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It is often the case when I wish to show a particular polynomial in $k[x_1, \ldots ,x_n]$ is irreducible. Assuming that the polynomial is sufficiently friendly (i.e. one I would encounter as part of a question that's designed to be answerable...), are there any pointers you could give me on how to approach it?

For example, consider the polynomial $f = X_0^8 + X_1^8 + X_2^8 \in k[X_0, X_1, X_2]$. Why is this irreducible? It looks it (i.e. I wouldn't know how to factorise it), but obviously this isn't a proof (or even almost a proof).

Any help would be greatly appreciated

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What assumptions are you making on $k$? If $k$ has characteristic $2$ then $f = (X_0 + X_1 + X_2)^8$. – Qiaochu Yuan May 4 '12 at 15:09
@QiaochuYuan Oops, didn't notice that. If that's the only problem then assume $\mathrm{char}(k) \neq 2$, but if needs be I guess just assume $\mathrm{char}(k) = 0$ – Jonathan May 4 '12 at 15:10
up vote 8 down vote accepted

First note that a reducible homogeneous polynomial necessarily splits into a product of homogeneous factors, so you can freely divide out by one of the variables. Thus your polynomial is irreducible if and only if $x^8 + y^8 + 1$ is irreducible.

Next you can use Eisenstein's criterion; think of $x^8 + y^8 + 1$ as a polynomial in $x$ with coefficients in $k[y]$. Since $\text{char}(k) \neq 2$ by assumption, the polynomial $y^8 + 1$ is squarefree, so it has a prime factor which divides it only once (exactly what this factor is depends on $k$), and by Eisenstein's criterion applied to this prime factor the conclusion follows.

For homogeneous polynomials in exactly three variables you can also use Bézout's theorem. If $k$ is algebraically closed and $f$ is reducible, then $f = 0$ defines at least two projective curves in $\mathbb{P}^2$, and by Bézout's theorem they intersect. At such an intersection point, all of the partial derivatives of $f$ will be equal to zero (exercise). So if you can show that there doesn't exist such a point, then $f$ must be irreducible. In this case, the partial derivatives are $8X_0^7, 8X_1^7, 8X_2^7$ which cannot simultaneously be zero if $\text{char}(k) \neq 2$.

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  1. In general, it is not easy at all. You can have a look at Eisenbud, D. (1995). Commutative algebra. There is some general strategies for proving irreducibility, using Serre's criterion for example, or simpler tools.

  2. Another good answer is to use computer algebra : it is efficient and reliable — in the sense that if you are guaranteed to find the correct decomposition, you do not depend on how lucky the algorithm is. Maple, Sage, Mathematica, Macaulay2, all of them provide factorization. However, there is some drawbacks : your polynomial to factor has to be concrete, no indetermination is tolerated in the exponents and the base field has to be fixed. Moreover, the absolute factorization — i.e. factorization over the algebraic closure of the coefficient field — is not widely implemented.

  3. The third answer is simple examples, simple tricks. Concerning your polynomial, the variety it represents is connected — over $\mathbb C$ for example —, because it is homogeneous, and its singular locus as codimension 2. So it has to be irreducible. Indeed, if the polynomial had two factor, then the associated variety would have two crossing components, and the intersection locus of two hypersurface is always empty or of codimenion $1$. (I assume that you are familiar with the correspondance algebra-geometry since you asked some questions about algebraic geometry.)

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