree terms of $f$ and $p$, $q$, one has $h$ also irreducible. It's a contradiction with the assumption. ∎
And the original prop,
Proposition 2 Let $f$ be a be a non-constant polynomial in $k[x_1, x_2, \dots, x_n]$ with the form $l+h$, where l is an irreducible non-constant homogeneous polynomial and $h$ is a homogeneous polynomial with $\operatorname{deg}h \neq \operatorname{deg}l$, and $l$ not dividing $h$. Then $f$ is irreducible.
Proof: From Proposition 1, one can suppose that $f$ is not irreducible, say $f = p q$, in which $p$ and $q$ are both non-constant polynomials.
Then from the argument on degree, the product of the lowest degree terms of $p$ and $q$ must be $l$. Since $l$ is irreducible, one of the lowest degree terms of $p$ and $q$ must be associated with $l$ and the other must be constant. Adjusting them with constants, one can assume that the lowest degree term of $p$ is $l$, and the lowest degree term of $q$ is $1$. Hence
$$f = (l + n) (1 + m) = l + n + l m + n m$$
, with $n$ zero or an polynomial of degree greater than $\operatorname{deg} l $, $m$ an polynomial of degree greater than $1$, and from this, $h = n + l m + n m$.
Since $l$ does not divide $h$, one knows that $n$ cannot be zero. Hence $\operatorname{deg} n m > \operatorname{deg}n$ and $\operatorname{deg}n m > \operatorname{deg}l m$. From this, $h$ cannot be homogeneous. Hence it's a contradiction.∎
One example of this proposition is $x-2y+z+x^2y^2z^3$. And $x+y+x^2+xy$ is a counter-example that showing $l$ not dividing $h$ is necessary, $(x+y^2)(1+x) = x+y^2+x^2+xy^2$ is a counter-example that showing $h$ being homogeneous is necessary. And $x+y+z^2+xyz$ is another non-example that is irreducible but not covered by this proposition.