Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.

share|cite|improve this answer
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 $p$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.