# Irreducibility of Polynomials in $k[x,y]$

I'm working through some Hartshorne problems and have noticed that in order to do certain problems properly one must prove a given polynomial $f\in k[x,y]$ is irreducible.

For example, in problem I.5.1(b) we are studying the polynomial $f=xy-x^6-y^6$. How can I show this is irreducible?

The standard tricks (Eisenstein, linearity in one variable, etc...) don't seem to work here.

Edit: Some of the other polynomials in the problem are also giving me trouble. They are $g=x^3-y^2-x^4-y^4$ and $h=x^2y+xy^2-x^4-y^4$. Any ideas for these?

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If all else fails and you need to determine if the polynomial is irreducible there is still Macaulay2. – user38268 Feb 14 '13 at 6:24
Yeah I threw it into Macaulay2 and it confirmed that it was irreducible but I was curious if there were other ways to do it. After all, Hartshorne's book was written in the 70s before computer algebra systems were prevalent so I figured there must be a way to do it by hand. – Brian Fitzpatrick Feb 14 '13 at 6:34
I suspect that one (ugly) way to do it is to show that there are no zero divisors in the ring $R = k[x, y]/f$. Note that because the only relation is $xy = x^6 + y^6$ the elements in $R$ have a canonical representative where each monomial $x^ny^m$ satisfies $n = 0$ or $m = 0$. It would be very messy with a few cases to do it this way but that's my only idea for how to do it by hand. – Jim Feb 14 '13 at 7:08

Assume $k$ is algebraically closed. Let $F = k((x))$. Consider the polynomial $g = xy - y^6$. Because

$$g \equiv f \mod x^6$$

then over $F$, factorization of $g$ should approximate the factorization of $f$. The factorization of $g$ is

$$g = y (x - y^5)$$

This is easy to check because its splitting field $E = F(x^{1/5})$ satisfies $[E : F] = 5$. The six roots of $g$ are $y=0$ and $y=\zeta^k x^{1/5}$ where $\zeta$ is a primitive fifth root of unity. (if $k$ has characteristic $5$, then instead $x^{1/5}$ is a root with multiplicity 5)

Now observe

$$f(x, x^{1/5}) \equiv g(x, x^{1/5}) = 0 \mod (x^{1/5})^{30}$$

and by Hensel's Lemma / Newton's algorithm, we can lift $x^{1/5}$ to an actual root $\alpha$ of $f$ in $E$ with leading term $x^{1/5}$; therefore $F(\alpha) = E$. (This might not work in characteristic 5; I haven't checked in detail)

Therefore, we conclude that $f$ has an irreducible factor of degree $5$ over $F$.

Therefore if $f$ were reducible over $k(x)$, then the only possibility is that it has an irreducible degree $5$ factor and a root. By inspection, we can see that such a root has to be $x^5$ plus higher powers of $x$, and from there it's easy to see that said root can't exist.

Thus $f$, as a polynomial in $y$ over $k[x]$, is irreducible and has content $1$. Therefore $f$ is irreducible in $k[x,y]$.

Now consider $k$ not algebraically closed, then the fact $f$ is irreducible over $\bar{k}[x,y]$ implies it is also irreducible over $k[x,y]$.

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This is exactly what I was looking for. Thanks! – Brian Fitzpatrick Feb 14 '13 at 9:02
In your first line did you mean $F=k[x]$? – JSchlather Feb 14 '13 at 15:16
@Jacob: I did mean $k[[x]]$. The power series ring is needed for the particular argument I made; e.g. $f$ doesn't have any roots for $y$ in the ring $k[x][x^{1/5}]$, let alone be totally split. – Hurkyl Feb 14 '13 at 17:11
So when you talk about $F(x^{1/5})$ this is $k((x))(x^{1/5})$? – JSchlather Feb 14 '13 at 17:14
@Jacob: Yeah. Actually I intended to write $k[[x]][x^{1/5}]$, but didn't particularly care about the distinction between the power series and laurent series fields. It's probably better to write the field version, though, so fixed! – Hurkyl Feb 14 '13 at 17:19

As a complement to other answers, I present here a very general method to see if a polynomial is irreducible of not. It needs a little more computation but it is not specific to the polynomial we consider.

There is an algorithm for absolute (i.e. over an algebraically closed field) factorisation of bivariate polynomials. It has been discovered by Picard (1906) and has been recently used by Ruppert (1999) and Gao (2003) to design efficient algorithms.

Let $f$ in $k[x,y]$ of bidegree $(m,n)$. Let assume that $gcd(f, \partial_xf) = 1$, if it is not the case, we have an obvious factor. Let $V$ be the vector space $$\left\{ (A,B)\in k[x,y]^2 \ \middle| \ \deg A \leq (m-1,n), \deg B \leq (m,n-1) \right \},$$ and $E$ the subspace of all $(A,B)$ in $V$ such that $$\frac{\partial}{\partial y}\left(\frac A f\right) = \frac{\partial}{\partial x}\left(\frac B f\right).$$

There is an obvious element in $E$: the couple $\left(\partial_x f,\partial_y f\right)$. More interestingly, if $g$ is a polynomial which divides $f$, then the couple $\left(\frac fg \partial_x g, \frac fg \partial_y g\right)$ is in $E$ as well. If $f = gh$, you can check that $$\left(\partial_x f,\partial_y f\right) = \left(\tfrac fg \partial_x g, \tfrac fg \partial_y g\right) + \left(\tfrac fh \partial_x h, \tfrac fh \partial_y h\right).$$

Now, the following fact should not surprise you : the dimension of E over $k$ is exactly the number of irreducible factors of $f$ over the algebraic closure of $k$.

So just with linear algebra over the rational numbers, I can prove that $xy+x^6+y^6$, as well as the two other polynomials you gave, are irreducible just by computing the rank of a matrix.

For the sake of completeness, here is a little piece of Maple code to perform the computation.

f := x*y+x^6+y^6:

# A and B are generic elements of V.

# This is the system of equation defining E.
collect(numer(diff(A/f,y)-diff(B/f,x)),[x,y],distributed):
eqs := [coeffs(%,[x,y])]:

# We compute a basis a generic element of E.
sol:=solve(eqs, indets(eqs)):

# The number of free variables in the generic
# element gives the dimension of E.
nops(indets(subs(sol, indets(eqs))));


This code give you the number of irreducible factors of $f$ !

• Picard, É. and Simart, G. (1906). Théorie des fonctions algébriques de deux variables indépendantes II
• Ruppert, W. M. (1999). Reducibility of polynomials $f(x,y)$ modulo $p$, J. Number Theory, 77(1), 62--70
• Gao, S. (2003). Factoring multivariate polynomials via partial differential equations, Math. Comp., 72(242), 801--822 (electronic)
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Yeah! This was exactly what I was looking for, without even knowing it existed. Thank you Lierre! (And thank you Google) – Vincent Oct 15 '14 at 13:31

I don't know about $f$ and $g$, but $h$ is irreducible because it is a sum of two forms, one of degree $4$ and one of degree $4-1=3$, with no common factors.

More generally, if you have two forms $F,G \in k[x_1,...,x_n]$ of degree $r,r-1$, with no common factors, then $F+G$ will be irreducible. See for example Fulton, Problem 2.34.

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This is a brute force outline to show that $xy - x^6 - y^6$ is irreducible. I am not sure what one can do in general.

Assume that it is reducible, being the product of $h(x,y)g(x,y)$, with $\deg h \ge \deg g$. We adopt the notation that $h_i$ (resp $h_i$) represents the degree $i$ part of $h$ (resp $g$).

By looking at the lowest degree term, we see that $h,g$ either starts with $(xy,1)$, $(1,xy)$, or $(x,y)$, $(y,x)$.

Case 1: $\deg g = 1$, then $g$ must be $x$ or $y$, since this is the degree 1 term for $g$. Impossible since $xy$ does not divide $x^6 + y^6$.

Case 2: $\deg g = 2$. $g$ can't be $xy$ by the same divisibility issue.

If $g$ starts with $x$ or $y$, check that the degree 5 term in the product $h(x,y)g(x,y)$ should be nonzero. But it is zero for $xy-x^6-y^6$.

If $g$ starts with 1. Check that the degree 1 term for $g$ is zero by looking at the degree 5 term in the product. Then look at the degree 4 term:

$$0 = h_2g_2 + h_3g_1 + h_4g_0 = xy g_2 + h_4$$

Degree 6 term says $h_4g_2 = -(x^6+y^6)$. This implies that $$h_4 = -xyg_2 \Rightarrow x^6+y^6 = -h_4g_2 = xyg_2^2$$ But $xy$ does not divide $x^6+y^6$, contradiction.

Case 3: $\deg g = 3$. If $g$ starts with $xy$, look at degree 5 term, contradiction.

If $g$ starts with $x$ or $y$, WLOG let $g$ starts with $x$. Degree 3 term says: $$0 = yg_2 + xh_2$$ Unique factorization forces $y|h_2$ and $x|g_2$. Now look at degree 4 term to conclude that $y|h_3$. But then $y | h$, so $y|x^6+y^6$ which is absurd.

If $g$ starts with 1, look at degree 3 term to conclude that $xy | h_3$. But then $xy | h$, so $xy | x^6+y^6$ which is absurd.

Thus $xy - x^6 - y^6$ is irreducible.

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+1 Most down to earth answer (by this I don't mean anything bad :D ). – user38268 Feb 14 '13 at 13:28

There is an easy way to see that all of $f,g,h$ are irreducible using Newton polytopes. The method is described in this math overflow answer.

For example, in the case of $h=x^2y+xy^2-y^4-x^4$ the polynomial has nonzero coefficients at $(2,1)$, $(1,2)$, $(0,4)$ and $(4,0)$. Going clockwise around convex hull of this, we move by vectors $$(-1,1), (-1,2), (4,-4), (-2,1).$$ This sequence cannot be partitioned into two such that each of them has sum zero, therefore $h$ is irreducible.

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