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I have a question about Exercise 2-34 from William Fulton's Algebraic Curves book. The exercise is as follows.

Suppose that $F, G \in k[X_1, \dots , X_n]$ are forms (i.e. homogeneous polynomials) of degree $r$ and $r+1$ respectively, without common factors (where $k$ is a field). Prove that $F + G$ is irreducible.

I'm having some trouble trying to prove this. First, since $F, G$ are arbitrary I can't think of a particular irreducibility criterion to apply here, so my only idea was to try by contradiction.


I use Fulton's notation for the homogenization and dehomogenization of a polynomial with respect to the variable $X_{n+1}$. Say if $f \in k[X_1, \dots , X_n]$ has degree $d$, then its homogenization with respect to $X_{n+1}$ is denoted by $f^* = X_{n+1}^d f\left ( \frac{X_1}{X_{n+1}} , \dots , \frac{X_n}{X_{n+1}} \right)$.

Similarly, if $F \in k[X_1, \dots , X_{n+1}]$ is homogeneous, then its dehomogenization with respect to $X_{n+1}$ is denoted by $F_* = F(X_1, \dots , X_n, 1)$.

My attempt

Thus I assume by contradiction that $F + G$ is reducible, then there are polynomials $H, R \in k[X_1, \dots , X_n]$ such that $F + G = HG$ and also $ 0 < \deg{H},\deg{G} < r + 1$. Then the homogenization is

$$(F + G)^* = X_{n+1} F + G = (HR)^* = H^* R^*$$

Now I'm somewhat stuck. I've tried to play around with this but without luck. I suppose that maybe I should try to get a contradiction to the assumption that $F$ and $G$ have no common factors, but I don't know how.

My questions

  1. How can this be proved?
  2. If my approach by contradiction is correct, how can the argument be finished?

As always, thank you for any help.

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up vote 3 down vote accepted

After you homogenize, the result is a linear polynomial in $X_{n+1}$ with coefficients in some field which just happens to be $k(X_1, ... X_n)$.

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Thank you very much for the answer Qiaochu. I was thinking about it. Let's see if I got it. Since a linear polynomial over a field is irreducible, then $(F+G)^* = X_{n+1} F + G$ is irreducible in $k(X_1 , \dots , X_n)[X_{n+1}]$ and since $F$ and $G$ have no common factors then the polynomial is primitive and therefore we can conclude that $(F+G)^*$ is irreducible in $k[X_1 , \dots , X_n][X_{n+1}]$. Am I right? – Adrián Barquero Jul 16 '12 at 0:38
@Adrián: yep. Equivalently, by looking at degrees as polynomials in $X_{n+1}$ we see that either $H^{\ast}$ or $R^{\ast}$ has degree $0$ and hence must lie in $k[X_1, ... X_n]$. – Qiaochu Yuan Jul 16 '12 at 0:43
Yes you're right, that was actually the first thing I wrote in my whiteboard when I was thinking about your answer, but then I thought about what I wrote above. Thanks for the help. – Adrián Barquero Jul 16 '12 at 1:04

Say that a non-zero polynomial $F$ has width $w\in\mathbb N_0$ if the degree of the monomial of highest degree appearing in $F$ with non-zero coefficient minus the degree of the monomial of lowest degree appearing in $F$ with non-zero coefficient is $w$. For example, if $F=F_{n+1}+F_n$ with $F_{n+1}$ and $F_n$ non-zero and homogeneous of degrees $n+1$ and $n$, the width of $F$ is $1$; notice that the polyonomials of width $0$ are exactly the non-zero homogeneous polynomials.

  • Show that if $F$ and $G$ are polynomials with widths $w_F$ and $w_G$, then the product $FG$ has width $w_F+w_G$.

  • Use that to prove that if a polynomial $F$ of width $1$, as in your case, is reducible it necessarily has an non-constant homogeneous factor.

  • Finally, use that to prove what you want.

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Doesn't this polynomial have width $1$? – Qiaochu Yuan Jul 16 '12 at 0:28
Yup, that was a typo :) – Mariano Suárez-Alvarez Jul 16 '12 at 0:29
Thank you very much for the answer Mariano. I'll prove it this way also. – Adrián Barquero Jul 16 '12 at 0:55

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