# Irreducible polynomials in 2 variables (over the integers)

I know a homogeneous polynomial $f(x,y)$ is irreducible if and only if $f(x,1)$ is. (Proof?)

I'm wondering if there's a similar criterion to check if $f(x,y)+c$ is irreducible, given that $f$ is homogeneous and irreducible.

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What do you mean by irreducible here? How could any monomial with total degree greater than 1 be irreducible? $xy^2$ seems like a counterexample, at my current (low) level of understanding. – rschwieb Sep 6 '12 at 12:38

For your first question, it is only true that if $P(X,1)$ is reducible then $P(X,Y)$ too. (rschwieb gave a counterexample.)

If $P(X,1)$ is reductible, there exist $Q,R \in k[X]$ such that $P(X,1)=Q(X)R(X)$. So $P(X,Y)=Y^d P(X/Y,1)= Y^d Q(X/Y) R(X/Y)$ in $k(X,Y)$. But $d= \text{deg}(P) = \text{deg}(R)+ \text{deg}(Q)$. So there exit $n,m \in \mathbb{N}$ such that $n+m=d$ and $Y^nQ(X/Y) \in k[X,Y]$ and $Y^mR(X/Y) \in k[X,Y]$. Thus, $P(X,Y)$ is reducible.

For your second question, I think there is no such criterium: for example, $XY$ is reducible but $XY+c$ is not for $c \neq 0$, and $X^2$ is reducible and $X^2-c^2$ is reducible too.

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The (irr)reducibility of $X^2-c$ depends on $c$. – user18119 Sep 6 '12 at 14:00
Exact, it depends on the existence of a square root. So I added a square to rectify this point. Thank you. – Seirios Sep 6 '12 at 14:30

Hi do you mean some thing like this?

$f(x,y) = x^{p-1} + yx^{p-2}+\cdots+ y^{p-1}$ over a field of characteristic $p$.

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This doesn't seem to be an answer to the question. – Gerry Myerson Sep 6 '12 at 13:24
Yea, this is not the answer, but I can't comment on the question due to lack of reputation, I will complete the answer tomorrow. – Ram Sep 6 '12 at 13:30