Suppose I have a homogeneous polynomial in at least 3 variables over some algebraically closed field (of characteristic 0, if need be). Question: How may I test — by hand — whether it is irreducible?

I understand that there aren't even yet algorithms which can decide whether a univariate polynomial is irreducible, but nonetheless there are tools like the Eisenstein criterion, reduction modulo $p$, and Gauss's lemma which come in useful.

Example: How can I show that $X^n + Y^n + Z^n$ is irreducible over a field of characteristic 0? The following argument works for $n$ a power of 2 and base field $\mathbb{Q}$: we know that $Y^n + Z^n$ is irreducible, since otherwise we could produce a $n$-th root of $-1$ in $\mathbb{Q}$. Thus $\mathfrak{p} = (Y^n + Z^n)$ is a prime ideal, and we can apply the generalised Eisenstein criterion. But this argument breaks down when $n$ is divisible by an odd number (because then we could factor out ($X^m + Y^m$, where $m$ is the highest power of 2 dividing $n$) or when there are $n$-th roots of $-1$. Alternatively, for $n = 2$, I can use an ad hoc approach: if $G(X, Y, Z) H(X, Y, Z) = X^2 + Y^2 + Z^2$, treating $Y$ and $Z$ as constants, we see that we must have a difference-of-squares factorisation, so we have $- (Y^2 + Z^2)$ a square, and iterating the argument, we see this cannot possibly be the case, even if the field is algebraically closed.

That kind of trick works well when there aren't any cross terms, but what about, say, the polynomial $X^2 Y^3 + Y^2 Z^3 + Z^2 X^3$? Here I'm quite lost. If the base field is algebraically closed, then substituting constants for $Z$ don't yield anything useful, since the resulting polynomial is always reducible. While it remains true that the polynomial is irreducible if if I can find an ideal $\mathfrak{I}$ such that the polynomial modulo $\mathfrak{I}$ is irreducible, it's not obvious to me whether any such ideals exist or not.

Context: Some of my exercises / exam problems ask me to show that a given hypersurface in projective space is irreducible. I'm not sure how much weight is placed on it, but it would be reassuring if I could do it properly.


In exactly $3$ variables, suppose a given homogeneous polynomial $F$ is reducible and suppose that the base field is algebraically closed. Then its factors describe projective curves which intersect by Bezout's theorem, and you can determine these intersection points by looking at the points in $\mathbb{P}^2$ where the partial derivatives of $F$ simultaneously vanish. If there are no such points, $F$ must be irreducible.

For example, $X^n + Y^n + Z^n$ must be irreducible by this criterion because the partial derivatives only simultaneously vanish at $(0 : 0 : 0)$.

  • $\begingroup$ I had a think about it... doesn't this imply every reducible projective plane curve has singular points? In which case it suffices to prove that a projective plane curve is smooth to show it's irreducible? $\endgroup$ – Zhen Lin May 17 '11 at 8:40
  • $\begingroup$ @Zhen: yes (over an algebraically closed field, and you can always pass to the algebraic closure). A short way to see this without doing any calculations is to change coordinates so that any point of intersection is at the origin. Locally around such a point the curve looks like $xy = 0$ or $x^2 = 0$. $\endgroup$ – Qiaochu Yuan May 17 '11 at 8:58
  • $\begingroup$ Yes, that was my intuition. And it seems to work for an arbitrary number of irreducible components, even with multiplicity. Excellent. $\endgroup$ – Zhen Lin May 17 '11 at 10:23
  • $\begingroup$ Dear @Qiaochu, in what sense does $x^3=y^4$ look locally around zero like $xy=0$ or $x^2=0$ ? $\endgroup$ – Georges Elencwajg May 21 '12 at 18:40
  • $\begingroup$ @Georges: $x^3 = y^4$ is not reducible, but your point is well taken. $\endgroup$ – Qiaochu Yuan May 21 '12 at 18:53

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