Is there a quicker way of telling if a polynomial of two variables is irreducible?

Question

Let's have a more concrete example, suppose we are considering $\mathbb{R}[X,Y]$ and $P(X,Y)=X^3+Y^3.$ Well I can tell that $X+Y$ is a factor straight away and I am left with $X^2-XY+Y^2$ but I cannot tell straightaway wether this is irreducible or not.

What I did was using quadratic formula by holding either $X$ or $Y$ constant to consider further its reducibility. However I am just a bit curious if there is a quicker way of doing this? Because had my polynomial being any higher 'degree' then I will be in a lot of trouble. Is there a trick where we can perhaps convert $\mathbb{R}[X,Y]$ into something we are more familiar about?

Many thanks in advance!

Answer 1

In general it's just hard but there are some methods that are easy when they work. Here are some exercises involving checking whether a polynomial in two variables is irreducible in roughly increasing order of difficulty, with spoilered solutions describing the techniques involved.

Exercise 1: When is the polynomial $ax^2 + bxy + cy^2$ irreducible?

This polynomial is homogeneous (every term has degree $2$), so if it factors it must factor into homogeneous polynomials (exercise). Hence any factorization must have the form $(px + qy)(rx + sy)$, and we see that as long as $a \neq 0$ the problem is the same as factoring the dehomogenized polynomial $at^2 + bt + c$ where $t = \frac{x}{y}$, and similarly as long as $c \neq 0$ the problem is the same as factoring the dehomogenized polynomial $a + bt + ct^2$ where now $t = \frac{y}{x}$. Either way the answer, over $\mathbb{R}$, is: if and only if the discriminant $\Delta = b^2 - 4ac$ is negative.

Exercise 2: When is the polynomial $y^2 - (x^3 + ax + b)$ irreducible?

(This is the (affine) equation of an elliptic curve in Weierstrass form.)

This polynomial is neither homogeneous nor quadratic. However, it is quadratic considered as a polynomial in $y$ only (with coefficients in $\mathbb{R}[x]$ or, if you prefer, the fraction field $\mathbb{R}(x)$), and considering it as a polynomial in $y$ only, if it has a nontrivial factorization it must have the form $(y - f(x))(y - g(x))$ where $f, g$ are polynomials in $x$. Expanding this out we get that $f(x) = - g(x)$ so in fact the factorization must have the form $(y - f(x))(y + f(x)) = y^2 - f(x)^2$; in other words, this polynomial factors iff $x^3 + ax + b$ is a square. But a square polynomial must have even degree! The conclusion is that this polynomial is always irreducible.

Exercise 3: Is the polynomial $(y^3 - y) - (x^6 - 1)$ irreducible?

(There's nothing special about this polynomial.)

This example is neither homogeneous nor quadratic in either variable so we'll need a new idea. We can do the following: it is cubic in $y$, so it factors as a polynomial in $y$ iff it has a linear factor as a polynomial in $y$, and we can test possible linear factors using a generalization of the rational root theorem, which here tells us that if $y - f(x)$ is a linear factor then $f(x)$ must divide $x^6 - 1$, whose irreducible factorization over $\mathbb{R}$ is $(x - 1)(x + 1)(x^2 - x + 1)(x^2 + x + 1)$. In order for $y - f(x)$ to be a linear factor, or equivalently for $f(x)$ to be a root of the above polynomial (as a polynomial in $y$), degree considerations show that $f$ must be quadratic, and now you can just check that none of the possible quadratics $f(x) = x^2 - 1, x^2 - x + 1, x^2 + x + 1$ work. So the polynomial is irreducible.

Answer 2

In two variables there is an irreducibility criterion called the Eisenstein-Dumas criterion, which generalises the Eisenstein criterion over $\mathbb Q$ and in which the role of the prime $p$ is played by the indeterminate $Y$ (which generates a prime ideal in $k[X,Y]$).

Let $F$ be any field and $f=f_0(Y)+f_1(Y)X+\cdots+f_n(Y)X^n$ such that $f_0(Y)\neq 0$ and such that $f_n(Y)$ is a nonzero constant in $F$. If the Newton polygon of $f$ has only one line segment from $(0,m)$ to $(n,0)$ and $\gcd(n,m)=1$, then $f$ is irreducible over $F$, as well as any algebraic extension of $F$.

The Newton polygon in question is the lower convex hull of the points $(0,d_0)$, $(1,d_1)$, $\dots$, $(n-1,d_{n-1})$ and $(n,0)$ where $d_i$ is the degree of $f_i(Y)$ with $d_0=m$ and $d_n=0$.

For example, if $f=aY^m+bX^vY^w+cX^n\in F[x,y]$ with $a,b,c,m,n,v,w\neq 0$ and $(v,w)$ is not on the line $y=-\frac mnx+m$, then the corresponding Newton polygon is the triangle with vertices $(0,m)$, $(v,w)$ and $(n,0)$. Since there is only one line segment from $(0,m)$ to $(n,0)$, the polynomial $f$ will be irreducible over $F$ if $\gcd (m,n)=1$.

The paper Absolute Irreducibility of Polynomials via Newton Polytopes by S. Gao gives more criteria and examples for absolute irreducibility (irreducibility over $\mathbb C$) for multivariate polynomials.